An Excursion through Elementary Mathematics, Volume I Real Numbers and Functions ( PDFDrive.com )

Problem Books in Mathematics
Antonio Caminha Muniz Neto
An Excursion
through Elementary
Mathematics,
Volume I
Real Numbers and Functions
Problem Books in Mathematics
Series Editor:
Peter Winkler
Department of Mathematics
Dartmouth College
Hanover, NH 03755
USA
More information about this series at http://www.springer.com/series/714
Antonio Caminha Muniz Neto
An Excursion through
Elementary Mathematics,
Volume I
Real Numbers and Functions
123
Antonio Caminha Muniz Neto
Mathematics
Universidade Federal do Ceará
Fortaleza, Ceará, Brazil
ISSN 0941-3502
Problem Books in Mathematics
ISBN 978-3-319-53870-9
DOI 10.1007/978-3-319-53871-6
ISSN 2197-8506 (electronic)
ISBN 978-3-319-53871-6 (eBook)
Library of Congress Control Number: 2017933290
© Springer International Publishing AG 2017
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To Gabriel and Isabela, my most beautiful
theorems.
To my teacher Valdenísio Bezerra, in
memorian
Preface
This is the first of a series of three volumes (the other ones being [4] and [5])
devoted to the mathematics of mathematical olympiads. Generally speaking, they
are somewhat expanded versions of a collection of six volumes, first published
in Portuguese by the Brazilian Mathematical Society in 2012 and currently in its
second edition.
The material collected here and in the other two volumes is based on course
notes that evolved over the years since 1991, when I first began coaching students
of Fortaleza to the Brazilian Mathematical Olympiad and to the International
Mathematical Olympiad. Some 10 years ago, preliminary versions of the Portuguese
texts also served as textbooks for several editions of summer courses delivered at
UFC to math teachers of the Cape Verde Republic.
All volumes were carefully planned to be a balanced mixture of a smooth and
self-contained introduction to the fascinating world of mathematical competitions,
as well as to serve as textbooks for students and instructors involved with math clubs
for gifted high school students.
Upon writing the books, I have stuck myself to an invaluable advice of the
eminent Hungarian-American mathematician George Pólya, who used to say that
one cannot learn mathematics without getting one’s hands dirty. That’s why, in
several points throughout the text, I left to the reader the task of checking minor
aspects of more general developments. These appear either as small omitted details
in proofs or as subsidiary extensions of the theory. In this last case, I sometimes
refer the reader to specific problems along the book, which are marked with an *
and whose solutions are considered to be an essential part of the text. In general, in
each section, I collect a list of problems, carefully chosen in the direction of applying
the material and ideas presented in the text. Dozens of them are taken from former
editions of mathematical competitions and range from the almost immediate to real
challenging ones. Regardless of their level of difficulty, we provide generous hints,
or even complete solutions, to virtually all of them.
This first volume concentrates on real numbers, elementary algebra, and real
functions. The book starts with a non-axiomatic discussion of the most elementary
properties of real numbers, followed by a detailed study of basic algebraic identities,
vii
viii
Preface
equations and systems of equations, elementary sequences, mathematical induction,
and the binomial theorem. These pave the way for an initial presentation of algebraic
inequalities like that between the arithmetic and geometric means, as well as those
of Cauchy, Chebyshev, and Abel. We then run through an exhaustive elementary
study of functions that culminates with a first look at implicitly defined functions.
This is followed by a second look on real numbers, focusing on the concept of
convergence for sequences and series of reals. We then return to functions, this
time to successively develop, in detail, the basics of continuity, differentiability,
and integrability. Along the way, the text stays somewhere between a thorough
calculus course and an introductory analysis one. Lots of interesting examples and
important applications are presented throughout. Whenever possible (or desirable),
the examples are taken from mathematical competitions, whereas the applications
vary from the proof and several applications of Jensen’s convexity inequality to
Lambert’s theorem on the irrationality of and Stirling’s formula on the asymptotic
behavior of nŠ. The text ends with a chapter on sequences and series of functions,
where, among other interesting topics, we construct an example of a continuous and
nowhere differentiable function, develop the rudiments of the generating function
method, and discuss Weierstrass’ approximation theorem and the rudiments of the
theory of Fourier series.
Several people and institutions contributed throughout the years for my effort
of turning a bunch of handwritten notes into these books. The State of Ceará
Mathematical Olympiad, created by the Mathematics Department of the Federal
University of Ceará (UFC) back in 1980 and now in its 36th edition, has since
then motivated hundreds of youngsters of Fortaleza to deepen their studies of
mathematics. I was one such student in the late 1980s, and my involvement with this
competition and with the Brazilian Mathematical Olympiad a few years later had a
decisive influence on my choice of career. Throughout the 1990s, I had the honor
of coaching several brilliant students of Fortaleza to the Brazilian Mathematical
Olympiad. Some of them entered Brazilian teams to the IMO or other international
competitions, and their doubts, comments, and criticisms were of great help in
shaping my view on mathematical competitions. In this sense, sincere thanks go to
João Luiz Falcão, Roney Castro, Marcelo Oliveira, Marcondes França Jr., Marcelo
C. de Souza, Eduardo Balreira, Breno Falcão, Fabrício Benevides, Rui Vigelis,
Daniel Sobreira, Samuel Feitosa, Davi Máximo Nogueira, and Yuri Lima.
Professor João Lucas Barbosa, upon inviting me to write the textbooks to the
Amílcar Cabral Educational Cooperation Project with Cape Verde Republic, had
unconsciously provided me with the motivation to complete the Portuguese version
of these books. The continuous support of Professor Hilário Alencar, president of
the Brazilian Mathematical Society when the Portuguese edition was first published,
was also of great importance for me. Special thanks go to professors Abdênago
Barros and Fernanda Camargo, my colleagues at the Mathematics Department of
UFC, who had made quite useful comments on the Portuguese editions, which
were incorporated in the text in a way or another; they had also read the entire
English version and helped me in improving it in a number of ways. If it weren’t for
my editor at Springer-Verlag, Mr. Robinson dos Santos, I almost surely would not
Preface
ix
have had the courage to embrace the task of translating more than 1500 pages from
Portuguese into English. I acknowledge all the staff of Springer involved with this
project in his name.
Finally, and mostly, I would like to express my deepest gratitude to my parents
Antonio and Rosemary, my wife Monica, and our kids Gabriel and Isabela. From
early childhood, my parents have always called my attention to the importance of
a solid education, having done their best for me and my brothers to attend the
best possible schools. My wife and kids filled our home with the harmony and
softness I needed to get to endure on several months of solitary nights of work
while translating this book.
Fortaleza, Brazil
December 2016
Antonio Caminha Muniz Neto
Contents
1
The Set of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
1.1
Arithmetic in R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
1.2
The Order Relation in R . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
1.3
Completeness of the Real Number System . . . .. . . . . . . . . . . . . . . . . . . .
1.4
The Geometric Representation . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
1
3
7
12
15
2
Algebraic Identities, Equations and Systems . . . . . . . .. . . . . . . . . . . . . . . . . . . .
2.1
Algebraic Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
2.2
The Modulus of a Real Number .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
2.3
A First Look at Polynomial Equations .. . . . . . . .. . . . . . . . . . . . . . . . . . . .
2.4
Linear Systems and Elimination . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
2.5
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
19
19
27
33
45
54
3
Elementary Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
3.1
Progressions .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
3.2
Linear Recurrences of Orders 2 and 3 . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
3.3
The † and … Notations . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
61
61
70
78
4
Induction and the Binomial Formula . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 87
4.1
The Principle of Mathematical Induction .. . . . .. . . . . . . . . . . . . . . . . . . . 87
4.2
Binomial Numbers .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 98
4.3
The Binomial Formula .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 104
5
Elementary Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
5.1
The AM-GM Inequality . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
5.2
Cauchy’s Inequality .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
5.3
More on Inequalities .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
111
111
123
128
6
The Concept of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
6.1
Definitions and Examples . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
6.2
Monotonicity, Extrema and Image . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
6.3
Composition of Functions . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
6.4
Inversion of Functions . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
143
143
153
163
172
xi
xii
Contents
6.5
6.6
6.7
Defining Functions Implicitly . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 176
Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 185
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 195
7
More on Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
7.1
Supremum and Infimum .. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
7.2
Limits of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
7.3
Kronecker’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
7.4
Series of Real Numbers.. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
201
201
208
221
228
8
Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
8.1
The Concept of Continuity . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
8.2
Sequential Continuity .. . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
8.3
The Intermediate Value Theorem . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
245
245
256
264
9
Limits and Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.1
Some Heuristics I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.2
Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.3
Basic Properties of Derivatives .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.4
Computing Derivatives . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.5
Rôlle’s Theorem and Applications .. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.6
The First Variation of a Function.. . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.7
The Second Variation of a Function.. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
9.8
Sketching Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
275
275
277
292
302
312
318
328
339
10 Riemann’s Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.1 Some Heuristics II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.2 The Concept of Integral . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.3 Riemann’s Theorem and Some Remarks . . . . . .. . . . . . . . . . . . . . . . . . . .
10.4 Operating with Integrable Functions .. . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.5 The Fundamental Theorem of Calculus . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.6 The Change of Variables Formula . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.7 Logarithms and Exponentials . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.8 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.9 Improper Integration .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
10.10 Two Important Applications .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
347
347
351
360
368
381
392
398
414
426
437
11 Series of Functions.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
11.1 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
11.2 Series of Functions.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
11.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
11.4 Some Applications .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
11.5 A Glimpse on Analytic Functions.. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
447
447
456
472
486
499
Contents
xiii
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 511
A
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 513
B
Hints and Solutions .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 515
Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 643
Chapter 1
The Set of Real Numbers
This first chapter recalls some definitions and results which are essential to all
further developments. We assume from the reader a modest acquaintance with the
most basic concepts of set theory; we also assume that he or she is familiar with the
sets of naturals,
N D f1; 2; 3; 4; : : :g;
integers,
Z D f0; ˙1; ˙2; ˙3; : : :g;
and rationals,
QD
na
b
o
I a; b 2 Z; b ¤ 0 ;
as well as with the elementary arithmetic operations within these sets.
In what concerns the integers1, given a; b 2 Z, with a ¤ 0, we say that a divides
b if there exists an integer c such that b D ac. Equivalently, to say that a divides b is
the same as to say that the rational number ba is an integer; for example, 13 divides
52, since 52
D 4.
13
If a divides b, we also say that a is a divisor of b, or that b is divisible by a; in
such a case, we denote a j b. If a doesn’t divide b (or, which is the same, if ba … Z),
we denote a − b. An integer n is even if 2 j n; otherwise, n is said to be odd. Hence,
0, ˙2, ˙4, ˙6, . . . are the even integers, while ˙1, ˙3, ˙5, . . . are the odd ones.
1
We refer the reader to Chap. 6 of [5] for a systematic discussion of what follows.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_1
1
2
1 The Set of Real Numbers
Given natural numbers a and b, it is well known that there exist unique integers
q and r satisfying the following conditions:
b D aq C r and 0 r < a:
(1.1)
The above relation is known as the division algorithm, and the integers q and r
are respectively called the quotient and the remainder of the division of b by a.
Conditions (1.1) are usually condensed in the diagram
b
r
a
q
In particular, for given naturals a and b and in the above notations, a j b is the
same as r D 0 and q D ba .
Two nonzero integers a and b always have a greatest common divisor, which
will be denoted gcd.a; b/; Moreover, a and b are said to be relatively prime if
gcd.a; b/ D 1; in particular, if b D ka C 1, then a and b are relatively prime.
If r D mn , with m and n integers, is a representation of the nonzero rational
r, then, by cancelling out factors common to m and n (i.e., cancelling gcd.m; n/
out of m and n), we get an irreducible representation for r. For example, since
2
gcd.12; 18/ D 6, the rational number 12
18 has 3 as an irreducible representation,
which was obtained by cancelling out a factor 6 from both 12 and 18.
An integer p > 1 is prime if 1 and p are its only positive divisors; in another way,
an integer p > 1 is prime if, for a 2 N, we have that
p
2 N ) a D 1 or p:
a
It is a well known fact (cf. Chap. 6 of [5]) that the set of prime numbers is infinite.
Below, we list all prime numbers less than 100:
2; 3; 5; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 53;
59; 61; 67; 71; 73; 79; 83; 89; 97:
An integer greater than 1 and which is not prime is said to be composite. It is also a
well known fact (cf. Chap. 6 of [5]) that every natural number n > 1 can be written
(or decomposed) as a product of a finite number of powers2 of prime numbers
(its prime factors); also, such a representation of n is unique, up to the order of the
powers. For instance, 9000 D 23 32 53 is the decomposition of 9000 as a product of
powers of primes. The existence and uniqueness of such a decomposition of n > 1
is known as the Fundamental Theorem of Arithmetic.
2
To recall the definition and the main properties of powers of numbers, we refer the reader to
Section 1.2.
1.1 Arithmetic in R
3
1.1 Arithmetic in R
We use to represent rational numbers in decimal notation. For the rational number
1
1
8 , for example, we write 8 D 0:125 as a shorthand for the equality
1
2
5
1
D
C 2 C 3;
8
10
10
10
and say that 0:125 is the decimal representation of 18 .
Some rational numbers have more complicated decimal representations. We take
1
as an example the rational 12
, for which we usually write
1
D 0:08333 : : : :
12
What does this equality mean? Arguing as in the case of 18 , we are tempted to say
that this equality is a shorthand for
8
1
3
3
3
D 2 C 3 C 4 C 5 C :
12
10
10
10
10
(1.2)
This is actually so, provided we correctly interpret the sum with an infinite number
of summands at the right hand side. Rigorously, (1.2) means that, if we fix the a
priori maximum error 101n D 0: „
00 ƒ‚
: : : 01
…, then we have
n
1
0<
12
3
3
3
8
C 3 C 4 CC k
2
10
10
10
10
1
10n
for every natural number k n; yet in another way, (1.2) means that all of the
1
numbers 1082 C 1033 C 1034 C C 103k , with k n, approximate 12
by defect with
error less than or equal to 0: „
00 ƒ‚
: : : 01
.
In
fact,
it
will
follow
from
Proposition
3.12
…
n
that
1
12
3
3
3
8
C 3 C 4 CC k
102
10
10
10
D
1
;
3 10k
1
equals
so that the error in the defect approximation 1082 C 1033 C 1034 C C 103k of 12
1
1
, which is always less than or equal to the maximum error 10n , whenever k n.
310k
1
It is precisely in this sense that we should think of the equality 12
D 0:08333 : : :.
4
1 The Set of Real Numbers
In the light of the above discussion, one question suggests itself naturally: for an
arbitrary sequence3 .a1 ; a2 ; a3 ; : : :/ of decimal digits, can we think of
0:a1 a2 a3 : : :
as the decimal representation of some rational number?
It is possible to prove (and we will do so in Problem 2, page 7) that the answer
to this question is yes if and only if the list .a1 ; a2 ; a3 ; : : :/ is periodic from a certain
point on, i.e., if and only if it is of the form
.a1 ; a2 ; : : : ; al ; b1 ; b2 ; : : : ; bp ; b1 ; b2 ; : : : ; bp ; b1 ; b2 ; : : : ; bp ; : : :/:
„ ƒ‚ … „ ƒ‚ … „ ƒ‚ …
p
p
(1.3)
p
1
(In particular, for the rational number 12
the list is .0; 8; 3; 3; 3; : : :/, which is clearly
periodic.) Therefore, if we are able to exhibit a list of digits that is not periodic from
any point on, we will conclude that the general answer to the question posed above
is no! We show an example of such a list now.
Example 1.1 The sequence of digits .0; 1; 0; 1; 1; 0; 1; 1; 1; 0; : : :/, with infinitely
many 0’s and such that the quantity of digits 1’s after each digit 0 equals the previous
quantity of digits 1 plus one, is not periodic from any point on.
Proof By the sake of contradiction, suppose that the list in the statement is periodic
from some point on, with, say, a block of p digits that repeats itself. The way we
defined the sequence assures that, from some point on, each occurrence of a block of
digits 1 would bring more than p digits 1. Therefore, the block of digits that repeats
itself should be composed only by digits 1, so that the list should be, from some
point on, equal to .1; 1; 1; 1; 1; 1; 1; : : :/. However, if this was so, then we could not
have an infinite number of digits 0, which is a contradiction.
t
u
In short, the previous discussion points to the following deficiency of the set
of rationals: every rational number admits a decimal representation, but not every
decimal representation represents a rational number. At this point, in order to fulfill
this deficiency, we postulate4 the existence of a set R, containing Q and having the
following properties:
3
As we shall see in Section 6.1, a sequence of real numbers is just a function f W N ! R; however,
for our purposes here, we can think of a sequence just as an ordered list of numbers, i.e., a list of
numbers in which we specify which is the first number, which is the second, third, and so on. We
will sistematically discuss some important elementary sequences of real numbers in Chap. 3.
4
An axiom or postulate in a certain theory is a property imposed as true. One of the fundamental
characteristics of Mathematics as a branch of human knowledge is the use of the axiomatic
method, i.e., the acceptance of the fact that not every mathematical property can be logically
deduced from previously established properties, being necessary the adoption of an adequate set
of axioms.
1.1 Arithmetic in R
5
(I) Addition, subtraction, multiplication and division in Q extend to similar
operations on R, in the sense that they have, on R, properties similar to those
on Q.
(II) The order relation within Q extends to an order relation within R, so that it has,
in R, the same properties its restriction has in Q; in particular, every element
of R is negative, zero or positive.
(III) In the sense of our previous discussions, to every sequence .a1 ; a2 ; a3 ; : : :/ of
digits there corresponds a unique element x 2 Œ0; 1, which will be denoted by
x D 0:a1 a2 a3 : : :. Conversely, to every x 2 Œ0; 1, there corresponds a (not necessarily unique) sequence .a1 ; a2 ; a3 ; : : :/ of digits, such that x D 0:a1 a2 a3 : : :.
The elements of R are called real numbers, and the set R as a whole is the set of
real numbers5 .
In the rest of this section and in the next two section we discuss in detail each of
the items (I), (II) and (III) above, showing what they really mean.
Beginning with (I), we postulate that R is furnished with two operations,
respectively denoted C and and called (by analogy with the corresponding
operations on Q) addition and multiplication. Such operations satisfy axioms (1)
to (7), quoted below:
(1) Consistency: for a; b 2 Q, the result a C b of the addition of a and b is the same,
whether we consider the usual addition on Q or the corresponding operation
on R. Analogously, the result a b of the multiplication of a and b is the
same, whether we consider the usual multiplication on Q or the corresponding
operation on R.
(2) Commutativity: the operations C and are commutative, i.e., they are such that
a C b D b C a and a b D b a, for all a; b 2 R.
(3) Associativity: the operations C and are associative, i.e., they are such that
a C .b C c/ D .a C b/ C c and a .b c/ D .a b/ c, for all a; b; c 2 R.
(4) Distributivity: Multiplication is distributive with respect to addition, i.e., it is
such that a .b C c/ D .a b/ C .a c/, for all a; b; c 2 R.
(5) The roles of 0 and 1 in R: the rational numbers 0 and 1 are such that 0 C a D a
and 1 a D a, for every a 2 R.
(6) Law of cancellation: if a; b 2 R are such that a b D 0, then a D 0 or b D 0.
(7) Existence of inverses: for every a 2 R, there exists b 2 R such that a C b D 0.
For every a 2 R n f0g, there exists b 2 R such that a b D 1.
The properties above have the following important consequences:
(i) Uniqueness of the additive inverse: for a given a 2 R, if b; b0 2 R are such that
a C b D 0 and a C b0 D 0, the associativity and commutativity of addition
5
There are more construtive (and rigorous) ways of introducing the real numbers, as the reader
can find in [6], for example. However, in these notes we chose to follow an approach that was as
close as possible to the previous experience of the medium reader, sacrificing rigor in the name of
understanding.
6
1 The Set of Real Numbers
give us
b D b C 0 D b C .a C b0 /
D .b C a/ C b0 D .a C b/ C b0
D 0 C b0 D b0 :
Hence, the real number a has a unique additive inverse, which from now on will
be denoted by a, as is usually done within Q. It follows from a C .a/ D 0
that a is the additive inverse of a; therefore, according to the notation just
established for additive inverses, we have .a/ D a.
(ii) Uniqueness of the multiplicative inverse: for a given a 2 R n f0g, if b; b0 2 R
are such that a b D 1 and a b0 D 1, then b D b0 . The proof of this fact
is completely analogous to that of the previous item (cf. Problem 3, page 7).
From now on, we will denote the multiplicative inverse of a 2 R n f0g by a1 ,
as is also usually done within Q.
(iii) For a 2 R, we have a 0 D 0: in order to check this, let a 0 D e. The
distributivity of multiplication with respect to addition gives
e D a 0 D a .0 C 0/ D a 0 C a 0 D e C e:
Hence,
e D e C 0 D e C .e C .e//
D .e C e/ C .e/
D e C .e/ D 0;
so that a 0 D e D 0.
In view of the above properties, we adopt the convention (as is usual within Q) to
omit the sign of multiplication, thus writing simply ab instead of ab. Now, observe
that the associativity and commutativity of addition and multiplication in R allow us
to add or multiply an arbitrary finite number of real numbers without worrying with
which summands or factors should be initially operated; the final result will always
be the same6 .
We also define the operations of subtraction () and division () in R as is
usually done in Q: for a; b 2 R, we set
a b D a C .b/ and a b D ab1 ;
6
Rigorously speaking, the validity of such a statement should be proved as a theorem, what can
be done with the aid of the principle of mathematical induction (cf. Chap. 5). However, in order
to have a less terse reading, we chose not to give a formal proof of this fact, just relying on the
previous experience of the reader.
1.2 The Order Relation in R
7
with b ¤ 0 in the last case; yet in the case of division, and whenever there is no
danger of confusion, we write a=b or ab as synonymous for a b.
Problems: Section 1.1
1. * Establish the following properties of proportions: if a, b, c and d are nonzero
reals, such that ab D dc , then
c
a˙c
a
D D
:
b
d
b˙d
2. * Given a sequence .a1 ; a2 ; a3 ; : : :/ of digits, prove that the real number
0:a1 a2 a3 : : : represents the decimal expansion of a rational if and only if the
sequence .a1 ; a2 ; a3 ; : : :/ is periodic from some point on, in the sense of (1.3).
3. * Prove the uniqueness of multiplicative inverses in R. More precisely, prove that
if a ¤ 0 is a real number and b; b0 2 R are such that a b D a b0 D 1, then
b D b0 .
4. Prove, from the axioms for addition and multiplication of reals, that a D .1/a,
for every a 2 R.
5. In each of the following items, decide whether the real number in question is
rational or not. In doing so, assume that the pattern of digits suggested up to the
dots is actually followed. Moreover, in case the number is rational, write it as an
irreducible fraction:
(a)
(b)
(c)
(d)
2:324444 : : :.
0:12121212 : : :.
2:1345454545 : : :.
0:1234567891011121314 : : :.
1.2 The Order Relation in R
We postulate the existence of an order relation on R, which amounts to a way
of comparing real numbers. By the sake of analogy with the corresponding order
relation on Q, we denote this order relation by , and also read it as greater than or
equal to. The order relation on R satisfies axioms (1’) to (5’) below:
(1’)
(2’)
(3’)
(4’)
(5’)
Consistency: for a; b 2 Q such that a b in Q, we have a b in R.
Reflexivity: a a, for every a 2 R.
Antisymmetry: if a; b 2 R are such that a b and b a, then a D b.
Transitivity: if a; b; c 2 R are such that a b and b c, then a c.
Dichotomy: for all a; b 2 R, one has either a b or b a.
8
1 The Set of Real Numbers
In all that follows, if a; b 2 R are such that a b and a ¤ b, we write a > b
and read a is greater than b. We also write a b (read a is less than or equal to b)
as a synonymous for b a, and a < b (read a is less than b) as a synonymous for
b > a. If a 2 R is such that a > 0, we say that a is positive; if a < 0, we say that a
is negative.
With respect to the order relation on R, we also impose axioms (6’) and (7’)
below, which guarantee – as they do in Q – its compatibility with the operations of
addition and multiplication:
(6’) a > b , a b > 0.
(7’) a; b > 0 ) a C b; ab > 0.
The next result collects some other useful properties of the order relation on R,
which can be deduced from axioms (1’) to (7’). From now on, we say that two
nonzero real numbers have equal signs if they are both positive or both negative.
Proposition 1.2 Let a; b; c; d 2 R.
(a) If a > 0, then a
< 0, and vice-versa.
b > 0 ) ab > 0
(b) If a > 0, then
.
b < 0 ) ab < 0
b > 0 ) ab < 0
(c) If a < 0, then
.
b < 0 ) ab > 0
(d) a > b ) a C c > b C c.
(e) a > b; c d )
a C c > b C d.
c > 0 ) ac > bc
(f) If a > b, then
.
c < 0 ) ac < bc
(g) a ¤ 0 ) a2 > 0.
(h) a > 0 , 1a > 0.
(i) If a and b have equal signs and a > b, then
1
a
< 1b .
Proof We prove just a few of the above items, letting the others as exercises (see
Problem 1, page 10).
(a) This follows from axiom (6’):
0 > a , 0 .a/ > 0 , a > 0:
(d) Again by axiom (6’), we have
a > b ) a b > 0 ) .a C c/ .b C c/ > 0 ) a C c > b C c:
(e) We use (d) and the transitivity of :
a>b ) aCc>bCc
cd ) bCcbCd
) a C c > b C d:
1.2 The Order Relation in R
9
(f) Suppose c > 0 (the case c < 0 can be treated analogously). It follows from (6’)
and (7’) that
a > b ) a b > 0 ) .a b/c > 0 ) ac bc > 0 ) ac > bc:
(h) Suppose a > 0. If we had 1a < 0, it would follow from (b) that 1 D a 1a < 0,
which is an absurd (for the order relation on R extends that on Q).
(i) Since a and b have equal signs, it follows from (b) and (c) that ab > 0. Thus,
1
item (h) gives ab
> 0. Since b a < 0, we get from either (b) or (c) that
ba
1
1 1
D
D .b a/ < 0:
a b
ab
ab
Finally, (6’) assures that this is equivalent to
1
a
< 1b .
t
u
Items (b) and (c) of the former proposition are known as the sign rules for
multiplication of real numbers.
Next, let r 2 R be given. The square of r, denoted r2 , is the real number r2 D rr;
the cube of r, denoted r3 , is the real number r3 D r r r. More generally, for a given
n 2 N, we define the nth power of r, denoted rn , as being r, if n D 1, or the real
number obtained by multiplying r by itself n times, if n > 1:
rn D „
r r ƒ‚
…r :
n
Once more we call the attention of the reader to the fact that the associativity of the
multiplication of reals, together with the principle of mathematical induction (cf.
Chap. 5) allows us to prove that the result of the right hand side of the equality above
does not depend on the order in which we multiply the n copies of r. Therefore, rn
is a well defined real number. Problem 3, page 10, lists some useful properties of
the powers of real numbers.
We collect below an important consequence of the properties of the order relation
on R, listed in the last proposition.
Corollary 1.3 Let r be a positive real number and m and n be natural numbers,
with m > n. Then:
(a) 0 < r < 1 ) rm < rn .
(b) r > 1 ) rm > rn .
Proof
(a) Since r is positive, multiplying both sides of the inequality r < 1 by r, we get
r2 < r. Multipliying both sides of this last inequality by r once more, it follows
that r3 < r2 and, hence, that r3 < r2 < r. Proceeding this way, we arrive at the
desired result, i.e.,
< r4 < r3 < r2 < r:
10
1 The Set of Real Numbers
(b) The proof of this item is essentially equal to that of (a), with the only difference
that, in this case, we have r > 1.
t
u
We next illustrate, in an example, how the above corollary can be useful in
comparing certain real numbers.
Example 1.4 In order to compare 2100 C 3100 and 4100 , for instance, it suffices to
see that
2100 C 3100 < 3100 C 3100 D 2 3100
D 2 33 397 D 54 397
< 64 497 D 43 497 D 4100 :
In Chapter 5, we will sistematically study some important inequalities involving
real numbers. For the time being, the following corollary – in spite of its simplicity
– will play an important role.
Corollary 1.5 For a; b 2 R, we have
a2 C b2 0;
(1.4)
with equality if and only if a D b D 0.
Proof Item (g) of Proposition 1.2 gives a2 ; b2 0. Therefore, item (d) of that
proposition gives a2 C b2 0. Now, let a ¤ 0. Then, it follows from item (g)
of the above mentioned proposition that a2 > 0. On the other hand, since we still
have b2 0, item (e) of that proposition guarantees that a2 C b2 > 0.
t
u
Problems: Section 1.2
1. * Prove items (b), (c) and (g) of Proposition 1.2.
2. * Generalize Corollary 1.5, showing that, if a; b; c 2 R, then
a2 C b2 C c2 0;
with equality if and only if a D b D c D 0.
3. * Given r; s 2 R and m; n 2 N, prove that7 :
(a) .rs/n D rn sn .
7
In order to prove properties (a) to (d) in a rigorous way, we have to rely on the principle of
mathematical induction (cf. Chap. 5). Thus, our intention here is simply to make the reader give
some arguments on their plausibility.
1.2 The Order Relation in R
11
(b) rmCn D rm rn .
(c) .rm/n D rmn .
n
n
(d) rs D rsn , if s ¤ 0.
4. * For r 2 R n f0g and n 2 N, we extend the notion of powers with natural
exponents to powers with integer exponents by defining rn D r1n . For example,
r1 D 1r , r2 D r12 etc. If we also set r0 D 1, prove that, for all m; n 2 N, one
m
has rrn D rmn .
5. Let a and b be nonzero integers, with b > 1. If the only prime divisors of b are
2 or 5, prove that the decimal representation of ab is finite.
6. * If x ¤ 0 is a real number and n 2 N, prove that xn is positive if n is even, and
has the same sign of x if n is odd.
1
1
7. Prove that 12 13 C 14 15 C 99
C 100
> 15 .
8. Given positive real numbers a and b such that a < b, compare (i.e., decide
aC2
which is the greatest of) the numbers aC1
bC1 and bC2 .
9. (TT) We are given ten real numbers such that the sum of any four of them is
positive. Show that the sum of all ten numbers is also positive.
10. Decide which of the numbers 3111 or 1714 is the greatest one.
11. * Let n 2 N and a; b be positive reals. Prove that:
(a) a < b if and only if a2 < b2 .
(b) a < b if and only if an < bn .
(c) an C bn < .a C b/n .
12. (EKMC - adapted) Let a, b and c be the lenghts of the sides of a right triangle,
c being the hypotenuse. Which one is the greatest: a3 C b3 or c3 ? Justify your
answer.
13. Show that, for every n 2 N, we have 12n C 22n C 32n 2 7n .
14. * Find all natural numbers a, b and c such that a b c and 1a C 1b C 1c is an
integer.
15. (IMO) Explain how to write 100 as a sum of naturals whose product is as large
as possible.
16. (Russia)8 The leftmost digit of the decimal representations of the natural
numbers 2n and 5n is the same. Prove that such digit is equal to 3.
17. (Russia) Let a, b, c and d be positive real numbers. Show that, among the
inequalities
a C b < c C d; .a C b/.c C d/ < ab C cd and .a C b/cd < ab.c C d/;
at least one is false.
8
For a converse to this problem, see Example 10.60.
12
1 The Set of Real Numbers
1.3 Completeness of the Real Number System
For the time being, we postulate that to every sequence .a1 ; a2 ; a3 ; : : :/ of digits,
there corresponds a unique x 2 R, in the following sense: for a fixed maximum
error 101n , with n 2 N, we have
0x
a
1
10
C
a2
1
ak
C C k n;
102
10
10
for every natural k n. In particular, it follows from this inequality that
x
a1
a2
ak
1
C 2 C C k C n;
10
10
10
10
for every k n. Taking k D n and recalling that aj 9 for every j, we get
a2
a1
C 2
10
10
9
9
C 2
10
10
9
9
C 2
D
10
10
9
9
C 2
D
10
10
x an1
an
1
C nC n
10n1
10
10
9
9
1
C C n1 C n C n
10
10
10
9
1
C C n1 C n1
10
10
1
C C n2
10
CC
D D 1;
Thus, 0 x 1.
As a shorthand to the above postulate, we say that the set R of real numbers is
complete. In this respect, we refer the reader to Section 7.1, as well as to Problem 9,
page 242, where the completeness of the real number system will be more rigorously
discussed.
Conversely, we also postulate that to every positive real number x there corresponds a nonnegative integer m and a sequence .a1 ; a2 ; a3 ; : : :/ of digits, such
that x D m C 0:a1 a2 a3 : : :, in the sense of the previous paragraph. If m > 0 and
m D bn : : : b1 b0 , with the bi ’s being their digits, we write
x D bn : : : b1 b0 :a1 a2 a3 : : :
and say that bn : : : b1 b0 :a1 a2 a3 : : : is the decimal representation or expansion of x.
As was seen in Problem 2, page 7, a real number is rational exactly when its
decimal representation is finite or infinite and periodic. On the other hand, a real
number which is not rational is said to be irrational. Thus, irrational numbers
are those real numbers which cannot be written as quotients of two integers,
or, in another way, those reals whose decimal representations are infinite and
nonperiodic.
1.3 Completeness of the Real Number System
13
Up to the present, the only irrational number we have met was the number
0;0101101110 : : : (cf. Example 1.1). In a sense, such an example is pretty dissatisfying, for it is difficult to manipulate this number (i.e., it is difficult to make
calculations with it). We will remedy this situation in what follows.
From an arithmetic point of view, one great advantage of the set of reals, in
comparison to the set of rationals, is the possibility of making root extractions of
positive real numbers. More precisely, given a real number x > 0 and a natural
number n, it is possible to prove (and we will do so, in two different ways, in
sections 7.1 and 8.3) that there exists a unique positive real number y such that
yn Dpx. From now on, we shall call this real number y the nth root of x, and write
y D n x; the natural n is the index of the root. In short,
p
y D n x , x D yn :
Indices n D 2 and n D 3 occur so frequently that deserve
and
p special names
p
2
notations.
When
n
D
2
(and
x
>
0),
we
write
simply
x,
instead
of
x,
p
p and
say that x is the square root of x; when n D 3 (and x > 0), we say that 3 x is the
cubic root of x.
The argument in the next example allows us to heuristically understand why roots
of positive reals do exist.
p 2
Example 1.6 By definition, we have p 2 D 2. Since 12 < 2 < 22 , the result
2
2
of Problem 11, page 11, gives 1 < 2 < 2. Now,
p since 1:4 < 2 < 1:5 , it
follows once again from that problempthat 1:4 < 2 < 1:5. Analogously, since
1:412 < 2 < 1:422 , we have 1:41 < 2 < 1:42. Continuing this way, we obtain
a unique list .a1 D 4; a2 D 1; a3 D 4; : : :/ of digits, such that 1:a1 a2 : : : an1 an <
p
2 < 1:a1 a2 : : : an1pa0n for all n 1, where a0n D an C 1. Thus, we have no option
but to conclude that 2 D 1:a1 a2 a3 : : : D 1:414 : : :.
At this point, we urge the reader to at least read the statements of Problems 1
and 2, page 14, to get an idea on how to (partially) extend the concept of nth root
to negative reals, as well as to take a look at the main properties of root extraction.
We now turn to powers of natural numbers. A perfect square is a natural number
which can be written in the form m2 , for some m 2 N; hence, the perfect squares
are the natural numbers 12 D 1, 22 D 4, 32 D 9, 42 D 16 etc. A perfect cube is a
natural number which can be written in the form m3 , for some m 2 N; the perfect
cubes are, then, the naturals 13 D 1, 23 D 4, 33 D 27, 43 D 64 etc. More generally,
a natural n is a perfect power if there exist natural numbers n and k, with k > 1,
such that n D mk . If this is the case, we say that n is a k-th perfect power, i.e., is
equal to one of the natural numbers 1k , 2k , 3k , 4k etc. Equivalently, to
psay that n 2 N
is a kth perfect power is the same as to say that its kth root, k n, is a natural
number.
The following proposition, which by now we take for granted, gives an infinite
supply of examples of irrational numbers, which, as we will see in a moment, are
more or less easy to handle. For a proof of it, we refer the reader to Chap. 6 of [5].
(Nevertheless, see Problems 7 and 8.)
14
1 The Set of Real Numbers
Proposition 1.7 p
Given natural numbers n and k, with k > 1, either n is a k-th
perfect power or k n is an irrational number.
p p p
According to the above proposition, numbers like 2, 3 3, 5 10 etc are all
examples of irrationals (for 2 is not a perfect square, 3 is not a perfect cube and
10 is not a perfect 5th power).
Now we can, at least formally (i.e., without worrying with approximations),
operate with several irrational numbers. Let us see an example where we (indirectly)
apply the last proposition to explain why a certain real number is irrational.
p
p
p
p
Example 1.8 The number 2 C 3 is irrational. In fact, if we let r D 2 C 3,
there are two possibilities: r 2 Q or r 62 Q. Suppose, for the sake of contradiction,
that r 2 Q. Then, since the set of rationals is closed with respect to multiplication,
we would have r2 2 Q. On the other hand, the distributivity of multiplication with
respect to addition gives
p
p p
p
r2 D . 2 C 3/. 2 C 3/
p
p p
p
p p
D 2. 2 C 3/ C 3. 2 C 3/
p
p
D .2 C 6/ C . 6 C 3/
p
D 5C2 6
p
p
2
so that 6 D r 25 . Therefore, 6 would be the quotient of the rational numbers
p
r2 5 and 2 and, as such, 6 would be itself a rational number. This contradicts the
result of Proposition 1.7, so that r … Q.
In order to finish our discussion on rational and irrational numbers, note that the
set of irrationals is not closed with respect to the ordinary arithmetic operations.
For instance, for a given irrational number r, although r is also irrational,
p we have
r C .r/ D 0, which is a rational. On the other hand, if we set r D 2, we get
r r D r2 D 2, which is also a rational. Finally, if r ¤ 0, then the quotient of r by
itself is equal to 1, again a rational number.
Problems: Section 1.3
p
1. * Given x < 0 real and n 2 N odd, let y D n x. Prove that yn D x (so that the
real number y is also called the nth root of x).
2. * Given m; n 2 N and x; y > 0, prove that:
p p
p
(a) n xy D n x n y.
pp
p
mn
(b) q
x D pm n x.
n x
(c) n xy D p
n y.
1.4 The Geometric Representation
3.
4.
5.
6.
7.
8.
15
Then, extend the above properties to all nonzero reals x and y, provided m and n
are odd naturals.
* Let a and b be rational numbers, and let r be an irrational one. If a C br D 0,
prove that a D b D 0.
Let a, b, c and d be rational numbers, and let r be an irrational one. If a C br D
c C dr, prove that a D c and b D d.
Let r be a positive real and let
pk be an integer greater than 1. If r is irrational,
prove that the numbers 1r and k r are also irrational.
p
p
(Canada) Let a; b and c be rational numbers, such that a C b 2 C c 3 D 0.
Prove that a; b and c are all equal to zero.
* Assuming the validity of the Fundamental Theorem of Arithmetic (according
p
to the last paragraph of the introduction to this chapter), prove that 2 is an
irrational number.
p
Let p be a prime number, and k > 1 be natural. Prove that k p is irrational.
1.4 The Geometric Representation
A quite useful way of thinking geometrically on the set of rational numbers is the
following: we take a line r, and mark on it a point O; then, we choose one of the halflines that O determines on r, call it positive (the other being called negative) and set
a line segment ` as unit of measure. Then, we associate to each rational number a
point of r, in the following way: first, we associate 0 to the point O; then (according
to Figure 1.1), given a rational number ab , with a; b 2 N, we mark, starting from O
and on the positive half-line, a line segment OA of length a` (i.e., OA is obtained by
juxtaposition of a segments equal to `). If b D 1, we associate a1 D a to the point A.
If b > 1, we divide OA into b equal line segments, by plotting b 1 points on OA;
from these b 1 points, we let B denote the one closest to O, and associate ab to the
point B. It is not difficult to show (cf. Problem 1) that this geometric construction is
consistent, in the sense that, by changing ab for another equivalent fraction, we get
the same point B on r. Moreover, an analogous construction can obviously be made
for the negative rationals, marked on the negative half-line.
By proceeding as described in the previous paragraph, it happens that there are
lots of points on r which are not associated to any rational number. In order to give a
simple example, let A be the point associated to the number 1, and construct a square
a
b
a
...
O B
Fig. 1.1 plotting rationals on a real line.
A
r
16
1 The Set of Real Numbers
C
O
B
A E
r
Fig. 1.2 a point that doesn’t represent any rational number.
OABC, as shown in Figure 1.2. By using a compass, mark on the positive half-line a
9
10
point E such that OE D
p OB. Since OA D 1,pit follows from Pythagoras Theorem
that OE D OB D 2. However, since 2 is irrational (cf. Proposition 1.7 or
Problem 7, page 15), we conclude that E is not associated to any rational number.
At this point, two natural questions pose themselves: is it possible to mark on r
all of the real numbers? If we assume that the answer to the previous question is
yes, then, after we mark all reals on r, will there be unmarked points on r? One of
the axioms of Plane Euclidean Geometry11 , stated below, assures that the answers
to these two questions are, respectively, yes and no.
Axiom 1.9 There is a one-to-one correspondence between the points on an
Euclidean line r onto the set of real numbers, which is completely determined
by the following choices:
(a) A point O on r, to represent the real number 0.
(b) A half-line, among those that O determines on r, where we mark the positive
reals.
(c) A point on the half-line of item (b), to represent the number 1.
If we fix choices on a line r as specified by the above axiom, we say that r is the
real line (cf. Figure 1.3).
For further use, we need the following definition.
Definition 1.10 For given real numbers a < b, we set12 :
(i) Œa; b D fx 2 RI a x bg.
9
Pythagoras of Samos was one of the greatest mathematicians of classical antiquity. The theorem
that bears his name was already known to babylonians, at least two thousand years before he was
born; nevertheless, Pythagoras
was the first one to prove it. It is also attributed to him the first proof
p
of the irrationality of 2.
10
We recall that Pythagoras’ Theorem, one of the most celebrated (and important) results of Plane
Euclidean Geometry, says that, in every right triangle, the square of the length of the hypotenuse
equals the sum of the squares of the lengths of the legs. For two different proofs of it, see chapters 4
and 5 of [4].
11
For an axiomatic construction of plane Euclidean Geometry, we refer the reader to [16].
12
We call the reader’s attention to the less common notations Œa; bŒ instead of Œa; b/, a; b instead
of .a; b, a; bŒ instead of .a; b/, Œa; C1Œ instead of Œa; C1/ and 1; aŒ instead of .1; a/.
1.4 The Geometric Representation
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
17
Œa; b/ D fx 2 RI a x < bg.
.a; b D fx 2 RI a < x bg.
.a; b/ D fx 2 RI a < x < bg.
Œa; C1/ D fx 2 RI a xg.
.a; C1/ D fx 2 RI a < xg.
.1; a D fx 2 RI x ag.
.1; a/ D fx 2 RI x < ag.
An interval in R is the set R itself, or a subset of R of any of the seven types
above. Observe that, in the real line, an interval corresponds to a line segment
(perhaps with the exclusion of one or both of its end points), to a half-line (perhaps
with the exclusion of its real end point), or even to the whole real line.
Remarks 1.11
i. We stress that the symbols C1 and 1 (one respectively reads plus infinite
and minus infinite) do not represent real numbers. They merely serve to point out
that, in each of the items (v), (vi), (vii), (viii) above, the corresponding intervals
do contain all reals greater than or equal to (resp. greater than), or less than or
equal to (resp. less than) a.
ii. According to the previous definition, we shall denote R D .1; C1/.
Given real numbers a < b, we say that a and b are the endpoints and that b a is
the length of each one of the intervals of items (i) to (iv), in the previous definition.
In this case, we also say that those intervals have finite length. Analogously, the
real number a is the (only) endpoint of each of the intervals of items (v) to (viii),
which have infinite (i.e., not finite) lengths. An interval in R is bounded provided
it has finite length; otherwise, the interval is said to be unbounded. In particular,
the bounded intervals of R are precisely those of items (i) to (iv), in the previous
definition.
If a bounded interval has endpoints a < b, we shall say that it is closed, closed
on the left, closed on the right or open when such an interval is respectively equal
to Œa; b, Œa; b/, .a; b or .a; b/. Alternatively, we say that Œa; b/ is open on the right
and .a; b is open on the left. Finally, infinite intervals are named accordingly, by
using obvious extensions of the above terminologies. Figure 1.3 shows the interval
Œa; b/,
0
a
Fig. 1.3 the open on the right interval Œa; b/.
b
R
18
1 The Set of Real Numbers
Problems: Section 1.4
1. * With respect to the geometric interpretation of rational numbers, discussed at
section 1.4, let the fractions ab and dc be given, where a; b; c; d 2 N. If ab D dc ,
explain why the construction given in the text associates both these fractions to
the same point of the real line.
Chapter 2
Algebraic Identities, Equations and Systems
The rest of this volume, up to Chap. 5, approaches and develops several tools
necessary for an adequate presentation of the material in volumes 2 and 3. We
start by studying, in this chapter, some important algebraic identities, equations and
systems of equations.
2.1 Algebraic Identities
Through the rest of these notes, we refer to a varying real number as a real
variable.1 In general, real variables will be denoted by lower case Latin letters,
for example a, b, c, x, y, z etc (an important exception to this usage is mentioned in
the next paragraph).
An algebraic expression, or simply an expression, is a real number formed
from a finite number of real variables, possibly with the aid of one or more
algebraic operations, i.e., additions, subtractions, multiplications, divisions, power
computations and root extractions (whenever the results of these operations make
sense in R). In particular, every real variable can be seen as an algebraic expression.
For another example,
xC
p
p
y x2 z
C 3 5 x2 yz3 x4
yz
is an algebraic expression which makes sense for all reals x; y; z, such that y > 0
and z ¤ 0 (recall that Problem 1, page 14, assures that we can extract roots of odd
index of any real number). We shall denote algebraic expressions by upper case latin
letters, as E, F etc.
1
In [5], we shall have the opportunity to consider complex variables.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_2
19
20
2 Algebraic Identities, Equations and Systems
We say that an algebraic expression E is a monomial if E is a product of
a given nonzero real number by powers of its variables, each of which having
nonnegative integer exponents. Thus, the monomials in the real variables x and y
are the expressions of the form axk yl , where a ¤ 0 is a given real number and
k; l 0 are nonnegative integers (here, we adopt the convention that xk D 1
whenever k D 0, and yl D 1 whenever l D 0—see Problem 4, page 11). For an
arbitrary monomial, the given nonzero real number that plays the role of a in axk yl
is called its coefficient. Hence, the monomials in x; y with coefficient 2 are those of
one of the forms
2; 2x 2y; 2x2 ; 2xy; 2y2 ; 2x3 ; 2x2 y; 2xy2 ; 2y3 etc:
A polynomial expression or simply a polynomial is (an expression that is) a
finite sum of monomials, as, for instance,
2 C 3xy p 2
5x yz:
The coefficients of a polynomial are the coefficients of its monomials.
Let E and F be algebraic expressions. We say that equality E D F is an algebraic
identity provided it is true for all possible values of the involved real variables.
In order to give a relevant example, let us consider the algebraic expression E D
.x C y/2 . The elementary properties of the operations of addition and multiplication
of real numbers (i.e., commutativity and associativity of addition and multiplication,
as well as distributivity of multiplication with respect to addition) give
E D .x C y/.x C y/ D x.x C y/ C y.x C y/
D .x2 C xy/ C .yx C y2 /
D x2 C 2xy C y2 ;
for all values of the real variables x and y. Therefore, by setting F D x2 C 2xy C y2 ,
we obtain the algebraic identity E D F, i.e.,
.x C y/2 D x2 C 2xy C y2 ;
(2.1)
to which we refer, from now on, as the formula for the square of a sum of two real
numbers.
The following proposition collects some important algebraic identities, which the
reader must keep for future use.
Proposition 2.1 For all x; y; z 2 R, we have:
(a) x2 y2 D .x y/.x C y/.
(b) .x ˙ y/2 D x2 ˙ 2xy C y2 .
(c) x3 ˙ y3 D .x ˙ y/.x2 xy C y2 /.
2.1 Algebraic Identities
21
(d) .x ˙ y/3 D x3 ˙ y3 ˙ 3xy.x ˙ y/.
(e) .x C y C z/2 D x2 C y2 C z2 C 2xy C 2xz C 2yz.
Proof We let the proofs of items (a)–(c) as exercises (see Problem 1), observing that
the identity of item (b), with the C sign, was established in (2.1). In item (d), let us
prove the identity for .x C y/3 ; that for .x y/3 is totally analogous: by invoking the
distributivity of the multiplication with respect to addition, as well as identity (2.1),
we get
.x C y/3 D .x C y/.x C y/2 D x.x C y/2 C y.x C y/2
D x.x2 C 2xy C y2 / C y.x2 C 2xy C y2 /
D .x3 C 2x2 y C xy2 / C .x2 y C 2xy2 C y3 /
D x3 C y3 C 3x2 y C 3xy2
D x3 C y3 C 3xy.x C y/:
In order to get the result of item (e), we apply that of item (b), with x C y in the
place of x and z in the place of y:
.x C y C z/2 D Œ.x C y/ C z2
D .x C y/2 C 2.x C y/z C z2
D .x2 C 2xy C y2 / C 2.xz C yz/ C z2
D x2 C y2 C z2 C 2xy C 2xz C 2yz:
t
u
The reader has certainly noticed that, in the previous proposition, one either has:
(i) an identity of the form E D F, where E is a product of (at least two) polynomials
and F is the sum of monomials we get from expanding the products in E (this is the
case of the identities of items (b), (d) and (e)); or else (ii) an identity of the form
E D F, where E is a polynomial and F is a product of (at least two) polynomials (as
in items (a) and (c) of the previous proposition). In case (ii), we shall sometimes say
that F is a factorisation of E, or that it is obtained by factoring out expression E.
The coming examples will give us an idea on how to apply the identities collected
in the previous proposition to solve several interesting problems.
Example 2.2 Let x; y; z be real numbers, not all zero, such that xCyCz D 0. Explain
why xy C xz C yz ¤ 0 and, then, compute all possible values of the expression
x2 C y2 C z2
:
xy C yz C zx
22
2 Algebraic Identities, Equations and Systems
Solution Squaring both sides of x C y C z D 0, it follows from item (e) of
Proposition 2.1 that x2 C y2 C z2 C 2.xy C xz C yz/ D 0. If xy C xz C yz D 0, we
would have x2 Cy2 Cz2 D 0, and a simple extension of Corollary 1.5 (cf. Problem 2,
page 10) would give us x D 0, y D 0 and z D 0, contradicting our hypotheses.
Therefore, xyCxzCyz ¤ 0, and it follows from x2 Cy2 Cz2 D 2.xyCxzCyz/ that
x2 C y2 C z2
D 2:
xy C yz C zx
t
u
Our next example shows how to use the algebraic identities we know so far to
prove inequalities.2
Example 2.3 (Poland) For given positive real numbers a and b, prove that 4.a3 C
b3 / .a C b/3 .
Proof By expanding the right hand side with the aid of item (d) of Proposition 2.1,
it is immediate to see that the inequality we want to prove is equivalent to a3 C b3 a2 b C ab2 . It now suffices to see that
a3 C b3 a2 b ab2 D a3 a2 b C b3 ab2 D a2 .a b/ b2 .a b/
D .a2 b2 /.a b/ D .a C b/.a b/.a b/
D .a C b/.a b/2 0;
for a C b > 0 and .a b/2 0.
t
u
We now generalize Example 1.8.
p
ab is irrational.
Example 2.4
(Austria)
Let
a
and
b
be
positive
rationals,
such
that
p
p
Prove that a C b is also irrational.
p
p
Proof By contraposition,
suppose that r D a C b were a rational number. Then,
p
r2 D a C b C 2 ab would also be rational. However, in such a case, we would have
p
r2 a b
;
ab D
2
which would be a rational number too, for, in the right hand side of the above
equality, both the numerator and the denominator are rational numbers.
t
u
Example 2.5 (Canada) For each natural number n, prove that
n.n C 1/.n C 2/.n C 3/
is never a perfect square.
2
We will undertake a thorough discussion of inequalities in Chap. 5 and Sects. 9.7 and 10.8.
2.1 Algebraic Identities
23
Proof Letting p D n.n C 1/.n C 2/.n C 3/, we have
p D Œn.n C 3/Œ.n C 1/.n C 2/
D .n2 C 3n/Œ.n2 C 3n/ C 2
D .n2 C 3n/2 C 2.n2 C 3n/
D Œ.n2 C 3n/2 C 2.n2 C 3n/ C 1 1
D Œ.n2 C 3n/ C 12 1:
If we set m D n2 C 3n C 1, we have m > 1 and, hence,
p D m2 1 > m2 2m C 1 D .m 1/2 :
Therefore, p is situated between the consecutive perfect squares .m 1/2 and m2 , so
that it cannot be, itself, a perfect square.
t
u
Apart from the algebraic identities collected in Proposition 2.1, another frequently useful one is that given by the equality
.x y/.x z/ D x2 .y C z/x C yz:
(2.2)
Observe that, at the right hand side of the above expression, both the sum S D y C z
and the product P D yz of y and z do appear. An expression of the form x2 Sx C P,
where S and P represent the sum and the product of two numbers or expressions, is
called a second degree trinomial in x. Hence, writing (2.2) backwards, we can also
see it as giving a factorisation for the second degree trinomial x2 Sx C P, where
S D y C z and P D yz:
x2 Sx C P D .x y/.x z/:
(2.3)
The above factorisation is sometimes called Viète’s formula, in honor of the French
mathematician François Viète.3
The following example shows us how to apply Viète’s formula.
Example 2.6 (Soviet Union) Let a; b and c be pairwise distinct real numbers. Show
that the number
a2 .c b/ C b2 .a c/ C c2 .b a/
is always different from zero.
3
François Viète, French mathematician of the XVI century. By his pioneerism in the usage of letters
to represent variables, Viète is sometimes called the father of modern Algebra.
24
2 Algebraic Identities, Equations and Systems
Proof Letting S denote the given number, we have
S D a2 .c b/ C b2 a b2 c C c2 b c2 a
D a2 .c b/ C .b2 a c2 a/ C .c2 b b2 c/
D a2 .c b/ C a.b C c/.b c/ C bc.c b/
D .c b/Œa2 a.b C c/ C bc
D .c b/.a b/.a c/;
where we used (2.3) in the last equality. Now, it follows from a ¤ b, b ¤ c and
c ¤ a that a b; c b; a c ¤ 0, so that S ¤ 0.
t
u
A useful variant of Viète’s formula is the factorisation for the expression x2 C
Sx C P, where, as before, S D y C z and P D yz:
x2 C Sx C P D .x C y/.x C z/:
(2.4)
If we change S, y and z in (2.3) respectively by S, y and z, we immediately see
that (2.4) is indeed equivalent to that factorisation.
The next example uses (2.4) to get yet another algebraic identity, which will be
further applied in a number of places, both in this volume as well as in [4] and [5].
Example 2.7 For all x; y; z 2 R, we have
.x C y C z/3 D x3 C y3 C z3 C 3.x C y/.x C z/.y C z/:
(2.5)
Proof Applying item (d) of Proposition 2.1 twice, first with x C y in place of x and
z in place of y, we successively get
.x C y C z/3 D Œ.x C y/ C z3
D .x C y/3 C z3 C 3.x C y/zŒ.x C y/ C z
D x3 C y3 C 3xy.x C y/ C z3 C 3.x C y/Œ.x C y/z C z2 /
D x3 C y3 C z3 C 3.x C y/Œxy C .x C y/z C z2 D x3 C y3 C z3 C 3.x C y/.y C z/.x C z/;
where, in the last equality, we have used the variant (2.4) of Viète’s formula.
Problems: Section 2.1
1. * Prove the other items of Proposition 2.1.
t
u
2.1 Algebraic Identities
25
2. If m C n C p D 6, mnp D 2 and mn C mp C np D 11, compute all possible
p
m
n
values of np
C mp
C mn
.
2
2
3. Let a and b be nonzero real numbers, such that a ¤ b; 1. If ba D 1b
,
1a
compute all possible values of 1a C 1b .
4. Given positive real numbers x and y, simplify the expression
2
1 xy
p
p
p :
. x y/2 C 2 xy
5. For x; y; z ¤ 0, such that y C z ¤ 0, simplify the expression
.x3 C y3 C z3 /2 .x3 y3 z3 /2
:
yCz
6. Let a and b be real numbers such that ab D 1 and a ¤ b. Simplify the
expression
a 1a b C 1b
:
a2 b2
7. Let x and y be natural numbers such that x2 C 361 D y2 . Find all possible
values of x.
8. Real numbers a and b are such that a C b D m and ab D n. Compute the value
of a4 C b4 in terms of m and n.
2
9. If a2 C b2 D 1, find all possible values of 13.ab/
.
a6 Cb6
10. (EKMC) Let p
a, b, c and d be real numbers such that a2 Cb2 D 1 and c2 Cd 2 D1.
If acCbd D 23 , compute the value of adbc, provided it is a positive number.
11. (Brazil) Find all natural numbers x and y such that x C y C xy D 120.
12. * Given positive distinct real numbers x and y, prove that the following
rationalisations4 are valid:
(a)
(b)
(c)
p
p
x y
xy .
p
p
3 2 p
x 3 xyC 3 y2
1
p
p
D
.
3 x˙ 3 y
x˙y
p
3 x˙ p
3
1
p D x˙y y .
p
3 2 p
x 3 xyC 3 y2
p 1p
x˙ y
D
4
In an informal way, one can think of a rationalisation as a way of clearing roots from
denominators.
26
2 Algebraic Identities, Equations and Systems
13. * For a natural number n > 1, show that
2
14. Rationalise
15. Rationalise
that
p
p
p
p
p
1
nC1 n < p <2
n n1 :
n
p1 p .
2C 2C 3
p 1p
. More
2C 3 3
precisely, obtain integers a, b, c, d, e, f and g such
p
p
p
p
1 p
3
3
p
D Œ.a 2 C b/ C .c 2 C d/ 3 C .e 2 C f / 9:
3
g
2C 3
1
16. Let x, y and z be nonzero real numbers, such that x C y C z D 0. Explain why
the sum of any two of them is also nonzero, and compute all possible values
of each of the following expressions:
(a)
(b)
x2
.yCz/2
x3
.yCz/3
C
C
y2
.xCz/2
y3
.xCz/3
C
C
z2
.
.xCy/2
z3
.
.xCy/3
17. Let a and b be distinct integers. Find, in terms of a and b, the quotient of the
division of a64 b64 by .a C b/.a2 C b2 /.a4 C b4 /.a8 C b8 /.a16 C b16 /.
18. * Given n > 1 integer and a; b 2 R, prove that the following factorizations are
valid:
(a) an bn D .a b/.an1 C an2 b C an3 b2 C C bn1 /.
(b) an C bn D .a C b/.an1 an2 b C an3 b2 C bn1 /, provided n is odd.
19. Write x4 C 4y4 as a product of two non constant polynomials in x and y, both
having integer coefficients.
20. (Canada) Let a; b; c 2 Z. Prove that 6 divides a C b C c if and only if 6 divides
a 3 C b 3 C c3 .
21. (Canada) If a, b and c are real numbers for which a C b C c D 0, show that
a3 C b3 C c3 D 3abc.
22. Prove the double radical formula, also known as Bhaskara’s formula5 : for all
positive real numbers a and b, such that a2 b, one has
q
p
a˙ bD
s
aC
s
p
p
a2 b
a a2 b
˙
:
2
2
5
In honor of the Indian mathematician of the XII century Bhaskara II, also known as
Bhaskaracharya (Bhaskara, the professor). The idea behind Bhaskara’s formula is that, if a and
b are
for which a2 b is a perfect square, then his formula provides a simpler expression
qnaturals
p
for a ˙ b.
2.2 The Modulus of a Real Number
27
23. Show that there do not exist nonzero real numbers x, y and z such that x C y C
z ¤ 0 and
1
1
1
1
D C C :
xCyCz
x
y
z
24. (Soviet Union) Let a, b and c be pairwise distinct rationals. Prove that
1
1
1
C
C
.b c/2
.c a/2
.a b/2
is the square of a rational.
p
p
p
25. (TT)
Let a, b and c be distinct rationals. If 3 a C 3 b 2 Q, prove that 3 a;
p
3
b 2 Q.
26. (TT) Let a, b, c, d, e and f be real numbers such that a C b C c C d C e C f D 0
and a3 C b3 C c3 C d3 C e3 C f 3 D 0. If no two of them are opposite to each
other, prove that
.a C c/.a C d/.a C e/.a C f / D .b C c/.b C d/.b C e/.b C f /:
27. (Poland) For positive integers a b, do the following items:
(a) Show that b3 < b3 C 6ab C 1 < .b C 2/3 .
(b) Find all such a and b for which both a3 C 6ab C 1 and b3 C 6ab C 1 are
perfect cubes.
2.2 The Modulus of a Real Number
We start this section by recalling the definition of modulus of a real number, a
concept which will be important in a number of places hereafter.
Definition 2.8 For x 2 R, the modulus of x, denoted jxj, is defined as
jxj D
x; if x 0
:
x; if x < 0
As
p sincep5 < 0, we have j 5j D .5/ D 5; analogously,
p an example,
j 3j D . 3/ D 3 etc. More generally, an immediate consequence of the
definition is that jxj 0 for all x 2 R, with equality if and only if x D 0. Moreover,
one always has
x jxj D j xj;
28
2 Algebraic Identities, Equations and Systems
with equality if and only if x 0. Note also that
jxj D
p
x2 D max fx; xg:
(2.6)
The simplest modular equation is the equation
jx aj D b;
where a and b are given real numbers. Since jx aj 0, such an equation does
not admit roots when b < 0. On the other hand, when b 0, it follows from the
definition of modulus that one must have either x a D b or x a D b, from
where we get the roots
x D a C b; a b:
The coming example shows how to solve a more elaborate equation in a single
variable, involving the concept of modulus of a real number.
Example 2.9 Solve equation jx C 1j C jx 2j C jx 5j D 7.
Solution First of all, note that
jx C 1j D
x C 1; if x 1
;
x 1; if x < 1
jx 2j D
x 2; if x 2
x C 2; if x < 2
and
jx 5j D
x 5; if x 5
:
x C 5; if x < 5
Now, the conjunction of the conditions x < 1 or x 1, x < 2 or x 2,
x < 5 or x 5 partitions the real line into the intervals .1; 1/, Œ1; 2/, Œ2; 5/
e Œ5; C1/. Hence, in order to simplify the left hand side of the given equation, we
separately consider x as varying in each one of these intervals. We thus obtain
8
3x C 6; if
ˆ
ˆ
<
x C 8; if
jx C 1j C jx 2j C jx 5j D
ˆ x C 4; if
:̂
3x 6; if
x < 1
1 x< 2
:
2x<5
x5
2.2 The Modulus of a Real Number
29
|x|
Fig. 2.1 Modulus of a real
number
x
0
R
Finally, note that
• 3x C 6 D 7 , x D 13 ; however, since the condition 13 < 1 is not satisfied,
there are no roots in this case.
• x C 8 D 7 , x D 1; since the condition 1 1 < 2 is satisfied, x D 1 is a
root of the equation.
• x C 4 D 7 , x D 3; since the condition 2 3 < 5 is satisfied, x D 3 is also a
root of the equation.
• 3x 6 D 7 , x D 13
; since the condition 13
5 is not satisfied, there are no
3
3
roots in this case.
Therefore, the solution set of the given equation is S D f1; 3g.
t
u
Back to the study of the properties of modulus, let us represent the real numbers
as points in the real line. It is easy to see that jxj is simply the distance from (the
point that represents) x to (the one representing) 0 (cf. Fig. 2.1). More generally,
given x; y 2 R, we can look at jx yj as the distance from the points x and y in the
real line. In fact, since jx yj D jy xj, we can suppose that x y. Then,
jx yj D y x D distance from x to y in the real line.
In the above reasoning, if we do not wish to consider which of x and y is the
greatest one, we can write
jx yj D maxfx; yg minfx; yg;
(2.7)
fx; yg D fmaxfx; yg; minfx; ygg:
(2.8)
for
There simple remarks suffice to consider the following hard example.
Example 2.10 (Yugoslavia) Let n 2 N and M D f1; 2; 3; : : : ; 2ng. Also, let M1 D
fa1 ; a2 ; : : : ; an g and M2 D fb1 ; b2 ; : : : ; bn g be subsets of M such that a1 < a2 <
< an and b1 > b2 > > bn . If M1 [ M2 D M and M1 \ M2 D ;, prove that
ja1 b1 j C ja2 b2 j C C jan bn j D n2 :
Proof It follows from (2.8) that
n
n
[
[
fmaxfai ; bi g; minfai ; bi gg D
fai ; bi g D f1; 2; 3; : : : ; 2ng:
iD1
iD1
30
2 Algebraic Identities, Equations and Systems
Also, (2.7) gives
ja1 b1 j C ja2 b2 j C C jan bn j D
D .maxfa1 ; b1 g C maxfa2 ; b2 g C C maxfan ; bn g/
.minfa1 ; b1 g C minfa2 ; b2 g C C minfan ; bn g/:
On the other hand, given integers 1 k; l n, with k ¤ l, we have
k > l ) maxfak ; bk g ak > al minfal ; bl g
and
k < l ) maxfak ; bk g bk > bl minfal ; bl g:
Therefore,
fmaxfa1 ; b1 g; maxfa2 ; b2 g; : : : ; maxfan ; bn gg D fn C 1; n C 2; : : : ; 2ng
and
fminfa1 ; b1 g; minfa2 ; b2 g; : : : ; minfan ; bn gg D f1; 2; : : : ; ng:
Finally, the above relations give
ja1 b1 j C ja2 b2 j C C jan bn j D
D ..n C 1/ C .n C 2/ C C 2n/ .1 C 2 C C n/
D n2 C .1 C 2 C C n/ .1 C 2 C C n/ D n2 :
t
u
We continue our study of the concept of modulus with the following important
result, which is known in mathematical literature as the triangle inequality.6
Proposition 2.11 For all real numbers a and b, we have
ja C bj jaj C jbj:
(2.9)
Moreover, if a; b ¤ 0, then equality holds if and only if a and b have the same sign.
6
At the end of Sect. 5.2, we will give an explanation of why (2.9), as well as the coming
inequality (4.6), are called triangle inequality.
2.2 The Modulus of a Real Number
31
Proof Since ja C bj and jaj C jbj are both nonnegative, we have
ja C bj jaj C jbj , ja C bj2 .jaj C jbj/2
, .a C b/2 jaj2 C jbj2 C 2jabj
, 2ab 2jabj;
which is clearly true. From the computations above it also follows that ja C bj D
jaj C jbj if and only if ab D jabj; in turn, this happens if and only if ab 0. Finally,
if a; b ¤ 0, then we have the equality if and only if ab > 0.
t
u
Corollary 2.12 For all real numbers a and b, we have
jjaj jbjj ja bj:
Moreover, if a; b ¤ 0, then the equality holds if and only if a and b have the same
sign.
Proof Applying the triangle inequality to a b in place of a, we get
jaj D j.a b/ C bj ja bj C jbj
and, hence, jaj jbj ja bj. Repeating the above argument with the roles of a
and b interchanged, it follows that jbj jaj ja bj.
Now, since ja bj jaj jbj; jbj jaj, we get
ja bj maxfjaj jbj; jbj jajg D jjaj jbjj;
where we used (2.6) in the last equality.
Equality happens if and only if we have equality in at least one of the triangular
inequalities jaj ja bj C jbj or jbj jb aj C jaj. If a; b ¤ 0 and, say, equality
holds in the first one of theses inequalities, i.e., if jaj D ja bj C jbj, then the
condition for equality in Proposition 2.11 assures that we must have .a b/b 0,
or, which is the same ab b2 . In particular, we must have ab > 0.
Conversely, suppose that ab > 0, and let us show that equality holds. There are
two possibilities: a; b > 0 or a; b < 0. Suppose that a; b < 0 (the other case can
be treated in a similar way). Then, jaj D a and jbj D b, so that jjaj jbjj D
j.a/ .b/j D jb aj D ja bj.
t
u
Given real numbers a, b and c and applying triangle inequality twice, we get
ja C b C cj ja C bj C jcj jaj C jbj C jcj;
(2.10)
Hence, we have the inequality
ja C b C cj jaj C jbj C jcj;
(2.11)
32
2 Algebraic Identities, Equations and Systems
which is the analogous of (2.9) for three real numbers, instead of two. Therefore,
we shall also refer to this last inequality as the triangle inequality.
If a; b; c ¤ 0 and we have equality in (2.11), then we should also have equality
in all inequalities in (2.10). In particular, we have ja C bj jaj C jbj, and it follows
from Proposition 2.11 that a and b have equal signs. Since we can also reach (2.11)
by writing
ja C b C cj jaj C jb C cj jaj C jbj C jcj;
we conclude, analogously, that b and c should also have equal signs.
Conversely, if a, b and c all have equal signs, say a; b; c < 0 (the case a; b; c > 0
is completely analogous), then a C b C c < 0, so that
ja C b C cj D .a C b C c/ D .a/ C .b/ C .c/ D jaj C jbj C jcj:
Hence, we have just shown that equality holds in (2.11) if and only if a, b and c
have equal signs.
As we shall see in Sect. 4.1 (cf. Problem 7, page 96), inequalities (2.9) and (2.10)
can be easily generalized for n real numbers. For the time being, we end this section
with the following
Example 2.13 Prove that, for every x 2 R, we have
jx 1j C jx 2j C jx 3j C C jx 10j 25:
Proof It follows from the triangle inequality that
jx aj C jx bj D jx aj C jb xj j.x a/ C .b x/j D jb aj:
Hence, grouping the summands at the left hand side in pairs, we get,
jx 1j C jx 10j j10 1j D 9I
jx 2j C jx 9j j9 2j D 7I
jx 3j C jx 8j j8 3j D 5I
jx 4j C jx 7j j7 4j D 3I
jx 5j C jx 6j j6 5j D 1:
Adding these inequalities, we obtain that of the statement.
t
u
2.3 A First Look at Polynomial Equations
33
Problems: Section 2.2
1. * Given real numbers a and b, show that
8
< ;; if b < 0
fx 2 RI jx aj < bg D fag; if b D 0
:
:
.a b; a C b/; if b > 0
Do the same for jx aj b, jx aj > b and jx aj b.
2. * Prove that, for all x; y 2 R, one has jxyj D jxj jyj.
3. Solve, for x 2 R, the following equations:
(a) jxj D x 6.
(b) jx C 1j C jx 2j C jx 5j D 4.
4. Solve, for x 2 R n f0; 1g, equation jxj
D jx1j
.
x
x1
5. Let a, b and c be given real numbers, with a < b. Discuss, in terms of a, b and c,
the number of solutions of the equation
jx aj C jx bj D c:
6. (Mexico) Let r be a nonnegative rational number. Prove that
ˇ
ˇ
p
ˇr C 2 p ˇ 1
ˇ
ˇ < jr 2j:
2
ˇr C 1
ˇ 2
(This inequality shows that the rational number
better as r does it.)
7. Prove that:
rC2
rC1
approximates
p
2 twice as
(a) If 0 x y, then
(b) If a; b 2 R, then
y
x
1Cy
.
1Cx
jaj
jbj
1Cjaj C 1Cjbj
jaCbj
1CjaCbj .
8. Let n > 1 be an integer. Prove that
jx 1j C jx 2j C C jx 2nj n2
for every real x, with equality for infinitely many values of x.
2.3 A First Look at Polynomial Equations
In this section we study some particular types of polynomial equations, postponing
a much deeper look to [5].
34
2 Algebraic Identities, Equations and Systems
In general, a polynomial equation of degree n is an equation of the form
an xn C an1 xn1 C C a1 x C a0 D 0;
(2.12)
where n 1 is an integer and a0 ; a1 ; : : : ; an are given real numbers,7 with an ¤ 0.
The simplest kind of such an equation is the first degree equation ax C b D 0,
where a and b are given real numbers and a ¤ 0. As the reader certainly knows, we
have
b
ax C b D 0 , ax D b , x D ;
a
so that ba is its only root.
The second simplest kind of polynomial equation is the second degree equation
ax2 C bx C c D 0;
(2.13)
where a, b and c are given real numbers, with a ¤ 0. For reasons that will soon
be clear, the left hand side of (2.13) is also known as the second degree trinomial
associated to Eq. (2.13).
In order to solve (2.13), we let (one reads delta) denote the real number
D b2 4ac;
and call it the discriminant of the equation (or of the associated trinomial). As we
shall see in a moment, the sign of discriminates whether or not the equation has
real roots. To this end, we need the following auxiliary result.
Lemma 2.14 Given a; b; c 2 R, with a ¤ 0, one has
#
"
b 2
ax C bx C c D a x C
2 :
2a
4a
2
(2.14)
This algebraic identity is called the canonical form of the second degree trinomial
ax2 C bx C c.
7
Here, we are using the concept of a sequence of real numbers. For further details in this respect,
we refer the reader to Chap. 3.
2.3 A First Look at Polynomial Equations
35
Proof It suffices to see that
c
b
ax2 C bx C c D a x2 C x C
a
a
b2
b
b2
D a x2 C x C 2 2 C
a
4a
4a
2 b
b2
b
D a x2 C x C 2 2
a
4a
4a
#
"
2
b
2 :
D a xC
2a
4a
c
a
C
4ac
4a2
t
u
Remark 2.15 The idea of adding and subtracting a certain summand out of a given
algebraic expression in order to complete a square, as was done right after the
second equality in the proof of the previous lemma, is very important and should
be learned as a kind of algebraic trick that will be useful in a number of places,
here as well as in [4] and [5]. Later in this chapter, we shall see other interesting
applications of such a technique.
Proposition 2.16 Let a, b and c be given real numbers, with a ¤ 0.
(a) The equation ax2 C bx C c D 0 has
real rootspif and only if 0. Moreover,
p
if this is so, then its roots are b2a and bC2a .
(b) If 0, then the sum S and the product P of the roots of ax2 C bx C c D 0 are
given by S D ba and P D ac .
Proof
(a) It follows from (2.14) that
b 2
D 2:
ax C bx C c D 0 , x C
2a
4a
2
Since x C
b 2
2a
(2.15)
0 for all x 2 R, if the equation has real roots, then one must
b
D˙
have 0. In this case, it transpires from (2.15) that x C 2a
(a) follows.
(b) It suffices to compute
p
p
b b C b
C
D
2a
2a
a
p
2a ,
and item
36
2 Algebraic Identities, Equations and Systems
and
b p b C p
.b/2 c
D
D :
2
2a
2a
4a
a
Remarks 2.17
t
u
p
i. When 0, formulae b˙2a for the roots of ax2 C bx C c D 0 are known as
Bhaskara’s formulae.
ii. The formulas of item (b) are also known as Viète’s formulae.
iii. In the notations of item (a), if D 0 we say that ax2 C bx C c D 0 has two
equal roots.
The coming example shows how one can reduce a seemingly complicated
equation to a simpler one by means of a suitable substitution of variable.
Example 2.18 (Brazil) Find all real numbers x such that x2 C x C 1 D
156
.
x2 Cx
Solution By performing the substitution y D x2 Cx, we get the equation yC1 D 156
y
or, which is the same, y2 C y 156 D 0. For this last equation, since D 12 4.156/ D 625 D 252 , it follows that y D 1˙25
D 13 or 12. Thus, we have
2
reduced the original equation to the second degree equations x2 C x D 13 and
x2 C x D 12. For the first one, we have D 51 < 0, so that there are no real
roots. For the second, D 49 and, hence, x D 1˙7
D 4 or 3.
t
u
2
Our next example shows that it is sometimes more useful to algebraically
manipulate a second degree equation than to solve it explicitly.
p
p
Example 2.19 Find the numerical value of .3C 2/5 C.3 2/5 without expanding
the powers involved.
p
p
Proof Letting u D 3 C 2 and v D 3 2, we have u C v D 6 and uv D 7, so
that u and v are the roots of the equation x2 6x C 7 D 0. Therefore, making x D u
and x D v in this equation gives us u2 6u C 7 D 0 and v 2 6v C 7 D 0, or,
equivalently, u2 D 6u 7 and v 2 D 6v 7. Multiplying the first equality by uk and
the second one by v k , where k 0 is an integer, we get
ukC2 D 6ukC1 7uk and v kC2 D 6v kC1 7v k I
adding both of these, we finally arrive at
ukC2 C v kC2 D 6.ukC1 C v kC1 / 7.uk C v k /:
2.3 A First Look at Polynomial Equations
37
Writing the previous relation for k respectively equal to 0, 1, 2 and 3, we
successively get
u2 C v 2 D 6.u C v/ 7 2 D 6 6 14 D 22I
u3 C v 3 D 6.u2 C v 2 / 7.u C v/ D 6 22 7 6 D 90I
u4 C v 4 D 6.u3 C v 3 / 7.u2 C v 2 / D 6 90 7 22 D 386I
u5 C v 5 D 6.u4 C v 4 / 7.u3 C v 3 / D 6 386 7 90 D 1686:
t
u
For the next example, recall that if the sum and the product of two real numbers
are positive, then both numbers are also positive.
Example 2.20 Let p and q be given real numbers. If the equation x2 C px C q D 0
has real, positive and distinct real roots, show that the same is true for the equation
qx2 C .p 2q/x C .1 p/ D 0.
Proof Note initially that q ¤ 0, for otherwise the first equation would reduce to
x2 C px D 0, which has 0 as one of its roots, thus contradicting our hypotheses.
Now, let and 0 be, respectively, the discriminants of x2 C px C q D 0 and
qx2 C .p 2q/x C .1 p/. Let us first show that 0 > 0, which will guarantee that
the second equation has distinct real roots. Since x2 C px C q D 0 has distinct real
roots, we have D p2 4q > 0. Therefore,
0 D .p 2q/2 4q.1 p/
D p2 4q C 4q2
D C 4q2 > 0:
Finally, according to the paragraph that immediately precedes this example, in
order to show that the roots of qx2 C .p 2q/x C .1 p/ D 0 are positive, it suffices
to show that the sum S0 and the product P0 of them are both positive. In order to do
this, we recall that the roots of x2 C px C q D 0 are known to be positive, so that (by
Viète’s formulae) p > 0 and q > 0. Hence, again by Viète’s formulae, we have
S0 D
p
1
p
2q p
1p
D2C
> 0 and P0 D
D C
> 0;
q
q
q
q
q
t
u
as we wished to show.
We finish our discussion of second degree equations with the following important
remark: if a ¤ 0 and ax2 CbxCc D 0 has real roots ˛ and ˇ (not necessarily ˛ ¤ ˇ),
then we have the factorisation
ax2 C bx C c D a.x ˛/.x ˇ/:
(2.16)
38
2 Algebraic Identities, Equations and Systems
In fact, it follows from item (b) of Proposition 2.16 that, for every real x,
a.x ˛/.x ˇ/ D aŒx2 .˛ C ˇ/x C ˛ˇ
c
b
D a x2 xC
a
a
D ax2 C bx C c:
It is instructive to compare (2.16) to (2.2). The right hand side of (2.16) is called the
factorised form of the second degree trinomial ax2 C bx C c.
In turning to more general polynomial equations, it is natural to try to look at
those of degrees n D 3 and n D 4. In these cases, formulas have been built, in
terms of the coefficients of the equations, to compute their real roots, if any.8 As
professor I. Stewart teaches us in Chap. 4 of his very interesting book [25], such
formulas derive from the works of the Italian mathematicians Scipione del Ferro,
Girolamo Cardano, Niccolò Fontana (conhecido como Tartaglia) and Lodovico
Ferrari. However, they are too complicated to be useful, and for this reason we
shall not discuss them here. In order to help convincing the reader, let us just
mention that Cardano’s formula for the roots of the third degree polynomial equation
ax3 C bx2 C cx C d D 0 is the following9:
q
3
q
p
p
3
2
2
3
q C q C .r p / C q q2 C .r p2 /3 C p;
b
c
, q D p3 C bc3ad
and r D 3a
(however, see Problems 16, 21 and 22).
where p D 3a
6a2
For polynomial Eq. (2.12) of degree n 5, the Norwegian mathematician Niels
H. Abel10 and the French mathematician Évariste Galois,11 both of the XIX century,
proved that there exists no similar formula, built on the coefficients of the given
polynomial equation, that gives its real (or even complex!) roots. Well understood,
it doesn’t matter how smart someone is; they proved that it is impossible to find such
a formula, simply because it doesn’t exist! For a beautiful and elementary account
of the ideas involved, we refer the reader to [14].
Some particular kinds of polynomial equations of degrees 4 and 6, however,
are sufficiently simple to deserve some attention, specially because appropriate
8
As it happens, these formulas also give the complex roots of the corresponding equations.
However, we shall postpone any considerations involving complex numbers to [5].
9
Cf. http://www.math.vanderbilt.edu/~schectex/courses/cubic.
10
In his 27 years of life, Abel made several deep contributions to Mathematics, among which the
most famous one is perhaps the impossibility of solving general polynomial equations of degree 5.
11
In spite of his premature death, when he was only 21 years old, Galois is considered to be one
of the greatest mathematicians the world has ever seen. His work on the connection between the
solvability of polynomial equations of degrees n 5 and Group Theory constitute the foundations
of what is known today as Galois’ Theory, a branch of modern Algebra with applications to several
distinct areas of Mathematics.
2.3 A First Look at Polynomial Equations
39
variable substitutions immediately reduce them to second degree equations. Let us
first examine biquadratic, i.e., equations of the form
ax4 C bx2 C c D 0;
(2.17)
with a; b; c 2 R and a ¤ 0. The variable substitution y D x2 transforms it into
the second degree equation ay2 C by C c D 0, which we already know how to
solve in R. Therefore, for each nonnegative root y D p
˛ of this last equation, solving
equation x2 D ˛ gives us the pair of real roots x D ˙ ˛ of the original biquadratic
equation. Conversely, if x D ˇ is a real root of the given biquadratic equation, it is
immediate to see that y D ˇ 2 is a nonnegative root of the second degree equation
ay2 C by C c D 0. We have, thus, proved the following result.
Proposition 2.21 Given real numbers a, b and c, with a ¤ 0, the real
p roots of the
biquadratic equation ax4 C bx2 C c D 0 are the reals of the form ˙ ˛, where ˛ is
a nonnegative root of the second degree equation ay2 C by C c D 0.
In order to actually compute the real roots of a specific biquadratic equation,
instead of invoking the last proposition, it is usually much better to recall our
previous discussion, remembering that the variable substitution y D x2 does the
job of reducing it to a second degree equation.
Example 2.22 Find the real roots of the biquadratic equation x4 C 5x2 7 D 0.
Solution The variable substitution y D x2 leads us to the second degree equation
y2 C 5y 7 D 0, for which
D 53. Hence, the roots of this last equation are
p
p
y D 5˙2 53 . Since 52 53 < 0, the real roots of the original biquadratic equation
q
p
p
t
u
are the solutions of x2 D 5C2 53 , i.e., are the real numbers ˙ 5C2 53 .
Given n 2 N, we point out that we can discuss the problem of finding the roots
of a polynomial equation of the form
ax2n C bxn C c D 0
(2.18)
in a way quite similar to that used to study biquadratic equations. We refer the reader
to Problem 16 for the corresponding details.
Let us now examine reciprocal polynomial equations of degree 4, i.e., polynomial equations of degree 4 having the form
ax4 C bx3 C cx2 C bx C a D 0;
where a, b and c are given real numbers, with a ¤ 0. As Problem 28 shows, the
name reciprocal applies to a larger class of polynomial equations, and comes from
the fact that x 2 R n f0g is a real root of it if and only if 1x also is.
Initially, note that 0 is not a root of the equation above, for a ¤ 0. Therefore, a
real number x is a root of it if and only if it is a root of the equation
40
2 Algebraic Identities, Equations and Systems
ax2 C bx C c C
a
b
C 2 D 0;
x
x
(2.19)
which is obtained from the original equation by dividing both sides of the equality
by x2 . Rewrite the left hand side of the last equation above as
1
1
2
C c D 0:
a x C 2 Cb xC
x
x
Now, the idea is to perform the variable substitution y D xC 1x . In order to implement
it, let us first of all note that, according to (2.1), one has
1 2
1
y2 D x C
D x2 C 2 C 2 :
x
x
Hence, x2 C x12 D y2 2, so that solving (2.19) amounts to solving the second degree
equation
a.y2 2/ C by C c D 0:
(2.20)
However, the above discussion hides a subtlety: it is sure that every real root
x D ˛ of (2.19) generates the real root ˇ D ˛ C ˛1 of a.y2 2/ C by C c D 0.
Nevertheless, the converse statement is not true: not every real root y D ˇ of this
last equation does generate real roots x D ˛ of the initial reciprocal equation. In
fact, once we get a real root y D ˇ of a.y2 2/ C by C c D 0, in order to obtain the
possible real roots of the reciprocal equation corresponding to ˇ, we have to solve
in R the equation
xC
1
D ˇ;
x
or, which is the same, x2 ˇx C 1 D 0. Since the discriminant of this last equation
is
D ˇ 2 4;
it will have real root only if ˇ 2 4 0, i.e., only if jˇj 2.
As was the case with biquadratic equations, in order to actually find the real roots
of a given reciprocal equation of degree 4, instead of memorizing the result of the
variable substitution y D x C 1x , it is more profitable to follow the steps that led us
from (2.19) to (2.20). Let us see an example in this respect.
Example 2.23 Find all real roots of the reciprocal equation
2x4 C 5x3 C 6x2 C 5x C 2 D 0:
2.3 A First Look at Polynomial Equations
41
Solution Dividing both sides by x2 and grouping summands, we obtain
1
1
2
C 6 D 0:
2 x C 2 C5 xC
x
x
Making the variable substitution y D x C 1x , it follows that y2 D x2 C x12 C 2, so that
the given equation is equivalent to
2.y2 2/ C 5y C 6 D 0:
Since this second degree equation has real roots y D 2 and y D 12 , it follows that
the real roots of the original equation are the real roots of the equations x C 1x D 2
and x C 1x D 12 . The first of these equations is equivalent to x2 C 2x C 1 D 0 and,
hence, has two real roots, both equal to 1. The second is equivalent to 2x2 CxC2 D
0, which has discriminant D 15 < 0; therefore, it has no real roots.
t
u
Problems: Section 2.3
1. Let b and c be given real numbers, such that the equation x2 C bjxj C c D 0 has
real roots. Prove that the sum of these roots is always equal to 0.
2. Solve, for x 2 R, the following equations:
p
(a) p
x C x C 2p
D 10.
p
(b) x C 10 2x C p
3 D 1 3x.
(c) x2 C 18x C 30 D 2 x2 C 18x C 45.
p
3. (IMO) In each of the cases (a) A D 2, (b) A D 1 and (c) A D 2, find all x 2 R
for which we have
q
q
p
p
x C 2x 1 C x 2x 1 D A:
4. A math teacher composed three different quizzes. In the first one, he put
a second degree equation. In the second, he put almost the same equation,
changing just the coefficient of the monomial of second degree. Finally, in the
third quiz, once more he put almost the same equation of the first, this time
changing just the constant coefficient. It is known that the roots of the equation
of the second quiz are 2 and 3, and that those of the third one are 2 and 7.
Decide whether the second degree equation of the first quiz has real roots and,
if this is so, compute them.
5. Let a and b be two distinct, nonzero real numbers. If a and b are the roots of the
equation x2 C ax C b D 0, find all possible values of a b.
42
2 Algebraic Identities, Equations and Systems
6. Let ˛ and ˇ be the roots of x2 13x C 9 D 0, and a and b be real numbers such
that the equation x2 C ax C b D 0 has roots ˛ 2 and ˇ 2 . Compute the value of
a C b.
7. Equation x2 Cx1 D 0 has roots u and v. Find a second degree equation whose
roots are u3 and v 3 .
8. The roots of the equation x2 Sx C P D 0 are the real numbers ˛ and ˇ. Find
a second degree trinomial whose coefficients are expressions built on S and P
and whose roots are the real numbers ˛S C P and ˇS C P.
9. Use the theory of second degree
developed in this section to compute
p equations p
the value of the sum .7 C 4 3/5 C .7 4 3/5 .
10. If ˛ is a root of x2 x 1 D 0, find all possible values of ˛ 5 5˛.
11. For which integer values of m does the equation x2 C mx C 5 D 0 have integer
roots?
12. Show that, for every a; b; c 2 R, with a ¤ 0, the equation
1
1
1
C
D 2
xb
xc
a
has exactly two distinct
q
q real roots.
13. Solve equation x D x 1x C 1 1x in the set of real numbers.
14. Show that, for every real number a ¤ 0, the equations
ax3 x2 x .a C 1/ D 0 and ax2 x .a C 1/ D 0
have a common root.
15. (Soviet Union) Do the following items:
(a) For real x, write the number x3 3x2 C 5x in the form
a.x 1/3 C b.x 1/2 C c.x 1/ C d;
with a; b; c; d 2 Z.
(b) If x and y are real numbers such that x3 3x2 C5x D 1 and y3 3y2 C5y D 5,
compute all possible values of x C y.
16. * Let n 2 N and a; b; c 2 R, with a ¤ 0. Elaborate, for the equation ax2n C
bxn C c D 0, a discussion analogous to that made on the text for biquadratic
equations, and which led us to Proposition 2.21.
17. * Consider the polynomial equation of third degree x3 C ax2 C bx C c D 0,
where a, b and c are given real numbers, with c ¤ 0. If ˛ is a (nonzero) real
root of it, prove that there exist real numbers b0 and c0 such that we have the
factorisation
x3 C ax2 C bx C c D .x ˛/.x2 C b0 x C c0 /:
2.3 A First Look at Polynomial Equations
43
Then, conclude that the original polynomial equation has at most three (not
necessarily distinct) real roots. Moreover, if this is the case, and ˛, ˇ and are
its three real roots, show that
x3 C ax2 C bx C c D .x ˛/.x ˇ/.x /:
This result generalizes the factorised form (2.16) of a second degree trinomial
12
and is a particular case of the division algorithm.
. p
p
p
p
3
3
18. (a) Show that the real number ˛ D 2 C 5 C 2 5 is a root of the
equation x3 C 3x 4 D 0.
(b) Conclude that ˛ is a rational number.
19. * Establish the following version of Girard’s relations13 between roots and
coefficients of a polynomial equation of third degree: if the real numbers a0 , a1 ,
a2 and a3 (a3 ¤ 0) are such that the equation
a 3 x3 C a 2 x2 C a 1 x C a 0 D 0
has (not necessarily distinct) real roots x1 , x2 and x3 , then
8
a2
ˆ
< x1 C x2 C x3 D a3
x1 x2 C x1 x3 C x2 x3 D
:̂ x x x D a0
1 2 3
a3
a1
a3
:
(2.21)
20. Assume that the equation x3 3x C 1 D 0 has three real roots, say ˛, ˇ and .
Compute the values of ˛ 2 C ˇ 2 C 2 , ˛ 3 C ˇ 3 C 3 and ˛ 4 C ˇ 4 C 4 .
21. Still concerning the third degree polynomial equation x3 C ax2 C bx C c D 0,
with a; b; c 2 R, prove that there exists a real number d such that the variable
substitution y D x d transforms the given equation into one of the form
y3 C py C q D 0, for certain real numbers p and q.
22. Concerning the equation x3 11x C 16 D 0, do the following items:
(a) Substitute x D uCv and get an equivalent equation in the two real variables
u and v.
(b) Impose that uv D 11
(i.e., make uv equals 13 times the coefficient of x in
3
the original equation) and conclude that the equation in u and v of item (a)
3
is equivalent to u6 C 16u3 C 11
D 0.
3
(c) Find u and v, and conclude that one of the roots of the given equation is
s
3
12
p
8 C
s
1191
C
9
3
p
1191
:
8 9
For more details concerning this point, as well as for the generalization of the result of
Problem 19, we refer the reader to [5].
13
After Albert Girard, French mathematician of the XVII century.
44
23.
24.
25.
26.
2 Algebraic Identities, Equations and Systems
The items above describe, by means of a specific example, the ideas behind
Cardano’s formula for the roots of a polynomial equation of third degree.
2
* Let x be a nonzero real number, such that x C 1x D 3. Compute all possible
values of x3 C x13 .
If x is a nonzero real number such that x C 1x D 4, calculate x4 C x14 .
2 2
If x2 x 1 D 0, compute all possible values of x 1x C x3 x13 .
(Singapore) If x2 4x C 1 D 0, compute all possible values of
6
x C 1x C 2
3 :
x3 C x13 x C 1x
x6 C
1
x6
27. Solve the reciprocal equation x4 7x3 C 14x2 7x C 1 D 0.
28. Given real numbers a0 ; a1 ; : : : ; an , with an ¤ 0, the polynomial equation
an xn C an1 xn1 C C a1 x C a0 D 0;
is said to be reciprocal if ak D ank , for 0 k n. If ˛ is a real root of
such an equation, show that ˛ ¤ 0 and that ˛1 is also a root of it. (Hence, the
name reciprocal justifies itself by the fact that reciprocal polynomial equations
have real roots which are pairwise reciprocal.14 In this case, one also says that
an xn C an1 xn1 C C a1 x C a0 is a reciprocal polynomial.).
29. Do the following items:
(a) * Given a real number x ¤ 0, write x3 C x13 in terms of y D x C 1x .
(b) Use the result of (a) to reduce the reciprocal equation ax6 C bx5 C cx4 C
dx3 C cx2 C bx C a D 0, of degree 6, to a polynomial equation of degree 3.
30. (Brazil—adapted) We are given nonzero integers a b, such that the quadratic
equation x2 C ax C b D 0 has nonzero integer roots c d. Then we form the
quadratic equation x2 CcxCd D 0 and check if it also has nonzero integer roots.
If this is so, we let e f be those roots and form the quadratic equation x2 C
exCf D 0. We proceed in a likewise manner until we reach a quadratic equation
with nonzero integer coefficients but with no integer roots. The purpose of this
problem is to find all a and b for which this process continues indefinitely. To
this end, do the following items:
(a) Show that if the process is to continue indefinitely, then we can assume that
a > 0 > b and that every subsequent equation x2 C ˛x C ˇ D 0 is also
such that ˛ > 0 > ˇ.
(b) Under the choices of (a), let x2 C mx C n D 0 and x2 C px C q D 0 be two
consecutive equations (i.e., such that p > 0 > q are the roots of the first),
14
We shall see in [5] that the same holds for the complex roots of this equation.
2.4 Linear Systems and Elimination
45
and let and 0 be their discriminants, respectively. Show that 0 < ,
unless n D 1 or m D 1, n D 2.
(c) Conclude that a D 1 and b D 2 is the only possible choice to begin with.
2.4 Linear Systems and Elimination
Let E and F be algebraic expressions in the real variables x1 ; : : : ; xn . By the
equation E D F in the real variables or unknowns x1 , . . . , xn , we mean the problem
of finding all sequences15 .a1 ; : : : ; an / of real numbers, such that E and F make sense
and the equality E D F holds when x1 D a1 , . . . , xn D an . In this case, each such
sequence .a1 ; : : : ; an / is said to be a solution of the equation E D F.
Now, let E1 , . . . , Em , F1 , . . . , Fm be algebraic expressions in the real variables
x1 ; : : : ; xn . The system of equations
8
E1 D F1
ˆ
ˆ
ˆ
< E2 D F2
::
ˆ
ˆ
:
:̂
Em D Fm
(2.22)
is the problem of finding all sequences .a1 ; : : : ; an / of real numbers, such that the
substitutions x1 D a1 , . . . , xn D an solve all of the equations Ej D Fj . As above,
each such sequence .a1 ; : : : ; an / is said to be a solution of the system (2.22). To
solve a system of equations as (2.22) means to find all of its solutions.
In this section and the next one, we shall learn how to solve some simple (though
useful) systems of equations. We are also going to see that, under certain conditions,
an equation E D F (in several real variables) is equivalent to a system of equations
like (2.22), and this will be a source of a number of interesting examples.
The simplest—and, for our purposes, also the most useful—systems of equations
are the linear systems with two (resp. three) equations in two (resp. three) real
unknowns, i.e., systems of equations of one of the forms
a 1 x C b 1 y D c1
a 2 x C b 2 y D c2
8
< a 1 x C b 1 y C c1 z D d 1
or
a x C b 2 y C c2 z D d 2 ;
: 2
a 3 x C b 3 y C c3 z D d 3
(2.23)
respectively. Here, the real numbers ai ; bi ; ci ; di are given and not all zero, being
called the coefficients of the linear system.
15
Although the reader is probably acquainted with the concept of sequence, we refer to Sect. 6.1
for a rigorous definition.
46
2 Algebraic Identities, Equations and Systems
The most efficient method for solving linear systems like those above is the
elimination algorithm,16 also known as gaussian elimination,17 which is based
in the following result.
Lemma 2.24 Let E and F be given algebraic expressions in the real variables x
and y (or x, y and z). For reals a, b and c, the systems of equations
EDa
and
FDb
E C cF D a C cb
FDb
have the same solutions.
Proof We shall prove the lemma in the case in which E and F are algebraic
expressions in the real variables x and y; the other case is completely analogous.
Suppose that x D x0 and y D y0 solve the system on the left, so that, when we
substitute x by x0 and y by y0 into E and F, both equalities E D a and F D b hold;
we shall denote such a situation by writing E.x0 ; y0 / D a and F.x0 ; y0 / D b. Then,
substituting x by x0 and y by y0 into E C cF gives us
.E C cF/.x0 ; y0 / D E.x0 ; y0 / C cF.x0 ; y0 / D a C cb;
so that x D x0 and y D y0 also solve the system on the right. Conversely, if x D x0
and y D y0 solve the system on the right, then, since E D .ECcF/cF, an argument
entirely analogous to the above shows that x D x0 and y D y0 do solve the system
on the left. Therefore, both systems have the same solutions.
t
u
Back to the analysis of the linear systems (2.23), let us show that a number of
careful applications of the previous lemma lead us to quick solutions of them (the
elimination algorithm consists exactly of this).
We start by the system on the left, which, for simplicity, we write as
ax C by D e
:
cx C dy D f
Since at least one of the coefficients a; b; c; d is nonzero, we can suppose, without
loss of generality, that a ¤ 0 (otherwise, it suffices to rewrite the system in one of
the forms
16
An algorithm is a finite sequence of precise procedures that, once followed, give an expected
result, also known as the output of the algorithm. Although algorithms will play a very modest
role in this volume, [5] brings a number of very interesting ones.
17
After Joanne Carl Friedrich Gauss, German mathematician of the XVIII and XIX centuries.
Gauss is generally considered to be the greatest mathematician of all times. In the several different
areas of Mathematics and Physics in which he worked, like Number Theory, Analysis, Differential
Geometry and Electromagnetism, there are always very important and deep results or methods that
bear his name. We refer the reader to [26] for an interesting biography of Gauss.
2.4 Linear Systems and Elimination
cx C dy D f
;
ax C by D e
47
by C ax D e
or
dy C cx D f
dy C cx D f
;
by C ax D e
according to whether c, b or d is nonzero, and change a by this number in the
following discussion).
Then, let a ¤ 0 and E D ax C by, F D cx C dy. Changing the equation F D f
by the equation
c
c
F E D f e;
a
a
we get the system
E
De
;
F ac E D f ac e
and Lemma 2.24 immediately assures that the solutions of this new system coincide
with those of the original one. Hence, in order to solve that system, it suffices to
solve this last one. On the other hand, since
c
c
bc
F E D .cx C dy/ .ax C by/ D d y;
(2.24)
a
a
a
the last system reduces to
ax C by D e
;
d0 y D f 0
c
0
where d 0 D d bc
a and f D f a e. Now, we shall consider three different cases:
• If d0 ¤ 0 (or, equivalently, ad bc ¤ 0), then the second equation above gives
0
y D df 0 , and the substitution of this value into the first equation gives x D 1a .e 0
by/ D 1a e bfd0 . In this case, the system is said to be determined, for it has a
unique solution.
• If d0 D 0 (or, equivalently, ad bc D 0) and f 0 ¤ 0, the system is impossible,
for the second equation reduces to 0y D f 0 , which has no roots at all.
• If d 0 D f 0 D 0, then the second equation reduces to 0y D 0, an equality which is
true for all real values of y. Therefore, the system as a whole consists only of the
equation ax C by D 0, which has infinitely many solutions (making y D ˛, with
˛ 2 R, we get x D b˛
). For this reason, the system is said to be undetermined.
a
As was the case in the previous chapter, we would like to stress that, in
specific examples, rather than memorizing the formulas obtained through the above
discussion, one should just execute the elimination process. The previous discussion
(more precisely, (2.24)) makes it clear that this process consists of subtracting an
48
2 Algebraic Identities, Equations and Systems
appropriate multiple of the first equation from the second, in order to eliminate the
variable x from it (and that’s where the name elimination comes from).
Example 2.25 Use Gaussian elimination to find all real values of a for which the
system equations
2x C ay D 3
ax C 2y D 32
is impossible, undetermined or determined.
Solution Subtracting a2 times the first equation from the second, we get the
equivalent system
(
2x C2 ay D 3
:
2 a2 y D 32 .1 a/
2
If 2 a2 ¤ 0, which is the same as a ¤ ˙2, then the second equation above
gives us y D 3.1a/
, so that the first equation furnishes a single value for x, namely,
4a2
xD
3.4 a/
1
.3 ay/ D
:
2
2.4 a2 /
Therefore, the system is determined whenever a ¤ ˙2.
If a D ˙2, then the second equation reduces to 0y D 32 .1 2/, which represents
an impossible equality. Therefore, the system is impossible.
t
u
For a geometric interpretation of Gaussian elimination for linear systems of two
equations in two real variables, see the problems of Sect. 6.2 of [4].
Let us now turn our attention to the linear system on the right, in (2.23). In order
to analyse it, let Ei D ai x C bi y C ci z, for 1 i 3, so that it reduces to
8
< E1 D d 1
E D d2 :
: 2
E3 D d 3
As was done for linear systems in two variables, we can suppose that a1 ¤ 0 (the
other cases being totally analogous). Changing equations E2 D d2 and E3 D d3
respectively by
E2 a2
a2
a3
a3
E1 D d2 d1 and E3 E1 D d3 d1
a1
a1
a1
a1
(in order to eliminate the variable x from the second and third equations), we get the
equivalent system
2.4 Linear Systems and Elimination
49
8
ˆ
< E1
E2 :̂ E 3
a2
E
a1 1
a3
a1 E1
D d1
D d2 D d3 a2
d
a1 1
a3
a1 d1
:
Since
E2 a2
a2
E1 D .a2 x C b2 y C c2 z/ .a1 x C b1 y C c1 z/
a1
a1
a2 b1
a 2 c1
y C c2 z
D b2 a1
a1
D b02 y C c02 z;
and, analogously, E3 a3
E
a1 1
D b03 y C c03 z, it suffices to solve the system
8
< a 1 x C b 1 y C c1 z D d 1
b02 y C c02 z D d20 ;
:
b03 y C c03 z D d30
where d20 D d2 aa21 d1 and d30 D d3 aa31 d1 .
In case all of the numbers b02 ; c02 ; b03 ; c03 are equal to 0, the last system above
reduces to
8
< a 1 x C b 1 y C c1 z D d 1
0 D d20 :
:
0 D d30
If d20 ¤ 0 or d30 ¤ 0, the system is impossible; if d20 D d30 D 0, the system is
undetermined, for it is equivalent to the single equation a1 x C b1 y C c1 z D 0, which
obviously has infinitely many solutions.
Suppose, then, that at least one of the numbers b02 ; c02 ; b03 or c03 is nonzero, say
0
b2 ¤ 0 (as before, the other cases can be treated quite analogously). Then, applying
Gaussian elimination to the system
b02 y C c02 z D d20
;
b03 y C c03 z D d30
we obtain a system of the form
8
< a 1 x C b 1 y C c1 z D d 1
b02 y C c02 z D d20 ;
:
c003 z D d300
50
2 Algebraic Identities, Equations and Systems
which is also equivalent to the original system. The discussion, then, goes on as
before, and we concentrate our attention in the third equation, c003 z D d300 . Also as
before, we have to distinguish three distinct cases:
• If c003 ¤ 0, then the third equation above gives a single value for z, say z D ;
since b02 ¤ 0, the substitution of this value into the second equation furnishes
y D b10 .d20 c02 z/ D b10 .d20 c02 /. Finally, since a1 ¤ 0, the substitution of the
2
2
values thus obtained for y and z into the first equation give us a single value for
x. Then, we conclude that the system is determined.
• If c003 D 0 and d300 ¤ 0, the system is impossible, for the third equation reduces to
0z D d300 .
• If c003 D d300 D 0, then the third equation reduces to the equality 0z D 0, and the
system as a whole reduces to
a 1 x C b 1 y C c1 z D d 1
b02 y C c02 z D d20
or, which is the same,
a 1 x C b 1 y D d 1 c1 z
:
b02 y D d20 c02 z
Since a1 and b02 are both nonzero, for each real value z D the system
corresponding to
a 1 x C b 1 y D d 1 c1 b02 y D d20 c02 is determined; therefore, the original system in x; y; z is undetermined.
We can summarize the above discussion by saying that the elimination algorithm
for linear systems of the form
8
< a 1 x C b 1 y C c1 z D d 1
a x C b 2 y C c2 z D d 2
: 2
a 3 x C b 3 y C c3 z D d 3
consists in performing, one by one, the following three steps:
1st. Eliminate the variable x from the second and third equations, by adding to
these equations appropriate multiples of the first one, thus obtaining a system
of the form
8
< a 1 x C b 1 y C c1 z D d 1
b02 y C c02 z D d20 :
:
b03 y C c03 z D d30
2.4 Linear Systems and Elimination
51
2nd. Eliminate the variable y (in case b02 ¤ 0 in the last system above) from the
third equation, by adding to it an adequate multiple of the second equation,
thus obtaining a system of the form
8
< a 1 x C b 1 y C c1 z D d 1
b02 y C c02 z D d20 :
:
c003 z D d300
3rd. Analyse equation c003 z D d300 and, after this, the other two equations in
succession, in order to decide whether the original system is determined,
undetermined or impossible.
As before, it is much more useful to keep the general steps above in mind than
to try to memorize any of the expressions obtained through the calculations in the
previous discussions. Let us see one more example to illustrate this point.
Example 2.26 Find all real values of a for which the system of equations
8
< x C y az D 1
2x C ay C z D 1
:
ax C y z D 2
is impossible.
Solution Multiplying the first equation respectively by 2 and by a, and subtracting
(also respectively) the results from the second and third equations, we get the
equivalent system
8
D 1
< x C y az
:
.a 2/y C .1 C 2a/z D 3
:
.1 a/y .1 a2 /z D 2 C a
If a D 1, the last equation reduces to the impossible equality 0 D 3, and the
original system is impossible. If a ¤ 1, add to the second equation a2
times the
1a
third one, thus obtaining the equivalent system
8
ˆ
D 1
< x C y az
2 3aC7
2
:
.a C a 1/z
D a 1a
:̂ .1 a/y .1 a2 /z D 2 C a
(Observe that we have slightly changed the second step, so that the roles of the
second and third equations of the last system are interchanged—it is the second
equation that, now, has just one variable. Obviously, this is perfectly right, and shows
that the elimination algorithm is quite a flexible one.)
52
2 Algebraic Identities, Equations and Systems
p
Now, if a2 C a 1 D 0, i.e., if a D 1˙2 5 , then the system is impossible,
for a2 3a C 7 ¤ 0 for these values of ap and, hence, the second equation
reduces to an impossible equality. If a ¤ 1˙2 5 , then the second equation gives
us a single possible value for z; in turn, upon substitution of this value of z into the
third equation, we find a single possible value for y; finally, putting these values for
y and z into the first equation, we find a single possible value for x, so that the system
is determined.
p
We conclude a D 1 and a D 1˙2 5 are the values of a for which the original
system is impossible.
t
u
We finish our discussion of linear systems by observing that Gaussian elimination
algorithm can easily be put to work to the analysis of the general linear system in m
equations and n unknowns
8
a11 x1 C a12 x2 C C a1n xn
ˆ
ˆ
ˆ
< a21 x1 C a22 x2 C C a2n xn
::
ˆ
ˆ
:
:̂
am1 x1 C am2 x2 C C amn xn
D b1
D b2
I
::
:
(2.25)
D bm
here, the aij and bi are given real numbers, such that at least one of the aij is nonzero.
Apart from an important particular case of general linear system, which will
make its appearance in Sect. 18:1 of [5] (and will be analysed there, by other
methods), we shall not make a systematic use of such systems along these notes;
hence, we shall not develop the general analysis of the application of the elimination
algorithm to them. If it is the case we have to solve such a linear system (as in the
example below), some clever reasoning, perhaps with the aid of Lemma 2.24, will
suffice.
Example 2.27 (OCM) Find all real solutions x1 , x2 , . . . , x100 of the linear system
8
ˆ
x 1 C x2 C x3
ˆ
ˆ
ˆ
ˆ
x 2 C x3 C x4
ˆ
ˆ
ˆ
<
::
:
ˆ
ˆ x98 C x99 C x100
ˆ
ˆ
ˆ
ˆ
ˆ x99 C x100 C x1
:̂
x100 C x1 C x2
D0
D0
::
:
D0
D0
D0
:
Solution Observe that each xj appears in exactly three of the given equations.
Therefore, adding all of them and dividing both sides by 3, we get
x1 C x2 C x3 C C x100 D 0:
2.4 Linear Systems and Elimination
53
In order to find x1 D 0, just note that
0 D x1 C .x2 C x3 C x4 / C C .x98 C x99 C x100 /
D x1 C 0 C 0 C C 0 D x1 :
For x2 D 0, write
0 D x2 C .x3 C x4 C x5 / C C .x99 C x100 C x1 /
D x 2 C 0 C 0 C C 0 D x2 :
Now, x1 C x2 C x3 D 0 implies x3 D 0. Then, x2 C x3 C x4 D 0 implies x4 D 0, and
so on, so that all of the xi ’s are equal to 0.
t
u
We refer the interested reader to Chap. 1 of [18] for quite a detailed exposition
of the elimination algorithm for general linear systems.
Problems: Section 2.4
1. Assume that, in the linear system
ax C by D e
, we have a, b, c, d, e, f ¤ 0.
cx C dy D f
Prove that:
(a)
(b)
(c)
a
c
a
c
a
c
¤
D
D
b
d
b
d
b
d
, the system is determined.
D ef , the system is undetermined.
¤ ef , the system is impossible.
2. Find all real values of a for which the system of equations
8
< x C 2y 3z D 4
3x y C 5z D 2
:
4x C y C .a2 14/z D a C 2
is impossible.
3. Solve the system of equations
82
3
ˆ
<xCy
3
5
x y C
:̂ 7 6 C
x
y
1
z
2
z
3
z
D8
D 1 :
D5
4. For 1 i; j 3, let aij be given real numbers, such that a2i1 C a2i2 C a2i3 D 1 for
1 i 3 and ai1 aj1 C ai2 aj2 C ai3 aj3 D 0, for 1 i; j 3 with i ¤ j. Given real
numbers b1 , b2 , b3 , solve the linear system
54
2 Algebraic Identities, Equations and Systems
8
< a11 x1 C a12 x2 C a13 x3 D b1
a x C a22 x2 C a23 x3 D b2 ;
: 21 1
a31 x1 C a32 x2 C a33 x3 D b3
writing x1 , x2 and x3 in terms of the aij and bi .
5. With respect to the linear system (2.25), do the following items:
(a) If b1 D b2 D D bm D 0, then the system always has at least one solution.
(b) If the system has at least two different solutions, then it has infinitely many
solutions.
(c) If the system has only one solution when b1 D b2 D D bm D 0, then, for
general values of b1 , b2 , . . . , bm , it has at most one solution.
2.5 Miscellaneous
Let us now turn our attention to the next simplest type of system of equations,
namely, second degree systems. Ultimately, such a system is just a rephrasing of
a second degree equation in terms of a system of two equations in two unknowns.
Nevertheless, the reader will be amazed on how they provide a number of interesting
applications.
Proposition 2.28 Let S and P be given real numbers. The system of equations
xCyDS
xy D P
(2.26)
has real solutions if and only if S2 4P. Moreover, in this case, the solutions are
given by x D ˛, y D ˇ or vice-versa, where ˛ and ˇ are the roots of the second
degree equation u2 Su C P D 0.
Proof First of all, if ˛ and ˇ are the roots of u2 Su C P D 0, then we know from
Viète’s formulas (see item (b) of Proposition 2.16) that ˛ C ˇ D S and ˛ˇ D P.
Therefore, the pairs x D ˛, y D ˇ and x D ˇ, y D ˛ do satisfy the system of
Eq. (2.26).
Conversely, let x D x0 , y D y0 be any solution of that system. Then, the first
equation gives y0 D S x0 , and the second equation then gives P D x0 y0 D x0 .S x0 /, or, which is the same, x20 Sx0 CP D 0. Therefore, x0 is a root of u2 SuCP D 0,
from where we get x0 D ˛ or x0 D ˇ. Since ˛ Cˇ D S, we have two possibilities:
• If x0 D ˛, then y0 D S x0 D S ˛ D ˇ.
• If x0 D ˇ, then y0 D S x0 D S ˇ D ˛.
t
u
2.5 Miscellaneous
55
The following examples show how we can sometimes reduce the search for
solutions of more complex systems of equations to that of simpler systems, of one
of the types we met so far.
Example 2.29 Find all real solutions of the system of equations
.x2 C 1/.y2 C 1/ D 10
:
.x C y/.xy 1/ D 3
Solution First of all, rewriting the left hand side of the first equation as
.x2 C 1/.y2 C 1/ D x2 y2 C x2 C y2 C 1
D .xy/2 C Œ.x C y/2 2xy C 1;
and letting x C y D a and xy D b, we get the system
b2 C a2 2b D 9
:
a.b 1/ D 3
Now, writing
b2 C a2 2b D a2 C .b2 2b C 1/ 1
D a2 C .b 1/2 1
and making the substitution b 1 D c, we reach the system
a2 C c2 D 10
:
ac D 3
By squaring the second equation, we transform this last system into one of the
form (2.26), whose unknowns are a2 and c2 , and such that S D 10 and P D 9. Since
the roots of the second degree equation u2 10u C 9 D 0 are 1 and 9, it follows that
a2 D 1 or 9 and, hence, a D ˙1 or ˙3. Then, we have the possibilities
.a; c/ D .1; 3/; .1; 3/; .3; 1/ or .3; 1/;
from where
.a; b/ D .1; 4/; .1; 2/; .3; 2/ or .3; 0/:
Finally, each of these pairs .a; b/ give us another system of the form (2.26), with
unknowns x and y. Solving the four systems thus obtained, we arrive at the solutions
of the original system:
56
2 Algebraic Identities, Equations and Systems
.x; y/ D .1; 2/; .2; 1/; .1; 2/; .2; 1/; .0; 3/ or .3; 0/:
t
u
Some seemingly complicated equations can be easily solved, provided we find a
way to transform them into systems of equations. Since there is no general procedure
that tells us when or how this can be done, each equation should be analysed
separately. In this respect, the following example shows that a frequently useful
algebraic trick is the introduction of new variables.
Example 2.30 (Israel) Find all real solutions of the equation
p
p
4
4
13 C x C 4 x D 3:
p
p
Solution Letting a D 4 13 C x and b D 4 4 x, we get a C b D 3 and a4 C b4 D
.13 C x/ C .4 x/ D 17. Hence, we have reduced the problem of solving the giving
equation to that of solving the system of equations
aCbD3
:
a4 C b4 D 17
Applying formula (2.1) for the square of a sum twice, we get
17 D a4 C b4 D .a2 C b2 /2 2a2 b2
D Œ.a C b/2 2ab2 2.ab/2
D .9 2ab/2 2.ab/2
D 81 36ab C 2.ab/2 ;
so that
.ab/2 18.ab/ C 32 D 0:
Solving for ab the second degree equation above, we find ab D 2 or ab D 16.
Therefore, there are two distinct possibilities:
.i/
aCbD3
aCbD3
or .ii/
:
ab D 2
ab D 16
Possibility (i) gives a D 1 and b D 2, or vice-versa. If a D 1, then 13 C x D 1
and, hence, x D 12. If a D 2, then 13 C x D 16 and, hence, x D 3. Possibility (ii)
does not give any real solutions, for, in this case, a and b would be real roots of the
second degree equation u2 3u C 16 D 0, which has none of them.
t
u
2.5 Miscellaneous
57
The following lemma states a relatively easy sufficient condition to transform the
search for the roots of an equation in one unknown into that of a system of equations.
Lemma 2.31 If E1 ; E2 ; : : : ; En are expressions involving one or more real variables, then
8
E1 D 0
ˆ
ˆ
ˆ
< E2 D 0
E12 C E22 C C En2 D 0 ,
:
:
ˆ
ˆ ::
:̂
En D 0
(2.27)
Proof This is an easy generalization of Corollary 1.5 and of Problem 2, page 10.
t
u
Example 2.32 Find all real roots of the equation
x4 y2 C y2 C 2 D 2y C 2x2 y:
Solution We can write the given equation as
.x4 y2 2x2 y C 1/ C .y2 2y C 1/ D 0;
or, which is the same,
.x2 y 1/2 C .y 1/2 D 0:
Therefore, by the previous lemma, the equation is equivalent to the system of
equations
x2 y 1 D 0
;
y1 D 0
whose solutions y D 1, x D ˙1 can be obtained without difficulty.
t
u
Problems: Section 2.5
1. (IMO) Find all real numbers x, y and z, such that the sum of any of them with
the product of the other two is always equal to 2.
2. Let a be a nonzero real constant. Find, in terms of a, all real numbers x and y
that solve the system of equations
(
1
xCy
x
xCy
Cx D aC1
:
Da
58
2 Algebraic Identities, Equations and Systems
3. Solve, for x; y 2 R, the system of equations
(
xCy
xy
xy
xCy
C
C
xy
xy
xy
xy
D5
:
D 56
4. (IMO—adapted) Consider the system of equations
8 2
< ax1 C bx1 C c D x2
ax2 C bx2 C c D x3 ;
: 22
ax3 C bx3 C c D x1
whose unknowns are x1 ; x2 ; x3 , where a; b and c are given real numbers, with
a ¤ 0. If D .b 1/2 4ac, do the following items:
(a) If < 0, then there is no real solution.
(b) If D 0, then there is exactly one real solution.
5. (TT) Find all real solutions of the system of equations
8 3
< x D 2y 1
y3 D 2z 1 :
: 3
z D 2x 1
6. Solve, for x; y 2 R, the equation x2 C 2xy C 3y2 C 2x C 6y C 3 D 0.
7. (IMO) Find all real values of a for which the system of equations
x2 C y2 D 4z
;
3x C 4y C z D a
with unknowns x, y and z, has a single solution.
8. (NMC) Find all real numbers x, y, z greater than 1, such that
xCyCzC
9.
10.
11.
12.
p
p
p
3
3
3
C
C
D2
xC2C yC2C zC2 :
x1
y1
z1
p
p
3
x C 5 C 3 11 x D 6.
Find all real roots of the equation p
p
Find all real roots of the equation 5 5 x D x.
x2
(Canada) Find all real roots of the equation x2 C .xC1/
2 D 3.
(Canada) Solve the system of equations
8 2
4x
ˆ
ˆ
2 D y
ˆ
< 1C4x
2
4y
2 D z
ˆ
ˆ 1C4y
:̂ 4z2 D x
1C4z2
2.5 Miscellaneous
59
13. (Romania) Let a, b, c and d be given real numbers, such that
8
a C b C c 3d
ˆ
ˆ
<
b C c C d 3a
:
ˆ c C d C a 3b
:̂
d C a C b 3c
Prove that a D b D c D d.
14. * Let E1 ; E2 ; : : : ; En and F1 ; F2 ; : : : ; Fn be given expressions in one or more real
variables, such that E1 F1 , E2 F2 , : : :, En Fn . Prove that the equation
E1 C C En D F1 C C Fn
is equivalent to the system of equations
8
E1 D F1
ˆ
ˆ
ˆ
< E2 D F2
:
::
ˆ
ˆ
:
:̂
En D Fn
15. (Romania) Find all real roots of the equation
p
4x2 x4 3 C
p
p
4y2 y4 C 4z2 z4 C 5 D 6:
16. Solve, in the set of positive reals, the system of equations
8
4
ˆ
<x C y D
y C 4z D
:̂ z C 4 D
x
5y
4
5z
4
5x
4
:
17. (Soviet Union—adapted)
p
p
(a) For x > 0, prove that x C 2x 2 2, with equality if and only if x D 2.
(b) Solve the system of equations
8
2
ˆ
< 2y D x C x
2z D y C 2y :
:̂ 2x D z C 2
z
Chapter 3
Elementary Sequences
An (infinite) sequence of real numbers is an infinite ordered list .a1 ; a2 ; a3 ; : : :/
of real numbers, i.e., an infinite list of real numbers, in which we specify who is
the first number of the list, who is the second, third and so on. It is customary to
denote an infinite sequence as above simply by .ak /k1 . A (finite) sequence of real
numbers is a finite ordered list .a1 ; a2 ; : : : ; an / of real numbers, i.e., a finite list of
real numbers, in which, as was the case with infinite sequences, we specify who is
the first number of the list, who is the second, the third, . . . , the n-th. Is it customary
to denote a finite sequence as above simply by .ak /1kn . In any of the cases above,
we say that ak is the k-th term of the sequence.1
Our aim in this chapter is to study some types of elementary sequences that
occur quite frequently in elementary Mathematics. Along this process, we shall
introduce several definitions and properties that will apply to general sequences as
well. Whenever there is no danger of confusion, we shall concentrate our discussion
on infinite sequences, letting to the reader the task of adapting all that comes to finite
sequences.
3.1 Progressions
We say that a sequence .ak /k1 is defined by a positional formula if the values
ak 2 R are given by means of an expression that depends on k.
1
As we will see in Chap. 6, an infinite (resp. finite) sequence of real numbers is, rigorously
speaking, a function f W N ! R (resp. f W f1; 2; : : : ; ng ! R, for some n 2 N). Thus, the above
notation comes from the shortcut of letting ak D f .k/, for every k 2 N (resp., every 1 k n).
However, for the time being, the above definition will be harmless and sufficient for all of our
purposes.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_3
61
62
3 Elementary Sequences
Example 3.1 The sequence .ak /k1 of the perfect squares is the infinite sequence
.12 ; 22 ; 32 ; : : :/. Therefore, we have a1 D 12 , a2 D 22 , a3 D 32 and, more generally,
ak D k2 , for every integer k 1.
An alternative to defining a sequence by means of a positional formula is to give
a recursive definition of it. This means that we should specify a (finite) number
of initial terms of the sequence, as well as a recipe to calculate the other terms by
recurring to those that precede it. Let us see an example.
Example 3.2 Consider the sequence .ak /k1 , defined recursively by a1 D 2, a2 D 5
and
ak D 2ak1 ak2 ; 8 k 3:
(3.1)
If we take k D 3 in the relation above, we get a3 D 2a2 a1 D 2 5 2 D 8; if
we now take k D 4, we find a4 D 2a3 a2 D 2 8 5 D 11, and so on. In the
above, relation (3.1) is the recurrence relation, or simply the recurrence satisfied
by the sequence .ak /k1 .
In the previous example, it is important to notice that we were only capable
to compute the value of a3 because we knew, beforehand, both the recurrence
relation (3.1) as well as the values of a1 and a2 . Knowing just the value of a1
would not suffice, for (3.1) computes each term as a function of the two ones that
immediately precede it. On the other hand, if we changed the values of a1 and a2 (but
maintained the recurrence relation), we would generally change the corresponding
values of subsequent terms (just make a1 D 1 and a2 D 2 in the given recurrence
relation, and compute the new value of the a3 ).
It is also worth observing that there are other equivalent ways of writing the
recurrence relation of the former example. For instance, letting k 2 D j, we get
k D j C 2 and, hence, ajC2 D 2ajC1 aj , 8 j 1 (since k 3 ) j 1). In other
words, the name we give to the index of a sequence (i.e., j, k etc.) is not relevant to
its definition; we could equally well write it as akC2 D 2akC1 ak , 8 k 1.
Sometimes, we will also need to list the terms of a sequence starting from zero,
denoting it as .ak /k0 . At first, this could look a little odd, for the first term of the
sequence would be a0 , the second would be a1 etc. Nevertheless, since this can be
a natural choice in a number of contexts, we urge the reader to get used to it. For
example, the sequence .1; 4; 9; 16; : : :/ of the perfect squares could be written as
ak D .k C 1/2 , for all k 0.
At this point, a natural question poses itself: if a certain sequence is defined
recursively, how can we find a positional formula for its terms? Unfortunately, this
question does not admit a simple answer. Instead of trying to attack this general
problem right now, we shall first discuss some simple examples, postponing a deeper
analysis to Chaps. 3 and 21 of [5].
We begin by looking at arithmetic progressions.
Definition 3.3 A sequence .ak /k1 of real numbers is an arithmetic progression
(we abbreviate AP) if there exists a real number r such that
3.1 Progressions
63
akC1 D ak C r;
(3.2)
for every k 1.
In the previous definition, we refer to the real number r as the common
difference of the AP. We also observe that, for an AP .ak /k1 to be completely
determined, one has to give both its first term a1 and its common difference r. For
example, the sequence .ak /k1 such that a1 D 2 and akC1 D ak C 3 for k 1, is the
AP with initial term 2 and common difference 3, and it is completely determined by
the recurrence relation it satisfies and by the value of its first term. However, if we
only knew that akC1 D ak C 3 for k 1, we would not have just one AP, for we
would not know how to start it.
Example 3.4 Let .ak /k1 be an AP of common difference r. Prove that the
sequences .bk /k1 and .ck /k1 , defined by bk D a2k and ck D a2k1 for k 1,
are also AP’s, with common differences equal to 2r.
Proof We look at the sequence .bk /k1 , the case of the sequence .ck /k1 being
totally analogous. By the definition of an AP, it suffices to show that bkC1 bk D 2r,
for k 1. Since .ak /k1 is an AP with common difference r, it follows from the
definition of bk that
bkC1 bk D a2.kC1/ a2k D a2kC2 a2k
D .a2kC2 a2kC1 / C .a2kC1 a2k /
D r C r D 2r;
t
u
as we wished to prove.
Another useful recursive characterization of AP’s is the one given in the
following result.
Proposition 3.5 A sequence .ak /k1 of real numbers is an AP if and only if
akC2 C ak D 2akC1 ; 8 k 1:
(3.3)
Proof By definition, a sequence is an AP if and only if a2 a1 D a3 a2 D ,
i.e., if and only if we have akC2 akC1 D akC1 ak for every integer k 1. This is
an equivalent way of writing (3.3).
t
u
The next result collects two useful formulas for AP’s.
Proposition 3.6 If .ak /k1 is an AP with common difference r, then
(a) ak D a1 C .k 1/r, for every k 1.
(b) a1 C a2 C C an D n.a12Can / , for every n 1.
64
3 Elementary Sequences
Proof
(a) Diagram
Cr
a1
Cr
/ a2
/ a3
Cr
/ Cr
/ ak1
Cr
/ ak
makes it clear that, in order to reach ak from a1 , it is necessary k1 steps, where
each step consists of adding r to a term. Therefore, in order to obtain ak from
a1 , we have to add .k 1/r to a1 , so that ak D a1 C .k 1/r.
(b) From diagram
Cr
a1
Cr
/ a2
/ a3
Cr
/ o
r
ak2 o
r
an1 o
r
an ;
we conclude that
a1 C an D .a2 r/ C .an1 C r/ D a2 C an1 ;
a2 C an1 D .a3 r/ C .an2 C r/ D a3 C an2
etc. Therefore, letting S D a1 C a2 C C an , we have
2S D 2.a1 C a2 C a3 C C an2 C an1 C an /
D .a1 C an / C .a2 C an1 / C .a3 C an2 / C C .an C a1 /
D .a1 C an / C .a1 C an / C .a1 C an / C C .a1 C an /
„
ƒ‚
…
n summands
D n.a1 C an /;
and the formula follows.
t
u
The formulas of items (a) and (b) of the previous proposition are respectively
known as the formulas for the general term and for the sum of the first n terms of
an AP. Let us see two simple applications of them.
Example 3.7 Compute, in terms of n, the sums of the first n positive integers and
the first n odd positive integers.
Solution The positive integers form the AP 1; 2; 3; 4; : : :, of common difference 1.
Since its n-th term is obviously equal to n, item (b) of the previous proposition gives
1C2C3CCn D
n.n C 1/
:
2
3.1 Progressions
65
The odd positive integers form the AP 1; 3; 5; 7; : : :, of common difference 2. Its
n-th term (i.e., the n-th odd positive integer) is, by the formula for the general term
of an AP, equal to 1 C .n 1/ 2 D 2n 1. Thus, again by item (b) of the previous
proposition, the sum of the first n odd positive integers is
nŒ1 C .2n 1/
D n2 :
2
Example 3.8 Let .ak /k1 be the sequence given by a1 D 1 and
1 C 3 C 5 C C .2n 1/ D
akC1 D
t
u
ak
;
1 C 2ak
for every integer k 1. Compute ak as a function of k.
Solution It is clear that all terms of the sequence .ak /k1 are positive, so that we
can define the sequence .bk /k1 by letting bk D a1k , for every k 1. Then, the
recurrence for .ak /k1 gives us
bkC1 D
1
1 C 2ak
1
D
D
C 2 D bk C 2;
akC1
ak
ak
so that .bk /k1 is the AP of initial term b1 D a11 D 1 and common difference
2. Therefore, this AP coincides with that of the odd positive integers, so that the
previous example gives bk D 2k 1, for every k 1. Hence,
ak D
1
1
:
D
bk
2k 1
t
u
Another useful class of sequences is the one of geometric progressions, according
to the next definition.
Definition 3.9 A sequence .ak /k1 of real numbers is a geometric progression (we
shall abbreviate GP) if there exists a real number q such that
akC1 D q ak ;
(3.4)
for every k 1.
The real number q that appears in the definition of a GP is its common ratio.
Observe that, if q D 0, then ak D 0 for every k > 1. On the other hand, if q D 1,
then ak D a1 for every k 1. As it happens with AP’s, a GP .ak /k1 is completely
determined provided we know its first term a1 and its common ratio q.
Example 3.10 For a fixed real number q ¤ 0, the sequence .ak /k1 such that ak D
qk for every k 1 (i.e., the sequence formed by the powers of q, with natural
exponents) is a GP of common ratio q. If q < 0, then Problem 6, page 11, shows
66
3 Elementary Sequences
that ak is positive if and only if k is even; if 0 < q < 1, then Corollary 1.3 gives
a1 > a2 > a3 > > 0; if q > 1, then, again by that result, we have 0 < a1 < a2 <
a3 < .
Another useful recursive characterization of (almost all) GP’s is the one given by
the following result.
Proposition 3.11 A sequence .ak /k1 of nonzero real numbers is a GP if and only if
akC2 ak D a2kC1 ;
(3.5)
for every k 1.
Proof By definition, .ak /k1 is a GP (of common ratio q) if and only if
a2
a3
a4
D
D
D D q:
a1
a2
a3
In other words, .ak /k1 is a GP if and only if for every integer k 1, we have
akC2
akC1
t
u
akC1 D ak , and this is an equivalent way of writing (3.5).
As we have done with AP’s, our next result brings formulas for the general term
and for the sum of the first k terms of a GP.
Proposition 3.12 If .ak /k1 is a GP of common ratio q, then:
(a) ak D a1 qk1 , for every k 1.
(b) If q ¤ 1, then a1 C a2 C C an D
anC1 a1
q1 ,
for every n 1.
Proof
(a) The proof we present parallels that for the general term of an AP (see the proof
of item (a) of Proposition 3.6): diagram
q
a1
/ a2
q
/ a3
q
/ q
/ ak1
q
/ ak
makes it clear that, in order to reach ak starting from a1 , it is necessary k 1
steps, where each step consists of multiplying a term of the sequence by q.
Hence, to reach ak we have to multiply a1 by q exactly k 1 times, so that
ak D a1 qk1 .
(b) Letting Sn D a1 C a2 C C an , it follows from (3.4) that
qSn D q.a1 C a2 C C an1 C an /
D qa1 C qa2 C C qan1 C qan
D a2 C a3 C C an C anC1 :
3.1 Progressions
67
Therefore,
.q 1/Sn D qSn Sn
D .a2 C a3 C C an C anC1 / .a1 C a2 C C an /
D .a2 C a3 C C an / C anC1 a1 .a2 C C an /
D anC1 a1 ;
where, in the last passage, we cancelled out the summand a2 C a3 C C an .
It now suffices to divide both members of the equality .q 1/S D anC1 a1
by q 1.
t
u
Example 3.13 Let .ak /k1 be an AP of natural numbers, with common difference
r > 0, and .bk /k1 be a GP of nonzero real numbers, with common ratio q. Consider
the sequence .ck /k1 , such that ck D bak for every integer k 1. Prove that .ck /k1
is a GP of common ratio qr .
Proof It suffices to show that the ratio of any two consecutive terms of the sequence
.ck /k1 is always equal to qr . For this, we use the formulas for the general terms of
AP’s and GP’s, as well as the result of Problem 4, page 11:
ba
b1 qakC1 1
ckC1
D kC1 D
D qakC1 ak D qr :
ck
bak
b1 qak 1
t
u
The purpose of the next example is to call the reader’s attention to the fact that
the idea used in the proof of item (b) of Proposition 3.12 is as important as the actual
result it collects. Hence, the reader should keep this idea in his/her mind.
Example 3.14 Compute the value of the sum
2 1 C 7 3 C 12 32 C 17 33 C C 497 399 C 502 3100 ;
where, from left to right, the k-th summand equals the product of the k-th term of
the AP 2; 7; 12; : : : ; 502 with the k-th term of the GP 1; 3; 32 ; : : : ; 3100 .
Solution Arguing exactly as in the proof of item (b) of Proposition 3.12, let S
denote the desired sum, and compute 3S (we use the factor 3, for it is the common
ratio of the GP 1; 3; 32 ; : : : ; 3100 ):
3S D 2 3 C 7 32 C 12 33 C 17 34 C C 497 3100 C 502 3101 :
68
3 Elementary Sequences
Hence,
2S D 3S S
D 2 3 C 7 32 C 12 33 C 17 34 C C 497 3100 C 502 3101
2 1 7 3 12 32 17 33 497 399 502 3100
D .502 3101 2/ 5.3 C 32 C 33 C 34 C C 3100 /
5
D .502 3101 2/ .3101 3/;
2
where we grouped multiples of equal powers of 3 in the third equality, and used the
formula of item (b) of Proposition 3.12 in the fourth one. Finally, grouping together
the summands of the last expression above, we get S D 14 .999 3101 C 11/.
t
u
Problems: Section 3.1
1. Compute the first four terms of the sequence .an /n1 , defined by a1 D 1 and
akC1 D a2k C ak C 1, for k 1.
2. Write a positional formula for each of the following sequences:
(a)
(b)
(c)
.1;
3; 4; 5; 6; 7; 8;
1 2;
: : :/.
2 3 4 5 6 7 8
;
;
;
;
;
;
;
;
:
:
:
.
2 13 4 15 6 17 8 1 9 1; 2 ; 3; 4 ; 5; 6 ; 7; 8 ; : : : .
3. Let .an /n1 be a sequence of positive real numbers satisfying the recurrence
k
relation akC1 D 3aak C1
, for k 1. Find a recurrence relation satisfied by the
sequence .bn /n1 , defined for n 1 by bn D a1n .
4. Write a recurrence relation satisfied for each of the following sequences:
(a) .1; 1; 1; 3; 5; 9; 17; 31; 57; 105; 183; : : :/.
2
22
(b) .1; 2; 22 ; 22 ; 22 ; : : :/.
5. Below, we show the first four lines of an infinite array of natural numbers, such
that, for i > 1, the i-th line starts with the number i and has two more entries
than the .i 1/-th line. Compute the sum of the entries in the n-th line.
1
2
3
4
3
4
5
4
5
6
7
6
7
8
9
10
3.1 Progressions
69
6. Show that the number 11 : : : 1 (n digits 1) is equal to 10 91 .
7. Compute the sum 1 C 11 C 111 C C 11
: : :…
1 as a function of n.
„ ƒ‚
n
n
8. If .ak /k1 is an AP with common difference r, prove that the sequence .bk /k1 ,
defined by bk D a2kC1 a2k for every k 1, is also an AP, and compute its
common difference.
9. Let .ak /k1 be an AP of nonzero common difference, and p, q, u, v be given
naturals. Prove that
ap C aq D au C av , p C q D u C v
10. Let .ak /k1 be an AP of common difference r ¤ 0. If ar1 is a nonnegative integer,
prove that the sum of any two terms of the AP is also a term of it.
11. Let .ak /k1 be an AP such that ap D ˛ and aq D ˇ, with p ¤ q. Compute apCq
in terms of p, q, ˛, ˇ.
12. (Romania) The sum of some (more than one) consecutive odd integers
equals 73 . Find these numbers.
13. The AP .ak /k1 is formed by pairwise distinct naturals. Prove that it contains
infinitely many composite naturals2 among its terms.
14. (Canada) Let an be the sum of the first n terms of the sequence
.0; 1; 1; 2; 2; 3; 3; 4; 4; : : : ; r; r; r C 1; r C 1; : : :/:
(a) Find a formula for an as a function of n.
(b) Prove that amCn amn D mn for every natural numbers m and n, with
m > n.
p p
p
15. (Canada) Show that the numbers 2, 3 and 5 cannot be terms of a
single AP.
16. Let .ak /k1 be a GP with common ratio q. Prove that, for an arbitrary integer
n 1, we have
a1 a2 : : : an D an1 q
n.n1/
2
:
17. Compute the value of the sum
3
5
99
1
C 2 C 3 C C 50 ;
2
2
2
2
in which the k-th summand from left to right equals the quotient between the
k-th term of the AP 1; 3; 5; : : : ; 99 and the k-th term of the GP 2; 22 ; 23 ; : : : ; 250 .
2
According to the introduction to Chap. 1, we recall that a natural n > 1 is composite if we can
write n D ab, for some naturals a; b > 1.
70
3 Elementary Sequences
18. The sequence .ak /k1 is an arithmetic-geometric progression if, for every
integer k 1, we have ak D bk ck , where the sequences .bk /k1 and .ck /k1
are respectively an AP and a GP. Let r be the common difference of the AP and
q be the common ratio of the GP. Compute, as a function of n, b1 , c1 , r and q,
the value of the sum of the first n terms of .ak /k1 .
19. (Macedonia) In a nonconstant AP of real numbers, the quotient between the
first term and the common difference is an irrational number. Prove that no
three distinct terms of this AP form a GP.
20. Compute, as a function of n, the n-th term of the sequence .ak /k1 , given by
a1 D 2 and akC1 D 2ak 1, for every integer k 1.
21. The sequence .an /n1 satisfies a1 D 1 and akC1 D 3ak 1, for every k 1. Do
the following items:
(a) If bn D an 12 , prove that bkC1 D 3bk for every k 1.
(b) Write down the first five terms of the sequence .bn /n1 and, then obtain a
general positional formula for it.
(c) Find a positional formula for an .
22. Prove that there does not exist a GP having the numbers 2, 3 and 5 as three of
its terms.
3.2 Linear Recurrences of Orders 2 and 3
At the end of last section, we saw that a sequence .an /n1 is a GP if and only if it
satisfies a recurrence relation of the form (3.4), i.e., if and only if
akC1 qak D 0;
(3.6)
for every integer k 1, where q is a real constant. For reasons that will be clear in
a few moments, we say that (3.6) is a first order linear recurrence relation with
constant coefficients. There, we also saw that a sequence .an /n1 is an AP if and
only if it satisfies a recurrence relation like (3.3).
In this section, our task is to study the more general class of sequences .an /n1
that satisfy recurrence relations like
akC2 C rakC1 C sak D 0;
(3.7)
where r and s are given real constants, not both zero. If s D 0, (3.7) essentially
reduces to (3.6); if s ¤ 0, it is said to be a second order linear recurrence relation
with constant coefficients.
Problem 5 explains what we mean by the property of linearity of sequences that
satisfy recurrence relations like (3.6) and, more generally, (3.7). For such a sequence
.an /n1 , Theorem 3.16 will explain how to compute an as a function of n. Before we
present it, it is instructive to show the idea behind its proof in a concrete example,
and we start by doing this.
3.2 Linear Recurrences of Orders 2 and 3
71
Example 3.15 Let .an /n1 be the sequence such that a1 D 1, a2 D 7 and akC2 D
8akC1 15ak , for every integer k 1. Find a positional formula for an .
Solution For every integer k 1, we have
akC2 3akC1 D 5.akC1 3ak /:
Thus, the sequence .bk /k1 , defined by bk D akC1 3ak for every k 1, is a GP of
common ratio 5 and initial term b1 D a2 3a1 D 4. Hence, bk D b1 5k1 D 45k1 .
Analogously, for every integer k 1, we have
akC2 5akC1 D 3.akC1 5ak /;
so that the sequence .ck /k1 , given by ck D akC1 5ak for every k 1, is a GP
of common ratio 3. Since its initial term is c1 D a2 5a1 D 2, it follows that
ck D c1 3k1 D 2 3k1 .
Therefore, for every integer k 1, we have the system of linear equations
akC1 3ak D 4 5k1
;
akC1 5ak D 2 3k1
which easily gives
ak D 2 5k1 3k1 ; 8 k 1:
t
u
The reasoning presented in the solution to the previous example can be easily
generalized to deal with a general second order linear recurrence relation with
constant coefficients. We do this now, referring the reader to Sect. 11.4 for a different
approach.
Theorem 3.16 Let .an /n1 be a sequence of real numbers such that
akC2 C rakC1 C sak D 0
for every integer k 1, where r and s are given real constants, not both zero. If the
second degree equation x2 Crx Cs D 0 has real roots3 ˛ and ˇ, then there exist real
constants A and B, completely determined by the values of a1 and a2 , such that:
(a) If ˛ ¤ ˇ, then an D A˛ n1 C Bˇ n1 , for every integer n 1.
(b) If ˛ D ˇ, then an D .A C B.n 1//˛ n1 , for every integer n 1.
Proof Recall from Proposition 2.16 that ˛ C ˇ D r and ˛ˇ D s. Thus, (3.7) can
be rewritten as
akC2 .˛ C ˇ/akC1 C .˛ˇ/ak D 0
3
The case of complex roots is dealt with, in a much more general setting, in Chap. 21 of [5].
72
3 Elementary Sequences
or, which is the same, as
akC2 ˛akC1 D ˇ.akC1 ˛ak /
or
akC2 ˇakC1 D ˛.akC1 ˇak /;
for every integer k 1.
Letting bk D akC1 ˛ak and ck D akC1 ˇak , it follows from the above relations
that .bk /k1 and .ck /k1 are GP’s of common ratios respectively equal to ˇ and ˛,
and initial terms respectively equal to b1 D a2 ˛a1 and c1 D a2 ˇa1 . Therefore,
the formula for the general term of a GP gives
bk D .a2 ˛a1 /ˇ k1 and ck D .a2 ˇa1 /˛ k1
or, which is the same,
akC1 ˛ak D .a2 ˛a1 /ˇ k1
:
akC1 ˇak D .a2 ˇa1 /˛ k1
(3.8)
Now, let us first consider the case ˛ ¤ ˇ. In (3.8), subtracting the first relation
from the second gives
ak D
a2 ˇa1 k1 a2 ˛a1 k1
˛
ˇ I
˛ˇ
˛ˇ
AD
a2 ˛a1
a2 ˇa1
and B D ˛ˇ
˛ˇ
letting
we get the formula of item (a).
In case ˛ D ˇ, relations (3.8) are equal and, at a first glance, it seems that we do
not have enough information to compute ak . Nevertheless, we can use the following
trick: it is immediate to verify that the sequences .uk /k1 and .vk /k1 , given by
uk D ˛ k1 and vk D .k 1/˛ k1 (the same ˛ as before, root of the second degree
equation x2 C rx C s D 0), are such that
ukC2 C rukC1 C suk D 0
;
vkC2 C rvkC1 C svk D 0
for every integer k 1 (see Problem 4); therefore, for fixed A; B 2 R, the sequence
zk D Auk C Bvk is also such that
zkC2 C rzkC1 C szk D 0
3.2 Linear Recurrences of Orders 2 and 3
73
(see Problem 5). Thus, the idea is to search for real numbers A and B such that, for
every integer k 1, we have ak D zk . Since both sequences .ak /k1 and .zk /k1
satisfy identical linear second order recurrence relations, Problem 13, page 96, will
assure that it suffices to find real numbers A e B for which a1 D z1 and a2 D z2 , i.e.,
such that
a1 D A
:
a2 D .A C B/˛
This can obviously be done, for ˛ D 2r ¤ 0.
t
u
In practice, we employ the formulas of the previous result in the following way:
let .an /n1 be a sequence of real numbers satisfying the recurrence relation (3.7) for
every integer k 1, where r and s are given real constants, with r ¤ 0. We start by
computing the real roots (if any) of the second degree equation
x2 C rx C s D 0;
which is called the characteristic equation of (3.7). If the characteristic equation
actually has real roots ˛ and ˇ, we then check whether ˛ ¤ ˇ or ˛ D ˇ, and use the
formulas of items (a) or (b) of Theorem 3.16, according to the case at hand. In order
to find the real constants A and B in case ˛ ¤ ˇ, we solve the system of equations
a1 D A C B
I
a2 D A˛ C Bˇ
in case ˛ D ˇ, we solve the system
a1 D A
:
a2 D .A C B/˛
In both of these cases, the systems we have to solve are obtained by taking k
respectively equal to 1 and 2 in the formulas of items (a) or (b) of the theorem.
Example 3.17 Let us execute the above procedure to get a positional formula for the
n-th term of the celebrated Fibonacci sequence,4 i.e., the sequence .Fn /n1 given
by F1 D 1, F2 D 1 and
FkC2 D FkC1 C Fk ;
(3.9)
4
After Leonardo di Pisa, also known as Fibonacci, Italian mathematician of the XII and XIII
centuries. As can be seen in Chap. 1 of [5], the Fibonacci sequence plays a relevant role in
Combinatorics.
74
3 Elementary Sequences
for every integer k 1. More precisely, let us show that
1
Fn D p
5
(
p !n
1C 5
2
p !n )
1 5
; 8 n 1:
2
(3.10)
Solution The recurrence relation (3.9) can be written as FkC2 Fp
kC1 Fk D 0 and,
p
1C 5
2
hence, has characteristic equation x x 1 D 0. Letting ˛ D 2 and ˇ D 12 5
be its roots, it follows from item (a) of Theorem 3.16 that
Fn D A˛ n1 C Bˇ n1 ;
with A and B chosen so that F1 D F2 D 1.
Making n D 1 and n D 2 in the above formula for Fn , we get the linear system
ACB D1
:
˛A C ˇB D 1
Multiplying the first equation by ˛ and subtracting the second equation
p from the
result, we obtain .˛ ˇ/B p
D ˛ 1. Now, since ˛ C ˇ D 1 and ˛ ˇ D 5, we can
write this last equation as 5B D ˇ, so that B D pˇ5 . Analogously, multiplying
the first equation by ˇ and subtracting the result from the second equation, we obtain
A D p˛ , so that
5
˛
ˇ
1
Fn D p ˛ n1 p ˇ n1 D p .˛ n ˇ n /:
5
5
5
t
u
Our next example shows a situation where Theorem 3.16 can be applied, albeit
not straightforwardly.
Example 3.18 Let .an /n1 be a sequence of real numbers satisfying the recurrence
relation akC1 D rak C s for every integer k 1, where r and s are given real
constants, with r ¤ 0; 1 (we need not consider the case r D 1, for in this case
.an /n1 is an AP). For an integer k 1, we have
akC2 rakC1 D s D akC1 rak
or, which is the same,
akC2 .r C 1/akC1 C rak D 0:
This is a second order linear recurrence relation with constant coefficients, whose
characteristic equation is
x2 .r C 1/x C r D 0:
3.2 Linear Recurrences of Orders 2 and 3
75
Since its roots are r and 1, it follows from item (a) of Theorem 3.16 that an D
Arn1 C B. If we now recall that a2 D ra1 C s, we see that A and B can be found by
solving the linear system
A C B D a1
:
Ar C B D ra1 C s
Therefore, A D a1 and B D s, so that
an D a1 rn1 C s:
Let us now turn our attention to third order linear recurrence relations with
constant coefficients. As the reader has probably guessed by now, in this case we
have a sequence .an /n1 satisfying a recurrence relation of the form
akC3 C rakC2 C sakC1 C tak D 0
for every integer k 1, where r, s and t are given real constants, not all zero.
As in the case of second order linear recurrence relations with constant coefficients, we define the characteristic equation of (3.11) as the third degree
polynomial equation
x3 C rx2 C sx C t D 0:
According to Problem 21, page 43, such an equation admits, at most, three real
roots. In what follows, we shall assume that it indeed has three such roots, let us say
˛, ˇ and . Then, we have the following result.
Theorem 3.19 Let .an /n1 be a sequence of real numbers such that
akC3 C rakC2 C sakC1 C tak D 0
(3.11)
for every integer k 1, where r, s and t are given real constants, not all zero. If the
characteristic equation x3 C rx2 C sx C t D 0 of (3.11) has real roots ˛, ˇ and ,
then there exist real constants A, B and C, completely determined by the values of
a1 , a2 and a3 , such that:
(a) If ˛ ¤ ˇ ¤ ¤ ˛, then an D A˛ n1 C Bˇ n1 C C n1 , for every n 1.
(b) If ˛ D ˇ ¤ , then an D .A C B.n 1//˛ n1 C C n1 , for every n 1.
(c) If ˛ D ˇ D , then an D .A C B.n 1/ C C.n 1/2 /˛ n1 , for every n 1.
Proof We shall only sketch the proof, leaving the details to the reader (see
Problem 6).
Let .bn /n1 be the sequence given by bn D ˛ n1 , for every integer n 1. Since
3
˛ C r˛ 2 C s˛ C t D 0, we also have ˛ kC2 C r˛ kC1 C s˛ k C t˛ k1 D 0 or, which
is the same,
bkC3 C rbkC2 C sbkC1 C tbk D 0;
76
3 Elementary Sequences
for every integer k 1. Hence, .bn /n1 satisfies the same recurrence relation as
.an /n1 does. Analogously, the sequences .cn /n1 and .dn /n1 , given for an integer
n 1 by cn D ˇ n1 and dn D n1 , also satisfy the same recurrence relation as
.an /n1 .
We now turn our attention to the cases (a)–(c):
(a) For all A; B; C 2 R, the sequence .un /n1 such that un D A˛ n1 C Bˇ n1 C
C n1 for every integer n 1 satisfies the same recurrence relation as .an /n1
does. On the other hand, since ˛ ¤ ˇ ¤ ¤ ˛, we can choose the constants
A, B and C in such a way that u1 D a1 , u2 D a2 and u3 D a3 . Therefore,
by invoking once more the result of Problem 13, page 96, we conclude that
un D an , for every integer n 1.
(b) The result of Problem 21, page 43, gives us
x3 C rx2 C sx C t D .x ˛/2 .x /:
With the aid of this identity, it is easy to show that bn D .n 1/˛ n1 also
satisfies the same recurrence relation as the sequence .an /n1 . Then, as in item
(a), the sequence .un /n1 such that un D .A C B.n 1//˛ n1 C C n1 for n 1
satisfies that same recurrence relation, for all A; B; C 2 R. Moreover, we can
again choose the real constants A, B and C such that u1 D a1 , u2 D a2 and
u3 D a3 . Thus, once more by the result of Problem 13, page 96, we conclude
that un D an for every integer n 1.
(c) This time, the result of Problem 21, page 43, gives
x3 C rx2 C sx C t D .x ˛/3 :
It is now easy to show that both sequences bn D .n 1/˛ n1 and cn D .n 1/2 ˛ n1 do satisfy the same recurrence relation as the sequence .an /n1 . Hence,
the same is true for the sequence un D .A C B.n 1/ C C.n 1/2 /˛ n1 , for all
A; B; C 2 R. The rest of the argument proceeds exactly as in items (a) and (b).
t
u
Example 3.20 Let .an /n1 be the sequence of real numbers given by a1 D 1, a2 D
4, a3 D 14 and
akC3 6akC2 C 12akC1 8ak D 0;
for every integer k 1. Compute an as a function of n.
Solution The characteristic equation of the given recurrence relation is x3 6x2 C
12x 8 D 0. Since x3 6x2 C 12x 8 D .x 2/3 , item (c) of the previous result
gives
an D .A C B.n 1/ C C.n 1/2 / 2n1 ;
3.2 Linear Recurrences of Orders 2 and 3
77
for every integer n 1, where A, B and C are real constants chosen so that a1 D 1,
a2 D 4 and a3 D 14. These initial conditions, in turn, give us the linear system of
equations
8
<A D 1
;
ACBCC D2
:
A C 2B C 4C D 7=2
whose solution is A D 1, B D 3=4, C D 1=4. Then, we obtain
1
3
2
an D 1 C .n 1/ C .n 1/ 2n1
4
4
D .n2 C n C 2/ 2n3 :
t
u
Problems: Section 3.2
1. Let .an /n1 be the sequence given by a1 D 1, a2 D 4 and akC2 D 5akC1 6ak ,
for every integer k 1. Compute an as a function of n.
2. Let .an /n1 be the sequence given by a1 D 3, a2 D 5 and akC2 D 3akC1 2ak ,
for every integer k 1. Prove that an D 2n C 1, for every n 2 N.
3. Let .an /n1 be the sequence given by a1 D 3 and akC1 D 2ak 1, for every
integer k 1. Compute an as a function of n.
4. * If the second degree equation x2 C rx C s D 0 has two real roots equal to ˛,
prove that the sequences uk D ˛ k1 and vk D .k 1/˛ k1 satisfy the recurrence
relations
ukC2 C rukC1 C suk D 0
:
(3.12)
vkC2 C rvkC1 C svk D 0
5. * The sequences .un /n1 and .vn /n1 satisfy the recurrence relations (3.12), for
every integer k 1. For any fixed real constants A and B, prove that the sequence
.zn /n1 , given by zk D Auk C Bvk , satisfies an analogous recurrence relation.
6. * Fulfill the details in the proof of Theorem 3.19.
7. * The Lucas sequence5 is the sequence .Ln /n1 given by L1 D 1, L2 D 3 and
LkC2 D LkC1 C Lk , for every integer k 1. Show that, for every integer n 1,
we have:
(a) Ln D ˛ n C ˇ n , where ˛ D
(b) L2n D L2n C 2.1/n1 .
5
p
1C 5
2
and ˇ D
p
1 5
2 .
After François Édouard Anatole Lucas, French mathematician of the XIX century.
78
3 Elementary Sequences
(c) L2n 5Fn2 D 4.1/n , where .Fn /n1 is the Fibonacci sequence (cf. Example 3.17).
8. (IMO shortlist—adapted) With respect to the sequence .an /n1 , such that a1 D 1
and
anC1 D
p
1
.1 C 4an C 1 C 24an /
16
for every integer n 1, do the following items:
p
(a) If bn D 1 C 24an, show that 2bnC1 D bn C 3 for every n 1.
(b) Show that bn D 3 C 24n for every n 1.
(c) Conclude that, for every n 1, we have
1
1
1
an D
1 C n1
1C n :
3
2
2
3.3 The † and … Notations
P
Q
In this section, we introduce the notations
(one reads sigma) for sums and
(one reads pi) for products, which will reveal themselves quite useful in the context
of sequences.
P
Definition 3.21 Given a sequence .ak /k1 , we write njD1 aj to denote the sum a1 C
a2 C C an , and read it as the sum of the aj ’s, from j D 1 to n. Hence,
n
X
jD1
aj D
; if n D 1
a1
:
a1 C a2 C C an ; if n > 1
As a particular case of the above definition, if .ak /k1 is a constant sequence
with, say, ak D c for every k 1, we clearly have
n
X
jD1
aj D
n
X
c D nc:
jD1
The main reason for the success of the † notation owes to the fact that it turns it
quite easy to manipulate sums with large numbers of summands, specially when
each such summand is itself a sum. For example, given sequences .ak /k1 and
.bk /k1 of real numbers, the associativity and commutativity of addition of reals
give
.a1 C a2 C C an / ˙ .b1 C b2 C C bn / D .a1 ˙ b1 / C .a2 ˙ b2 / C C .an ˙ bn /I
3.3 The † and … Notations
79
with the aid of the † notation, the equality above can be written in the much more
compact form
n
X
aj ˙
jD1
n
X
bj D
jD1
n
X
.aj ˙ bj /:
(3.13)
jD1
On the other hand, given c 2 R, the distributivity of the multiplication of reals with
respect to addition gives
c.a1 C a2 C C an / D ca1 C ca2 C C can ;
an equality that can be written, again with the aid of the † notation, as
c
n
X
aj D
jD1
n
X
caj :
(3.14)
jD1
The following example shows, in a simple case, how one could apply the two
identities above in order to simplify the task of calculating the value of a sum.
P
Example 3.22 Compute the value of nkD1 .2k C 1/ in terms of n 2 N.
Solution Successively applying (3.13), (3.14) and the first part of Example 3.7, we
get
n
X
.2k C 1/ D
kD1
n
X
2k C
kD1
D2
n
X
1D2
kD1
n
X
kCn
kD1
n.n C 1/
C n D n2 C 2n:
2
t
u
P
The
notation is particularly useful to perform cancellations in sums. More
precisely, given a sequence .ak /k1 , and performing the possible cancellations in
the sum
.a2 a1 / C .a3 a2 / C .a4 a3 / C C .an1 an2 / C .an an1 /;
we get an a1 as result. With the aid of the
just get as
n1
X
jD1
P
notation, we can write the equality
.ajC1 aj / D an a1 :
(3.15)
80
3 Elementary Sequences
An equivalent formula (obtained from the above by writing n C 1 in place of n),
which will sometimes be used in the place of (3.15), is
n
X
.ajC1 aj / D anC1 a1 :
(3.16)
jD1
Any one of (3.15) or (3.16) is known as a telescoping sum. The idea behind
the name is the following: as a telescope shortens the immense distance between
a celestial body and our eyes, the above formulas shorten the way between certain
given sums and their results.
P
Telescoping sums are the greatest advantage of having the
notation at our
disposal. Let us justify this claim by examining two interesting examples.
Example 3.23 Deduce the formula for the general term of an AP with the aid of
telescoping sums.
Solution Let the sequence .ak /k1 be an AP of common difference r. Then, ajC1 aj D r for each integer j 1, so that (3.16) gives
an a1 D
n1
X
.ajC1 aj / D
jD1
n1
X
r D .n 1/r
jD1
or, which is the same, an D a1 C .n 1/r.
t
u
To the next example, we say that a sequence .ak /k1 is a second order AP
provided the sequence .bk /k1 , given for k 1 by bk D akC1 ak , is a
nonconstant AP. In order to build a second order AP .ak /k1 , we can start with a
nonconstant AP, as
.3; 7; 11; 15; 19; 23; 27; : : :/;
and stipulate the initial term of the second order AP, say a1 D 2; then, we
successively compute the values of a2 ; a3 ; : : : from the relations a2 a1 D 3,
a3 a2 D 7, a4 a3 D 11 etc. Proceeding this way we obtain, from the AP
given above, the second order AP
.2; 5; 12; 23; 38; 57; 80; : : :/:
The discussion of the previous paragraph makes it clear that a second order AP
is completely determined only if we know its first three terms. In fact, it is only by
knowing its first three terms that we will know the first two terms of the nonconstant
AP formed by the differences of the consecutive terms of the second order AP.
We are now in position to use telescoping sums to deduce a formula for the
general term of a general second order AP.
3.3 The † and … Notations
81
Example 3.24 Given a second order AP .ak /k1 , prove that
an D a1 C .a2 a1 /.n 1/ C
.n 1/.n 2/r
;
2
(3.17)
where r D a3 2a2 C a1 is the common difference of the nonconstant AP formed
by the differences of the consecutive terms of .ak /k1 .
Solution Let bk D akC1 ak for every integer k 1. The formulas for telescoping
sums and for the sum of the terms of a finite AP give us
an a1 D
n1
n1
X
X
.n 1/.b1 C bn1 /
:
.ajC1 aj / D
bj D
2
jD1
jD1
On the other hand, by applying the formula for the general term of an AP, we get
bn1 D b1 C .n 2/r D .a2 a1 / C .n 2/r
and, thus,
.n 1/.b1 C bn1 /
2
.n 1/.2.a2 a1 / C .n 2/r/
D a1 C
2
.n 1/.n 2/r
:
D a1 C .n 1/.a2 a1 / C
2
an D a1 C
Finally, it suffices to observe that
r D b2 b1 D .a3 a2 / .a2 a1 / D a3 2a2 C a1 :
t
u
In spite of the previous P
examples, if we wish to use the formula for telescoping
sums to compute the sum njD1 bj of the first n terms of a given sequence .bk /k1 ,
we first need to find a way to write the summands bj as the differences ajC1 aj ,
between consecutive terms of some other sequence .ak /k1 . The difficulty with this
reasoning relies upon the fact that we do not know, beforehand, who the sequence
.ak /k1 is. The discussion of some additional examples will help to clarify this point.
Example 3.25 Consider the sequence .ak /k1 given by a1 D 1 and
akC1 D
ak
;
1 C kak
for every integer k 1. Compute an as a function of n.
82
3 Elementary Sequences
Solution Since the terms of the sequence .ak /k1 are all positive (why?), we can
define the sequence .bk /k1 by setting bk D a1k . The given recurrence relation, then,
gives,
bkC1 D
1
akC1
D
1 C kak
1
D
C k D bk C k:
ak
ak
Thus, by applying the formula for telescoping sums, together with the formula for
the sum of the terms of a finite AP, we get
n1
n1
X
X
n.n 1/
:
.bkC1 bk / D
kD
bn b1 D
2
kD1
kD1
Therefore,
bn D b1 C
and, hence, an D
1
bn
D
n.n 1/
n2 n C 2
n.n 1/
D1C
D
;
2
2
2
2
.
n2 nC2
t
u
For the next example define, for each positive integer n, the factorial of n,
denoted nŠ, as the product of all natural numbers from 1 to n (by convention, 1Š D 1),
i.e., 2Š D 2, 3Š D 6, 4Š D 24, 5Š D 120 etc. Observe that, in general, we have
.k C 1/Š D .k C 1/ kŠ.
P
Example 3.26 (Canada) Compute the value of the sum njD1 j.jŠ/ in terms of n.
Solution In order to use the formula for telescoping sums, we have first to succeed
in writing j.jŠ/ as a difference ajC1 aj , of consecutive terms of a single sequence
.ak /k1 . We do this by observing that
j.jŠ/ D Œ.j C 1/ 1jŠ D .j C 1/ jŠ jŠ D .j C 1/Š jŠ:
Hence, by defining ak D kŠ for k 1, we get
n
X
jD1
j.jŠ/ D
n
X
jD1
Œ.j C 1/Š jŠ D
n
X
.ajC1 aj /
jD1
D anC1 a1 D .n C 1/Š 1Š:
t
u
As anticipated at the beginning of this section, we now present a very useful
notation for products of real numbers.
Q
Definition 3.27 Given a sequence .ak /k1 , we write njD1 aj to denote the product
a1 a2 : : : an , and read the product of the aj ’s, from j D 1 to j D n. Thus,
3.3 The † and … Notations
83
n
Y
aj D
jD1
; if n D 1
a1
:
a1 a2 : : : an ; if n > 1
Q
With the aid of the
notation, we can write the factorial of n 2 N (see the
paragraph that precedes Example 3.26) by writing
nŠ D
n
Y
j:
(3.18)
jD1
P
Q
As was the case with the
notation, the usefulness of the
notation comes
from the fact that it formally commutes with multiplications and divisions. In fact,
given a real number c and sequences .ak /k1 and .bk /k1 , we have
.a1 a2 : : : an /.b1 b2 : : : bn / D .a1 b1 /.a2 b2 / : : : .an bn /;
a1 a2
an
a1 a2 : : : an
D
b1 b2 : : : bn
b1 b2
bn
and
cn .a1 a2 : : : an / D .ca1 /.ca2 / : : : .can /
second equality above). Writing both
(provided all of the bj ’s are nonzero, in the Q
sides of the identities above by means of the notation, we obtain the relations
0
10
1
n
n
n
Y
Y
Y
@ aj A @ bj A D
.aj bj /;
jD1
Qn
jD1
aj
jD1
bj
Qn
D
n
Y
jD1
jD1
aj
bj
and cn
jD1
n
Y
jD1
aj D
n
Y
.caj /:
jD1
Below, we present an example of application of such formulas.
Q
Example 3.28 Compute, in terms of n, the value of nkD1 2 C 2k .
Q
Solution We apply the properties of the notation listed above:
Qn
n n
Y
Y
2
.k C 1/
kC1
n
Qn
D
D 2 kD1
2C
2
k
k
kD1 k
kD1
kD1
D 2n
.n C 1/Š
.n C 1/ nŠ
D 2n
nŠ
nŠ
D 2n .n C 1/:
t
u
84
3 Elementary Sequences
P
Q
Analogously to the case of the
notation, the notation is particularly useful
when performing cancellations in certain products. This is the content of the formula
for telescoping products, collected in the following result.
Proposition 3.29 If .ak /k1 is a sequence of nonzero real numbers, then
n
Y
ajC1
jD1
aj
D
anC1
:
a1
(3.19)
Proof As with telescoping sums, it suffices to observe that the intermediate factors
of the product at the left hand side all cancel. In symbols,
n
Y
ajC1
jD1
aj
D
a2 a3 a4
an anC1
anC1
D
:
a1 a2 a3
an1 an
a1
t
u
Example 3.30 We take another look at the previous example, this time with the
formula for telescoping products at our disposal. To this end, we first of all note that
n n Y
Y
2
kC1
D 2n
:
2C
k
k
kD1
kD1
Now, defining the sequence .ak /k1 by ak D k, it follows from (3.19) that
n n Y
Y
2
anC1
akC1
n
D2
2C
D 2n D 2n .n C 1/:
k
a
a
k
1
kD1
kD1
Problems: Section 3.3
1. Prove that a sequence .ak /k1 is a second order AP if and only if akC2 2akC1 C
ak ¤ 0 and akC3 3akC2 C 3akC1 ak D 0, for every integer k 1.
2. Let .ak /k1 be the sequence defined by a1 D 1 and anC1 D an C 3n 1, for
every positive integer n. Compute the n-th term of this sequence in terms of n.
3. The sequence .an /n1 is given by a1 D 1 and anC1 D an C 8n, for n 1.
Compute an in terms of n.
P
1
4. * Compute the value of the sum nkD2 .k1/k
as a function of n 2 N.
5. The sequence .ak /k1 is an AP. Prove that, for every positive integer n, we have
n1
X
kD1
n1
1
D
:
ak akC1
a1 an
3.3 The † and … Notations
85
6. Prove that, for every positive integer n, we have
1
1
1
1
1
C 2 C 2 CC 2 < 2 :
2
1
2
3
n
n
P
1
7. Compute the value of the sum n1
kD1 .4k1/.4kC3/ as a function of n 2 N.
8. (Romania) Let k and n be positive integers. Prove that:
(a) .2kC1/3 .2k1/3 can always be written as a sum of three perfect squares.
(b) .2n C 1/3 2 can always be written as a sum of 3n 1 perfect squares.
P
k
as a function of the integer n > 1.
9. Compute the value of the sum nkD1 .kC1/Š
10. (Brazil—adapted) Do the following items:
(a) For k 2 N, write .k C 1/2 C k2 C k2 .kqC 1/2 as a perfect square.
P
1
1
C .kC1/
(b) Compute the value of the sum 99
2 C 1.
kD1
k2
11. * (Brazil) Let .Fk /k1 be the Fibonacci sequence, i.e., the sequence given by
F1 D 1, F2 D 1 and FkC2 D FkC1 C Fk , for every integer k 1. Compute the
P
F
in terms of n 2 N.
value of the sum nkD1 Fk FkC1
kC2
12. The sequence .ak /k1 is an AP. Prove that, for every positive integer n, we have
n1
X
kD1
n1
1
Dp
p
p :
p
ak C akC1
a1 C an
2
C 1.
13. (a) Factorise the expression x4 C xP
(b) Compute the value of the sum nkD1 k4 Ckk2 C1 in terms of n 2 N.
14. (Australia) Compute the value of the sum
999;999
X
kD1
1
p
p
p
:
3 2
3
2
.k C 1/ C k 1 C 3 .k 1/2
15. Compute, in terms of n 2 N, the value of the sum
n
X
kD1
1
p
:
p
.k C 1/ k C k k C 1
16. (Germany) Given a natural
number n > 1, compute, as a function of n, the value
Qn 1
of the product jD2 1 j2 .
17. For 0 k 101, let xk D
k
.
101
Compute the value of the sum
101
X
kD0
x3k
:
1 3xk C 3x2k
86
3 Elementary Sequences
18. (Leningrad) Compute, as a function of n 2 N, the value of the product
n1 3
Y
k C1
:
k3 1
kD2
For the next problem, we recall that the arithmetic mean A of a finite
collection a1 , a2 , . . . , ak of real numbers is the real number
AD
a1 C a2 C C ak
:
k
19. (Macedonia) Find all values of n 2 N for which we can write the set A D
f1; 2; 3; : : : ; 4ng as the union of n pairwise disjoint 4-element subsets, such that,
in each of them, one element is the arithmetic mean of the other three.
20. (China—adapted) Use the result of Problem 13, page 26, to compute the value
of the greatest integer that is less than or equal to
1
1
1
1
S D p C p C p CC p
:
10;000
1
2
3
21. * Given positive integers n and p, prove that:
< .k C 1/p , for every k 2 N.
(a) kp < .kC1/ pC1k
Pn1 p npC1 Pn
(b) kD1 k < pC1 < kD1 kp .
pC1
pC1
Chapter 4
Induction and the Binomial Formula
With the algebraic background of the previous chapters at our disposal, we devote
the first section of this chapter to the development of a very important topic in
elementary Mathematics, namely, the principle of mathematical induction. It will
considerably improve our ability of elaborating proofs of mathematical statements
that depend on natural numbers. In the other two sections, we shall apply this
principle to deduce Newton’s formula for binomial expansion, as well as some
important related results.
4.1 The Principle of Mathematical Induction
Generally speaking, there are several ways of proving something. For instance,
we can make a direct proof, or a proof by contradiction (for more on Logic and
proof techniques, see the first chapters of [21]). In this sense, the principle of
mathematical induction will provide us yet another way of proving certain types
of mathematical statements.
To understand how this principle works, let’s consider a subset A N such that
1 2 A. Also, suppose we know that, whenever a certain natural number k is in A,
then, so k C 1 also is in A. Therefore, 1 2 A assures that 2 D 1 C 1 2 A; accordingly,
2 2 A allows us to conclude that 3 D 2 C 1 2 A. Continuing this way, we conclude
that A contains all natural numbers, or, which is the same, that A D N. This intuitive
discussion can be formalized as the following axiom, known as the principle of
mathematical induction.
Axiom 4.1 Let A N be a set satisfying the following conditions:
(a) 1 2 A.
(b) k 2 A ) k C 1 2 A.
Then A D N.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_4
87
88
4 Induction and the Binomial Formula
At this point, a natural question is: how can we apply the principle of mathematical induction to prove something in Mathematics? In order to answer it, suppose
there is given a property P.n/ of the natural number n, which we wish to prove to
be true for every n 2 N. Then, we let
A D fk 2 NI P.k/ is trueg
and observe that
A D N , .P.n/ is true for every n 2 N/:
Thus, in order to prove that P.n/ is true for every n 2 N, it suffices to prove that
A D N; yet in another way, by invoking the principle of mathematical induction, it
suffices to prove that:
• 1 2 A;
• k 2 A ) k C 1 2 A.
In turn, the definition of the set A assures that, showing the validity of the two items
above is the same as showing that
• P.1/ is true;
• P.k/ true ) P.k C 1/ true.
The previous discussion can be summarized as the following recipe for a proof
by induction.
Proposition 4.2 If P.n/ is a property of the natural number n, then P.n/ is true for
every n 2 N if and only if the following two conditions are satisfied:
(a) P.1/ is true;
(b) P.k/ true ) P.k C 1/ true.
To understand how a proof by induction works in practice, let’s start by
discussing two examples, the first of which was already considered in the previous
chapter.
Example 4.3 Prove that, for every n 2 N, the sum of the first n odd natural numbers
equals n2 .
Proof Since the kth odd natural number is 2k1, the property P.n/ is, in this case,
the same as
P.n/ W
n
X
.2j 1/ D n2 :
jD1
As was said before, in order to prove by induction that P.n/ is true for every n 2 N,
we have to verify that:
4.1 The Principle of Mathematical Induction
89
i. P.1/ is true.
ii. P.k/ true ) P.k C 1/ true.
Checking i. is immediate: the first odd natural number is 1, which is the same
as 12 . In order to verify ii., we have to assume that P.k/ is true, i.e., that
k
X
.2j 1/ D k2 ;
(4.1)
jD1
and find a way to deduce from this that P.k C 1/ is also true, i.e., that
kC1
X
.2j 1/ D .k C 1/2 :
(4.2)
jD1
Since we are assuming that P.k/ is true, we can substitute (4.1) into the left hand
side of (4.2) to get
kC1
X
jD1
.2j 1/ D
k
X
.2j 1/ C .2k C 1/ D k2 C .2k C 1/ D .k C 1/2 :
jD1
Hence, it follows by induction that P.n/ is true for every n 2 N.
t
u
Example 4.4 Prove that, for every n 2 N, the sum of the first n perfect squares is
equal to
1
n.n C 1/.2n C 1/:
6
Proof Since the kth perfect square is the number k2 , the property P.n/ is, in this
case,
P.n/ W
n
X
j2 D
jD1
1
n.n C 1/.2n C 1/:
6
As before, in order to compose a proof by induction, we have to verify that:
i. P.1/ is true.
ii. P.k/ true ) P.k C 1/ true.
Again, verifying i. is immediate: 12 D
P.k/ is actually true, i.e., that
k
X
jD1
j2 D
1.1C1/.21C1/
.
6
To verify ii. let’s assume that
1
k.k C 1/.2k C 1/;
6
(4.3)
90
4 Induction and the Binomial Formula
and find a way to deduce that P.k C 1/ is also true, i.e., that
kC1
X
j2 D
jD1
1
.k C 1/Œ.k C 1/ C 1Œ2.k C 1/ C 1:
6
(4.4)
Substituting (4.3) into the left hand side of (4.4), we obtain
kC1
X
j2 D
jD1
k
X
j2 C .k C 1/2
jD1
1
k.k C 1/.2k C 1/ C .k C 1/2
6
1
D .k C 1/Œk.2k C 1/ C 6.k C 1/
6
1
D .k C 1/.k C 2/.2k C 3/:
6
D
Hence, it follows by induction that P.n/ is true for every n 2 N.
t
u
Below, we state (also as an axiom) a slightly more general form of the principle
of mathematical induction, which gives it greater flexibility in applications.
Axiom 4.5 Let a 2 N and A fa; a C1; a C2; : : :g be a set satisfying the following
conditions:
(a) a 2 A.
(b) k 2 A ) k C 1 2 A.
Then, A D fa; a C 1; a C 2; : : :g.
As the reader may be guessing right now, the prototype application of this form
of the principle of mathematical induction is to prove that a certain property P.n/ is
true for every natural number n a, where a is a natural number given in advance.
This can be done by letting
A D fk 2 NI P.k/ is trueg
and observing that
A D fa; a C 1; a C 2; : : :g
m
P.n/ is true for every natural number n a:
This way, we obtain the following more general recipe for proofs by induction.
4.1 The Principle of Mathematical Induction
91
Proposition 4.6 Let a 2 N. If P.n/ is a property of the natural number n, then P.n/
is true for every natural number n a if and only if the following two conditions
are satisfied:
(a) P.a/ is true;
(b) P.k/ true ) P.k C 1/ true.
The next example illustrates the fact that, sometimes, this more general form of
proof by induction is really necessary.
Example 4.7 Prove that nŠ > 2n , for every natural number n 4.
Proof First of all, observe that we really need to start from n D 4, for the inequality
nŠ > 2n is not true when n D 1, 2 or 3.
The property P.n/ we wish to prove is
P.n/ W nŠ > 2n :
In order to prove it by means of mathematical induction, we have to prove that
P.4/ is true and that P.k/ true ) P.k C 1/ true. The validity of P.4/ follows from
4Š D 24 > 16 D 24 . Let’s then suppose that, for some k 2 N, the property P.k/ is
actually true, i.e., that
kŠ > 2k :
From this, we wish to deduce the trueness of P.k C 1/, i.e., that .k C 1/Š > 2kC1 .
To this end, we first use the trueness of P.k/ to get
.k C 1/Š D .k C 1/ kŠ > .k C 1/ 2k I
subsequently, we note that
.k C 1/ 2k 2kC1
for every integer k 1. Hence, by combining the two inequalities got above, we
arrive at .k C 1/Š > 2kC1 , thus concluding that P.k C 1/ is indeed true.
Therefore, it follows by induction that the property P.n/ is true for every integer
n 4.
t
u
Before we present another example, it is appropriate to make some comments
on terminology: in a proof by induction, the step P.k/ ) P.k C 1/ is generally
called the induction step. In order to execute it, we assume that the property P.k/
is indeed true (and this assumption constitutes our induction hypothesis) and,
then, use this trueness, possibly together with other arguments, to deduce that the
property P.k C 1/ is also true. Thus, a proof by mathematical induction, as stated in
Proposition 4.6, is usually done by executing the following steps:
92
•
•
•
•
4 Induction and the Binomial Formula
identification: isolate the property P.n/ to be proved;
initial case: verify the validity of P.a/;
induction hypothesis: assume the trueness of P.k/.
induction step: deduce the validity of P.k C 1/, with the aid of the induction
hypothesis and (perhaps) other arguments.
The statement of each problem generally makes it quite clear what the property
P.n/ is. Therefore, a proof by induction is usually centered around the three last
steps of the above scheme. Moreover, it is also customary not to explicitly state
either the property P.k/ or the induction step P.k/ ) P.k C 1/. The discussion of
the next example is performed in this shortened way.
Example 4.8 (Brazil) For every integer n > 2, show that there exist n pairwise
distinct natural numbers whose sum of inverses equals 1.
Proof Let’s make induction on n 3, the initial case n D 3 following from
1
1
1
C C D 1:
2
3
6
Now suppose, by induction hypothesis, that, for a certain natural k 3, there
exist natural numbers x1 < x2 < < xk such that
1
1
1
C
CC
D 1:
x1
x2
xk
Multiplying both sides of the equality above by
the resulting equality, we obtain
1
2
and adding
1
2
to both sides of
1
1
1
1
C
C
CC
D 1:
2
2x1
2x2
2xk
Since 2 < 2x1 < 2x2 < < 2xk , we have got k C 1 pairwise distinct
natural numbers whose sum of inverses is also equal to 1, and this completes the
induction step.
t
u
When applying the principle of mathematical induction to prove something,
one must be very careful in the execution of the induction step, to avoid absurd
conclusions. This point is better illustrated with the following classical example and
the subsequent discussion on the reasoning we shall present.
Example 4.9 In a farm, if one horse is white, show that all horses are white.
Proof Let’s “prove” the statement above by induction on the number n of horses.
To the initial step, note that if a farm has just one horse and at least one of its
horses is white, then surely all horses in the farm are white.
By induction hypothesis, assume that in a farm with k horses, if at least one of
them is white, then all are white. Then, consider a farm with k C 1 horses, also with
4.1 The Principle of Mathematical Induction
93
at least a white one. Take one horse from the farm, one that is not that we knew a
priori to be white. Since we are left with k horses in the farm, at least one of which
is white, it follows from the induction hypothesis that all of the k horses are white.
Now, bring back the horse who was taken from the farm, and take from it one that
was left at the first time. Again, we are left with k horses, at least one of which is
white, so that, applying the induction hypothesis one more time, we conclude that
all of these k horses are also white. However, since the horse we took from the
farm this second time was already white, we conclude that all of the k C 1 horses
are white.
t
u
It is quite evident that there is some absurd in the “proof” presented above, for the
statement of the example does not reflect the reality. The point is that we did not
succeed in executing the induction step, for our reasoning does not work in a farm
with two horses, as you can readily verify.
There is yet another important form of the principle of mathematical induction,
generally referred to as the strong principle of mathematical induction, or simply
as strong induction. We describe it next.
Axiom 4.10 Let A N be a set satisfying the following conditions:
(a) 1 2 A.
(b) f1; : : : ; kg A ) k C 1 2 A.
Then, A D N.
At this point, the reader should probably find it clear how to use of the strong
principle of induction to compose proofs. Nevertheless, in order to further illustrate
it, we take a look at a couple of examples.
Example
4.11 (OCS
p
p - adapted) Show that, for every n 2 N, the number .7 C
4 3/n C .7 4 3/n is a positive even integer.
p
p
Proof Let u D 7C4 3 and v D 74 3. Then, uCv D 14 and uv D 1, from where
it follows that u and v are the roots of the second degree equation x2 14x C 1 D 0.
It follows from this that u2 D 14u 1 and v 2 D 14v 1, and hence, for every
integer k 2,
uk D 14uk1 uk2 and v k D 14v k1 v k2 :
Thus, letting sj D uj C v j and adding both relations above, we get, for every integer
k 2, that
sk D 14sk1 sk2 :
Now, observe that s0 D 2 and s1 D u C v D 14, both integers. Suppose, by
induction hypothesis, that sk 2 Z for every integer 1 k < n. Then, the recurrence
relation above gives
sn D 14sn1 sn2 ;
from where we conclude that sn , being the sum of two integers, is also integer.
94
4 Induction and the Binomial Formula
To what was left to prove, note that u; v > 0 guarantee that sn D un C v n is
positive for every n 2 N. On the other hand, again from the recurrence relation for
the sequence .sk /k1 , we conclude that sk and sk2 always have the same parity, i.e.,
they are either both even or both odd. However, since s0 and s1 are both even, it
follows again by induction that sn is an even integer, for every natural n.
t
u
Example 4.12 Show that every natural number n can be written, in a unique way,
as a sum of powers of 2 with nonnegative and pairwise distinct integer exponents.
This way of writing n is called its binary representation or expansion.
Proof Let’s make a proof by strong induction. For n D 1, we have 1 D 20 , and
this is obviously the only possible binary representation of 1. Now, suppose that the
desired result is true for every natural number less than a certain natural n > 1.
We first show that n has at least one binary representation. To this end, take the
greatest power of 2 which is less than or equal to n, say 2k . Then,
2k n < 2kC1 ;
so that 0 n 2k < 2k . If n 2k D 0, there is nothing left to do. Otherwise,
1 n 2k < n and, by induction hypothesis (here, we are assuming that every
natural number less than n has a binary representation; therefore, we are making a
strong induction argument), there exist nonnegative integers 0 a0 < a1 < < al
such that
n 2k D 2a0 C 2a1 C C 2al :
Using once more the inequality n 2k < 2k , it follows that 2a0 C 2a1 C C 2al < 2k
and, thus, al < k. Hence,
n D 2a0 C 2a1 C C 2al C 2k ;
with 0 a0 < a1 < < al < k.
We then show that the binary representation of n is unique. So, let’s assume that
n D 2a0 C 2a1 C C 2aj D 2b0 C 2b1 C C 2bl ;
with 0 a0 < a1 < < aj and 0 b0 < b1 < < bl . Then,
2aj 2a0 C 2a1 C C 2aj
D n D 2b0 C 2b1 C C 2bl
20 C 21 C C 2bl
D 2bl C1 1;
4.1 The Principle of Mathematical Induction
95
where in the last passage we used the formula for the sum of the terms of a finite
GP. It follows from the above that 2aj < 2bl C1 and, hence, aj < bl C 1, i.e., aj bl .
Changing the roles of aj and bl in the above computations, we analogously get bl aj and, thus, aj D bl . Letting aj D bl D k, say, we obtain
n 2k D 2a0 C 2a1 C C 2aj1 D 2b0 C 2b1 C C 2bl1 :
(4.5)
Since n 2k < n, if we now invoke the uniqueness part of the induction hypothesis
(i.e., if we assume that the binary representation of every natural number less than
n is unique), then (4.5) gives j 1 D l 1 and a0 D b0 , a1 D b1 , . . . , aj1 D bl1 ,
as we wished to prove.
t
u
Before you go on and try the proposed problems, there are two remarks that
ought to be made. Firstly, if we want to prove the validity of a certain property P.n/
of the natural number n, then using induction will not always be the best possible
choice. For example, try Problem 1 and, after you are done, compare your proof by
induction to the one given in Example 3.7. Secondly, apart from Problems 8 to 12,
we do not present any applications of induction (and there are lots of them) to either
Combinatorics or Number Theory. Instead, we refer the interested reader to [5].
Problems – Section 4.1
1. Use the principle of induction to prove that the sum of the first n natural
numbers equals n.nC1/
.
2
2. Prove that, for every natural number n, one has
n.n C 1/ 2
13 C 23 C C n3 D
:
2
3. Prove that, for every natural number n, one has
1
1
1
1
1
1
1
C C
D C
CC
:
2
3
2n 1
n
nC1
2n 1
4. (Canada) For n 2 N, let h.n/ D 1 C
1
2
C
1
3
C C 1n . Prove that
n C h.1/ C h.2/ C h.3/ C C h.n 1/ D nh.n/:
5. Show that, for every integer n > 1, one has
1 2 C 2 3 C C .n 1/n D
1
.n 1/n.n C 1/:
3
6. Prove that, for every integer n > 1, one has
12 C 32 C 52 C C .2n 1/2 D
1
.2n 1/2n.2n C 1/:
6
96
4 Induction and the Binomial Formula
7. Given n 2 N and real numbers a1 ; a2 ; : : : ; an , prove the general version of the
triangle inequality:
ja1 C a2 C C an j ja1 j C ja2 j C C jan j:
8.
9.
10.
11.
12.
13.
14.
(4.6)
Also, show that if a1 ; a2 ; : : : ; an ¤ 0, then equality holds in (4.6) if and only if
a1 , a2 , . . . , an have equal signs.
Prove that 9 divides 4n C 15n 1, for each natural n.
Prove that 3 divides n3 n, for each natural number n.
n1
Prove that 3n divides 43 1, for every integer n 1.
The result of the next problem is known as the fundamental principle of
counting or as the multiplicative principle.
* Show that, if we have n1 ways of choosing an object of type 1, n2 ways of
choosing an object of type 2, . . . , nk ways of choosing an object of type k, then
the number of ways of choosing simultaneously one object of each of the types
from 1 to k is n1 n2 : : : nk .
* Prove that a set with n elements has exactly 2n subsets.
* Let .an /n1 and .bn /n1 be two sequences satisfying the second order linear
recurrence relation with constant coefficients ukC2 C rukC1 C suk D 0, for
every k 1. If a1 D b1 and a2 D b2 , prove that an D bn , for every n 1. Then, extend this result to the case in which both sequences .an /n1 and
.bn /n1 satisfy identical third order linear recurrence relations with constant
coefficients.
Let .an /n1 be a sequence of nonzero real numbers such that, for every n 2,
we have
n1
X
jD1
n1
1
D
:
aj ajC1
a1 an
Prove that the sequence is an AP.
15. (Bulgaria - adapted) * The sequence .an /n1 is defined by a1 D 1 and, for k 1
integer, akC1 D a2k ak C 1. Prove that, for every integer n 1, we have:
(a) P
anC1 D a1 : : : an C 1.
n
1
1
(b)
jD1 aj D 2 a1 a2 :::an .
16. Prove that, for every natural number n, we have 22 > nn .
17. (Macedonia) Let x be a nonzero real number, such that x C x1 2 Z. Prove that
xn C xn 2 Z, for every integer n.
18. (Brazil) Let .xn /n1 and .yn /n1 be sequences of real numbers such that, for
every integer k 1, we have xkC1 D x3k 3xk e ykC1 D y3k 3yk . If x21 D y1 C 2,
show that x2n D yn C 2 for every integer n 1.
19. (Putnam) Let .xn /n0 be a sequence of nonzero real numbers satisfying, for
every integer n 1, the recurrence relation x2n xn1 xnC1 D 1. Prove that there
exists a real number ˛ such that xnC1 D ˛xn xn1 , for every n 2 N.
n
4.1 The Principle of Mathematical Induction
97
The next four problems concern the Fibonacci sequence. We refer the reader
to Example 3.17, for a review of its definition.
20. Let .Fn /n1 be the Fibonacci sequence. Prove that, for every n 2 N, the
following identities are true:
(a)
(b)
(c)
(d)
(e)
F1 C F2 C C Fn D FnC2 1.
F12 C F22 C C Fn2 D Fn FnC1 .
F1 C F3 C C F2n1 D F2n .
F2 C F4 C C F2n D F2nC1 1.
2
FnC1
Fn FnC2 D .1/n .
21. Let .Fn /n1 be the Fibonacci sequence. Prove that, for every m; n 2 N, with
m > 1, we have FmCn D Fm FnC1 C Fm1 Fn .
22. Let .Fn /n1 be the Fibonacci sequence and .Ln /n1 be the Lucas sequence (cf.
Problem 7, page 77). Given m; n 2 N, prove that:
(a) 2FmCn D Fm Ln C Fn Lm .
(b) 2LmCn D 5Fm Fn C Lm Ln .
23. Let .Fn /n1 be the Fibonacci sequence.
(a) Prove thatpFn D n2 if and only if n D 1 or 12.1
(b) If ˛ D 1C2 5 , do the following items:
i. Prove that ˛ n D Fn ˛ C Fn1 , for every integer n 1.
ii. Find all n 2 N such that ˛ n n2 ˛ is an integer.
24. (TT) Let .an /n1 be a sequence of pairwise distinct natural numbers. Prove that
there are infinitely many naturals k such that ak > k.
25. (France) Let .an /n1 be a sequence of positive real numbers such that a1 D 1
and
a31 C a32 C C a3n D .a1 C a2 C C an /2 ;
for every integer n 1. Show that an D n, for every integer n 1.
26. * For a fixed real number a > 1, let .xn
/n1 be a sequence of real numbers such
p
p
1
that a < x1 < a C 1 and xkC1 D 2 xk C xak , for each integer k 1. Prove
that, for every integer n 1, we have
p
a < xn p
1
a C n1 :
2
27. (France) Prove that, for every integer n > 5, there exist n positive integers such
that the sum of the inverses of their squares is equal to 1.
More generally, it can be shown that Fn is a perfect square if and only if n D 1 or 12. For a proof
of this fact, see the problems of Section 12.3 of [5].
1
98
4 Induction and the Binomial Formula
28. Given m; n 2 N, with m > 1, prove that there exist unique nonnegative integers
k, a0 , a1 , : : :, ak such that 0 ai < m for 0 i k, ak ¤ 0 and
n D ak mk C ak1 mk1 C C a2 m2 C a1 m C a0 :
In this case, we write n D .ak ak1 : : : a1 a0 /m and say that the right hand side
above is the representation of n in base m, or the madic representation of n.
29. (Sweden) Prove that, for every natural number n, there exists a unique sequence
.aj /j1 of integers, such that 0 aj j for every j 1 and
n D a1 1Š C a2 2Š C a3 3Š C :
30. * Prove Zeckendorf’s theorem2 : every natural number can be uniquely written
as a sum of Fibonacci numbers of nonconsecutive and greater than 1 indices.
4.2 Binomial Numbers
We start by recalling the definition of fatorial, extended to the nonnegative integers.
Definition 4.13 The fatorial of a given nonnegative integer n is the number nŠ,
given by
(
nŠ D
1; if n D 0
Qn
:
jD1 j; if n 1
One could guess that it would be much more reasonable to define 0Š as being
equal to 0. Nevertheless, the reasons for letting 0Š be equal to 1 will soon become
evident.
Definition
4.14 Given integers n and k, with 0 k n, we define the binomial
number nk by
!
nŠ
n
D
:
k
kŠ.n k/Š
It is easy to verify, directly from the definition, that
!
n
D 1;
0
2
!
n
D n;
1
!
n.n 1/
n
D
2
2
After Édouard Zeckendorf, belgian mathematician of the XX century.
4.2 Binomial Numbers
99
for every nonnegative integers n for which the binomial numbers above are defined.
On the other hand, for every integers n and k such that 0 k n, we have
!
!
n
n
D
:
(4.7)
k
nk
Indeed, this follows at once from
!
!
nŠ
nŠ
n
n
D
D
D
:
k
kŠ.n k/Š
.n k/Š.n .n k//Š
nk
Notice that the binomial numbers n0 , n1 and n2 , whenever defined, are all
naturals (in the case of n2 D n.n1/
, this is due to the fact that the product of
2
any
two
consecutive
integers
is
an
even
number). This is also true in the case of
n
n.n1/.n2/
D
,
for,
among
any
three
consecutive integers, there is always a
6
3
multiple of 3 and an even integer, so that n.n
1/.n
n 3/ isnalways
n a multiple
of 6. On the other hand, (4.7) assures that nn D n0 , n1
D 1 , n2
D n2 and
n n
D 3 , whenever defined, are also natural numbers.
n3
The discussion of the previous paragraph makes it natural to ask whether nk is
a natural number for every choice of integers n and k such that 0 k n. This is
indeed the case, and will be a straightforward consequence of relation (4.8) below,3
which is known as Stifel’s relation.4
Proposition 4.15 Given integers n and k such that 1 k < n, we have
!
!
!
n
n1
n1
D
C
:
k
k
k1
(4.8)
Proof It suffices to apply the definition of binomial number to the right hand side
of the equality above, together with some algebraic manipulations:
!
!
.n 1/Š
n1
n1
.n 1/Š
C
D
C
k
k1
kŠ.n 1 k/Š
.k 1/Š.n k/Š
1
.n 1/Š
1
C
D
.k 1/Š.n 1 k/Š k
nk
.n 1/Š
n
.k 1/Š.n 1 k/Š k.n k/
!
nŠ
n
D
:
D
kŠ.n k/Š
k
D
t
u
3
4
Another proof is the object of Problem 8.
After Michael Stifel, german mathematician of the XVI century.
100
4 Induction and the Binomial Formula
With the binomial numbers we construct an infinite triangular numerical table,
known as Pascal’s triangle,5 in the following way: we count lines and columns
starting from 0, lines being labelled from top to bottom and columns from left to
right; for 0 k n, the .n; k/th entry, i.e., the number written
on the crossing of
the nth line and the kth column, is the binomial number nk . More precisely:
• The entries of column
0, read
from
top to bottom, are respectively equal to the
binomial numbers 00 ; 10 ; 20 ; 30 ; : : :. As we have already seen, all such numbers
are equal to 1.
• Line 0 is formed solely by the binomial number 00 D 1. Line 1 is composed by
the binomial numbers 10 and 11 , both of which are also equal to 1.
• In general, the entries ofline
n,
read from
left to right, are respectively equal to
the binomial numbers n0 ; n1 ; n2 ; : : : ; nn .
According to the above recipe, we show below the first few lines of Pascal’s
triangle:
Pascal’s Triangle
0
0
1
0
1
1
2
0
2
1
2
2
3
0
3
1
3
2
3
3
4
0
4
1
4
2
4
3
4
4
5
0
5
1
5
2
5
3
5
4
5
5
6
0
6
1
6
2
6
3
6
4
6
5
6
6
···
···
···
···
···
···
···
···
With respect to Pascal’s triangle, Stifel’s relation says that, whenever we add, in
line n 1, the entries at columns k 1 and k, we get the entry situated at line n and
column k. This is more difficult to say than to understand and verify,
and allows us
to recursively get the numerical values of the binomial numbers nk . The table below
shows the numerical values of the binomial numbers nk for 0 n 6, obtained
with the aid of Stifel’s relation.
5
After Blaise Pascal, french mathematician of the XVII century. Besides the triangle that bears his
name, there is an important Pascal’s theorem in the theory of conics, which has greatly motivated
the developments of Projective and Algebraic Geometry.
4.2 Binomial Numbers
101
1
1
1
1
1
1
1
1
2
3
4
5
6
1
3
6
10
15
1
4
10
20
1
5
15
1
6
1
···
···
···
···
···
···
···
···
More gerally, since n0 D 1 and nn D 1 whenever these numbers are defined,
it is not difficult for the reader to convince himself or herself that nk 2 N for all
integers n and k such that 0 k n. We give a formal proof of this fact in the
corollary below.
Corollary 4.16 For all integers n and k such that 0 k n, we have nk 2 N.
Proof Let’s make induction
on n 0, the case n D 0 being obvious, for the only
such binomial number is 00 D 1.
Now suppose, by induction hypothesis, that n1
is a natural number for every
j
0 j n 1, and consider a binomial number of the form nk . There are two cases
to consider:
(i) If k D 0 or k D n, then we have already observed that n0 D nn D 1.
(ii) If 1 k n 1, then it follows from the induction hypothesis that both n1
k
n1
and k1 are natural numbers. Hence, Stifel’s relation gives
!
!
!
n
n1
n1
D
C
2 N:
k
k
k1
t
u
Later (cf. Theorem 4.20), we shall see that one of the main utilities of Pascal’s
triangle is that it provides an easy and quick way to write down the expansion of
the binomial .x C y/n for small values of n. For the time being, let’s establish some
useful identities relating the entries of a row, column or diagonal of it.
The formula of the following proposition is known as the columns’ theorem of
Pascal’s triangle.
Proposition 4.17 In the column n of Pascal’s triangle, the sum of the entries in
rows n, n C 1, . . . , n C k 1 is equal to the entry located at column n C 1 and row
n C k. In symbols,
!
!
k1
X
nCj
nCk
D
:
n
nC1
jD0
(4.9)
102
4 Induction and the Binomial Formula
Proof Let’s make
a proof
by induction on k 1. The initial case k D 1 reduces to
verifying that nn D nC1
nC1 , which is immediate.
By induction hypothesis, suppose that, when k D l 1, (4.9) is true for all
nonnegative integer values of n. Then, for k D l C 1 and every integer n 0,
we have
!
!
!
l
l1
X
X
nCj
nCj
nCl
D
C
n
n
n
jD0
jD0
!
!
nCl
nCl
D
C
nC1
n
!
nClC1
D
;
nC1
where we used Stifel’s relation in the last equality above.
Therefore, it follows by induction that (4.9) is true for every k 2 N and every
integer n 0.
t
u
In the coming example, we employ the columns’ theorem to compute the sum of
the first n perfect squares.
Example 4.18 Given n 2 N, compute the value of 12 C 22 C C n2 in terms of n.
P
Solution Letting S D njD1 j2 , we have
!
!
n
n
X
X
j
j
C
:
SD
j.j 1/ C
jD2
2
1
jD1
jD1
jD2
jD1
n
X
n
X
Therefore, it follows from (4.9) that
!
!
1
nC1
nC1
SD2
C
D n.n C 1/.2n C 1/:
6
3
2
t
u
Given an integer n 0, the diagonal n of Pascal’s triangle is formed by the
binomial numbers
!
!
!
!
n
nC1
nC2
nC3
;
;
;
;::::
0
1
2
3
With respect to them, the following corollary is the diagonals’ theorem of Pascal’s
triangle.
4.2 Binomial Numbers
103
Corollary 4.19 In the diagonal n of Pascal’s triangle, the sum of the entries in rows
0; 1; : : : ; k1 is equal to the entry located at row nCk and column k1. In symbols,
!
!
k1
X
nCj
nCk
D
:
j
k1
jD0
Proof Since
nCj
j
D
(4.10)
nCj
, it follows from the columns’ theorem that
n
! k1
!
!
k1
X
X nCj
nCj
nCk
D
D
D
j
n
nC1
jD0
jD0
!
nCk
:
k1
t
u
Problems – Section 4.2
is even, for every n 2 N.
1. Prove that 2n
n
2. Use the results of this section to compute the value of 13 C 23 C C n3 , for
every n 2 N.
3. * For n 2 N, prove that
!
!
n
n
<
< <
0
1
n
!
>
n
2
n
2
!
!
n
n
> >
C1
n
if n is even, and
!
!
n
n
<
< <
0
1
!
n
n1
2
D
n
!
nC1
2
!
n
> >
n
if n is odd.
5n
4. Prove that, for every integer n 2, one has 2 4 < 2n
n .
5. Given natural numbers k and m, with m > k, prove that
n k1
nnC1
nDk k
k
m
X
1
D 1 mC1 :
k
6. Let n and k be given integers, such that 0 k n. Prove that
!
!
!
!
!
n
n
n
n
n
1
C
C .1/k
D .1/k
:
0
1
2
k
k
104
4 Induction and the Binomial Formula
7. Let .Fk /k1 be the Fibonacci sequence (cf. Example 3.17). Show that, for every
n 2 N, one has
n1
!
bX
2 c
nj1
;
Fn D
j
jD0
where b n1
2 c denotes the greatest integer less than or equal to
n1
2
D
n1
;
2
n3
2 ;
n1
2 ,
i.e.,
if n is odd
:
if n is even
8. * Givenintegers
n and k such that 0 k n, prove that the set f1; 2; : : : ; ng has
exactly nk subsets of k elements each.
4.3 The Binomial Formula
We shall now obtain Newton’s binomial formula,6 i.e., we shall explicitly write
.x C y/n as a sum of monomials of the form xk yl .
Theorem 4.20 Given n 2 N, we have
!
n
X
n nj j
x y:
.x C y/ D
j
jD0
n
(4.11)
Proof For n D 1, we have
!
!
1 1
1 1
.x C y/ D x C y D
x C
y:
0
1
1
Suppose, by induction hypothesis, that (4.11) is true for n D k, i.e., that
!
k
X
k kj j
x y:
.x C y/ D
j
jD0
k
6
Sir Isaac Newton, english mathematician and physicist of the XVII century, is considered to be
one of the greatest scientists ever. Actually, it is difficult to properly address Newton’s contribution
to the development of science. Known as the father of modern Physics, Newton also created,
together with G. W. Leibniz, the Differential and Integral Calculus. His masterpiece, Philosophiae
Naturalis Principia Mathematica, is one of the most influential books ever written and contains
the cornerstones of both Calculus and Physics.
4.3 The Binomial Formula
105
For n D k C 1, we thus have
.x C y/
kC1
!
k
X
k kj j
x y
D .x C y/.x C y/ D .x C y/
j
jD0
k
!
!
k
k
X
k kC1j j X k kj jC1
x
x y
D
y C
j
j
jD0
jD0
Dx
kC1
!
!
k
k1
X
k kC1j j X k kj jC1
x
x y C ykC1 :
C
y C
j
j
jD1
jD0
P In the last line above, let’s perform
P the following changes of indices: in the first
,
change
j
per
l
and,
in
the
second
, change j C 1 per l; in this way, in the second
P
we have j D l 1 and 0 j k 1 , 1 l k. Then, we get
.x C y/
kC1
!
!#
k
k
C
xkC1l yl C ykC1
Dx
C
l
l
1
lD1
!
k
X
k C 1 kC1l l
kC1
x
Dx
C
y C ykC1 ;
l
lD1
kC1
k
X
"
where we used Stifel’s
relation
kC1 in the last equality.
Finally, since kC1
D
0
kC1 D 1, we can write the last line above as
!
!
!
k
k C 1 kC1 X k C 1 kC1l l
k C 1 kC1
x
x
C
y C
y
l
0
kC1
lD1
or, which is the same,
!
kC1
X
k C 1 kC1l l
x
y:
l
lD1
This is exactly the expression we would like to reach, so that, by induction, we
conclude that (4.11) is true for every n 2 N.
t
u
Corollary 4.21 Given n 2 N, we have
!
n
.x y/n D
xnj yj :
.1/j
j
jD0
n
X
Proof It suffices to apply (4.11), writing y in place of y.
t
u
106
4 Induction and the Binomial Formula
It is customary to write
!
n nj j
x y
Tj D
j
and say that such a monomial Tj is the general term of the expansion of .x C y/n .
Hence,
.x ˙ y/n D T0 ˙ T1 .˙1/n Tn :
In what follows, we collect some examples of application of the binomial
formula, as well as some important consequences of it.
n
p
Example 4.22 Find the least n 2 N for which the expansion of x x C x14 has a
summand not depending on x. For such an n, compute this summand.
Solution In expanding the given expression, the general term is
!
!
!
p nk 1 k
n
n 3 .nk/ 4k
n 3n11k
.x x/
D
x2
x
D
x 2 :
k
x4
k
k
Therefore, there is a summand not depending on x if and only if there exists 0 k n such that 3n 11k D 0. Hence, n D 11k
, so that the least possible n 2 N is
3
n
D
11,
obtained
for
k
D
3.
In
this
case,
the
summand
that doesn’t depend on x is
11
D
165.
t
u
3
p
Example 4.23 Let k 2 N and a; b; r 2 Q, with r > 0 such that r is irrational.
Prove that:
p k
p
(a) There exist
p c;k d 2 Q such
p that .a C b r/ D c C d r.
p
p
(b) If .a C b r/ D c C d r, with c; d 2 Q, then .a b r/k D c d r.
Proof
p
(a) By expanding .a C b r/k , we get
X
p
.a C b r/k D
0jk
2jj
!
!
k kj j p j p X k kj j p j1
a b r C r
a b r :
j
j
0jk
2−j
Making
cD
X
0jk
2jj
!
!
X k
p
k kj j p j
a b r and d D
akj bj rj1 ;
j
j
0jk
it is immediate that c; d 2 Q (since
an odd j).
(4.12)
2−j
p
p
rj 2 Q for an even j, and rj1 2 Q for
4.3 The Binomial Formula
107
p
(b) It sufficespto note that the expansion of .a b r/k is essentially equal to that
of .a C b r/k , the difference lying on the signs of half of the summands. More
precisely, in the notations of (4.12), we get
X
p
.a b r/k D
0jk
2jj
!
!
k kj j p j p X k kj j p j1
a b r r
a b r
j
j
0jk
p
D c d r:
2−j
t
u
Example 4.24 Use Newton’s binomial formula to prove Lagrange’s identity:7
!2
!
n
X
n
2n
D
:
j
n
jD0
(4.13)
Proof Writing .1Cx/2n D .1Cx/n .1Cx/n and applying Newton’s binomial formula
to both sides, we get
!
! !0 n
! 1
! !
2n
n
n
X
X
2n k
n i @X n j A X n n iCj
x D
x
x D
x :
k
i
j
i
j
kD0
iD0
jD0
i;jD0
Comparing the coefficients of xn in the first and last expressions above and
using (4.7), we get
!
! !
!
!
!2
n
n
X n n
X
X
2n
n
n
n
D
D
D
:
n
i
j
i
ni
i
iCjDn
iD0
iD0
t
u
Item (a) of the following corollary is known as the lines’ theorem of the Pascal’s
triangle.
Corollary 4.25 Given n 2 N, we have:
Pn n
D 2n .
(a)
PjD0 jn P
(b)
D 0jn nj D 2n1 :
0jn
j
2jj
2−j
Proof For item (a), it suffices to set x D y D 1 in the formula for .x C y/n .
7
Joseph Louis Lagrange, french physicist and mathematician of the XVIII century. Lagrange was
one of the greatest scientists of his time, with notable contributions to Physics and Mathematics.
In particular, he was a pioneer in the fields of Calculus of Variations and Celestial Mechanics.
108
4 Induction and the Binomial Formula
In what concerns item (b), we start by setting x D 1 and y D 1 in the binomial
formula to get (check it!)
X
0D
0jn
2jj
Now, letting A D
P
0jn
2jj
above relation that
n
j
!
!
X n
n
:
j
j
0jn
and B D
2−j
P
0jn
2−j
n
, it follows from (a) and the
j
A C B D 2n
:
ABD0
Hence, we have A D B D 2n1 .
t
u
Remark 4.26 It is possible to generalize the formulas of item (b) of the previous
corollary in the following way: given integers 0 r < k < n, we can compute the
value of the sum
!
!
!
n
n
n
C
C
C
r
kCr
2k C r
in terms of n, k andr.For instance,
for kD 3 we
can
compute, interms
of
n2 N, the
values of the sums n0 C n3 C n6 C , n1 C n4 C n7 C and n2 C n5 C n8 C .
The corresponding deduction uses the multisection formula, which will be obtained
in Chapter 15 of [5], with the aid of complex numbers.
We finish this small chapter with an example that illustrates the use of the lines’
theorem of Pascal’s triangle to the computation of sums.
Example 4.27 Given n 2 N, compute the value of
!
!
!
!
n
n
n
n
C2
C3
CCn
:
1
2
3
n
Solution First of all, note that for j 2 N we have
n
j
j
!
nŠ
nŠ
D
jŠ.n j/Š
.j 1/Š.n j/Š
!
.n 1/Š
n1
Dn
:
Dn
.j 1/Š.n j/Š
j1
Dj
4.3 The Binomial Formula
109
Hence, it follows from the lines’ theorem that
!
!
n
n
X
X
n1
n
D n 2n1 :
j
Dn
j
1
j
jD1
jD1
t
u
Problems – Section 4.3
P 1. Given n 2 N, compute the value of the sum njD0 nj 3j .
Pn n k
Pn1 n1 k
B
38
2. Let A D
kD0 k 3 and B D
kD0
k 11 . If A D 4 , compute the
value of n.
1
3. Show that .1; 1/n 200
.n2 C 19n C 200/.
4. (OCM) Without directly evaluating all summands, compute the value of the
sum
1
1
1
1
1
1
C
C
C
C
C
:
10Š
3Š8Š
5Š6Š
7Š4Š
9Š2Š
11Š
65
5. Find the maximal term in the expansion of 1 C 13 .
6. (Baltic Way). Let a, b, c and d be real numbers for which a2 C b2 C .a C b/2 D
c2 C d2 C .c C d/2 . Prove that a4 C b4 C .a C b/4 D c4 C d4 C .c C d/4 .
22n1
7. Given n 2 N, prove that 2n
n .
n >
8. In the expansion of x.1Cx/n , divide the coefficient of each term by the exponent
nC1
of x in that term. Prove that the sum of all these numbers equals 2 n 1 .
Pn n j
9. Given a 2 Z n f0g and n 2 N, compute the value of the sums jD0 j ja and
Pn n 2
jD0 j j in terms of a and n.
10. The sequence .ak /k1 is an AP. Prove that, for every integer n > 1, the following
identities are valid:
Pn n
(a)
.1/jC1 aj D 0.
j
PjD0
n n
jC1 2
(b)
ajC1 D 0.
jD0 j .1/
11. * If 0 < q < 1 and n 2 N, prove that qn <
q
qCn.1q/ .
p
p
12. Given a; b; n 2 N, with n > 1, prove that the number a C n b is the root of a
polynomial equation with integer coefficients and degree 2n.
13. (Croatia) Let x, y and z be nonzero real numbers, such that x C y C z D 0. Prove
that the value of
x5 C y5 C z5
xyz.xy C yz C zx/
does not depend on the particular values of x, y and z.
110
4 Induction and the Binomial Formula
14. * Given n 2 N and j; k; l nonnegative
integers such that j C k C l D n, we define
n the trinomial number j;k;l
by
!
nŠ
n
:
D
jŠkŠlŠ
j; k; l
Prove the trinomial expansion formula
.x C y C z/ D
n
X
jCkClDn
!
n
xj yk zl ;
j; k; l
(4.14)
where the sum in the right hand side extends over all possible choices of
nonnegative integers j; k; l such that j C k C l D n.
15. Use the result of the previous problem to expand the trinomials .x C y C z/3 and
.x C y C z/4 .
10
16. Compute, in the expansion of 1 C x C 6x , the term that doesn’t depend on x.
17. Prove the following identities involving trinomial numbers:
n
P
n
(a)
jCkClDn j;k;l D 3 .
P
l n
(b)
jCkClDn .1/ j;k;l D 1.
18. * Prove that, for every n 2 N, we have
!
n
X
n
F2nC1j D F2nC1 1;
j
jD1
where Fk is the kth Fibonacci number (cf. Example 3.17).
19. (IMO shortlist) Prove that
!
995
X
1
.1/k
1991 k
D
:
1991
k
k
1991
kD0
Chapter 5
Elementary Inequalities
This chapter is an invitation to the systematic study of algebraic inequalities. More
precisely, our main purpose here is to discuss some interesting examples of inequalities, for whose derivation we can use the simple mathematics we developed so far.
Later, when we have the tools of Calculus at our disposal, we shall return to the study
of algebraic inequalities, largely generalizing some of those we will study here.
5.1 The AM-GM Inequality
As will be clear in this section, what allows us to develop a systematic study of
inequalities is the basic fact that the square of any real number is nonnegative, and
it is zero if and only if the number in question is also equal to zero.
To start with, for x; y 2 R we know that .jxj jyj/2 0, with equality if and only
if jxj D jyj. If we expand the left hand side, we get the inequality jxj2 C jyj2 2jxyj
or, which is the same,
x2 C y2
jxyj;
2
(5.1)
with equality if and only if jxj D jyj.
p
If we p
now start with two positive real numbers a and b, and make x D a 0
and y D b 0, it follows from (5.1) that
aCb p
ab;
2
p
p
with equality if and only if a D b, i.e., if and only if a D b.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_5
(5.2)
111
112
5 Elementary Inequalities
The simplicity of the previous reasoning hides its importance. Actually, (5.2)
is a particular case of a much more general inequality, known as the arithmeticgeometric mean inequality (see Theorem 5.7). For the time being, let’s see how to
deduce other interesting inequalities from (5.2).
Example 5.1 Given positive real numbers x and y, we have:
(a) x C 1x 2, with equality if and only if x D 1.
4
(b) 1x C 1y xCy
, with equality if and only if x D y.
Proof
(a) Applying (5.2) with a D x and b D 1x , we get
r
1
1
x C 2 x D 2;
x
x
with equality if and only if x D 1x , i.e., if and only if x D 1 (for, x > 0 by
hypothesis).
4
(b) Since x C y > 0, we have 1x C 1y xCy
if and only if .x C y/ 1x C 1y 4 or,
which is the same, if and only if xy C yx 2. In turn, this last inequality follows
directly from that of item (a), with xy in the place of x. It also follows from (a)
that equality holds if and only if xy D yx , i.e., if and only if x2 D y2 . However,
since x; y > 0, such condition is clearly equivalent to x D y.
For another proof of item (b), apply (5.2) twice and multiply the results:
1
1
C
.x C y/
x
y
p
2 xy 2
s
1 1
D 4:
x y
It follows directly from (5.2) that we have equality if and only if x D y.
t
u
Example 5.2 Given positive real numbers x, y and z, prove that
x2 C y2 C z2 xy C xz C yz;
(5.3)
with equality if and only if x D y D z.
Proof Applying (5.1) three times, we get the inequalities
x2 C z2
y2 C z2
x 2 C y2
xy;
xz;
yz:
2
2
2
It now suffices to add these three inequalities to reach (5.3).
If x D y D z, then (5.3) clearly becomes an equality. Conversely, if x ¤ y (the
2
2
cases x ¤ z and y ¤ z can be treated analogously), then x Cy
> xy. Therefore,
2
an obvious generalization of item (e) of Proposition 1.2 assures that, adding the
inequalities
5.1 The AM-GM Inequality
113
x2 C z2
y2 C z2
x2 C y2
> xy;
xz;
yz;
2
2
2
we get x2 C y2 C z2 > xy C xz C yz. Hence, in order to have equality in (5.3), we
must have x D y D z.
t
u
The next two examples extend (5.2) to three and four positive reals. We start with
the case of four positive numbers.
Example 5.3 Given positive real numbers a, b, c and d, we have
aCbCcCd p
4
abcd;
4
with equality if and only if a D b D c D d.
p
ab and
Proof We already know that aCb
2
aCbCcCd
D
4
aCb
2
C
2
cCd
2
cCd
2
(5.4)
p
cd. Therefore,
p
p
q
p p
p
ab C cd
4
ab cd D abcd;
2
where the last inequality above follows from yet another application of (5.2).
Equality happens if and only if we have equality inpall of the inequalities
p above.
cCd
For the first one of them, we must have aCb
D
ab
and
D
cd, and,
2
p 2 p
hence, a D b and c D d. For the second one,
pwe must
p have ab D cd; however,
since a D b and c D d, this is the same as a2 D c2 , i.e., a D c. Therefore, we
have equality if and only if a D b D c D d.
t
u
The previous example raises the natural question of whether a similar inequality
is true for three positive real numbers. As we have already anticipated, this is indeed
the case, albeit its deduction will not be as immediate as the one just discussed
(nevertheless, see Problem 11).
Example 5.4 Given positive real numbers a, b and c, we have
aCbCc p
3
abc;
3
(5.5)
with equality if and only if a D b D c.
ProofpWe apply the inequality of Example 5.4 to the four positive reals a, b, c and
d D 3 abc to get
p
q
p
p
p
a C b C c C 3 abc
4
4
3
3
abc abc D d3 d D d D abc:
4
p
p
p
3 abc.
Then, a C b C c C 3 abc 4 3 abc or, which is the same, aCbCc
3
114
5 Elementary Inequalities
Fig. 5.1 Rectangular
parallelepiped with
dimensions x, y and z
y
z
x
From the above computations it is clear that equality holds if and only if it also
happens in the inequality
aCbCcC
4
p
3
abc
q
4
p
3
abc abc:
Therefore,
it follows from Example 5.3 that equality holds if and only if a D b D
p
c D 3 abc, i.e., if and only if a D b D c.
u
t
At this point, the reader is probably formulating in his/her mind a natural
generalization of inequalities (5.2), (5.4) and (5.5), which we anticipated to be true.
Before we formally state and prove it, let’s see an interesting application of (5.5).
Example 5.5 A box (i.e., a rectangular parallelepiped1) is such that the sum of its
edge lengths equals 48 cm. If the box encloses the largest possible volume, find its
dimensions.
Solution Figure 5.1 shows a rectangular parallelepiped with dimensions x, y and
z (measured in centimeters), such that the sum of the lengths of its twelve edges
equals 48 cm. Since each dimension comprises four equal and parallel edges, this is
the same as saying that x C y C z D 12.
It is a well known fact that the volume V of such a parallelepiped is given by
the formula V D xyz. Hence, algebraically our problem reduces to the one of
maximizing the product xyz, under the restrictions that x; y; z > 0 and xCyCz D 12.
To this end, we apply inequality (5.5) to get
V D xyz xCyCz
3
3
D
12
3
3
D 64;
thus concluding that V is at most 64 cm3 . Example 5.4 also shows that the volume
equals 64 cm3 (i.e., that equality holds in the above inequality) if and only if x D
y D z or, which is the same, if and only if x D y D z D 4 cm. Therefore, the box of
largest possible volume is a cube with edge length equal to 4 cm.
t
u
1
For the elementary facts on the notion of volume of solids, we refer the reader to [4].
5.1 The AM-GM Inequality
115
Before we state the promised generalization of (5.2), (5.4) and (5.5), we need
some terminology.
Definition 5.6 Given n > 1 positive real numbers a1 ; a2 ; : : : ; an , we define their:
(a) Arithmetic mean as the number 1n .a1 C a2 C C an /.
p
(b) Geometric mean2 as the number n a1 a2 : : : an .
In the context of the previous definition, what we did in (5.2) and in Examples 5.3
and 5.4 was to show that the arithmetic means of two, three or four positive reals
are always greater than or equal to the respective geometric means, with equality in
each case if and only if the given numbers are all equal. We establish the general
case in the coming result, whose proof can be omitted in a first reading. As the title
of this section suggests, inequality (5.6) below is known as the arithmetic meangeometric mean inequality.
Theorem 5.7 Given n > 1 positive reals a1 ; a2 ; : : : ; an , their arithmetic mean is
always greater than or equal to their geometric mean. In symbols:
a1 C a2 C C an p
n a1 a2 : : : ; an ;
n
(5.6)
with equality if and only if a1 D a2 D D an .
Proof Firstly, let’s prove by induction that the desired inequality is true whenever n
is a power of 2, with equality if and only if a1 D a2 D D an . To this end, we have
to verify the initial case n D 2 (which was already done along the discussion that
established (5.2)), formulate the induction hypothesis (for n D 2j , say) and execute
the induction step (i.e., deduce the case n D 2jC1 from the case n D 2j ). However,
since 2jC1 D 2 2j , it suffices to suppose the inequality to be true for any k positive
reals (with equality if and only if these k numbers are all equal) and, from this, to
deduce that it is also true for any 2k positive reals (with equality if and only if these
2k positive reals are also all equal). Hence, to perform the induction step, consider
2k positive reals a1 ; a2 ; : : : ; a2k . The induction hypothesis, together with the case
n D 2, give
1
0
2k
k
k
1 @1 X
1X
1 X
aj D
aj C
akCj A
2k jD1
2 k jD1
k jD1
1 p
p
k
a1 : : : ak C k akC1 : : : a2k
2
q
p
p
k a1 : : : ak k akC1 : : : a2k
p
D 2k a1 : : : ak akC1 : : : a2k :
2
The reader maybe find it useful if we observe that the adjective geometric attached to this number
comes from the case n D 2. In this case, inequality (5.2) has a simple geometric interpretation, for
which we refer to the problems of Sect. 4:2 of [4].
116
5 Elementary Inequalities
To have equality, we must have it in all passages above. Then, it must be that
akC1 C C a2k
a1 C C ak
p
p
D k a1 : : : ak ;
D k akC1 : : : a2k
k
k
and
p
p
q
k a :::a C k a
p
p
1
k
kC1 : : : a2k
D k a1 : : : ak k akC1 : : : a2k :
2
For the first two equalities, it follows from the induction hypothesis that a1 D
p
D ak and akC1 D D a2k . For the third one, we must have k a1 : : : ak D
p
k a
kC1 : : : a2k ; this last condition, together with the two former ones, implies that
we must have a1 D D ak D akC1 D D a2k . It is also evident (verify,
anyhow!) that, if the 2k positive given numbers are equal, then equality must happen.
Therefore, (5.6) is true by induction, with the stated condition for equality, whenever
n is a power of 2.
Let’s now prove, by strong induction, that the inequality is true in general, with
equality holding if and only if all of the numbers are equal. To this end, let n > 1 be
natural and a1 ; a2 ; : : : ; an be given positive reals. Take k 2 N such that 2k > n, and
apply the arithmetic mean-geometric mean inequality to the numbers a1 ; a2 ; : : : ; an ,
p
together with 2k n copies of the number a D n a1 a2 : : : an (thus, to a total of
k
k
n C .2 n/ D 2 numbers, in which case we already know that the inequality is
true). We get
a1 C C an C a C C a
2k
D
q
2k
a1 : : : an a2k n
p
p
2k
an a2k n D
a2k D a:
2k
Therefore, it follows that a1 C a2 C C an C .2k n/a 2k a and, hence,
p
a1 C a2 C C an
a D n a1 a2 : : : an :
n
For the equality to hold, we already know that we must have a1 D a2 D D
an D a D D a; in particular, all of the numbers a1 ; a2 ; : : : ; an must be equal.
Finally, it’s immediate to see that, if all of the numbers a1 , a2 , . . . , an are equal, then
equality holds.
t
u
The following corollary generalizes item (b) of Example 5.1. For an alternative
proof, see Problem 21.
Corollary 5.8 Given n > 1 positive reals a1 ; a2 ; : : : ; an , we have
.a1 C a2 C C an /
1
1
1
C
CC
a1
a2
an
n2 ;
(5.7)
5.1 The AM-GM Inequality
117
with equality if and only if a1 D a2 D D an .
Proof Applying (5.6) twice, we get
1
1
1
C
CC
.a1 C a2 C C an /
a1
a2
an
s
!
p
1
1
n 1
n
D n2 :
n a1 a2 : : : an n
a1 a2
an
In order to have equality in (5.7), we must have equality in (5.6), and we know that
this implies a1 D a2 D D an . Conversely, it is immediate to verify that, if all of
the n given numbers are equal, then we have equality in (5.7).
t
u
Remark 5.9 The inequality (5.7) is sometimes referred to as the arithmetic meanharmonic mean inequality. This is due to the fact that it can be written as
a1 C a2 C C an
n
1=a1 C 1=a2 C C 1=an
n
1
;
and that the right hand side above (i.e., the inverse of the arithmetic mean of the
inverses of a1 , a2 , . . . , an ) is known as the harmonic mean of a1 , a2 , . . . , an .
We now illustrate the use of the inequality on the arithmetic and geometric means
in three examples.
Example 5.10 (Israel-Hungary) Let k and n be positive integers, with n > 1. Prove
that
!
r
1
1
1
n k C 1
C
CC
>n
1 :
kn
kn C 1
kn C n 1
k
Proof It suffices to see that
n1
X
jD0
X
n1 n1
X
1
kn C j C 1
1
Cn D
C1 D
kn C j
kn
C
j
kn C j
jD0
jD0
v
r
un1
uY kn C j C 1
n k C 1
n
> nt
Dn
;
kn
C
j
k
jD0
where we’ve applied the inequality on the arithmetic and geometric means once.
Note that, since the numbers knCjC1
are pairwise distinct, equality never holds. u
t
knCj
118
5 Elementary Inequalities
Example 5.11 (APMO) If a, b and c are positive reals, prove that
b c
aCbCc
a
1C
1C
2 1C p
1C
:
3
b
c
a
abc
Proof Expanding the left hand side, we obtain
b c
aCc
bCc
aCb
a
1C
1C
D2C
C
C
;
1C
b
c
a
b
a
c
and it suffices to prove that
bCc
aCb
2.a C b C c/
aCc
C
C
:
p
3
b
a
c
abc
Letting S denote the left hand side of the last expression above, it follows
from (5.6) and (5.7) that
1
1
1
C C
3
a
b
c
1
1
1
1
1
2
1
1
C C
C .a C b C c/
C C
3
D .a C b C c/
3
a
b
c
3
a
b
c
2
1
3
.a C b C c/ p
C 93
3
3
3
abc
S D .a C b C c/
D
2.a C b C c/
p
:
3
abc
t
u
Example 5.12 Gabriel has a sheet of cardboard of 2 m by 3 m. In order to assemble
an open box, he cuts four equal squares from the corners of the sheet, folds it along
the cuts and glues the lateral faces of the box along their common edges. If the box
is to have the largest possible volume, what should be the length of the sides of the
squares he cut? Justify your answer.
Solution If x is the common length of the sides of the cut squares, then one must
clearly have 0 < x < 1. Since the box has height x and its bottom is a rectangle of
side lengths 2 2x and 3 2x, the volume Gabriel wants to maximize depends on
x and equals .2 2x/.3 2x/x.
One possibility for him is to try to apply the inequality between the arithmetic
and geometric means to get rid of x and, then, see what the condition for equality
says about the size of x. However, he cannot do this directly, for, although
.2 2x/.3 2x/x .2 2x/ C .3 2x/ C x
3
the expression at the right hand side still depends on x.
3
;
5.1 The AM-GM Inequality
119
Nevertheless, the following trick does the job: he starts by choosing positive reals
a, b and c such that a.22x/Cb.32x/Ccx doesn’t depend on x and such that there
exists at least one value of x 2 .0; 1/ for which a.2 2x/ D b.3 2x/ D cx. This
amounts to finding a positive solution .a; b; c/ for the linear equation 2aC2bc D 0,
such that this solution, in turn, gives equal solutions for the first degree equations
2.a b/x D 2a 3b and .2a C c/x D 2a, which should belong to the interval .0; 1/.
Hence, we should have
2a C 2b D c and
2a 3b
2a
D
2 .0; 1/:
2.a b/
2a C c
If we succeed in finding a; b; c > 0 satisfying the given equations, we will
2a
automatically have 2aCc
2 .0; 1/. To what is left to do, substitute c D 2a C 2b into
2a3b
a
the second equation to get 2.ab/
D 2aCb
or, which is the same, 2a2 2ab3b2 D 0.
a
Therefore, b is a positive solution of the second degree equation 2u2 2u 3 D 0,
p
p
so that ab D 1C2 7 . It thus suffices to choose b D 2, a D 1 C 7 and c D 2a C 2b D
p
6 C 2 7.
With these choices of a; b; c at hand, Gabriel can successfully implement the
heuristic reasoning of the second paragraph of the proof. Writing V for the volume
and recalling that 2a C 2b D c, we have
abc V D a.2 2x/b.3 2x/cx
a.2 2x/ C b.3 2x/ C cx
3
2a C 3b 3
D
:
3
3
Equality holds if and only if a.22x/ D b.32x/ D cx. However, we already know
2a
that these two equations have x D 2aCc
as common solution, so that the maximal
possible volume is attained only for x D x0 D
1 2a C 3b
abc
3
3
2a
2aCc
D
p
5 7
,
6
and equals
p
10 C 7 7
:
D x0 .2 2x0 /.3 2x0 / D
27
t
u
Later, when we have the methods of Calculus at our disposal, the previous
example will fall into the general framework of the analysis of the first and second
variations of the function V.x/ D .2 2x/.3 2x/x; as such, it will have a
straightforward solution. Nevertheless, the solution we presented above, of choosing
adequate weights prior to applying the inequality between the arithmetic and
geometric means, albeit quite tricky, is an instructive one for several other situations.
120
5 Elementary Inequalities
Problems: Section 5.1
1. * Generalize item (a) of Example
ˇ
ˇ 5.1. More precisely, prove that, if x is a
nonzero real number, then ˇx C 1x ˇ 2, with equality if and only if jxj D 1.
3
3
2. Given positive reals a and b, prove that ab C ba a2 C b2 . When does equality
hold?
3. Given real numbers a < b < c, prove that the equation
1
1
1
C
C
D0
xa
xb
xc
has exactly two distinct real roots.
4. (Brazil) Let a, b and c be positive real numbers. Prove that
p
.a C b/.a C c/ 2 abc.a C b C c/:
5. (USA) Prove that, for every positive real numbers a, b and c, one has
a3
1
1
1
1
C 3
C 3
:
3
3
3
C b C abc
b C c C abc
c C a C abc
abc
For the next two problems, the reader shall need some Euclidean Geometry,
which we review next.3 More precisely (cf. Fig. 5.2), if a D BC, b D AC and
c D AB are the lengths of the sides of a triangle ABC, then there exist x; y; z > 0
such that a D y C z, b D x C z and c D x C y. In fact, it suffices to take x, y
and z as being equal to the lengths of the line segments determined on the sides
of ABC by the points of tangency of its incircle. In the context of inequalities
involving the lengths of the sides of a triangle, the substitution of them by y C z,
x C z and x C y is frequently referred to as Ravi’s transformation.
6. (IMO) If a, b and c are the lengths of the sides of a triangle, prove that
abc .a C b c/.b C c a/.c C a b/:
Fig. 5.2 Ravi’s
transformation
y
x
A
3
B
y
z
I
x
For a thorough discussion of the facts that follow, see Sect. 3:4 of [4], for instance.
z
C
5.1 The AM-GM Inequality
121
7. Let a, b and c be the lengths of the sides of a triangle. Prove that
b
c
a
C
C
3:
bCca
cCab
aCbc
8. (Baltic Way) Let a, b, c and d be given positive reals. Prove that
aCc
bCd
cCa
dCb
C
C
C
4:
aCb
bCc
cCd
dCa
9. If 0 < x ¤ 1 and n is a positive integer, prove that
1 x2nC1
.2n C 1/xn :
1x
10. (England) Prove that 3a4 C b4 4a3 b, for all nonzero real numbers a and b,
with equality if and only if jaj D jbj and ab > 0.
11. * Prove directly (i.e., without appealing to (5.5)) that a3 C b3 C c3 3abc.
12. (Soviet Union) Let a, b and c be positive reals. Prove that
.ab C ac C bc/2 3abc.a C b C c/:
13. (Soviet Union) If x; y; z > 0, prove that
z
x
y2
z2
y
x2
C 2 C 2 C C :
2
y
z
x
x
y
z
14. Let a and b be given positive reals. Prove that
9.a3 C b3 C c3 / .a C b C c/3 :
15. Given positive reals a, b and c, prove that
a4 .1 C b4 / C b4 .1 C c4 / C c4 .1 C a4 / 6a2 b2 c2 ;
with equality if and only if a D b D c D 1.
16. Let a1 , a2 , . . . , an be given positive reals. Prove that
a2
a3
an1
an
a1
C
C
CC
C
n:
a2
a3
a4
an
a1
17. Let n > 1 be an odd integer and a1 , a2 , . . . , an be negative reals. Show that
a1 C a2 C C an p
n a1 a2 : : : an ;
n
with equality if and only if all of the ai ’s are equal.
122
5 Elementary Inequalities
18. (BMO) Prove that, for every natural n, one has:
(a) .n C 1/n 2n nŠ.
(b) .n C 1/n .2n C 1/n 6n .nŠ/2 .
19. (Slovenia) Let x be a positive real number and m be a natural number. Prove that
1
1
1
x.x C 1/.x C 2/ : : : .x C m 1/ mŠ x1C 2 C 3 CC m :
20. (Poland) If x1 , x2 , . . . , xn are positive reals whose sum equals S, prove that
S
S
n2
S
;
C
CC
S x1
S x2
S xn
n1
with equality if and only if all of the xi ’s are equal.
21. The purpose of this problem is to present an alternative proof of inequality (5.7),
one that doesn’t make use of (5.6). To this end, do the following items:
P
P
P ai
aj
n
n
1
(a) Show that
D
n
C
a
i
iD1
iD1 ai
i<j aj C ai .
(b) Apply item (a) of Example 5.1 to each sum
ai
aj
C
aj
,
ai
with i < j, to get (5.7).
22. Prove the weighted arithmetic-geometric mean inequality: let a1 , a2 , . . . , an
be positive reals and k1 , k2 , . . . , kn be positive integers whose sum of inverses
equals 1. Prove that
ak11
a k2
a kn
C 2 C C n a1 a2 : : : an ;
k1
k2
kn
with equality if and only if a1 D a2 D D an . (Note that the case k1 D
k2 D D kn D 1n corresponds to the usual inequality between the arithmetic
and geometric means.)
23. (Romania) Let n > 1 be an integer and 0 < a1 < a2 < < an be given real
numbers. Prove that
12
22
n2
n
n1
n2
1
C
CC
C
C
CC
:
a1
a2
an
a1
a2 a1
a3 a2
an an1
Under what conditions does equality occur?
24. (China) Given positive reals a, b and c, prove that
aC
s p
p
ab C 3 abc
aCbCc
aCb
3
a
:
3
2
3
5.2 Cauchy’s Inequality
123
5.2 Cauchy’s Inequality
As a further application of the ideas of the previous section, let n > 1 be an
integer and a1 , a2 , a3 , b1 , b2 , b3 be real numbers such that a21 C a22 C a23 D 1
and b21 C b22 C b23 D 1. Since x2 C y2 2jxyj for all x; y 2 R, with equality if and
only if jxj D jyj, we have
a21 C b21 ja1 b1 j;
a22 C b22 ja2 b2 j;
a23 C b23 ja3 b3 j;
(5.8)
with equality if and only if ja1 j D jb1 j, ja2 j D jb2 j and ja3 j D jb3 j. Adding the left
and right hand sides of the above inequalities, we get
.a21 C a22 C a23 / C .b21 C b22 C b23 / D .a21 C b21 / C .a22 C b22 / C .a23 C b23 /
2.ja1b1 j C ja2 b2 j C ja3 b3 j/
2ja1 b1 C a2 b2 C a3 b3 j;
where in the last step we applied the triangle inequality for three real numbers, (2.11).
Hence, it follows from a21 C a22 C a23 D 1 and b21 C b22 C b23 D 1 that
ja1 b1 C a2 b2 C a3 b3 j 1:
Equality holds if and only if it holds in the three inequalities (5.8), as well as in
the triangle inequality. Therefore, equality holds if and only if ja1 j D jb1 j, ja2 j D
jb2 j, ja3 j D jb3 j and either a1 b1 , a2 b2 , a3 b3 0 or a1 b1 , a2 b2 , a3 b3 0. However,
it is immediate to check that such conditions are equivalent to a1 D b1 , a2 D b2 and
a3 D b3 .
Now, consider arbitrary real numbers a1 ; a2 ; a3 and b1 ; b2 ; b3 , such that at least
one of a1 ; a2 ; a3 and at least one of b1 ; b2 ; b3 are nonzero. For a positive real number
c, let xi D aci for 1 i 3. Since
x21 C x22 C x23 D
a21 C a22 C a23
;
c2
we have
x21 C x22 C x23 D 1 , c D
Analogously, letting yi D
bi
d
q
a21 C a22 C a23 :
for 1 i 3, with d D
q
b21 C b22 C b23 , we have
y21 C y22 C y23 D 1. Therefore, it follows from our previous discussion that
124
5 Elementary Inequalities
jx1 y1 C x2 y2 C x3 y3 j 1;
with equality if and only if xi D yi for 1 i 3.
Substituting the definitions of xi and yi in the above inequality and recalling that
c; d > 0, we conclude that the last inequality above is equivalent to
ja1 b1 C a2 b2 C a3 b3 j
1;
cd
q
q
with c D a21 C a22 C a23 and d D b21 C b22 C b23 . Moreover, equality holds if
and only if ai D dc bi for 1 i 3. Finally, note that the last inequality above is
equivalent to
ja1 b1 C a2 b2 C a3 b3 j cd D
q
q
a21 C a22 C a23 b21 C b22 C b23 :
Up to this point, what we have done was to establish, for n D 3, the famous
Cauchy’s inequality.4 We now turn to the general case.
Theorem 5.13 (Cauchy) Let n > 1 be an integer and a1 , a2 , . . . , an , b1 , b2 , . . . , bn
given real numbers. Then,
ˇ v
ˇ
v
ˇ u n
ˇ n
u n
ˇ uX 2 uX
ˇX
ˇ
ˇ
t
t
a
b
a
b2j ;
j jˇ
j
ˇ
ˇ
ˇ jD1
jD1
jD1
(5.9)
with equality if and only if the ai ’s and bi ’s are respectively proportional, i.e., if and
only if there exists a nonzero real number such that
a1 D b1 ; a2 D b2 ; : : : ; an D bn :
Proof If all of the ai ’s or all of the bi ’s are equal to zero, there is nothing to do.
Otherwise, in order to establish (5.9), it suffices to follow the steps of the particular
case n D 3 discussed above. The only difference is that, whenever convenient, we
have to use (4.6) instead of (2.11) (see Problem 2, page 127).
t
u
Later (see Example 6.18), we shall give another proof of Cauchy’s inequality
as an application of the theory of maxima and minima of quadratic functions. For
a geometric interpretation of Cauchy’s inequality for n D 2, see the problems of
Sect. 6:3 of [4].
The coming two examples illustrate how one can apply Cauchy’s inequality.
4
Augustin Louis Cauchy, one of the greatest mathematicians of the XIX century, and maybe of
History. Cauchy was one of the precursors of Mathematical Analysis, an extremely important area
of higher Mathematics. He also has his name attached to several important results in Differential
Equations and Mathematical Physics.
5.2 Cauchy’s Inequality
125
Example 5.14 Let a, b and c be given real numbers. Show that the system of
equations
3.x2 C y2 C z2 / C a2 C b2 C c2 D 6
ax C by C cz D 2
doesn’t have any real solutions x; y; z.
Proof By contradiction, if there existed a real solution x; y; z, we would have, by
Cauchy’s inequality,
4 D .ax C by C cz/2 .a2 C b2 C c2 /.x2 C y2 C z2 /:
Thus, letting u D a2 C b2 C c2 and v D x2 C y2 C z2 , we would have u C 3v D 6 and
uv 4. At this point, the inequality on the arithmetic and geometric means would
give us
p
p
p
p
6 D u C 3v 2 u 3v D 2 3uv D 2 3 4 D 4 3;
t
u
which is an absurd.
Example 5.15 (Romania) Let x1 ; x2 ; : : :, xnC1 be positive reals satisfying x1 C x2 C
C xn D xnC1 . Prove that
p
p
x1 .xnC1 x1 / C C xn .xnC1 xn / p
xnC1 .xnC1 x1 / C C xnC1 .xnC1 xn /:
Proof For 1 j n, let yj D xnC1 xj . By Cauchy’s inequality, we have
p
p
p
p
x1 y 1 C C xn y n x1 C C xn y1 C C yn
p
p
D xnC1 .xnC1 x1 / C C .xnC1 xn /:
t
u
For future use, we collect the following corollary of Cauchy’s inequality.
Corollary 5.16 Given real numbers a1 ; : : : ; an and b1 ; : : : ; bn , we have
v
v
v
uX
uX
u n
u n
u n 2 uX
t .aj C bj /2 t
aj C t
b2j ;
jD1
jD1
(5.10)
jD1
with equality holding if and only if a1 , . . . , an and b1 , . . . , bn are positively
proportional, i.e., if and only if there exists a positive real number such that
ai D bi for 1 i n.
126
5 Elementary Inequalities
Proof For the sake of clarity, let’s prove the corollary for n D 3, the general case
being totally analogous. Since both sides of (5.10) are nonnegative real numbers,
it suffices to show that the square of the left hand side is less than or equal to the
square of the right hand side. In symbols,
2
2
2
.a1 C b1 / C .a2 C b2 / C .a3 C b3 / q
a21
C
a22
C
a23
C
q
b21
C
b22
C
b23
2
:
Expanding .ai C bi /2 , it follows that the square of the left hand side equals
.a21 C 2a1 b1 C b21 / C .a22 C 2a2 b2 C b22 / C .a23 C 2a3 b3 C b23 /:
Analogously, the square of the right hand side equals
q
q
.a21 C a22 C a23 / C 2 a21 C a22 C a23 b21 C b22 C b23 C .b21 C b22 C b23 /:
Cancelling the summand .a21 C a22 C a23 / C .b21 C b22 C b23 / from both sides, we
conclude that (5.10) is equivalent to
q
q
2.a1 b1 C a2 b2 C a3 b3 / 2 a21 C a22 C a23 b21 C b22 C b23 ;
which is precisely Cauchy’s inequality.
The analysis of the conditions for equality will be left as an exercise to the reader.
t
u
As will be seen in Chaps. 8 and 13 of [4], for n D 2 and 3 inequality (5.10)
has the following geometric interpretation: in a cartesian coordinate system, if O D
.0; : : : ; 0/, A D .a1 ; : : : ; an / and B D .b1 ; : : : ; bn /, and we set C D A C B, then
C D .a1 C b1 ; : : : ; an C bn / and (5.10) amounts to the inequality
OC OA C OB;
where XY stands for the length of the line segment XY. This is the same as the
(geometric) triangle inequality for the (possibly degenerated) triangle OAC, and for
this reason (5.10) is also known as the triangle inequality.
p Note also that,
p if npD 1 and we make a1 D a and b1 D b, then (5.10) reduces to
.a C b/2 a2 C b2 or, which is the same, jaCbj jajCjbj. This is the reason
why (5.10), as well as its generalization (4.6), are known as triangle inequality.
5.2 Cauchy’s Inequality
127
Problems: Section 5.2
1. Given real numbers x and y such that 3x C 4y D 12, compute the minimum
possible value of x2 C y2 .
2. Prove the general case (5.9) of Cauchy’s inequality.
3. Given positive reals a1 , a2 , . . . , an , we define its quadratic mean as the number
s
a21 C a22 C C a2n
:
n
Prove the inequality between the quadratic and arithmetic means:
s
a1 C a2 C C an
a21 C a22 C C a2n
;
n
n
(5.11)
with equality if and only if a1 D a2 D D an .
4. (IMO shortlist) Let a1 , a2 , a3 , a4 be positive reals. Prove that
X
a2i C a2j C a2k
1i<j<k4
ai C aj C ak
a1 C a2 C a3 C a4 ;
with equality if and only if a1 D a2 D a3 D a4 .
5. Use Cauchy’s inequality to give a third proof of (5.7).
6. (Leningrad) Given positive reals a, b, c and d, prove that
1
1
4
16
64
C C C
:
a
b
c
d
aCbCcCd
7. (OIM) Let x, y and z be positive reals whose sum equals 3. Prove that
p
p
p
p
p
3
3 9 < 2x C 3 C 2y C 3 C 2z C 3 3 5:
When does the right hand inequality become an equality?
8. Let n > 2 be an integer. Prove that
v !
v ! v !
u
u
u
u n
u n
u n
p
t
t
C
CCt
< n.2n 1/:
1
2
n
9. (Soviet Union—adapted) Given x; y; z > 0, apply Cauchy’s inequality to prove
that
z
x
y2
z2
y
x2
C
C
C C ;
y2
z2
x2
x
y
z
with equality if and only if x D y D z.
128
5 Elementary Inequalities
10. (APMO) Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be given positive reals, such that
a1 C a2 C C an D b1 C b2 C C bn . Show that
n
X
kD1
1X
a2k
ak :
ak C bk
2 kD1
n
11. (TT) Let a1 , a2 , . . . , an be given positive reals. Prove that
a2
a2
a2
1C 1
1 C 2 : : : 1 C n .1 C a1 /.1 C a2 / : : : .1 C an /:
a2
a3
a1
12. (APMO) Let a, b and c be the lengths of the sides of a triangle. Show that
p
aCbcC
p
p
p
p
p
bCcaC cCab aC bC c
and explain when equality occurs.
13. Let a1 , a2 , . . . , an , b1 , b2 , . . . , bn and c1 , c2 , . . . , cn be given positive real
numbers. Show that
n
1 X
a k b k ck
n kD1
!4
n
X
kD1
!
a4k
n
X
kD1
!
b4k
n
X
!
c4k
:
kD1
14. (IMO) Let a, b and c be positive reals such that abc D 1. Prove that
1
1
1
3
C 3
C 3
:
C c/
b .a C c/
c .a C b/
2
a3 .b
5.3 More on Inequalities
This section is devoted to the study of other important elementary inequalities. The
first of them is generally attributed to the Bernoulli brothers,5 being known as Bernoulli’s inequality. In spite of its simplicity, we will see that it is quite useful in
applications.
Proposition 5.17 (Bernoulli) If n is a natural number and x > 1 is a real number,
then
.1 C x/n 1 C nx;
with equality holding for n > 1 if and only if x D 0.
5
Jacob and Johann Bernoulli, Swiss mathematicians of the XVIII century.
5.3 More on Inequalities
129
Proof Let’s make induction on n, the case n D 1 being immediate. Suppose, by
induction hypothesis, that .1 C x/k 1 C kx; since 1 C x > 0, we have
.1 C x/kC1 D .1 C x/.1 C x/k .1 C x/.1 C kx/
D 1 C .k C 1/x C kx2 1 C .k C 1/x;
with equality if and only if .1 C x/k D 1 C kx and kx2 D 0, i.e., if and only if x D 0.
t
u
Example 5.18 Given a natural number n and positive reals a and b, show that
a
1C
b
n
b n
C 1C
2nC1 ;
a
with equality if and only if a D b.
Proof Dividing both sides of the inequality by 2n , we see that it suffices to prove
that
1
a n
b n
1
1 C
C 1 C
2:
2
2b
2
2a
a
b
> 1 and 12 C 2a
> 1, if we apply Bernoulli’s inequality to both
Since 12 C 2b
summands at the left hand side above and add the results, we get
1
a n
b n
b
1
a
1 C
C
1 :
C 1 C
2Cn
2
2b
2
2a
2b
2a
Now, it suffices to apply the inequality between the arithmetic and geometric means
to obtain
r
b
a
a b
C
12
1 D 0;
2b
2a
2b 2a
with equality if and only if
a
2b
D
b
,
2a
i.e., if and only if a D b.
t
u
We continue by presenting an inequality known in the literature as Chebyshev’s
inequality.6
Theorem 5.19 (Chebyshev) If a1 , a2 , . . . , an and b1 , b2 , . . . , bn are real numbers
such that
a1 a2 an and b1 b2 bn ;
6
After Pafnuty Chebyshev, Russian mathematician of the XIX century.
130
5 Elementary Inequalities
then
1X
ai
n iD1
n
!
1X
bi
n iD1
n
!
1X
ai bi ;
n iD1
n
with equality holding if and only if a1 D a2 D D an ou b1 D b2 D D bn .
Proof We have to show that
n
n
X
ai bi iD1
n
X
!
ai
iD1
n
X
!
bi
0;
iD1
and for this it suffices to observe that the expression at the left hand side equals
n
X
.ai aj /.bi bj /;
(5.12)
i;jD1
which, in turn, is nonnegative (for, the ai ’s and bi ’s are equally ordered).
Now, if a1 D a2 D D an or b1 D b2 D D bn , it is immediate to check that
equality holds in Chebyshev’s inequality. Conversely, suppose that equality holds
in such an inequality. Then, the argument of the previous paragraph assures that the
expression in (5.12) must be equal to zero. Since .ai aj /.bi bj / 0 for all indices
1 i; j n, we must have .ai aj /.bi bj / D 0 for all 1 i; j n. If there exists
1 k n such that bk < bkC1 , then b1 bk < bkC1 bn , and the
condition .ai akC1 /.bi bkC1 / D 0 for every 1 i n gives ai D akC1 for 1 i k. Hence, a1 D a2 D D ak D akC1 . If we now start from .ai ak /.bi bk / D 0
for k < i n, we conclude in a similar way that akC1 D D an . Therefore, all of
the ai ’s must be equal.
t
u
The following corollary collects an important consequence of Chebyshev’s
inequality.
Corollary 5.20 If k is a natural number and a1 , a2 , . . . , an are positive reals, then
ak1 C ak2 C C akn
n
a1 C a2 C C an
n
k
:
(5.13)
Moreover, if k > 1, then equality holds if and only if all of the ai ’s are equal.
Proof Let’s prove the inequality by induction on k 1, noting that (5.13) is trivially
true for k D 1 and all positive reals a1 , . . . , an . By induction hypothesis, let l > 1 be
a natural number such that (5.13) is true for k D l 1 and all positive reals a1 , a2 ,
. . . , an .
Since both sides of the inequality we wish to prove are invariant under permutations of the indices 1, 2, . . . , n, we can suppose, without any loss of generality,
5.3 More on Inequalities
131
that a1 a2 an . Then, a1l1 a2l1 anl1 , and it follows from
Chebyshev’ inequality that
1X l
a n iD1 i
n
1X
ai
n iD1
n
!
!
n
1 X l1
:
a
n iD1 i
(5.14)
On the other hand, induction hypothesis gives
1 X l1
a n iD1 i
n
1X
ai
n iD1
n
!l1
;
(5.15)
and if we combine these two inequalities we get
1X l
a n iD1 i
n
1X
ai
n iD1
n
!
1X
ai
n iD1
n
!l1
1X
ai
n iD1
n
D
!l
:
Finally, let l > 1 and suppose we have equality in (5.13) when k D l. Then,
the argument of the previous paragraph assures that we must have equality in (5.14)
and (5.15). However, by the condition for equality in Chebyshev’s inequality, the
only way to have equality in (5.14) is having a1 D D an .
t
u
The two coming examples illustrate how one can apply Chebyshev’s inequality.
Example 5.21 (Poland) Let a1 ; a2 ; : : : ; an be positive reals with sum equal to s.
Prove that
a1
a2
an
n
:
C
CC
s a1
s a2
s an
n1
Proof Suppose, without loss of generality, that a1 a2 an . Then, s a1 s a2 s an and, since s ai > 0 for every i, it follows that
1
1
1
sa
sa
. Therefore, Chebyshev’s inequality gives
sa1
2
n
n
X
iD1
! n
!
n n
X
X 1
1 X
1
ai
ai D
ai
s ai
s ai
n iD1
s ai
iD1
iD1
!
n
s X 1
D
:
n iD1 s ai
On the other hand, it follows from Corollary 5.8 that
!
n
X
.s ai /
n
X
iD1
iD1
1
s ai
!
n2 :
(5.16)
132
However, since
5 Elementary Inequalities
Pn
iD1 .s
ai / D .n 1/s, this last inequality yields
n
X
iD1
n2
1
:
s ai
.n 1/s
(5.17)
If we now combine (5.16) and (5.17), we arrive at the desired inequality.
t
u
Example 5.22 (Turkey)
Let n > 1 be a natural number and x1 ; x2 ; : : : ; xn be positive
P
reals such that niD1 x2i D 1. Find the least possible value of
n
X
iD1
x5i
;
x1 C Cb
xi C C xn
where the hat over xi at the denominator of the i-th summand indicates that it
contains all of x1 ; x2 ; : : : ; xn , except xi .
Proof Let s denote the sum of the xi ’s and suppose, without loss of generality, that
1
1
1
x1 x2 xn . Then, x21 x22 x2n and sx
sx
sx
.
1
2
n
The expression to be minimized can be written as
SD
n
X
iD1
1
x5 :
s xi i
Hence, if we apply Chebyshev’s inequality twice, together with the inequality
between the arithmetic and quadratic means (cf. Problem 3, page 127) and
Corollary 5.8, we get
n
1 X 1
S n iD1 s xi
1
2
n
1
3
n
s
3
n
D
n
X
iD1
n
X
iD1
n
X
iD1
!
1
s xi
1
s xi
1
s xi
!
!
!
n
X
!
x5i
iD1
n
X
!
xi
iD1
n
X
!
xi
iD1
n
X
!
x4i
iD1
n
X
!2
x2i
iD1
!1
n
X
s
3 n2
.s xi /
n
iD1
1
1
s
D 2
:
n n.n 1/s
n .n 1/
5.3 More on Inequalities
133
1
In order to guarantee that n2 .n1/
is the least possible value, we ought to show
that
attained. To this end, it is enough to see that, in view of the constraint
Pn it is
2
iD1 xi D 1, the condition for equality in Chebyshev’ inequality gives x1 D x2 D
D xn D p1n if equality holds. However, computing with these values for x1 ,
. . . , xn , we see that all of the inequalities above become equalities, so that
indeed, the least possible value.
1
n2 .n1/
is,
t
u
Next, we present an inequality known as the rearrangement inequality. We
recall that a sequence .x1 ; x2 ; : : : ; xn / is a permutation or a rearrangement of
.a1 ; a2 ; : : : ; an / if these two sequences differ only by the order of their terms.7
Proposition 5.23 Let a1 < a2 < < an be given positive real numbers. If
.x1 ; x2 ; : : : ; xn / is any permutation of .a1 ; a2 ; : : : ; an /, then
n1
X
ai ani iD1
n1
X
iD1
a i xi n1
X
a2i ;
iD1
with equality in the left (resp. right) inequality above if and only if xi D ani (resp.
xi D ai ), for 1 i n.
Proof Let us show how to maximize the sum a1 x1 Ca2 x2 C Can xn . (The argument
to minimize it is completely analogous.)
Since the number of permutations8 .x1 ; x2 ; : : : ; xn / of .a1 ; a2 ; : : : ; an / is finite,
there is at least one of them which maximizes the sum a1 x1 C a2 x2 C C an xn .
If .b1 ; b2 ; : : : ; bn / is such a permutation, we want to show that bi D ai for
1 i n. To this end, it suffices to show that b1 < b2 < < bn . By contradiction,
suppose that there exist indices 1 i < j n for which bi > bj . Define a
permutation .b01 ; b02 ; : : : ; b0n / of the ai ’s by setting
8
< bk ; se k ¤ i; j
b0k D bi ; se k D j :
:
bj ; se k D i
Then,
n
X
iD1
ai b0i n
X
ai bi D .ai b0i C aj b0j / .ai bi C aj bj /
iD1
D .ai bj C aj bi / .ai bi C aj bj /
D .ai aj /.bj bi / > 0:
7
In the more precise language of functions (cf. Chap. 6), we say that .x1 ; x2 ; : : : ; xn / is a
permutation of .a1 ; a2 ; : : : ; an / if there is a bijection ' W f1; : : : ; ng ! f1; : : : ; ng, so that
xi D a'.i/ , for 1 i n.
8
Actually, it is easy to show that there are exactly nŠ such permutations. For a proof, see [5], or
provide one yourself, by making an induction argument.
134
5 Elementary Inequalities
This is the same as
a1 b01 C a2 b02 C C an b0n > a1 b1 C a2 b2 C C an bn ;
which, in turn, contradicts the fact that .b1 ; b2 ; : : : ; bn / is a permutation of
.a1 ; a2 ; : : : ; an / that maximizes the sum a1 x1 C a2 x2 C C an xn . Therefore,
b1 < b2 < < bn .
t
u
With essentially the same argument as above, one can easily extend the rearrangement inequality to the case in which a1 a2 an . In this case, if
.x1 ; x2 ; : : : ; xn / is any permutation of .a1 ; a2 ; : : : ; an /, then
n1
X
ai ani iD1
n1
X
iD1
a i xi n1
X
a2i ;
iD1
with equality in the left (resp. right) inequality above if (and no more if, and only
if ) xi D ani (resp. xi D ai ), for 1 i n. The coming example explores this more
general form of the rearrangement inequality.
Example 5.24 Given positive reals a, b and c, show that:
(a) a3 C b3 C c3 a2 b C b2 c C c2 a.
1
1
1
(b) aCbCc
abc a2 C b2 C c2 .
Proof
(a) Suppose, without loss of generality, that a b c. (Other orderings of a,
b and c would give rise to analogous arguments.) A direct application of the
rearrangement inequality gives
a3 C b3 C c3 D a2 a C b2 b C c2 c a2 b C b2 c C c2 a:
(b) The symmetry of both sides with respect to a, b and c allows us to suppose again
that a b c. The inequality to be proved is equivalent to
a2 bc C ab2 c C abc2 .ab/2 C .bc/2 C .ca/2 :
In order to get this, let’s first observe that the condition 0 < a b c implies
ab ac bc. Therefore, upon applying the general form of the rearrangement
inequality, we obtain
a2 bc C ab2 c C abc2 D ab ac C ab bc C ac bc
.ab/2 C .bc/2 C .ca/2 :
t
u
5.3 More on Inequalities
135
A very useful idea in certain types of problems involving inequalities is to try to
use arguments similar to the one of the proof of the rearrangement inequality. Let’s
see an example along these lines.
Example 5.25 (Taiwan) Let n > 2 be an integer. Compute the greatest possible
value of the expression
X
xi xj .xi C xj /;
1i<jn
when .x1 ; x2 ; : : : ; xn / varies over all sequences of nonnegative reals such that x1 C
x2 C C xn D 1.
Solution Let
En .x1 ; : : : ; xn / D
X
xi xj .xi C xj /
1i<jn
and observe that the symmetry of the right hand side allows us to suppose that
x1 xn 0.
Let’s first of all show that
En .x1 ; : : : ; xn2 ; xn1 C xn ; 0/ En .x1 ; : : : ; xn /:
(5.18)
To this end, set yj D xj for 1 j < n 1, yn1 D xn1 C xn and yn D 0, and note
that y1 C y2 C C yn D 1. For the sake of notation, let En denote the difference
En D En .y1 ; : : : ; yn / En .x1 ; : : : ; xn /:
Then, we get
X
En D
yi yj .yi C yj / C
1i<jn2
C
n1
X
X
1i<jn2
yi yn1 .yi C yn1 /
iD1
yi yn .yi C yn / iD1
n2
X
X
1i<jn2
xi xn1 .xi C xn1 / xi xj .xi C xj /
X
1i<jn1
xi xn .xi C xn /:
136
5 Elementary Inequalities
Taking the definition of the yi ’s into account, we obtain
En D
n2
X
xi .xn1 C xn /.xi C xn1 C xn /
iD1
n2
X
n1
X
xi xn1 .xi C xn1 / iD1
D
xi xn .xi C xn /
iD1
n2
X
Œxi .xn1 C xn /.xi C xn1 C xn / xi xn1 .xi C xn1 /
iD1
xi xn .xi C xn / xn1 xn .xn1 C xn /
D
n2
X
2xi xn1 xn xn1 xn .xn1 C xn /
iD1
D xn1 xn .2x1 C C 2xn2 xn2 xn /:
Since x1 x2 xn 0 and n 3, it perspires that
2x1 C C 2xn2 xn1 xn 2x1 xn1 xn
D x1 xn1 C x1 xn 0;
which, in turn, establishes (5.18).
Thus, in order to maximize En .x1 ; : : : ; xn /, we can restrict our attention to the
sequences .x1 ; x2 ; : : : ; xn / of nonnegative reals for which x1 C C xn1 C xn D 1
and xn D 0. In this case, it is immediate to see that
En .x1 ; : : : ; xn1 ; 0/ D En1 .x1 ; : : : ; xn1 /;
with x1 ; : : : ; xn1 nonnegative reals for which x1 C C xn1 D 1. Then, we can
repeat the same argument and, doing this several times, we conclude that we can
suppose xi D 0 for i > 2. Therefore, it is sufficient to maximize
E2 .x1 ; x2 / D x1 x2 .x1 C x2 /;
for nonnegative x1 and x2 satisfying x1 C x2 D 1. This, in turn, is immediate:
x1 x2 .x1 C x2 / D x1 x2 x1 C x2
2
2
D
1
:
4
Finally, we have
En .x1 ; : : : ; xn / E2
1 1
1
;
D :
2 2
4
t
u
5.3 More on Inequalities
137
The last inequality we wish to consider is due to the XIX century Norwegian
mathematician N. H. Abel. For this reason, it is known as Abel’s inequality.
Theorem 5.26 (Abel) Let n > 1 be a natural number and a1 , a2 , . . . , an , b1 , b2 ,
. . . , bn be given real numbers, with a1 a2 an 0. If M and m respectively
denote the maximum and minimum elements of the set of sums fb1 ; b1 Cb2 ; : : : ; b1 C
b2 C C bn g, then
ma1 a1 b1 C a2 b2 C C an bn Ma1 :
Proof Let’s prove the right hand side inequality, the proof of the left hand side one
being totally analogous.
Make s0 D 0 and si D b1 C C bi for 1 i n. Then,
n
X
n
X
ai bi D
iD1
ai .si si1 / D
iD1
n
X
iD1
ai si n1
X
aiC1 si
iD0
n1
X
.ai aiC1 /si C an sn
D
iD1
n1
X
M.ai aiC1 / C Man D Ma1 :
t
u
iD1
For future reference, and in the notations of the proof of Abel’s theorem, the
identity
n
X
iD1
ai bi D
n1
X
.ai aiC1 /si C an sn
(5.19)
iD1
is usually referred to as Abel’s identity. As we shall see here and later (see
Problem 18, page 243), it is almost as useful as the inequality itself.
We finish this section by presenting a beautiful application of Abel’s inequality.
In the course of the proof, we will use the fact that every set with k elements
has exactly 2k distinct subsets. For a proof of this last fact, we refer the reader to
Problem 12, page 96, or to Chap. 1 of [5].
Example 5.27 (Romania) Do the following items:
(a) Let n > 1 be an integer and x1 ; : : : ; xn , y1 ; : : : ; yn positive reals such that x1 y1 <
x2 y2 < < xn yn and, for 1 k n, x1 C C xk y1 C C yk . Prove that
1
1
1
1
1
1
C
CC
C
CC :
x1
x2
xn
y1
y2
yn
138
5 Elementary Inequalities
(b) Let A D fa1 ; a2 ; : : : ; an g be a set of positive integers with the following
property: the sums of the elements of any two nonempty subsets of A are always
distinct. Prove that
1
1
1
C
CC
< 2:
a1
a2
an
Proof
(a) First of all, observe that
n
n
n
X
X
1 X1
xi y i
D
:
y
x
xi yi
iD1 i
iD1 i
iD1
(5.20)
On the other hand, the condition x1 C C xk y1 C C yk for 1 k n
can be rewritten as
.x1 y1 / C C .xk yk / 0
for 1 k n. Thus, making ai D xi1yi and bi D xi yi for 1 i n, we have
i
a1 > a2 > > an > 0, b1 C C bi 0 and xxi y
D ai bi , for 1 i n.
i yi
Applying Abel’s inequality to (5.20), we obtain
n
n
X
1 X1
an minfb1 C C bi I 1 i ng 0:
y
x
iD1 i
iD1 i
(b) Suppose, without loss of generality, that a1 < a2 < < an , and let Bk D
fa1 ; : : : ; ak g for 1 k n. The hypothesis on the set A assures that all of the
2k 1 nonempty subsets of Bk have distinct sums of elements. However, since
each of these sums is a natural number and a1 C C ak is the greatest of them,
we conclude that
a1 C C ak 2k 1:
Now, observing that 2k 1 D 20 C 21 C C 2k1 , we have
a1 C a2 C C ak 20 C 21 C C 2k1
for 1 k n. On the other hand, it is obvious that
20 a1 < 21 a2 < < 2n1 an ;
5.3 More on Inequalities
139
so that the inequality of item (a) gives
1
1
1
1
1
1
C
CC
0 C 1 C C n1 < 2:
a1
a2
an
2
2
2
t
u
For another proof of item (b) of the previous example, see Chap. 3 of [5].
Problems: Section 5.3
n 1. Given n 2 N, prove that 1 C 1n < 1 C
2. (USA) Given naturals m and n, let a D
1 nC1
.
nC1
mmC1 CnnC1
. Prove
mm Cnn
that
a m C a n mm C n n :
3. Let a, b and c be positive reals. Prove that
1
1
a 8 C b 8 C c8
1
C C :
a
b
c
a 3 b 3 c3
4. (OIM) Find all real positive solutions of the system of equations
x1 C x2 C C x1994 D 1994
:
x41 C x42 C C x41994 D x31 C x32 C C x31994
5. (IMO shortlist) Let a1 , a2 , a3 , a4 be positive reals. Prove that
X
a3i C a3j C a3k
1i<j<k4
ai C aj C ak
a21 C a22 C a23 C a24 ;
with equality if and only if a1 D a2 D a3 D a4 .
6. Let n > 1 be an integer and a1 , a2 , . . . , an , b1 , b2 , . . . , bn be given real numbers,
with a1 a2 an and b1 b2 bn . If 1 2 n are
positive reals whose sum is equal to 1, prove that
n
X
iD1
!
i a i
n
X
iD1
!
i b i
n
X
i a i b i
iD1
and give necessary and sufficient conditions for the equality. To which particular
case there corresponds Chebyshev’s inequality?
140
5 Elementary Inequalities
7. (IMO shortlist) Let a, b, c and d be nonnegative reals for which ab C bc C cd C
da D 1. Prove that
b3
c3
d3
1
a3
C
C
C
:
bCcCd
aCcCd
aCbCd
aCbCc
3
8. For positive reals a, b, c and n 2 N, prove that
bn
cn
an1 C bn1 C cn1
an
C
C
:
bCc
aCc
aCb
2
9. (IMO shortlist) Let x, y and z be positive reals such that xyz D 1. Prove that
y3
z3
3
x3
C
C
:
.1 C y/.1 C z/
.1 C x/.1 C y/
.1 C x/.1 C y/
4
10. (India) Let n > 1 be an integer and x1 , x2 , . . . , xn be given positive reals whose
sum is equal to 1. Prove that
n
X
iD1
xi
p
1 xi
r
Xp
1
n
p
xi :
n1
n 1 iD1
n
11. (Slovenia) Given 2n positive reals a1 , a2 , . . . , a2n , how should we arrange them
in pairs, such that the sum of the n products of the numbers of each pair is
maximal?
12. (IMO) Let .ak /k1 be a sequence of pairwise distinct positive integers. Prove
that, for every n 2 N, we have
n
X
ak
kD1
k2
n
X
1
kD1
k
:
13. Do the following items:
(a) If x < y are given positive reals and a D xCy
2 , prove that a.x C y a/ xy.
(b) Use item (a) to furnish another proof for the inequality between the
arithmetic and geometric means.
14. (TT) Let a1 ; a2 ; : : : ; an be given positive reals. Prove that
Y
n
a2
a2
a2
1 C 2 ::: 1 C n .1 C ak /:
1C 1
a2
a3
a1
kD1
15. (Taiwan) Let n 3 be an integer and x1 , x2 , . . . , xn be nonnegative reals whose
sum equals 1. Prove that
x21 x2 C x22 x3 C x23 x4 C C x2n x1 4
:
27
5.3 More on Inequalities
141
16. If a, b, c and d are nonnegative reals such that a 1, a C b 5, a C b C c 14
and a C b C c C d 30, use Abel’s inequality to prove that
p
p
p
p
a C b C c C d 10:
17. Let a1 , . . . , an and b1 , . . . , bn be real numbers such that a1 a2 an > 0
and b1 a1 , b1 b2 a1 a2 , . . . , b1 b2 : : : bn a1 a2 : : : an . Show that
b1 C b2 C C bn a1 C a2 C C an :
Chapter 6
The Concept of Function
With the algebraic background of the previous chapters at our disposal, we now
begin the study of real functions of a single variable. After presenting the basic
concepts of domain, codomain and image, some relevant examples of functions
are given. Then, we introduce the notions of monotonicity and extremal values,
discussing various examples that, in spite of their elementary character, will reveal
themselves to be very useful. The chapter continues with the study of the operations
of composition and inversion of functions, and culminates with the orchestration of
the whole of it to a first study of implicitly defined functions. We end the chapter by
discussing, in its last two sections, graphs of elementary functions.
6.1 Definitions and Examples
Let X and Y be two given nonempty sets. Informally, a function f from X to Y is a
rule that associates to each x 2 X a unique y 2 Y. It is sometimes useful to visualize
a function f W X ! Y in a more concrete way, by means of diagrams as that of
Fig. 6.1, where each arrow indicates which element y 2 Y is associated to a given
x 2 X.
We write f W X ! Y to denote that f is a function from X to Y. In this case,
the element y 2 Y associated to x 2 X via f is denoted by y D f .x/, and is called
the image of x 2 X by f . In the example of Fig. 6.1, we have X D f1; 2; 3g,
Y D fa; b; c; dg and f .1/ D a, f .2/ D a, f .3/ D c. Thus, a is the image of 1 and 2
by f , and c is the image of 3 by f .
The discussion of the previous paragraph makes it clear that the definition of
function allows, in the corresponding diagram, that one or more elements of Y do
not receive arrows, or that one or more elements of Y receive more than one arrow
(observe that both these possibilities are present in Fig. 6.1).
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_6
143
144
6 The Concept of Function
Fig. 6.1 Example of a
function f from X to Y
X
Y
f
1
a
2
c
3
d
b
Fig. 6.2 No arrow starts at
12X
X
Y
a
1
b
2
c
3
d
Fig. 6.3 More than one
arrow starts at 1 2 X
X
Y
1
a
2
c
3
d
b
Note, however, that the diagrams in Fig. 6.2 and 6.3 do not correspond to
functions. The situation of Fig. 6.2 is forbidden because there is no arrow departing
from 1 2 X. The situation of Fig. 6.3 is forbidden because more than one arrows
departs from 1 2 X.
The next three definitions isolate some quite useful types of functions.
Definition 6.1 Let X and Y be given nonempty sets. For a fixed element c 2 Y, the
constant function c from X to Y is the function f W X ! Y such that f .x/ D c for
every x 2 X.
6.1 Definitions and Examples
145
Thus, in the extreme case of the constant function equal to c, defined as above,
every x 2 X is associated to a single y 2 Y, namely, y D c. Nevertheless, the
conditions required by the “definition” of function are fully satisfied, i.e., every x 2
X is associated to a unique y 2 Y.
Definition 6.2 Let X be a given nonempty set. The identity function of X is the
function IdX W X ! X, such that IdX .x/ D x, for every x 2 X.
As in the previous definition, it is immediate to see that the conditions required
by the “definition” of function are satisfied, so that IdX is indeed a function from X
to itself.
For the next definition, given an arbitrary n 2 N, we let In be the set whose
elements are the first n natural numbers, i.e., I1 D f1g, I2 D f1; 2g, I3 D f1; 2; 3g
and, more generally,
In D fk 2 NI 1 k ng:
Definition 6.3 An infinite sequence of real numbers is a function f W N ! R. A
finite sequence of real numbers is a function f W In ! R, for some n 2 N.
As was anticipated in Chapter 3, given a sequence f W N ! R (resp. f W In ! R)
and k 2 N (resp. k 2 In ), one uses to write ak , instead of f .k/, to denote the image of
k by f . In this case, one also writes .ak /k1 (resp. .ak /1kn ) to denote the sequence
as a whole, and says that ak is its k –th term.
From a mathematically rigorous viewpoint, a function is a particular type of
relation, according to Definition 6.4 below. In order to state it properly, we begin by
recalling two simple facts from elementary set theory1.
Given nonempty sets X and Y, and elements x 2 X and y 2 Y, the ordered pair
.x; y/ is defined by
.x; y/ D ffxg; fx; ygg:
From this, it is immediate to prove that, if x; x0 2 X and y; y0 2 Y, then
.x; y/ D .x0 ; y0 / , x D x0 and y D y0 :
The cartesian product of X and Y (in this order), denoted X
X
Y, is the set
Y D f.x; y/I x 2 X and y 2 Yg:
Definition 6.4 Given nonempty sets X and Y, a relation from X to Y (or between
X and Y, in this order) is a subset R of the cartesian product X Y, i.e., R is a set of
ordered pairs .x; y/, with x 2 X and y 2 Y. If R is a relation from X to X, we simply
say that R is a relation on X.
1
For a more thorough account of set theory, we refer the reader to the classic [15].
146
6 The Concept of Function
Example 6.5 Let X D f1; 2; 3g and Y D f2; 3; 4; 5g. The set
R D f.x; y/ 2 X
YI x yg
is a relation from X to Y, given by R D f.2; 2/; .3; 2/; .3; 3/g. Indeed, these are the
only ordered pairs .x; y/, with x 2 f1; 2; 3g, y 2 f2; 3; 4; 5g and such that x y.
The example above illustrates an obvious procedure we can use to define a
specific relation R between given nonempty sets X and Y (in this order): it suffices
to declare, somehow, a subset of the cartesian product X Y; those ordered pairs of
X Y that satisfy the given prescription will be precisely the elements of R.
If R is a relation from X to Y, then R X Y, by definition. Conversely, once
we choose an ordered pair .x; y/ 2 X Y, then either .x; y/ 2 R or .x; y/ … R. In
the former case, we say that x and y are related by R, and write x R y; in the latter
case, we say that x and y are not related by R, and write x R y. Thus, in symbols,
x R y , .x; y/ 2 R:
(6.1)
Therefore, with respect to the relation of Example 6.5, we have 3 R 2 but 2 R 3,
since 2 3 is false.
Among all types of relations we can consider, one of the most important ones is
given by the coming definition.
Definition 6.6 Given nonempty sets X and Y, a relation R from X to Y is a function
if the following condition is satisfied:
8 x 2 X; 9 a unique y 2 YI x R y:
As in the beginning of this section, given nonempty sets X and Y, it is customary
to use smallcase latin letters like f , g, h etc to denote functions from X to Y. Also
as before, we write f W X ! Y if f is such a function, and write f .x/ D y if the
ordered pair .x; y/ 2 X Y belongs to f , i.e., if it is such that x f y. Notice that such a
notation makes sense, for the definition of function assures that, if .x; y1 / and .x; y2 /
are ordered pairs in X Y such that x f y1 and x f y2 , then y1 D y2 . On the other
hand, a moment’s thought shows that the formal definition of function given above
is simply the correct way of validating the informal definition given at the beginning
of this section.
We will usually work with functions f W X ! Y such that X; Y R. In these
cases, we will generally indicate which element f .x/ 2 Y is associated to a generic
element x 2 X by means of a formula in x that ought to be seen as a rule to get
f .x/ out of x. For instance, we can say “consider the function f W R ! R given
by f .x/ D x2 ” to mean that the function f associates, to each x 2 R, its square x2 .
Observe that the defining requisites of a function are fulfilled, for, to each x 2 R,
2
we have associated a single
yet with respect to
p f .x/ 2
pR,2 namely, x . In particular,
this example, we have f . 2/ D . 2/ D 2, f .3/ D 32 D 9 etc.
6.1 Definitions and Examples
147
When X; Y R and f W X ! Y is a function such that the element f .x/ 2 Y
associated to x 2 X is given by a formula in x, we will sometimes denote such a
correspondence by writing
f W X ! Y
:
x 7! f .x/
This way, the function of the previous paragraph, that associates to each x 2 R its
square x2 , could have been denoted in the following manner:
f W R ! R
:
x 7! x2
Let’s see another example.
Example 6.7 Consider the function f W Q ! R, given by
f .x/ D
p
x2 C 1; if x 0
:
x C 1; if x > 0
We surely have defined a function, for the expressions that define f .x/ have sense
in R and, although we must apply different formulas, according to the rational x
satisfies x 0 or x > 0,peach rational x p
has a single, well defined image f .x/. Thus,
for example, f .1/ D .1/2 C 1 D 2 (since 1 0), but f .2/ D 2 C 1 D 3
(since 2 > 0).
Observe that we could have defined f .x/ by writing
p
x2 C 1; if x 0
f .x/ D
:
x C 1; if x 0
If that were the case, then, albeit conditions x 0 and x 0 cover all rationals, they
wouldn’t be mutually exclusive: x D 0 would satisfy both. However, the formulas
that should
p be applied in one case or the other would give the same result for x D 0
(since 02 C 1 D 0 C 1), thus avoiding any possibility of inconsistence.
We sometimes say that a function f like that of the previous example is defined
by parts. This expression alludes to the fact that there is a formula to compute
f .x/ when x 2 .1; 0, and another one to compute it when x 2 .0; C1/ (or
x 2 Œ0; C1/, as one may wish).
When dealing with a function f W X ! Y, it’s frequently useful to refer to the sets
X and Y as the domain and codomain of the function, respectively; in this context,
we will write X D Dom .f /. For instance, for the function f of Example 6.7, the
domain and codomain are, respectively, Q and R.
We will often work with functions f W X ! R, with X R. In cases like
these, we shall say that f is a real function (alluding to the fact that it assumes real
148
6 The Concept of Function
values – i.e., that its codomain is R) of a real variable (now alluding to the fact
that a generic element x of the domain X of f – the variable of the function – is a
real number).
In cases like those of the previous paragraph, if f .x/ is defined by a formula on
x, then, unless explicitly stated otherwise, we will stick to the usage of taking X as
the largest possible domain. In other words, in such cases we will take X to be the
largest possible subset of R in which the mathematical operations that define the
expression f .x/ have sense. We shall, then, say that X is the maximal domain of
definition, or simply the maximal domain of f .
Example 6.8 Find the maximal domain of definition X R of the function f W X !
1
R, given by f .x/ D px.x1/
.
1
Solution For px.x1/
to be a real number, it’s necessary and sufficient that x.x 1/
be positive. Thus,
n
o
1
X D x 2 RI p
2R
x.x 1/
D fx 2 RI x.x 1/ > 0g:
A product of two real numbers is positive only if both factors have the same sign.
Therefore, we must have x > 0 and x 1 > 0, or else x < 0 and x 1 < 0. Hence,
we must have x > 1 or x < 0, so that
X D .1; 0/ [ .1; C1/:
t
u
In the context of real functions of a real variable, there are standard ways of
building new functions out of other known ones: we just have to use the arithmetic
operations of their codomain R. More precisely, given a nonempty set X R, a
real number c and functions f ; g W X ! R (with the same domain!), we define the
functions f C g; f g; c f W X ! R by setting
.f C g/.x/ D f .x/ C g.x/;
.f g/.x/ D f .x/ g.x/;
.c f /.x/ D c f .x/;
for every x 2 X.
Some remarks are in order. First of all, notice that the addition signs in the first
equality above have distinct meanings: in the left hand side, the C sign is used in
the very definition of the function f C g, while in the right hand side f .x/ C g.x/
stands for the usual addition of the real numbers f .x/ and g.x/. Analogous remarks
are valid for the multiplication signs used in the definitions of the functions f g and
c f . Secondly, as is usual with real numbers, we shall generally omit multiplication
signs, writing fg and cf instead of f g and c f . This usage will not be source of any
kind of confusion.
6.1 Definitions and Examples
149
It is evident that f C g, fg and cf are indeed functions from X to R. On the other
hand, by analogy with real numbers, f Cg and fg are called the sum and the product
of f and g, respectively. Also, note that the product cf of the real number c by the
function f can be seen as a particular case of the product of two functions: taking
g W X ! R to be the function identically equal to c, we have fg D cf ; yet for such a
g, we shall denote f C g simply by f C c, so that
.f C c/.x/ D f .x/ C c;
for every x 2 X.
Example 6.9 Let f and g be functions from R to R given by f .x/ D
g.x/ D x C 3, for every x 2 R. Then,
.f C g/.x/ D f .x/ C g.x/ D
D
x2
x
x2 C1
and
x
C .x C 3/
C1
x C .x2 C 1/.x C 3/
x3 C 3x2 C 3
D
;
2
x C1
x2 C 1
.fg/.x/ D f .x/g.x/ D
x2 C 3x
x
.x
C
3/
D
x2 C 1
x2 C 1
and
p
p
p
. 3f /.x/ D 3f .x/ D 3 p
x 3
x
D 2
:
x2 C 1
x C1
We leave to the reader the task of verifying that the operations of addition and
multiplication of functions, defined as above, satisfy properties analogous to those
satisfied by the corresponding operations with real numbers. More precisely, for
functions f ; g; h W X ! R and real numbers a, b, we have:
• Commutativity: f C g D g C f ; fg D gf .
• Associativity: f C .g C h/ D .f C g/ C h; f .gh/ D .fg/h; a.bf / D .ab/f .
• Distributivity: f .g C h/ D fg C fh; .a C b/f D af C bf .
Finally, note that the associativity property of the operations of addition and
multiplication of two functions allows us to define, in entirely analogous ways, the
sum and the product of an arbitrary finite number of functions from X to R. For
example, given functions f ; g; h W X ! R, we define f C g C h W X ! R as either
.f C g/ C h or f C .g C h/, since these are equal functions; analogously, fgh W X ! R
can be defined as either .fg/h or f .gh/.
For an additional extension of the discussion of operations on functions, we refer
the reader to Problem 8.
Turning our attention to the codomain of a general function f W X ! Y, it’s
important for the reader to realize that Y doesn’t generally coincide with the set of
the images of the elements of X (which, anyhow, is a subset of Y). Let’s illustrate this
150
6 The Concept of Function
point using again the function f of Example 6.7. We have already observed that its
codomain is the set R of real numbers. On the other hand, it’s certain that its subset
ff .x/I x 2 Qg doesn’t contain real numbers less than 1. Indeed, since x2 C 1 1 for
any real x, we have
f .x/ D
p
p
x2 C 1 1 D 1;
whenever x < 0 is rational; on the other hand, for a rational x > 0, we have f .x/ D
x C 1 > 1. In any case,
ff .x/I x 2 Qg Œ1; C1/;
which is a proper subset of the codomain R of f .
More generally, given a function f W X ! Y, the image of f is the set Im.f /,
whose elements are the images f .x/ 2 Y of the elements x 2 X:
Im.f / D ff .x/ 2 YI x 2 Xg:
In particular, we always have Im.f / Y.
In the example just discussed, we showed that the image of a function can be a
proper subset of its codomain. Nevertheless, we didn’t get an explicit description of
the image of the function under consideration. As further developments will show,
in specific situations this can be a somewhat hard task. For the time being, let’s
consider an illustrative example.
Example 6.10 Find an explicit description of the image of the function f W Qnf0g !
Q given by f .r/ D 1b , if the nonzero rational r is written as r D ab , with a 2 Z, b 2 N
and gcd.a; b/ D 12 .
Solution The function f is well defined, in the sense that its definition is not
ambiguous. Indeed, it’s a standard fact that every nonzero rational number admits
a unique representation as the quotient of two relatively prime integers,
4 the
denominator being a natural number; for example, 46 D 2
,
so
that
f
6 D 13 .
3
On the other hand, in the notations of the statement of the example, since b 2 N, we
immediately see that
Im.f / n1
o n 1 1 1
o
I b 2 N D 1; ; ; ; : : : :
b
2 3 4
Moreover,
it is clear that all elements of the last set above do belong to the image of
f , for f 1b D 1b , for every b 2 N. Hence, we conclude that
2
For further details on the gcd of two nonzero integers, read again the introduction to Chapter 1 or,
alternatively, read Section 6:2 of [5].
6.1 Definitions and Examples
151
o
n 1 1 1
Im.f / D 1; ; ; ; : : : :
2 3 4
t
u
Unfortunately, there exists no algorithm that allows us to explicitly describe the
image of an arbitrarily given function. Nevertheless, in subsequent chapters we will
solve this problem for several important classes of functions.
We finish this section discussing the important concept of equality of functions.
p
With respect
to the function f of Example 6.7, it makes no sense to consider f . 2/,
p
since 2 … Q and the domain of f is Q. What we could do would be to consider,
instead of f , the function g W R ! R, given by
p 2
x C 1; if x 0
:
g.x/ D
x C 1; if x > 0
Although the formulas that define f .x/ and g.x/ are the same, for f they can be
applied only to x 2 Q, while for g they can be applied to every real x. Therefore, it
makes no sense to think about f and g as being equal functions, just denoted in two
different ways.
In the positive direction, we have the following
Definition 6.11 Functions f W X ! Y and g W W ! Z are equal if X D W, Y D Z
and f .x/ D g.x/, for every x 2 X.
If f W X ! Y and g W W ! Z are equal functions, we write f D g. We also stress
that, according to the above definition, the equality of f and g doesn’t merely reduce
to f .x/ D g.x/; it also means the equality X D W of the domains of f and g, as well
as the equality Y D Z of the codomains of these two functions. If functions f and
g as above are not equal, we will write f ¤ g and say that f and g are different or
distinct functions.
Problems: Section 6.1
p
1. Find the maximal domain of definition of the function f , such that f .x/ D px1
.
3x
2. Find
maximal domain of definition of the function f , such that f .x/ D
r the
q
p
p
1
3
3 x.
2
2
2 b2 j
3. Let f W QC ! QC be the function defined by f ab D jaa2 Cb
2 , if a; b 2 N are
relatively prime.
(a) Compute f .1/, f .10/ and f 24
.
36
32
101
(b) Among the rationals 55
,
and
73 257
89 , which do belong to the image of f ?
152
6 The Concept of Function
4. Consider the function f W R ! R given by f .x/ D x3 2x2 C 5x, for every
x 2 R. Prove that f .x/ has the same sign of x,pfor every real number x ¤ 0.
5. Function f W R ! R is such that f .1/ D 2, f . 2/
pD 4 and f .x C y/ D f .x/f .y/,
for every x; y 2 R. Compute the value of f .3 C 2/.
6. Let f W R ! R be a function such that f .x C y/ D f .x/ C f .y/, for every reals
x and y. If .ak /k1 is an AP of common difference r, prove that the sequence
.f .ak //k1 is an AP of common difference f .r/.
7. Let f W R ! R be a function such that f .x C y/ D f .x/f .y/, for every reals
x and y. If .ak /k1 is an AP of common difference r and such that f .a1 / ¤ 0,
prove that the sequence .f .ak //k1 is a GP of common ratio f .r/.
8. * Given a nonempty set X R and functions f ; g W X ! R, extend the
discussion in the text, furnishing adequate definitions to the difference f g
and the quotient gf of the functions f and g.
9. * The integer part of a real number x, denoted bxc, is defined to be the greatest
integer which is less than or equal to x. For instance, bc D 3, b 32 c D 2 and
b1c D 1. Find the image of the integer part function
b c W R ! R
;
x 7! bxc
(6.2)
that associates to each x 2 R its integer part bxc.
10. * The fractional part of a real number x, denoted fxg, is defined as fxg D
x bxc, where bxc stands for the integer part of x (see the previous problem).
For example, fg D 3, f 32 g D 32 .2/ D 12 and f1g D 1 1 D 0.
Find the image of the fractional part function
f g W R ! R
;
x 7! fxg
(6.3)
that associates to each x 2 R its fractional part fxg.
11. (TT) Prove
p that the nth natural number which is not a perfect square is equal
to bn C n C 12 c, where bc is defined as in Problem 9.
12. * Let f W Q ! Q be a function such that f .x C y/ D f .x/ C f .y/, for all x; y 2 Q.
Prove the following identities, for all x; y 2 Q and m; n 2 Z, with n ¤ 0:
(a)
(b)
(c)
(d)
(e)
f .0/ D 0 and f .x/ D f .x/.
f .x y/ D f .x/ f .y/.
f .mx/
D mf .x/.
f 1n D f .1/
.
n
f mn D mn f .1/.
6.2 Monotonicity, Extrema and Image
153
6.2 Monotonicity, Extrema and Image
We start this section concentrating ourselves in the problem of finding the image of
a given function. To this end, recall that the image of a function f W X ! Y is the set
Im.f / D ff .x/ 2 YI x 2 Xg
D fy 2 YI y D f .x/ for some x 2 Xg:
This second way of declaring Im.f / is particularly useful in case f is a real function
of a real variable, i.e., if f W X ! R, with X R. Indeed, for such an f , if the values
f .x/ are given by a formula on x 2 X, we can look at the problem of finding the
image of f as that of finding the y 2 R for which the equation f .x/ D y has at least
one solution x 2 X. Let’s see some examples.
Example 6.12 An affine function is a function f W R ! R such that f .x/ D ax C b,
for every real x, where a and b are given real numbers, with a ¤ 0. A linear function
is an affine function f as above, such that b D 0.
According to the previous paragraph, the image of an affine function f as above
can be found by searching for the y 2 R for which the equation ax C b D y has
at least one solution x 2 R. Since a ¤ 0, this equation always admits the solution
x D yb
a . Therefore, we conclude that every y 2 R belongs to the image of f and,
hence, Im .f / D R.
Example 6.13 The function of inverse proportionality is the function f W R n
f0g ! R n f0g given by f .x/ D 1x , for every x 2 R n f0g.
In order to find its image, it suffices to find all y 2 R for which there exists a
real number x ¤ 0 (i.e., x belonging to the domain of f ), such that f .x/ D y, i.e.,
such that 1x D y. If y D 0, it is clear that this equation doesn’t admit solutions; on
the other hand, if y ¤ 0, this same equation admits the solution x D 1y ¤ 0. Hence,
Im .f / D R n f0g.
It is time we define one of the most important classes of elementary real functions
of a real variable.
Definition 6.14 A quadratic, or second degree function is a function f W R ! R
such that f .x/ D ax2 C bx C c for every real x, where a, b and c are given real
numbers, with a ¤ 0. The discriminant of f is the discriminant of the associated
second degree trinomial ax2 C bx C c, so that D b2 4ac.
The problem of finding the image of a quadratic function is sufficiently important
to be collected in the following
Proposition 6.15 With respect to the quadratic function f .x/ D ax2 C bx C c, we
have that:
(a) If a > 0, then Im .f / D 4a
; C1 .
(b) If a < 0, then Im .f / D 1; 4a
.
154
6 The Concept of Function
Moreover, in any of the above cases,
f .x/ D b
,xD :
4a
2a
Proof Following the general procedure described at the beginning of this section,
it suffices to find all y 2 R for which the equation ax2 C bx C c D y, i.e., the
second degree equation ax2 C bx C .c y/ D 0, has at least one real solution.
As we already know from Section 2.3, a necessary and sufficient condition for the
existence of such (a) root(s) is that this last equation has nonnegative discriminant,
i.e., that b2 4a.c y/ 0. Since we agreed to let b2 4ac D , the y’s we’re
looking for are precisely the solutions of the first degree inequality
C 4ay 0:
Now, we separately consider the cases a > 0 and a < 0. If a > 0, then
4ay C 0 , y ;
4a
and it follows that
Im .f / D y 2 RI y D ; C1 I
4a
4a
if a < 0, then
4ay C 0 , y 4a
and, hence,
Im .f / D y 2 RI y 4a
D
1; :
4a
To what was left to do, notice that, for y 2 Im .f /, the solutions of the equation
ax2 C bx C c D y (, ax2 C bx C .c y/ D 0) are
xD
b ˙
p
p
b ˙ C 4ay
b2 4a.c y/
D
:
2a
2a
(6.4)
Therefore, equation f .x/ D y admits a single solution if and only if C 4ay D 0,
or, which is the same, if and only if y D 4a
; this being the case, we have from (6.4)
b
that x D 2a .
t
u
6.2 Monotonicity, Extrema and Image
155
To what comes next, we make the convention of saying that the quadratic
function f .x/ D ax2 C bx C c has constant sign if f .x/ 0 for every x 2 R, or
f .x/ 0 for every x 2 R.
Corollary 6.16 The quadratic function f .x/ D ax2 CbxCc has constant sign if and
only if 0. In this case, we have af .x/ 0 for every x 2 R. In other words:
(a) If 0 and a > 0, then f .x/ 0 for every x 2 R.
(b) If 0 and a < 0, then f .x/ 0 for every x 2 R.
Proof Let’s look at the case a > 0, the case a < 0 being totally analogous. If 0,
it follows from the previous proposition that
f .x/ 0; 8 x 2 R:
4a
Conversely, suppose that a > 0 and that f has constant sign. Again from the
previous proposition, the image of f contains positive numbers, so that we must
have f .x/ 0, for every x 2 R. In particular,
b
Df 0:
4a
2a
Therefore, 0.
t
u
Remark 6.17 A simple modification of the argument presented in the proof of the
previous corollary allows us to conclude that
i. If < 0 and a > 0, then f .x/ > 0 for all x 2 R.
ii. If < 0 and a < 0, then f .x/ < 0 for all x 2 R.
From now on, we shall use these results whenever needed, without further comments.
The previous corollary can also be used to give a much simpler proof of Cauchy’s
inequality (cf. Theorem 5.13).
Example 6.18 Let n > 1 be an integer and a1 , a2 , . . . , an , b1 , b2 , . . . , bn be real
numbers, such that at least one of the ai ’s and at least one of the bi ’s is nonzero.
Consider the quadratic function
f .x/ D .a1 x b1 /2 C .a2 x b2 /2 C C .an x bn /2
D Ax2 2Bx C C;
where A D a21 C a22 C C a2n > 0 (for, at least one of the ai ’s is nonzero), B D
a1 b1 C a2 b2 C C an bn and C D b21 C b22 C C b2n .
Since f .x/ is a sum of squares, we have f .x/ 0 for every x 2 R. On the other
hand, since A > 0, Corollary 6.16 gives D 4.B2 AC/ 0. Hence, B2 AC,
156
6 The Concept of Function
p p
or, which is the same, jBj A C. Substituting the values of A, B and C, we get
Cauchy’s inequality.
According to the above reasoning, equality in Cauchy’s inequality is equivalent,
for the function f , to D 0. In turn, this is equivalent to the existence of a single
˛ 2 R such that f .˛/ D 0. However, since f .˛/ is a sum of squares, the only way
we can have f .˛/ D 0 is if each one of these squares vanishes, i.e., if
a1 ˛ b1 D a2 ˛ b2 D D an ˛ bn D 0:
Finally, since at least one of the bi ’s is nonzero, we have ˛ ¤ 0 and, writing D ˛1 ,
we get
a1 D b1 ; a2 D b2 ; : : : ; an D bn
as a necessary and sufficient condition for equality.
In order to continue in our study of function, we need a piece of terminology.
Definition 6.19 Let I R be an interval. A function f W I ! R is said to be:
(a)
(b)
(c)
(d)
increasing, if f .x1 / < f .x2 /, for all x1 < x2 in I.
decreasing, if f .x1 / > f .x2 /, for all x1 < x2 in I.
nondecreasing, if f .x1 / f .x2 /, for all x1 < x2 in I.
nonincreasing, if f .x1 / f .x2 /, for all x1 < x2 in I.
Moreover, in any of the cases above, we say that the function f is monotonic in I 3 .
Regarding the above definition, an interesting (and, as we shall see, important)
problem is the one of finding the monotonicity intervals of a function f W I ! R,
where I R is an interval. By that we mean to find the intervals J I such that f
is increasing (resp. decreasing, nondecreasing or nonincreasing) in J. Here we shall
see some elementary examples, postponing a more general analysis to Chapter 9.
Example 6.20 The affine function f W R ! R, given by f .x/ D ax C b, is increasing
if a > 0 and decreasing if a < 0.
Let’s verify this claim in the case a > 0, the analysis of the case a < 0 being
totally analogous. Letting x1 < x2 be two real numbers, it follows from a > 0 that
f .x2 / f .x1 / D .ax2 C b/ .ax1 C b/ D a.x2 x1 / > 0:
Therefore, f is increasing.
3
In the notations of this definition, it is worth observing that, for some authors, a function f
satisfying the condition of item (a) (resp. of item (b), (c) or (d)) is said to be strictly increasing
(resp. strictly decreasing, increasing or decreasing).
6.2 Monotonicity, Extrema and Image
157
2
x
Example 6.21 The function f W Œ0; C1/ ! R, given by f .x/ D xC2
, is increasing
in the whole interval Œ0; C1/. To check this, take real numbers 0 a < b. Then,
b2
a2
bC2 aC2
1
Œb2 .a C 2/ a2 .b C 2/;
D
.a C 2/.b C 2/
f .b/ f .a/ D
and, since .a C 2/.b C 2/ > 0, it suffices to show that b2 .a C 2/ a2 .b C 2/ > 0.
To this end, start by writing
b2 .a C 2/ a2 .b C 2/ D b2 a a2 b C 2.b2 a2 /
D ab.b a/ C 2.b a/.b C a/
D .b a/Œab C 2.b C a/:
Now, since 0 a < b, both factors in the last product above are positive, so that
b2 .a C 2/ a2 .b C 2/ > 0.
Example 6.22 Function f W R ! R, given by f .x/ D x3 C 2x, for every x 2 R, is
increasing. Indeed, for any real numbers a < b, we have
f .b/ f .a/ D b3 a3 C 2b 2a
D .b a/.b2 C ba C a2 / C 2.b a/
D .b a/.b2 C ab C a2 C 2/:
Since b a > 0, it suffices to show that a2 C ab C b2 C 2 > 0; in order to do this,
one possibility is to use the inequality between the arithmetic and geometric means
for two numbers:
a2 C b2 C ab C 2 2jabj C ab C 2 jabj C 2 > 0;
where, in the next to last inequality, we used the fact that j˛j C ˛ 0, for every
˛ 2 R.
The coming proposition solves, for quadratic functions, the problem of finding
the monotonicity intervals.
Proposition 6.23 Let a, b, c 2 R, with a ¤ 0, and f .x/ D ax2 C bx C c for every
x 2 R.
b
b
(a) If a > 0, then f is decreasing in 1; 2a
and increasing in 2a
; C1 .
b
b
and decreasing in 2a
; C1 .
(b) If a < 0, then f is increasing in 1; 2a
158
6 The Concept of Function
Proof Let’s do the proof of item (a), the proof of item (b) being totally analogous.
b
For x2 > x1 2a
, we have
f .x2 / f .x1 / D a.x22 x21 / C b.x2 x1 /
b
D a.x2 x1 / x2 C x1 C
> 0;
a
b
gives x2 x1 > 0, as well as x2 C x1 C
since x2 > x1 2a
b
a
> 0.
t
u
The next definition is, in a certain sense, complementary to Definition 6.19.
Definition 6.24 Let I R be an interval and f W I ! R be a given function. We
say that y0 2 R is the minimum value of f in I if the two following conditions are
satisfied:
(a) Im .f / Œy0 ; C1/.
(b) y0 2 Im .f /.
In this case, the real numbers x0 2 I such that f .x0 / D y0 are called the minimum
points of the function f .
Similarly, we define what one means by the maximum value and the maximum
points of a function f W I ! R (I R being an interval). The maximum and
minimum points of a given function (provided they exist) are collectively called its
extreme points; accordingly, the values the function takes at those points are its
extreme values.
In Section 9.6, we shall see how to search extreme points for differentiable
functions, i.e., functions possessing derivatives. For the time being, we shall content
ourselves in analysing some elementary examples, the first of which being an
immediate consequence of Proposition 6.15.
Proposition 6.25 With respect to the quadratic function f .x/ D ax2 C bx C c, if
b
a > 0 (resp. a < 0), then 2a
is the only minimum (resp. maximum) point of f .
Moreover, the minimum (maximum) value of f is 4a
.
The proposition above has several interesting applications, two of which are
collected below, for the sake of illustrating its use. For the necessary geometric
background, we refer the reader to [4].
Example 6.26 There is given a semicircle of 1cm of radius. A rectangle is so
situated that one of its sides lies on the diameter of the semicircle, whereas its other
two vertices lie on the semicircle itself. Compute the largest possible value for the
area of the rectangle.
Solution Let Fig. 6.4 account for the described situation, so that AB is the diameter
of the semicircle, O is its center and PQRS is the given rectangle, with PQ AB.
Setting OQ D x and QR D y, the area of PQRS equals 2xy. On the other hand,
applying Pitagoras’ theorem to triangle OQR, we get x2 C y2 D 1 and, hence,
6.2 Monotonicity, Extrema and Image
159
Fig. 6.4 Maximizing the
area of rectangle PQRS
R
S
1
A
P
O x Q
y
B
p
p
p
2xy D 2x 1 x2 D 2 x2 .1 x2 / D 2 x2 x4 :
Now, making the substitution z D x2 , it follows from the last expression above for
the area that it suffices to maximize the quadratic function f .z/ D z z2 , subjected
to the condition 0 < z < 1 (for, x < OR D 1). By Proposition 6.25, such a
function (without any further restrictions) admits z D 12 as its only maximum point.
Moreover, since 12 2 .0; 1/, it follows that the desired maximum value is f 12 D 14 .
q
Therefore, the maximum value for the area is 2 14 D 1.
t
u
Example 6.27 Given a triangle ABC in the plane, show that its baricenter is the
2
2
2
only point P in the plane of ABC for which the sum AP C BP C CP attains its
minimum possible value.
Proof Fix a cartesian system of coordinates in the plane, with respect to which
A.x1 ; y1 /, B.x2 ; y2 / and C.x3 ; y3 /. If P.x; y/, then the formula for the distance
between two points in the plane furnishes
2
2
2
AP C BP C CP D f .x/ C g.y/;
where
f .x/ D
3
X
.x xi /2 D 3x2 2.x1 C x2 C x3 /x C .x21 C x22 C x23 /
iD1
and, analogously, g.y/ D 3y2 2.y1 C y2 C y3 /y C .y21 C y22 C y23 /.
2
2
Since x and y are independent variables, in order to minimize AP C BP C
2
CP it suffices to minimize the quadratic functions f and g. To this end, invoking
Proposition 6.25 we conclude that f and g attain their minimum values only at the
points x D 13 .x1 C x2 C x3 / and y D 13 .y1 C y2 C y3 /, respectively. However, it is
a well known fact (cf. Chapter 6 of [4], for instance) that these are precisely the
coordinates of the baricenter of triangle ABC.
t
u
To address the problem of finding the maximum and/or minimum values of a
given function, another elementary strategy which is sometimes useful is to resort
to inequalities. In what follows we shall see two examples along these lines.
160
6 The Concept of Function
2
C1
Example 6.28 Let f W Œ0; C1/ ! R be the function given by f .x/ D xxC1
, for
every x 2 R. What is the minimum value of f ? Does f attain a maximum value?
Solution First of all, note that
x2 C 1
x2 1 C 2
D
xC1
xC1
2
2
D .x C 1/ C
2:
Dx1C
xC1
xC1
f .x/ D
Therefore, applying (5.2) (with a D x C 1 and b D
.x C 1/ C
2
),
xC1
we get
r
p
2
2
2 .x C 1/ D 2 2;
xC1
xC1
2
with equality if and only if x C 1 D xC1
, i.e., if and only if x2 C 2x 1 D 0. Since
x p
0, we conclude that equality takes place in the above inequality if and only if
x D 2 1. Thus, for x 0 we have
f .x/ D .x C 1/ C
p
2
2 2 2 2;
xC1
p
so that
p 2 2 2 is the minimum possible value for f , and it is attained only at
x D 2 1.
2
To what is left, just notice that, for n 2 N, we have f .n/ D n 1 C nC1
n 1.
Therefore, f does not attain a maximum value.
t
u
Example 6.29 Find the maximum value
and the maximum point(s) of the function
p
x
f W Œ0; C1/ ! R, given by f .x/ D x2 C16 .
Solution It follows from the inequality between the arithmetic and geometric
means that
16
16
16
C
C
x2 C 16 D x2 C
3
3
3
r
32 p
16 16 16
4
D p
4 x2 x;
4
3 3 3
27
so that
p
p
4
x
27
:
f .x/ D 2
x C 16
32
6.2 Monotonicity, Extrema and Image
161
Equality holds if and only if x2 D
Thus,
p
4
27
32
16
3 ,
which is the same as x D
p4
3
is the maximum value of f and it is attained only at x D
(since x 0).
p4 .
3
t
u
Problems: Section 6.2
1. Find the image of the function f W R ! R, given by f .x/ D x2 1C1 , for every
x 2 R.
2. * Find the image of the function f W R ! R, given by f .x/ D x C 1x , for every
x 2 R .
3. Let I R be an interval, let a 2 I and f W I ! R be a given function. If f is
increasing (resp. decreasing) in .1; a \ I and decreasing (resp. increasing)
in Œa; C1/ \ I, prove that a is the only maximum (resp. minimum) point of f
in I.
4. * Let X R be a nonempty set, f W X ! R be a given function and c 2 R.
Relate the images of the functions f and f C c. More precisely, if Y D Im .f /,
prove that Im .f C c/ D Y C c, where Y C c denotes the set
Y C c D fy C cI y 2 Yg:
5. * Let X R be a nonempty set, f W X ! R be a given function and c 2 R .
Relate the images of the functions f and cf . More precisely, if Y D Im .f /,
prove that Im .cf / D cY, where cY denotes the set
cY D fcyI y 2 Yg:
6. Motivated by the canonical form of the second degree trinomial ax2 C bx C c,
from now on we shall say that
(
f .x/ D a
b
xC
2a
2
2
4a
)
(6.5)
is the canonical form of the quadratic function f .x/ D ax2 C bx C c. Use this
canonical form to give another proof of Proposition 6.15.
7. Let f .x/ D ax2 C bx C c be a quadratic function for which > 0, and x1 < x2
be the roots of f .x/ D 0. Prove the following items:
(a) If a > 0, then f .x/ < 0 , x 2 .x1 ; x2 /.
(b) If a < 0, then f .x/ < 0 , x … Œx1 ; x2 .
8. Let f .x/ D ax2 C bx C c be a quadratic function. If there exists a real number
x0 for which af .x0 / < 0, prove that > 0 and x0 2 .x1 ; x2 /, where x1 < x2 are
the roots of the equation f .x/ D 0.
162
6 The Concept of Function
9. Among all rectangles of a given perimeter, prove that the one of largest possible
area is the square.
10. The cross section of a tunnel has the shape of a semicircle of radius 5m. The
tunnel has two lanes of traffic, which go in opposite directions and are separated
one from the other by a narrow median. The trucks of a transportation firm are
to cross the tunnel to take goods from one city to another. If the trucks are 18m
long, what should be their widths and heigths, so that they can carry the largest
possible load volume per travel?
11. Compute the maximum value of the function f W R ! R given by f .x/ D x5x1
2 C1 ,
for every x 2 R.
12. Let ˛1 < ˛2 < < ˛n be given reals, and f W R ! R be given by
f .x/ D jx ˛1 j C jx ˛2 j C C jx ˛n j;
for every x 2 R. Prove that f attains a minimum value and compute it in terms
of ˛1 , ˛2 , . . . , ˛n .
13. In each of the following items, use the inequality between the arithmetic and
geometric means to compute the maximum value of the given function:
(a) f W R ! R given by f .x/ D
x
, for every x 2 R.
2x2 C3
2x2 C5xC2
, for every x 2
x2 C1
3
R.
(b) f W R ! R given by f .x/ D
(c) f W Œ0; 1 ! R given by f .x/ D x.1 x /, for every x 2 Œ0; 1.
14. In each of the following items, use the inequality between the arithmetic and
geometric means to compute the minimum value of the given function:
(a) f W .0; C1/ ! R given by f .x/ D .xC10/.xC2/
, for every x 2 .0; C1/.
xC1
(b) f W .0; C1/ ! R given by f .x/ D x2 C ax , for every x 2 .0; C1/, where a
is a positive real constant.
2
(c) f W .0; C1/ ! R given by f .x/ D x3xCa , for every x 2 .0; C1/.
(d) f W .0; C1/ ! R given by f .x/ D 6x C 24
, for every x 2 .0; C1/.
x2
15. (Romania) Let x; y 2 R be such that x2 xy C y2 2. Show that x4 C y4 8
and explain when the equality holds.
16. (TT) Find all reals x, y, z and t such that
8
y D x3 C 2x
ˆ
ˆ
<
z D y3 C 2y
:
ˆ t D z3 C 2z
:̂
x D t3 C 2t
For the next problem, the reader might find it helpful to recall the discussion
on the gcd of two nonzero integers, at the beginning of Chapter 1. alternatively,
look at Chapter 6 of [5].
6.3 Composition of Functions
163
17. (OCM) Let f W N ! N be a function such that f .mn/ D f .m/ C f .n/ whenever
m and n are relatively prime. A natural number m is said to be a strangulation
point of f if n < m ) f .n/ < f .m/ and n > m ) f .n/ > f .m/. If f has
infinitely many strangulation points, show that it is an increasing function.
6.3 Composition of Functions
Functions f W X ! Y and g W Y ! Z give well defined rules for, departing from
x 2 X via f , get y D f .x/ 2 Y and then, via g, get z D g.y/ 2 Z. It thus seems
reasonable that we may form a new function that allows us to go directly from X
to Z. This is indeed the case, and the corresponding function is called the composite
of f and g, according to the following
Definition 6.30 Given functions f W X ! Y and g W Y ! Z, the composite
function of f and g (in this order) is the function g ı f W X ! Z, defined for
x 2 X by
.g ı f /.x/ D g.f .x//:
Roughly speaking, the above definition means that, in order to find the image of
x 2 X by g ı f , it suffices to find the image of f .x/ 2 Y by g. On the other hand,
it is easy to verify that g ı f , as defined above, is indeed a function. Also, observe
that to form the composite of f and g, it is necessary that the domain of g equals the
codomain of f . Let’s see some examples.
Example 6.31 Let X and Y be nonempty sets and f W X ! Y an arbitrary function.
If IdX W X ! X and IdY W Y ! Y are the identity functions on X and Y, respectively,
then
f ı IdX D f and IdY ı f D f :
Let’s check the equality f ı IdX D f , the other one being totally analogous. To this
end, it suffices to note that f ı IdX is a function from X to Y such that, for evey x 2 X,
.f ı IdX /.x/ D f . IdX .x// D f .x/:
Example 6.32 Let f ; g W R ! R be the functions given by f .x/ D x2 and g.x/ D
1
, for every x 2 R. Then, g ı f and f ı g are functions from R to R, with
x2 C1
.g ı f /.x/ D g.f .x// D
1
1
1
D 2 2
D 4
.f .x//2 C 1
.x / C 1
x C1
164
6 The Concept of Function
and
.f ı g/.x/ D f .g.x// D .g.x//2 D
1
2
x C1
2
D
x4
1
:
C 2x2 C 1
The preceding example show an interesting phenomenon: it must happen that
g ı f ¤ f ı g. More precisely, it may happen that we can form g ı f but cannot form
f ı g (or vice-versa); it suffices to have, for instance, f W X ! Y and g W Y ! Z, with
X ¤ Z. Nevertheless, even if we can form both g ı f and f ı g, it may well be the
case that g ı f ¤ f ı g.
Example 6.33 Let f ; g W .0; C1/ ! .0; C1/ be functions given by
f .x/ D
x2 C 1
3x2
and .f ı g/.x/ D
xC2
;
3
for every x 2 .0; C1/. Find an expression for g.x/ in terms of x.
Solution The definition of composite function gives
g.x/2 C 1
xC2
D .f ı g/.x/ D f .g.x// D
;
3
3g.x/2
so that
g.x/2 C1
3g.x/2
D
xC2
3
or, which is the same,
3g.x/2 C 3 D 3.x C 2/g.x/2 :
Looking at this expression as a first degree equation in g.x/2 , we obtain g.x/2 D
1
and, hence, g.x/ D ˙ pxC1
, for each x > 0. However, since g has positive
1
xC1
image, we must have g.x/ D
p1 ,
xC1
for every x > 0.
t
u
The operation of composition of functions, albeit not commutative, is associative,
as the next proposition teaches us.
Proposition 6.34 Given functions f W X ! Y, g W Y ! Z and h W Z ! W, we have
h ı .g ı f / D .h ı g/ ı f :
Proof First of all, both h ı .g ı f / and .h ı g/ ı f are functions from X to W. Hence,
we only need to check that they associate, to each x 2 A, a single element of W. To
see this, just note that
.h ı .g ı f //.x/ D h..g ı f /.x// D h..g.f .x///
D .h ı g/.f .x// D ..h ı g/ ı f /.x/:
t
u
6.3 Composition of Functions
165
The previous proposition is quite important, for, it guarantees that if functions f ,
g and h (in this order) can be composed, then the composite function can be safely
denoted by h ı g ı f , and we need not worry about which composition to perform
first. As an immediate generalization of this remark, suppose functions f , g, h and l
(in this order) can be composed. Then, the previous proposition gives
l ı .h ı .g ı f // D .l ı h/ ı .g ı f / D ..l ı h/ ı g/ ı f D ;
so that the order in which we insert parentheses does not alter the composition.
It is not difficult to see that the situation described in the last paragraph remains
true for the composition (whenever possible) of any finite number of functions. In
particular, given a nonempty set X, a function f W X ! X and a natural number
n, this last fact allows us to unambiguously define the nth composite function
f .n/ W X ! X, by setting
f .n/ D f ı f ı ı f :
„ ƒ‚ …
n
Example 6.35 Let f W R n f1; 1g ! R n f1; 1g be the function given by f .x/ D
1x
1Cx , for every x ¤ ˙1. For each n 2 N, find the expression that defines the nth
composite function f .n/ .
Solution Firstly, note that f .n/ W R n f1; 1g ! R n f1; 1g. Now, since
f .2/ .x/ D .f ı f /.x/ D f .f .x// D
1x
1 1Cx
1 f .x/
D
D x;
1x
1 C f .x/
1 C 1Cx
we have f .2/ D IdX , the identity function of X D R n f1; 1g. This gives us
f .3/ D f ı f .2/ D f ı IdX D f and f .4/ D f ı f .3/ D f ı f D IdX :
In general, suppose (by induction hypothesis) that we have already proved that
f .2k1/ D f and f .2k/ D IdX , for some integer k 1. Then,
f .2kC1/ D f ı f .2k/ D f ı IdX D f
and
f .2kC2/ D f ı f .2kC1/ D f ı f D IdX :
Therefore, we conclude that f .n/ D f whenever n is odd, and f .n/ D IdX whenever n
is even.
u
t
Given a function f W X ! Y, we’ve already seen some examples that illustrate
the fact that the image of f is not necessarily equal to its codomain Y. On the other
166
6 The Concept of Function
hand, we can also have two distinct elements of X with the same image via f . For
an example, consider the quadratic function f .x/ D x2 , with x 2 R; for every x 2 R,
we have f .x/ D x2 D .x/2 D f .x/. We attach special names to functions whose
images coincide with their codomains, or which associate distinct images to distinct
elements of their domains. This is set in the coming
Definition 6.36 A function f W X ! Y is said to be:
(a) Injective, or one-to-one, or an injection, if, for every y 2 Y, there exists at
most one x 2 X such that f .x/ D y.
(b) Surjective, or onto, or a surjection, if, for every y 2 Y, there exists at least one
x 2 X such that y D f .x/. In other words, this is the same as saying that the
image of f is all of Y.
(c) Bijective, or one-to-one onto, or a bijection, if it is simultaneously injective
and surjective.
An efficient way of verifying whether a function f W X ! Y is injective or not is
to verify whether the implication
f .x1 / D f .x2 / ) x1 D x2
(6.6)
is true for all x1 ; x2 2 X. Accordingly, in order to prove that f is surjective, one must
be capable of, for each y 2 Y, guarantee the existence of at least one solution x 2 X
for the equation f .x/ D y. Let’s see some examples.
Example 6.37 If X R is a nonempty set and f W X ! X is a function such that
f .f .x// D x for every x 2 X, then f is a bijection.
Proof Let x1 and x2 be elements of X for which f .x1 / D f .x2 /. According to (6.6),
in order to show that f is injective it suffices to prove that x1 D x2 . To this end,
observe that f .x1 / D f .x2 / implies f .f .x1 // D f .f .x2 // and, then (by using the given
hypothesis), x1 D x2 .
The surjectivity of f also follows from the hypothesis: for a fixed y 2 X, taking
x D f .y/ 2 X we get f .x/ D f .f .y// D y, so that y 2 Im.f /.
t
u
As a particular case of the previous example, the function of inverse proportionality (cf. Example 6.13) is a bijection from R n f0g into itself.
Example 6.38 Let f W Œ0; 1 ! Œ0; 1 be a surjective function, such that jf .x1 / f .x2 /j jx1 x2 j for all x1 ; x2 2 Œ0; 1. Prove that there are only two possibilities:
either f .x/ D x for every x 2 Œ0; 1, or f .x/ D 1 x for every x 2 Œ0; 1.
Proof Let a; b 2 Œ0; 1 be chosen in such a way that f .a/ D 0 and f .b/ D 1 (that it
is possible to make such a choice follows from the surjectivity of f ). Then, the given
hypothesis allows us to write
1 D j1 0j D jf .b/ f .a/j jb aj 1;
6.3 Composition of Functions
167
so that jb aj D 1. However, the only a; b 2 Œ0; 1 such that jb aj D 1 are a D 0
and b D 1, or vice-versa. Suppose that a D 0 and b D 1 (the other case can be
haldled similarly), and take any c 2 .0; 1/. Triangle inequality (5.10), together with
the hypothesis on f , give
1 D jf .1/ f .0/j
jf .1/ f .c/j C jf .c/ f .0/j
j1 cj C jc 0j
D .1 c/ C c D 1:
Hence, we must have jf .c/ f .0/j D jc 0j and, since c; f .c/ 0, we get f .c/ D c.
Finally, this is true for every c 2 Œ0; 1, so that f .x/ D x for every x 2 Œ0; 1.
t
u
The coming proposition teaches us the way injective, surjective and bijective
functions behave with respect to composition.
Proposition 6.39 Let f W X ! Y and g W Y ! Z be given functions. Then:
(a)
(b)
(c)
(d)
(e)
g ı f injective ) f injective, but the converse is not necessarily true.
g ı f surjective ) g surjective, but the converse is not necessarily true.
g and f injective ) g ı f injective.
g and f surjective ) g ı f surjective.
g and f bijective ) g ı f bijective.
Proof
(a) For x1 ; x2 2 X, we have
f .x1 / D f .x2 / ) g.f .x1 // D g.f .x2 //
) .g ı f /.x1 / D .g ı f /.x2 /
) x1 D x 2 ;
where, in the last passage, we used the fact that g ı f is injective. We now have
to give an example in which f is injective but g ı f is not. To this end, take
X D Y D Z D R, f .x/ D x and g.x/ D x2 .
(b) Choosing z 2 Z arbitrarily, the surjectivity of g ı f guarantees the existence of
at least one x 2 X such that z D .g ı f /.x/. Then, z D g.f .x//, so that g is also
surjective. For the second part, take X D Y D Z D R, g.x/ D x and f .x/ D x2 ;
then, g is surjective, but g ı f is not.
(c) Let x1 ; x2 2 X. By using the injectivity of g and, then, that of f , we obtain
.g ı f /.x1 / D .g ı f /.x2 / ) g.f .x1 // D g.f .x2 //
) f .x1 / D f .x2 /
) x1 D x2 ;
Hence, g ı f is also injective.
168
6 The Concept of Function
(d) Let z 2 Z be arbitrarily choosen. The surjectivity of g guarantees the existence
of y 2 Y such that z D g.y/. On the other hand, the surjectivity of f assures the
existence of x 2 X for which f .x/ D y. Then, we have
.g ı f /.x/ D g.f .x// D g.y/ D z;
so that g ı f is surjective, too.
(e) It follows from items (c) and (d) that
g and f bijective ) g and f injective and surjective
) g ı f injective and surjective
) g ı f bijective:
t
u
Let’s revisit Example 6.37 in light of the previous proposition.
Example 6.40 Let X be a nonempty set. If f W X ! X is a function such that
f ı f D IdX , then f is a bijection.
Proof Indeed, since the identity function IdX W X ! X is a bijection, it follows from
items (a) and (b) of the previous proposition that f is injective and surjective, hence,
bijective.
t
u
The material of this section also allows us to study the important concept of
countably infinite sets, according to the following
Definition 6.41 An infinite set A is said to be countable if there exists an injective
function f W A ! N. In this case, we shall also say that A is a countably infinite set.
If A is countably infinite and B A is infinite, then B is also countably infinite.
Indeed, by composing an injective function f W A ! N with the inclusion W B ! A
(that sends each x 2 B to itself), we get the injective function f ı W B ! N. In
particular, every infinite subset of N is countable.
The set Z of integers is also countably infinite, for one can easily check that the
function f W Z ! N given by
f .x/ D
2x; if x < 0
2x C 1; if x 0
is a bijection. Perhaps a little more surprising is the following
Example 6.42 The cartesian product N
Proof It suffices to define f W N
N is a countably infinite set.
N ! N by setting
f .m; n/ D 2m1 .2n 1/; 8 m; n 2 N:
6.3 Composition of Functions
169
A straightforward argument on odd and even numbers shows that f is injective.
(Actually, the Fundamental Theorem of Arithmetic – cf. introduction to Chapter 1
or Chapter 6 of [5] – assures that f is bijective.)
t
u
With a little more effort (see Problem 21), we can prove that Q is countably infinite. However, as will be seen in Section 7.4 (cf. Example 7.46), R is uncountable,
i.e., not countable. In turn, this implies (cf. Problem 22) that I D R n Q is also
uncountable. For yet another example of an uncountable set, see Problem 23.
The following lemma shows that the elements of every countably infinite set can
be written as terms of a sequence.
Lemma 6.43 If A is a countably infinite set, then there exists a bijection f W N ! A.
In particular, letting an D f .n/, we get A D fa1 ; a2 ; a3 ; : : :g.
Proof By definition, there exists an injective function g W A ! N. Then, g induces
a bijection (which we will also denote by g) from A to B D Im.g/ N. If we
construct a bijection h W N ! B, then f D g ı h W N ! A will also be a bijection.
To what is left to do, we start by letting b1 D min B and setting h.1/ D b1 . Since
B is infinite, we can let b2 D min.Bnfb1 g/ and set h.2/ D b2 . By the same token, let
b3 D min.B n fb1 ; b2 g/ and set h.3/ D b3 . Continuing this way, we define a function
h W N ! B such that, letting h.k/ D bk , we have b1 D min B and
bk D min B n fb1 ; : : : ; bk1 g
for every natural k > 1. In particular, b1 < b2 < b3 < and h is injective.
If h was not surjective, there would exist b 2 B such that b ¤ b1 ; b2 ; b3 ; : : :. If
bk < b for every k 2 N, then B f1; 2; : : : ; bg, a contradiction to the fact that B
is infinite. Hence, there would exist a natural m > 1 such that bm1 < b < bm .
However, since
h.m/ D min B n fh.1/; : : : ; h.m 1/g
D min B n fb1 ; : : : ; bm1 g ;
we should have defined h.m/ to be b, instead of bm . Since this is a contradiction, we
conclude that h is indeed surjective.
t
u
The concept and properties of countable sets give rise to interesting results, one
of which we collect in the following
Example 6.44 Show that it is possible to partition the set of natural numbers in two
sets A and B satisfying the following conditions:
(a) Neither A nor B contains the terms of an infinite and nonconstant AP.
(b) For all distinct x; y 2 A, we have jx yj 2016.
Proof An infinite and nonconstant AP of naturals is characterized by the ordered
pair .a; r/, where a is its first term and r is its common difference. Since N N
170
6 The Concept of Function
is countable, the same is true of the family P of infinite and nonconstant AP’s of
naturals. Hence, we can write P D fs1 ; s2 ; s3 ; : : :g.
Now, let Ak be the set whose elements are the terms of sk , and let A be defined in
the following way: take x1 2 A1 , x2 2 A2 , . . . such that x2 x1 2016, x3 x2 2017, x4 x3 2018, . . . (that this is possible follows from the fact that each Ak is
an infinite set). Then, let B D N n A.
It is clear that A does not contain the terms of an infinite AP of common
difference r 2 N, for the differences xj xj1 are eventually all greater than r.
On the other hand, B does not contain the terms of an infinite and nonconstant AP
either, for, if it did, we should have B Ak for some k 2 N; however, xk 2 Ak n B. u
t
Problems: Section 6.3
1. Let f and g be real functions of a real variable, given by f .x/ D x 72 and
g.x/ D x2 14 . Find the solution set of the inequality j.g ı f /.x/j > .g ı f /.x/.
2. Let f and g be real functions of a real variable, such that f .x/ D 2x C 7 and
.f ı g/.x/ D x2 2x C 3, for every x 2 R. Find the expression that defines g.x/
in terms of x.
3. Let f ; g W R ! R be such that g.x/ D 2x 3 and .f ı g/.x/ D 2x2 4x C 1.
Find the expression that gives f .x/ in terms of x.
4. Let f and g be real functions of a real variable, given by f .x/ D ax C b and
g.x/ D cx C d for every x 2 R, with ac ¤ 0. Show that
f ı g D g ı f , .a 1/d D .c 1/b:
5. Let f W R ! R be the function defined by
(
f .x/ D
xCa
;
xCb
if x ¤ b
:
1; if x D b
If f .f .x// D x for every real x, compute the possible values of b.
6. * Let I R be an interval and f W I ! R be an increasing or decreasing
function. Prove that f is injective.
7. Let I; J R be intervals and f W I ! J and g W J ! R be given functions. If f
and g are increasing (resp. if f is increasing and g is decreasing, or vice-versa),
prove that g ı f is also increasing (resp. decreasing).
8. Let f W R n f0g ! R be a function such that
1
1
1
f .x/f
D 1 and f x C
D f .x/ C f
;
x
x
x
2
2
for every
1 real x ¤ 0. If u and v are nonzero
u real numbers such that u C v D 1
and f uv D 2, compute the value of f v .
6.3 Composition of Functions
171
9. Given a function f W X ! Y, we define its graph4 as the subset Gf of the
cartesian product X Y given by
Gf D f.x; y/ 2 X
10.
11.
12.
13.
14.
15.
YI y D f .x/g:
Let F W X ! Gf be the function defined by F.x/ D .x; f .x//, for every x 2 X.
Prove that F is a bijection.
* Let ; ¤ X R be a union of intervals, which is symmetric with respect to
0 2 R. We say that a function f W X ! R is even (resp. odd) if f .x/ D f .x/
(resp. f .x/ D f .x/), for every x 2 X. If ; ¤ X R is as above, prove that
every function f W X ! R can be written, in a unique way, as a sum of an even
and an odd function with domain X.
Let f W R n f0g ! R be a function such that f ab D f .a/ f .b/, for every
nonzero real numbers a and b. Prove that f is an even function.
Let f W R ! R be an odd function. Decide whether the function f ı f is even,
odd, or not even nor odd.
Let g W R ! R be an odd function, such that g.x/ > 0 whenever x > 0. Show
that there exists a function f W R ! R for which g D f ı f .
Find all real values of k for which the image of the function f W R n f1g ! R,
2 CkxCk
given by f .x/ D 4x xC1
, equals R n .L; L/, for some positive real L.
* A function f W R ! R is periodic if there exists a smallest positive real
number p, called the period of f , such that f .x C p/ D f .x/ for every x 2 R.
Given a periodic function f W R ! R, of period p > 0, do the following items:
(a) Let g W R ! R be also periodic of period p. If f .x/ D g.x/ for every
x 2 Œ0; p/, prove that f D g.
(b) Given a 2 R n f0g, prove that the function g W R ! R, given by g.x/ D
p
f .ax/, is periodic of period jaj
.
16. (Italy) Let f W R ! R be a function such that f .10 C x/ D f .10 x/ and
f .20 C x/ D f .20 x/, for every real x. Prove that f is odd and find p > 0
such that f .x C p/ D f .x/, for every x 2 R.
17. (IMO - adapted) Let f W R ! Œ0; 1 be a function such that, for a certain a 2 R,
we have
f .x C a/ D
1 p
C f .x/ f .x/2 ;
2
for every x 2 R. Prove that f is periodic.
18. (Brazil) The function f W Z ! R is such that f .x/ D x 10 for x > 100 and
f .x/ D f .f .x C 11// for x 100. Find the image of f .
4
Graphs of real functions of a real variable will be one of the main objects of study along these
notes, starting from Section 6.6.
172
6 The Concept of Function
19. (Hungary) Let f W N ! N be a function satisfying the following conditions:
(a) f .1/ D 1.
(b) f .2n/ D 2f .n/ C 1.
(c) f .f .n// D 4n C 1.
Compute f .1993/.
20. Give an example of a surjective function f W N ! N such that, for every n 2 N,
the set fx 2 NI f .x/ D ng is infinite.
21. * The purpose of this problem is to show that Q is countably infinite. To this
end, do the following items:
(a) If sets A and B are either finite or countably infinite sets, show that A B is
either finite or countably infinite.
(b) Construct a surjective function from Z N into Q.
(c) Conclude that there exists a surjection f W N ! Q and, then, that Q is
countable.
22. * The purpose of this problem is to show that I D R n Q is uncountable,
assuming that R itself is uncountable (this will be proved in Example 7.46).
To this end, do the following items:
(a) Let A1 ; A2 ; A3 ; : : : be a countably infinite collection of sets. Show
Sthat there
exist
B
A
such
that
B
;
B
;
B
;
:
:
:
are
pairwise
disjoint
and
k
k
1
2
3
k1 Ak D
S
B
.
k1 k
(b) In the notations
S of (a), if each of A1 ; A2 ; A3 ; : : : is finite or countably infinite,
show that k1 Ak is also finite or countably infinite.
(c) Show that I is uncountable.
23. * Let F be the family5 of infinite subsets A of N such that N n A is also infinite,
i.e.,
F D fA NI A and N n A are infinite:g:
Show that F is an uncountable set.
6.4 Inversion of Functions
Among all functions f W X ! Y, the case of a bijection is, in a certain sense, the
best possible. Indeed, in this case the elements of X and Y are in a one-to-one onto
correspondence, so that to each element of X there corresponds a single element of
Y via f , and vice-versa. When this happens, we can form a function g W Y ! X by
asking that
f .x/ D y , g.y/ D x:
5
A family is a set whose elements are sets themselves.
6.4 Inversion of Functions
173
At this point, a natural question is this: why cannot we use the same declaration
above to define g when f is not bijective? From an intuitive point of view, if f is not
surjective, then there exists an element y of Y which is not the image of any element
of X via f ; therefore, there is no natural way of using f and y to define g.y/. On the
other hand, if f is not injective, then there exist distinct elements x1 ; x2 2 X with the
same image y 2 Y via f ; if we were to try to define g by using f , there would also
be no natural way of deciding which of x1 and x2 should be equal to g.y/.
Back to the case in which f is bijective, it is not difficult to see that g, defined
as above, is indeed a function, moreover such that .g ı f /.x/ D x, for every x 2 X,
and .f ı g/.y/ D y, for every y 2 Y. Yet in another way, we have g ı f D IdX
and f ı g D IdY . Conversely, if f W X ! Y and g W Y ! X are functions such
that g ı f D IdX and f ı g D IdY , then Proposition 6.39 guarantees that f must
be a bijection, and Problem 1 assures that g is the only function satisfying these
compositions.
We summarize the above discussion in the following
Definition 6.45 Let f W X ! Y be a given bijection. The inverse function of f is
the function g W Y ! X such that, for every y 2 Y,
g.y/ D x , y D f .x/:
From now on, whenever there is no danger of confusion, we shall denote the
inverse (function) of a bijection f W X ! Y by f 1 W Y ! X. Notice that the exponent
1 in the notation of inverse function has no arithmetic meaning; it simply calls the
reader’s attention to the fact that f 1 does the opposite of f , i.e., applies Y into X
instead of X into Y, and reverses the arrows of the correspondences made by f .
Now, it naturally emerges the question of how to effectively compute the inverse
of a given bijection. Although such a computation is generally more complicated
than that of composite functions, in the case of real functions of a real variable f W
X ! Y we can reason in the following simple way: since f 1 .y/ D x , f .x/ D y,
in order to find f 1 .y/ D x it suffices to solve, for x 2 X, the equation f .x/ D y.
If we find a single solution x 2 X for every y 2 Y, then f will be a bijection and
x (computed in terms of y) will give the sought for expression for f 1 .y/. Let’s
implement such a reasoning in some relevant examples.
Example 6.46 Let a and b be given reals, with a ¤ 0, and let f W R ! R be the
affine function given by f .x/ D ax C b, for every x 2 R. Show that f is a bijection
and find the expression of f 1 .
Proof Initially, note that
f .x/ D y , ax C b D y , x D
yb
:
a
174
6 The Concept of Function
Hence, f is a bijection and, by the definition of f 1 , this value of x is precisely equal
to f 1 .y/, so that
f 1 .y/ D
yb
:
a
t
u
Example 6.47 If X is a nonempty set, then, analogously to the former example, the
inverse function of the identity function IdX is this function itself, so that we can
write . IdX /1 D IdX . Nevertheless, the inverse of a bijective function can be the
function itself, even if it is not the identity function of some nonempty set; an
example is furnished by the inverse proportionality function f W R n f0g ! R n f0g
(which we already know to be bijective), for which f .x/ D 1x , for every x 2 R n f0g.
Indeed, since
f .x/ D y ,
1
1
Dy,xD ;
x
y
we conclude that
f 1 .y/ D x D
1
y
and, hence, f 1 D f .
Example 6.48 Let a, b and c be given real numbers, with a > 0. Propositions 6.15
and 6.23 guarantee that the quadratic function f .x/ D ax2 CbxCc, seen as a function
f W b
; C1
2a
! ; C1 ;
4a
is a bijection. Compute the expression of its inverse.
Solution According with the discussion
that
6.46, in order
preceded
b Example
to get the expressão of f 1 W 4a
; C1 ! 2a
; C1 , it suffices to fix
b
y 2 4a
; C1 and solve, for x 2 2a
; C1 , the equation f .x/ D y, i.e.,
ax2 C bx C c y D 0. Since this is a second degree equation in y, the condition
b
x 2a
gives (remember that a > 0)
xD
b C
p
p
b C C 4ac
b2 4a.c y/
D
;
2a
2a
where D b2 4ac is the discriminant of f . Therefore,
f 1 .y/ D
b C
p
C 4ay
:
2a
t
u
6.4 Inversion of Functions
175
Example 6.49 As a particular case of the previous example6 , the function f W
Œ0; C1/ ! Œ0; C1/, given by f .x/ D x2 for every nonnegative real x, is a bijection
having as its inverse the square root function
f 1 W Œ0; C1/ ! Œ0; C1/
p
:
x
x
7!
We finish this section obtaining a useful formula relating composition and
inversion of bijections.
Proposition 6.50 If f W X ! Y and g W Y ! Z are bijections, then g ı f W X ! Z is
also a bijection and
.g ı f /1 D f 1 ı g1 :
Proof We already know, from item (e) of Proposition 6.39, that g ı f is bijective. On
the other hand, since .g ı f /1 and f 1 ı g1 are both functions from Z to X, in order
to check that .g ı f /1 D f 1 ı g1 it suffices, by the uniqueness of the inverse (cf.
Problem 1), to show that
.f 1 ı g1 / ı .g ı f / D IdX and .g ı f / ı .f 1 ı g1 / D IdZ :
These verifications are quite straightforward and will be left for the reader as
exercises.
u
t
Problems: Section 6.4
1. * Let f W X ! Y be a given function.
(a) If g W Y ! X is such that g ı f D IdX and f ı g D IdY , prove that f is a
bijection.
(b) Prove that there exists at most one function g W Y ! X for which gıf D IdX
and f ı g D IdY .
2. * Complete the proof of Proposition 6.50, showing (in the notations of its
statement) that
.f 1 ı g1 / ı .g ı f / D IdX e .g ı f / ı .f 1 ı g1 / D IdZ :
6
Here and in the previous example, we are relying on previous knowledge from the reader, namely,
the existence of square roots of positive real numbers. Therefore, the present discussion will be
rigorously justified only after Theorem 7.9, where we prove the existence of such roots. In this
respect, the reader might also want to take a look at Example 8.36.
176
6 The Concept of Function
3. Let f W Π12 ; C1/ ! Π34 ; C1/ be the function given by f .x/ D x2 x C 1, for
every x 12 . Show that f is a bijection and obtain the defining expression of its
inverse.
4. Let f W R n f2g ! R n f3g be the function given by f .x/ D 3x5
. Show that f is
x2
a bijection and obtain the defining expression of its inverse.
5. Let a; b; c; d 2 R and f W R n f dc g ! R be the function given by f .x/ D
axCb
. Generalize the previous problem, showing that f defines a bijection from
cxCd
R n f dc g onto R n f ac g. Moreover, conclude that, if d D a, then f 1 D f .
6. * Let n be a natural number and f W Œ0; C1/ ! Œ0; C1/ be the function
given by f .x/ D xn . Admitting that Im.f / D Œ0; C1/ (a fact to be proved in
Example 8.36), show that f is an increasing bijection and get the expression for
f 1 , thus defining the n–th root function.
7. Give an example of a bijection f W R ! R for which both f C f 1 and f f 1
are also bijections.
8. * Let I; J R be intervals and f W I ! J be a bijection. If f is increasing (resp.
decreasing), prove that f 1 W J ! I is also increasing (resp. decreasing).
9. Let f W R ! R be a bijection and g W R R ! R R be the function defined
by g.x; y/ D .x3 ; x f .y//. Prove that g is bijective and find the expression that
defines its inverse in terms of f 1 .
10. (IMO) Let G be a (nonempty) set of affine functions possessing the folllowing
properties:
(a) If f ; g 2 G, then f ı g 2 G.
(b) If f 2 G, then f 1 2 G.
(c) For every f 2 G, there exists xf 2 R such that f .xf / D xf .
Prove that there exists a real number x0 such that f .x0 / D x0 for every f 2 G.
11. (France) Let f W N ! N be a bijection. Prove that there exist natural numbers
a < b < c such that f .a/ C f .c/ D 2f .b/.
6.5 Defining Functions Implicitly
A function can be defined implicitly by a set of properties. For example, letting
g.x/ D x C 1 and h.x/ D x 1 for every x 2 R, the function f W R ! R given by
f .x/ D x2 satisfies
f .g.x// D g.x/2 and f .h.x// D h.x/2 ;
i.e., it is such that
f .x C 1/ D .x C 1/2 D x2 C 2x C 1 and f .x C 1/ D .x 1/2 D x2 2x C 1I
therefore, for every x 2 R, we have
f .x C 1/ f .x 1/ D 4x:
6.5 Defining Functions Implicitly
177
Now, we can try to reverse the steps above, asking which functions f W R ! R are
such that
f .x C 1/ f .x 1/ D 4x; 8 x 2 R:
(6.7)
Clearly, f .x/ D x2 is not the only one, for, fc .x/ D x2 Cc also satisfies (6.7), whatever
the real constant c.
Now, let f W R ! R be any real function satisfying (6.7), so that f is not explicitly
given by its values. Since we merely know that f must satisfy that relation, we say
that f is implicitly defined.
On the other hand, since we know that (6.7) must be satisfied, it allows us to find
other relations f must also satisfy. For instance, if g W R ! R is given by g.x/ D x2 ,
then
f .g.x/ C 1/ f .g.x/ 1/ D 4g.x/
or, which is the same,
f .x2 C 1/ f .x2 1/ D 4x2 ; 8 x 2 R:
(6.8)
Thus, any function f W R ! R satisfying (6.7) will also satisfy (6.8). Nevertheless,
relation (6.8) may not be very useful to the task of finding all functions f W R ! R
that satisfy (6.7).
It is our purpose in this section to approach the problem of finding all functions
implicitly defined by a certain (finite) set of relations. Since there is no general
theory to accomplish this, we shall content ourselves in analysing a number of
interesting examples. In turn, these examples will provide us with several useful
techniques to be employed in such problems.
Let’s start by looking at two simple examples.
Example 6.51 (Canada) Find all increasing functions f W N ! N, such that f .2/ D
2 and f .mn/ D f .m/f .n/, for every m; n 2 N.
Solution From 1 f .1/ < f .2/ D 2 we get f .1/ D 1; also, f .4/ D f .2/f .2/ D 4
and f .8/ D f .4/f .2/ D 8. Suppose, by induction hypothesis, that f .2k / D 2k for a
certain natural number k. Then,
f .2kC1 / D f .2k /f .2/ D 2k 2 D 2kC1 ;
and it follows that f .2n / D 2n for every nonnegative integer n.
Now, let n be a fixed natural number. Since f is increasing, we have
2n D f .2n / < f .2n C 1/ < < f .2nC1 1/ < f .2nC1 / D 2nC1 :
178
6 The Concept of Function
However, f .2n C 1/, f .2n C 2/, : : :, f .2nC1 1/ are natural numbers too, so that the
only possibility is
f .2n C 1/ D 2n C 1; f .2n C 2/ D 2n C 2; : : : ; f .2nC1 1/ D 2nC1 1:
Finally, since such a reasoning is valid for whatever natural n, we conclude that
f .m/ D m, for every m 2 N.
t
u
Example 6.52 (OIM) If D D R f1; 0; 1g, find all functions f W D ! R such
that, for every x 2 D, we have
f .x/2 f
1x
1Cx
D 64x:
1x ¤ 0 for every
Solution First of all, note that, since x ¤ 0, we have f .x/2 f 1Cx
1x
x 2 D. In particular, f .x/ ¤ 0 for every x 2 D. Now, let g.x/ D 1Cx
for x 2 D. The
definition of D easily gives that g.D/ D, so that we can compose f and g. This
way, for every x 2 D, we have
2
f .g.x// f
1 g.x/
1 C g.x/
D 64g.x/:
(6.9)
Substituting the defining formula for g in the above relation, we arrive at the equality
f
1x
f
1Cx
1
1C
1x
1Cx
1x
1Cx
!
1x
D 64
1Cx
or, which is the same,
f
1x
1Cx
2
1x
;
f .x/ D 64
1Cx
for every x 2 D. Squaring both sides of the relation given in the statement of the
problem and dividing the result by the last relation above, we find
f .x/3 D 64x2
1x
;
1Cx
q 1x
so that f .x/ D 4 3 x2 1Cx
.
Up to this point, we have only shown that, if there exists a function f W D ! R
satisfying the stated relation, then it must be given by the last expression above.
Hence, it is necessary that we verify that such an f does satisfy that relation, for
every x 2 D. Since such a verification is straightforward, we leave it to the reader.
t
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6.5 Defining Functions Implicitly
179
Yet in respect to the previous example, a little more practice would have allowed
us to get rid of formally defining g to, then, compose it with f in order to get (6.9).
Instead, we could have just said
Substituting x by
1x
in the given relation; we obtain : : : ;
1Cx
keeping in mind that this substitution is actually a composition of functions. From
now on, whenever there is no danger of confusion, we shall adopt this language
shortcut, which already appears in the next example. When reading it, try to identify
the compositions that correspond to the employed substitutions.
Example 6.53 (Poland) Find all functions f W R ! R such that, for every x; y 2 R,
we have
.x y/f .x C y/ .x C y/f .x y/ D 4xy.x2 y2 /:
Solution Since this relation is valid for every x; y 2 R, it must be valid if we make
ab
x D aCb
2 and y D 2 , with a; b 2 R. Substituting these values for x and y, we get
the relation
bf .a/ af .b/ D .a2 b2 /ab;
which must be satisfied for every a; b 2 R. In particular, when ab ¤ 0, dividing
both sides of the last relation above by ab we get
f .a/ f .b/
D a2 b2 ;
a
b
for all a; b 2 R n f0g. Hence, letting g W R n f0g ! R be the function given by
g.x/ D f .x/
x2 , it follows from the above that g.a/ D g.b/, for all a; b 2 R n f0g. In
x
other words, g must be constant, so that there must exist a real number k such that
g.x/ D k, for every x 2 R n f0g. But this is the same of saying that f .x/ D x3 C kx,
for every x 2 R n f0g.
On the other hand, making x D y D 1 in the relation given in the statement of
the problem, we get f .0/ D 0, for every function f W R ! R that satisfies those
conditions. Since 03 C k 0 D 0, we conclude that every such function must be of
the form f .x/ D x3 C kx, for every x 2 R and some real constant k.
As in the previous example, we have to verify that every such function does
satisfy the stated conditions. This amounts to straightforward algebrism, which will
be left to the reader.
t
u
For the next example, we need the following
Definition 6.54 Let X be a nonempty set and f W X ! X be a given function. An
element x0 2 X is said to be a fixed point of f if f .x0 / D x0 .
180
6 The Concept of Function
If I R is an interval, then a decreasing function f W I ! I admits at most one
fixed point. Indeed, if x1 ; x2 2 I were fixed points of f , with x1 < x2 , it would follow
from the fact that f is decreasing that
x1 D f .x1 / > f .x2 / D x2 ;
a contradiction to x1 < x2 .
Example 6.55 (Argentina) Let f W R ! R be a decreasing function such that
f .x C f .x// D x C f .x/, for every real number x. Prove that f .f .x// D x, for every
real x.
Proof The hypotheses on f guarantee that x C f .x/ is a fixed point of f , for every
x 2 R. On the other hand, according to the previous discussion, the decreasing
character of f assures the existence of at most one fixed point for it. Therefore, there
must exist a 2 R such that x C f .x/ D a, for every x 2 R, so that f .x/ D a x for
every x 2 R. Hence,
f .f .x// D f .a x/ D a .a x/ D x;
for every x 2 R.
t
u
The next example develops a body of ideas which will reveal themselves to be
useful in a number of other situations involving implicitly defined functions. In
particular, the first part of the presented reasoning solves Problem 12, page 152.
Example 6.56 Find all functions f W R ! R such that f .1/ D 1 and, for all
x; y 2 R:
(a) f .x C y/ D f .x/ C f .y/.
(b) f .xy/ D f .x/f .y/.
Solution Let f be a function satisfying the stated conditions. Making x D y D 0 in
(a), we get
f .0/ D f .0 C 0/ D f .0/ C f .0/ D 2f .0/;
so that f .0/ D 0. Making y D x in (a), we obtain
f .2x/ D f .x C x/ D f .x/ C f .x/ D 2f .x/;
for every x 2 R. Now, making y D 2x in (a), it comes that
f .3x/ D f .x C 2x/ D f .x/ C f .2x/ D f .x/ C 2f .x/ D 3f .x/;
6.5 Defining Functions Implicitly
181
for every x 2 R. Repeating the above reasoning we easily conclude, by induction
on n 2 N, that
f .nx/ D nf .x/; 8 n 2 N; x 2 R
(6.10)
In particular, making x D 1 in (6.10), we get f .n/ D n, for every n 2 N. Letting
x D 1n in (6.10), it follows that
1
1
D nf
1 D f .1/ D f n n
n
and, hence, f
1
n
D 1n . Finally, x D
f
1
m
in (6.10), with m 2 N, furnishes
1
1
n
1
Df n
D nf
Dn D :
m
m
m
m
m
n
In order to see what happens with negative rationals, let y D x in item (a), to get
0 D f .0/ D f .x C .x// D f .x/ C f .x/;
or, which is the same,
f .x/ D f .x/; 8 x 2 R:
(6.11)
In particular, if x < 0 is rational, it follows from (6.11), together with the fact that
x is a positive rational, that f .x/ D f .x/ D .x/ D x. Therefore, f .x/ D x,
for every x 2 Q.
Since f .x/ D x for every x 2 Q, we suspect that the identity function of R is the
only one satisfying the stated conditions. To confirm that, we now turn our attention
to item (b). First of all, let’s show that, if f .x/ D 0 for some x 2 R, then x D 0.
Indeed, if x ¤ 0, then, making y D 1x in item (b), we would have
1
1
Df x
D f .1/ D 1;
0 D f .x/f
x
x
which is a contradiction. Now, letting y D x ¤ 0 in (b), we obtain
f .x2 / D f .x x/ D f .x/ f .x/ D f .x/2 > 0I
(6.12)
hence, if x; y 2 R, with x < y, and a ¤ 0 is such that y x D a2 , then, successively
applying (a), (6.11) and (6.12), we get
f .y/ f .x/ D f .y/ C f .x/ D f .y x/ D f .a2 / D f .a/2 > 0;
so that f is an increasing function.
182
6 The Concept of Function
Finally, suppose that there exists a 2 R such that f .a/ < a. We invoke the
result of Problem 4, page 206, according to which there exists a rational number
between any two real numbers. This allows us to choose a rational number r such
that f .a/ < r < a, and the increasing character of f gives
r D f .r/ < f .a/;
which is a contradiction. Analogously, we cannot have f .a/ > a, so that the only
possibility is f .a/ D a. However, since a 2 R was arbitrarily chosen, we conclude
that f .x/ D x for every x 2 R.
t
u
Our last example shows that, for implicitly defined functions f W N ! N,
elementary divisibility arguments are sometimes useful.
Example 6.57 (Lituania) Find all functions f W N ! N such that, for all natural
numbers m and n, we have
f .f .m/ C f .n// D m C n:
Solution Let’s first prove that f is injective. To this end, let m and n be natural
numbers such that f .m/ D f .n/ D k. Then, f .2k/ D f .f .n/ C f .n// D 2n and,
analogously, f .2k/ D 2m, so that m D n.
Now, let k > 1 be a natural number. From .k 1/ C 2 D k C 1, it follows that
f .f .k 1/ C f .2// D k C 1 D f .f .k/ C f .1//:
By the injectivity of f , we then get f .k 1/ C f .2/ D f .k/ C f .1/ or, which is the
same, f .k/ f .k 1/ D f .2/ f .1/, for every k > 1. Writing this relation for
k D 2; 3; : : : ; n and adding the corresponding equalities, we arrive at
f .n/ D .n 1/.f .2/ f .1// C f .1/;
(6.13)
for every natural n > 1. Letting n D 2f .1/ in the above relation, it comes that
2 D f .f .1/ C f .1// D f .2f .1// D .2f .1/ 1/.f .2/ f .1// C f .1/
and, hence,
f .2/ f .1/ D
2 f .1/
:
2f .1/ 1
Since f .2/ f .1/ is an integer, we conclude that 2f .1/ 1 divides 2 f .1/. In
particular, 2f .1/ 1 j2 f .1/j, and from this inequality it is easy to conclude that
the only possibility is f .1/ D 1, so that f .2/ D 2. It thus follows from (6.13) that
f .n/ D n, for every natural n.
t
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6.5 Defining Functions Implicitly
183
Problems: Section 6.5
1. * Generalize the discussion of the paragraph that precedes Example 6.55,
showing that, if I R is an interval and f ; g W I ! R are functions such
that f is decreasing and g is increasing, then there exists at most one x0 2 I for
which f .x0 / D g.x0 /.
p
p
2. Find all positive reals x for which 2 C x D 8x .
3. Find all functions f W Q ! Q such that
f
xCy
2
D
f .x/ C f .y/
;
2
for all x; y 2 Q.
4. (Austria) Find all functions f W Z n f0g ! Q such that, for all x; y 2 Z n f0g for
which x C y is a multiple of 3, we have
f
xCy
3
D
f .x/ C f .y/
:
2
5. (Vietnam) Find all functions f W R ! R such that
1
1
1
f .xy/ C f .xz/ f .x/f .yz/ ;
2
2
4
for all x; y; z 2 R.
6. (Spain) Find all increasing functions f W N ! N such that f .n C f .n// D 2f .n/,
for every n 2 N.
7. (OIM shortlist) Find all functions f W R ! Z which satisfy the following set of
conditions:
(a) f .x C a/ D f .x/ C a, for every x 2 R and every a 2 Z.
(b) f .f .x// D 0, for x 2 Œ0; 1/.
8. (Austrian-Polish) Prove that there doesn’t exist a function f W Z ! Z such that,
for all x; y 2 Z, we have
f .x C f .y// D f .x/ y:
9. (Romania) Find all functions f W Z ! Z such that f .0/ D 1 and
f .f .k// C f .k/ D 2k C 3;
for every k 2 Z.
10. (Romania) Let k > 1 be an odd integer and A D fx1 ; x2 ; : : : ; xk g be a set of k
real numbers. Find all injective functions f W A ! A such that
jf .x1 / x1 j D jf .x2 / x2 j D D jf .xk / xk j:
184
6 The Concept of Function
11. (BMO) Let a be a given real number and f W R ! R be a function such that
f .0/ D 12 and
f .x C y/ D f .x/f .a y/ C f .y/f .a x/;
for all x; y 2 R. Prove that f is constant.
12. Find all functions f W Q ! QC such that f .x C y/ D f .x/f .y/, for all x; y 2 Q.
13. Find all functions f W Œ0; 1 ! Œ0; 1 such that f .0/ D 0, f .1/ D 1 and
f .x C y/ C f .x y/ D 2f .x/;
for all x; y 2 Œ0; 1 such that x y; x C y 2 Œ0; 1.
14. (Lituania) Let f W Z ! Z be such that f .m2 C f .n// D f .m/2 C n, for all
m; n 2 Z.
(a) Prove that f .0/ D 0 and f .1/ D 1.
(b) Find all such functions.
15. (OIM) Find all increasing functions f W N ! N such that f .yf .x// D x2 f .xy/,
for every x; y 2 N.
16. (IMO) Find all functions f W Œ0; C1/ ! R satisfying the following set of
conditions:
(a) f .xf .y//f .y/ D f .x C y/, for all x; y 2 Œ0; C1/.
(b) f .2/ D 0 and f .x/ ¤ 0, for 0 x < 2.
17. (IMO) Let S D fx 2 RI x > 1g. Obtain all functions f W S ! S satisfying the
following conditions:
(a) f .x C f .y/ C xf .y// D y C f .x/ C yf .x//, for every x; y 2 S.
(b) f .x/
x is increasing in each of the intervals .1; 0/ and .0; C1/.
18. (IMO) Decide whether there exists a function f W N ! N satisfying the
following set of conditions:
(a) f .1/ D 2.
(b) f .n/ < f .n C 1/, for every n 2 N.
(c) f .f .n// D f .n/ C n, for every n 2 N.
19. (Iran) Find all functions f W R ! R such that
f .f .x C y// D f .x C y/ C f .x/f .y/ xy;
for all x; y 2 R.
20. (Poland) Find all functions f W QC ! QC satisfying, for every positive rational
number x, the following set of conditions:
(a) f .x C 1/ D f .x/ C 1.
(b) f .x3 / D f .x/3 .
6.6 Graphs of Functions
185
6.6 Graphs of Functions
Given a function f W X ! Y, we define its graph to be the subset Gf of the cartesian
product X Y, defined by
Gf D f.x; y/ 2 X
YI y D f .x/g:
(6.14)
If f W X ! R is a real function of a real variable, so that X R is a finite
union of intervals (possibly X D R), then the graph of f has considerable geometric
importance, for
Gf X
YR
R;
and this last set can be identified with the euclidean plane, furnished with a fixed
cartesian coordinate system7 .
Our purpose in these last two sections of the chapter is to examine some
elementary examples and study some simple properties of graphs of functions
f W X ! R, when X R is a finite union of intervals. We postpone the discussion
of the main properties of graphs of continuous and differentiable functions to
chapters 8 and 9.
In all that follows, we fix a cartesian coordinate system is in the plane.
The first point worth observing is that not every subset of the cartesian plane
can be the graph of a function. Indeed, suppose f W X ! R is a real function of a
real variable, X being a finite union of intervals. If .x0 ; y0 / 2 Gf , then, by the very
definition of graph, x0 2 X and y0 D f .x0 /. On the other hand, for a fixed x0 2 X, if
A1 .x0 ; y1 / and A2 .x0 ; y2 / are points on the graph of f , then, again from the definition,
we have
y1 D f .x0 / D y2 ;
so that A1 D A2 . In short, for a given x0 2 X, the vertical line x D x0 of the
cartesian coordinate system intersects the graph of f if and only if x0 2 X; moreover,
in this case such a line intersects the graph in exactly one point. Thus, the subset C
depicted in Fig. 6.5 as a continuous curve doesn’t represent the graph of any function
f W Œ3; 3 ! R, since every vertical line contained in the gray strip intersects C in
more than one point.
The graph of a real function of a real variable gives a quite simple geometric
interpretation for the image of the function. To discuss it, let’s consider the cartesian
plane of Fig. 6.6, in which the graph of a function f W Œa; b/ ! R is drawn and its
points of intersection with the horizontal line y D y0 are marked. (There are three
such points in Fig. 6.6, whose abscissas are denoted by ˛, ˇ and .) Let .x0 ; y0 / be
7
We refer the reader to Chapter 6 of [4] for an adequate presentation of cartesian coordinate
systems.
186
6 The Concept of Function
y
(3, 0)
x
O
(−3, 0)
C
Fig. 6.5 A subset of the cartesian plane which is not the graph of a function
Gf
y
y = y0
a α
β
γ b
x
Fig. 6.6 Image graph
an intersection point of the line and the graph. By pertaining to the graph of f , the
point .x0 ; y0 / must be such that x0 2 Œa; b/ and f .x0 / D y0 , so that x0 D ˛, ˇ or .
Conversely, let be given a point .x0 ; y0 / of the cartesian plane, with x0 2 Œa; b/. It is
clear that .x0 ; y0 / belongs to the horizontal line y D y0 ; moreover, if x0 is a solution
of the equation f .x/ D y0 , i.e., if f .x0 / D y0 , then x0 D ˛, ˇ or and we also
have .x0 ; y0 / 2 Gf . Therefore, the horizontal line y D y0 intersects the graph of f
exactly when y0 belongs to the image of f . The reasoning for an arbitrary function
f W X ! R, with X R, is entirely analogous and allow us to conclude that
The image of f is precisely the set of y0 2 R for which the horizontal line
y D y0 intersects the graph of f :
6.6 Graphs of Functions
187
y
Gf
x0
a
b
x
y = y0
Fig. 6.7 Minimum point of f W I ! R
Now, if I R is an interval, then the monotonicity of a function f W I ! R also
says a lot about the behavior of its graph. For instance, if we suppose that f increases
(resp. decreases) in I, we conclude that, as long as the x increases in I, the values
f .x/ increase (resp. decrease) in R, so that the graph of f rises (resp. falls).
On the other hand, if y0 2 R is the minimum value of f W I ! R and x0 2 I is a
minimum point of f (cf. Definition 6.24), then, for every x 2 I, the point .x; f .x// is
above or coincides with the point .x; y0 / (cf. Fig. 6.7). Yet in another way, the graph
of f is entirely contained in the closed upper halfplane determined by the horizontal
line y D y0 , touching it at the point .x0 ; y0 /.
Notice that the concepts of maximum value and maximum point of a function
f W I ! R admit geometric interpretations analogous to those discussed above.
In the rest of this section we shall examine some important examples of graphs
of functions.
Example 6.58 Let f W R ! R be a constant function, with f .x/ D c for every x 2 R.
The graph of f is the set
Gf D f.x; y/I x 2 R and y D cg D f.x; c/I x 2 Rg;
i.e., it is the horizontal line y D c, which crosses the vertical axis at the point .0; c/
(cf. Fig. 6.8, where we consider the case c > 0).
Example 6.59 Recall (cf. Definition 6.2) that the identity function IdR W R ! R is
given by IdR .x/ D x, for every x 2 R. Therefore, its graph is the set
G IdR D f.x; y/I x 2 R and y D xg D f.x; x/I x 2 Rg:
It is an easy exercise in euclidean geometry to verify that the points of the cartesian
plane of the form .x; x/ or .x; x/ are precisely those situated on the bisectors of the
angles formed by the coordinate axes, those of the form .x; x/ belonging to the first
188
6 The Concept of Function
y
Gf
(0, c)
x
O
Fig. 6.8 Graph of the constant function f .x/ D c; 8 x 2 R
y
G IdR
(0, a)
(a, a)
(a, 0)
x
Fig. 6.9 Graph of the identity function IdR
or third quadrants. Hence, the graph of the identity function IdR is the line shown
in Fig. 6.9, and, from now on, will be called the bisector of odd quadrants. Observe
that the set of points of the form .x; x/, i.e., the bisector of even quadrants, is the
graph of the function f W R ! R given by f .x/ D x.
Example 6.60 The modular function is the function f W R ! R given by f .x/ D
jxj. It follows immediately from the definition of modulus of a real number that
Gf D f.x; jxj/I x 2 Rg
D f.x; jxj/I x 2 RC g [ f.x; jxj/I x 2 R g
D f.x; x/I x 2 RC g [ f.x; x/I x 2 R g:
Since the points .x; x/ and .x; x/ are symmetric with respect to the horizontal axis,
the graph of the modular function is obtained by reflecting, along the horizontal
axis, the portion of the graph of the function IdR situated in the third quadrant (cf.
Fig. 6.10).
6.6 Graphs of Functions
189
y
G|x|
(x, −x)
x
(x, x)
Fig. 6.10 Graph of the modular function f .x/ D jxj
y
B(0, b)
A(−b/a, 0)
O
x
Fig. 6.11 Graph of the affine function f .x/ D ax C b
Example 6.61 If f .x/ D ax C b is an affine function, then its graph is the subset of
the cartesian plane given by
Gf D f.x; y/I x; y 2 R and y D ax C bg:
According to basic analytic geometry (cf. Chapter 6 of [4], for instance), the graph
of f is the
line of equation y ax b D 0, with slope a and passing through the
points A ba ; 0 and B D .0; b/. Fig. 6.11 depicts the graph of f .x/ D ax C b in the
case a; b > 0.
For what comes next, recall (cf. Chapter 6 of [4], for instance) that, given in the
plane a point F and a line d, with F … d, the parabola of focus F and directrix d
(cf. Fig. 6.12) is the locus of the points P in the plane for which
PF D dist.P; d/;
190
6 The Concept of Function
Fig. 6.12 Parabola of focus
F and directrix d
P
F
V
d
where dist.P; d/ stands for the distance from P to d. The axis of the parabola is the
line that passes through F and is perpendicular to d, while its vertex is the point V
where it intersects its axis.
We now show that the graph of every quadratic function is a parabola. More
precisely, we have the following result.
Theorem 6.62 Given a; b; c 2 R, with a ¤ 0, the graph of the quadratic function
b
f .x/ D ax2 CbxCc is the parabola whose axis is the line x D 2a
and whose vertex
b
is V 2a ; 4a . Moreover, it “opens upwards” if a > 0, and “opens downwards”
if a < 0.
Proof Let’s look for x0 ; y0 ; k 2 R such that y0 ¤ k and, letting F.x0 ; y0 / and d be
the line y D k, we have
P 2 Gf ” PF D dist.P; d/:
To this end, set P.x; y/. We first notice that
P 2 Gf ” y D ax2 C bx C cI
also, the formula for the distance between two points of the cartesian plane gives
PF D dist.P; d/ ”
p
.x x0 /2 C .y y0 /2 D jy kj:
Therefore, we want to find x0 , y0 and k such that
y D ax2 C bx C c , .x x0 /2 C .y y0 /2 D .y k/2
,yD
x2 C y20 k2
1
x0
x2 xC 0
:
2.y0 k/
y0 k
2.y0 k/
6.6 Graphs of Functions
191
Hence, it is natural to try to solve, with respect to x0 , y0 and k, the system of
equations
1
D a;
2.y0 k/
x0
D b;
y0 k
x20
1
C .y0 C k/ D c:
2.y0 k/
2
b
; then,
This, in turn, is immediate: the first two equations readily give x0 D 2a
substituting the first equation and the value of x0 into the third equation, it comes
that
b2
1
D2 c 2 a D I
y0 C k D 2 c x20 2.y0 k/
4a
2a
finally, by solving the system of equations
y0 k D
1
; y0 C k D ;
2a
2a
1C
we get y0 D 1
4a and k D 4a .
To what was left to prove, since the vertex of the parabola is the intersection of
b
the line x D 2a
with the graph, its ordinate y equals
b 2
b
yDa CcD :
Cb 2a
2a
4a
t
u
We finish this section with the coming proposition, which establishes an important relation between the graphs of a bijection and its inverse. Then, we use it to
sketch the graphs of two important functions.
Proposition 6.63 Let I; J R be finite unions of interval. If f W I ! J is a bijection,
then the graphs of f and f 1 are symmetric with respect to the bisector of the odd
quadrants of the cartesian plane.
Proof Fix a 2 I and b 2 J. From the definition of inverse function, we have
.a; b/ 2 Gf , b D f .a/
, a D f 1 .b/
, .b; a/ 2 Gf 1 :
However, since the points .a; b/ and .b; a/ are symmetric with respect to the line
y D x, there is nothing left to do.
t
u
192
6 The Concept of Function
y
g
f
O
x
Fig. 6.13 Sketching the graph of x 7!
p
x
Example 6.64 Sketch the graph of the square root function
f W Œ0; C1/ ! Œ0; C1/
p
:
x
7!
x
Solution We saw at Example 6.49 that f is the inverse of g W Œ0; C1/ ! Œ0; C1/,
given by g.x/ D x2 for every x 2 Œ0; C1/. Since we already know the graph of g,
it follows from the previous proposition tha the graph of f is the symmetric of the
graph of g with respect to the line y D x (cf. Fig. 6.13).
t
u
Example 6.65 Recall that the inverse proportionality function is f W R n f0g !
R n f0g, such that f .x/ D 1x , for every x 2 R n f0g. With the previous discussion
at our disposal, we can accurately sketch its graph. Indeed, f is clearly decreasing
in .0; C1/; also, we already know from the Problem 10, page 171, that f is odd,
so that (this time by the Problem 5, page 193) its graph is symmetric with respect
to the origin of the cartesian plane; on the other hand, since f is its own inverse,
the previous proposition assures that its graph is also symmetric with respect to the
bisector of the odd quadrants; finally, we will see later that its graph is a continuous
curve (i.e., one with no interruptions) which “opens upwards” in .0; C1/.
Since it is intuitively clear (and will be formalized later) that f .x/ D 1x becomes
more and more close to zero as long as x increases without bound, we arrive at
the sketch shown
in Fig. 6.14. There, besides the above remarks, we plotted the
auxiliary points n; 1n , for 1 n 4 integer.
6.6 Graphs of Functions
193
y
y=x
O
x
Fig. 6.14 Sketching the graph of f .x/ D
1
x
Problems: Section 6.6
1. If f W Œ2; C1/ ! Œ1; C1/ is given by f .x/ D x2 4x C 5, show that f is a
bijection and find the intersection points of the graphs of f and f 1 .
For the next problem, given a nonempty set X and a function f W X ! R,
we say that f is bounded if there exists M > 0 such that jf .x/j M, for every
x 2 X.
2. * Let I R be an interval and f W I ! R be a given function. If f is bounded,
prove that its graph is contained in a horizontal strip of the cartesian plane
bounded by two parallel lines.
3. * If I R is an interval and f W I ! I is a given function, show that the fixed
points of f are the abscissas of the intersection points of the graph of f with the
bisector of the odd quadrants.
4. * Let I R be an interval and f ; g W I ! R be given functions. Explain how to
analytically identify the intersection points of the graphs of f and g.
For the next problem, we think the reader will find it useful to read again the
statement of Problem 10, page 171.
5. * Let I R be a union of intervals, symmetric with respect to 0 2 R, and
f W I ! R be a given function.
(a) If f is even, prove that Gf is symmetric with respect to the vertical axis.
(b) If f is odd, prove that Gf is symmetric with respect to the origin.
6. In each of the following items, sketch the graphs of the given real functions of
a real variable in a single cartesian system:
194
6 The Concept of Function
(a) f1 .x/ D x2 , f2 .x/ D x4 and f3 .x/ D x3 .
(b) f1 .x/ D x, f2 .x/ D x3 and f3 .x/ D x5 .
p
7. Sketch the graph of the function f W R ! R such that f .x/ D 3 x, for every
x 2 R.
8. Sketch the graph of the integer part function bc W R ! R (cf. Problema 9,
página 152).
For the next problem, we think the reader may find it useful to read the
statement of Problem 15, page 171 again.
9. * Do the following items:
(a) If f W R ! R is periodic of period p > 0, explain how to draw the graph of
f by knowing the portion of it in the interval 0 x < p.
(b) Use item (a) to sketch the graph of the fractional part function f g W R ! R
(cf. Problem 10, page 152).
10. * Let f W R ! R be a given function and a ¤ 0 be a given real number. Prove
that the graph of:
(a) g.x/ D f .x C a/ is obtained by translating the graph of f of a units in the
direction of the horizontal axis.
(b) g.x/ D f .x/ C a is obtained by translating the graph of f of a units in the
direction of the vertical axis.
(c) g.x/ D f .x/ is obtained by reflecting the graph of f along the horizontal
axis.
(d) g.x/ D f .x/ is obtained by reflecting the graph of f along the vertical axis.
(e) g.x/ D af .x/ is obtained by vertically stretching (resp. compressing) the
graph of f by a factor a, if a > 1 (resp. 0 < a < 1).
(f) g.x/ D f .ax/ is obtained by horizontally stretching (resp. compressing) the
graph of f by a factor a, if a > 1 (resp. 0 < a < 1).
11. * Use the results of the previous problem to sketch the graph of the function
x
f W R n f2g ! R given by f .x/ D 2x
, for every x 2 R n f2g.
12. Let I be an interval and f W I ! R be a given function. Which relation does exist
between the graphs of f and of the function g W I ! R, given by g.x/ D jf .x/j?
Apply your conclusion, together with the result of Problem 10, to sketch the
graphs of the functions listed below:
(a)
(b)
(c)
(d)
1
g.x/ D jxC1j
, for every x 2 R n f1g.
2
g.x/ D jx 4j, for every x 2 R.
g.x/ D jx2 jx C 2j C 2j, for every x 2 R.
1
g.x/ D 1 .x2/
2 , for every real x ¤ 2.
13. Prove that the graph of the inverse proportionality function is obtained by a
trigonometric (i.e., counterclockwise) rotation, of 4 radians, of the hyperbola8
of equation x2 y2 D 2.
8
For the necessary background on the equation of hyperbolas, we refer the reader to Chapter 6 of
[4], for instance.
6.7 Trigonometric Functions
195
6.7 Trigonometric Functions
The sine function is the function sin W R ! R, that associates to each x 2 R the
sine of an arc of x radians9 :
sin W R ! R
:
x 7! sin x
Analogously, we define the cosine function by
cos W R ! R
;
x 7! cos x
where cos x stands for the cosine of an arc of x radians.
Some basic properties of the sine and cosine functions are collected in the
following proposition, for which the reader may find it useful to recall the statements
of Problem 15, page 171, and Problem 10, page 171.
Proposition 6.66 Sine and cosine functions have image Œ1; 1 and are periodic of
period 2. Furthermore, sine function is odd, whereas cosine function is even.
Proof Immediate from basic Trigonometry.
t
u
According to the above proposition and the discussion contained in Problem 15,
page 171, in order to get an accurate sketch of the graph of the sine function, it
suffices to make it in the interval Œ; , then copying this portion of the graph to
each interval of the form ΠC 2k; C 2k, where k 2 Z.
On the other hand, since the sine function is odd, in order to get its graph in
the interval Œ; , it suffices to sketch it in the interval Œ0; ; once this is done,
then, reflecting it across the origin of the cartesian plane, we obtain (from item (c)
of Problem 10, page 171) the graph in the interval Œ; .
We shall prove in Section 8.1 (cf. Example 8.8) that the graph of the sine
function is a continuous curve, i.e., a curve with interruptions. On the other hand, in
Section 9.7 we shall show (cf. Example 9.66) that such a graph “ opens downwards”
in the interval Œ0; . For the time being, by assuming the validity of these facts, we
can finally sketch the graph of the sine function.
Example 6.67 Gathering together the above information on the behavior of the sine
function in the inteval Œ0; (image, periodicity,
together
continuity and
concavity),
with the fact that it is increasing in 0; 2 , decreasing in 2 ; and such that
sin. x/ D sin x for every x 2 R, we conclude that, in order to get a reasonably
accurate sketch of its graph
in that interval, we just need to compute the values of
sin x for some values x 2 0; 2 . This is done in the table below:
For the necessary background on Trigonometry for this section, we refer the reader to Chapter 7
of [4].
9
196
6 The Concept of Function
Fig. 6.15 Graphs of the sine
and cosine functions
y
1
−π
− π2
π
2
0
π
x
−1
6
0
0
4
p
2=2
1=2
3
p
3=2
2
1
With all this at hand, we immediately get Fig. 6.15, first on the interval Œ0; ,
then on the interval Œ; and finally on R.
We now recall one more piece of basic Trigonometry, which assures that
cos x D sin x C
2
for every x 2 R. Hence, item (a) of Problem 10, page 194, guarantees that, once the
graph of the sine function is drawn, we can get a corresponding sketch for the graph
of the cosine function by translating the graph of the sine function of 2 in the
direction of the horizontal axis. In Fig. 6.15, the graph of the cosine function in the
interval Œ; is given by the dashed curve contained in the strip of the cartesian
plane bounded by the (also dashed) lines y D 1 and y D 1.
Let us now look at a relevant application of the Ptolemy’s formulae for the sine
and cosine of the sum and difference of two arcs, which tells us how to proceed to
study a function given as a linear combination of the sine and cosine functions.
Example 6.68 Given positive integers a and b, let f W R ! R be the function such
that
f .x/ D a cos x C b sin x;
for every x 2 R. Writing
a cos x C b sin x D
p
b
a
a2 C b2 p
cos x C p
sin x ;
a2 C b2
a2 C b2
we notice that
p
a
a2 C b2
2
C
b
p
2
a C b2
2
D 1:
6.7 Trigonometric Functions
197
Fig. 6.16 Defining angle ˛
α
√
a2 + b 2
a
b
Therefore, point P p 2a 2 ; p 2b 2 belongs to the portion of the trigonometric
a Cb
a Cb
circle situated in the first quadrant, so that there exists a real number ˛ 2 .0; 2 / for
which
a
cos ˛ D p
a2 C b2
and
sin ˛ D p
b
a2 C b2
(cf. Fig. 6.16). Now, it follows from the formula for the cosine of a difference that
f .x/ D a cos x C b sin x
p
D a2 C b2 .cos ˛ cos x C sin ˛ sin x/
p
D a2 C b2 cos.x ˛/:
(6.15)
In particular, since j cos.x ˛/j 1, we get
jf .x/j D
p
p
a2 C b2 j cos.x ˛/j a2 C b2 ;
and it’s not difficultp
to use (6.15) to prove that the image of f is precisely the interval
Œc; c, where c D a2 C b2 (in this respect, see Problem 3).
We close this section with an elementary study of the tangent function, i.e., the
sin x
function that associates, to each real x in its domain, the real number tan x D cos
x.
Since
cos x D 0 , x D
C k; 9 k 2 Z;
2
the (maximal) domain of definition of the tangent function is the set
DDRn
n
2
o
C kI k 2 Z ;
so that the function we wish to study is
tan W D ! R
:
x 7! tan x
198
6 The Concept of Function
For x 2 D, we have
tan.x C / D
sin x
sin.x C /
D
D tan x;
cos.x C /
cos x
and it’s immediate to verify that there doesn’t exist a real number 0 < p < such
that tan.x C p/ D tan x for every x 2 D. Therefore, the tangent function is periodic
of period . Furthermore, since D is a subset of R which is symmetric with respect
to 0 and
tan.x/ D
sin.x/
sin x
D
D tan x
cos.x/
cos x
for every x 2 D, we conclude that the tangent function is odd.
From what was collected above, in order to sketch
the graph of the tangent
function it suffices to do it on the interval 0; 2 . Indeed, the odd character of
the tangent function assures that its graph on the interval 2 ; 0 is obtained
by reflecting,
the origin of the cartesian system, the portion of it on
around
the interval 0; 2 . On the other hand, since we have sketched the desired graph
function will allows us to
on the interval 2 ; 2 , the periodicity of the tangent
sketch the graph of it on all intervals of the form 2 C k; 2 C k , with k 2 Z:
it suffices to translate the graph on the interval 2 ; 2 of k units in the direction
of the horizontal axis, for every k 2 Z.
In Section 8.1 (cf. Problem 5, page 254),
we shall
prove that the graph of the
tangent function, restricted to the interval 2 ; 2 , is a continuous curve, i.e., a
curve without interruptions. On the other hand, in Section 9.7 (cf. Example
9.66) we
shall establish the fact that the graph “opens upwards” on the interval 0; 2 . For the
time being, assuming the validity of these statements and in view of the discussion
on the previous paragraphs, we can sketch the graph of the tangent function by
tabulating some values of it on the interval 0; 2 . The result is approximately that
of Fig. 6.17.
Problems: Section 6.7
1. * The purpose of this problem is to introduce the inverse trigonometric
functions. To this end, do the following items:
(a) Show that the restriction of the sine function to the interval 2 ; 2 defines
an increasing bijection (which we also
denote by sin, whenever there is
no danger of confusion) sin W 2 ; 2 ! Œ1; 1; its inverse is the arc
sine, denoted arcsin W Œ1; 1 ! 2 ; 2 , which is also increasing. Then,
compute arcsin 12 , arcsin 1 and arcsin.1/.
6.7 Trigonometric Functions
199
Fig. 6.17 Graph of the
tangent function
y
1
− π2 − π4
π
4
0
π
2
x
−1
(b) Show that the restriction of the cosine function to the interval Œ0; defines
a decreasing bijection (which we also denote by cos, whenever there is no
danger of confusion) cos W Œ0; ! Œ1; 1; its inverse is the arc-cosine,
denoted arccos W Œ1; 1 ! Œ0; , which is also decreasing. Then, compute
arccos 12 , arcsin 1 and arcsin.1/.
(c) Show that the restriction of the tangent function to the interval 2 ; 2
defines an increasing bijection (which
we also
denote by tan, whenever
there is no danger of confusion) tan W 2 ; 2 !
R; its inverse is the arctangent function, denoted arctan W R ! 2 ; 2 , which is also increasing.
p
Then, compute arctan 1, arctan 3 e arctan p13 .
For item (a) of the next problem, recall that the cotangent of x ¤ k (k 2 Z)
x
1
is the real number cot x, defined by cot x D cos
sin x . In particular, cot x D tan x .
2. In each of the items below, sketch the graph of the given function:
(a)
(b)
(c)
(d)
f
f
f
f
W R n fkI k 2 Zg !
R, given by f .x/ D cot x.
W Œ1; 1 ! 2 ; 2 , given by f .x/ D arcsin x.
W Œ1; 1 ! Œ0;, given by f .x/ D arccos x.
W R ! 2 ; 2 , given by f .x/ D arctan x.
3. * Let a and b given real numbers, not both zero, and f W R ! R be the function
given by
f .x/ D a cos x C b sin x:
(a) Find the image set of f .
(b) Prove that f is periodic of period 2.
(c) Sketch the graphs of f and of the sine function in a single cartesian
coordinate system.
200
6 The Concept of Function
4. * Let a and b be given real numbers, at least one of which is nonzero, and
f W R ! R be the function given by
f .x/ D a cos.x/ C b sin.x/;
where is a nonzero real number. Show that f is periodic and compute its
period.
5. Let f W R ! R be the function given by f .x/ D 2 sin x C cos.2x/.
(a) Compute the maximum and minimum values of f , as well as the real
numbers x for which f attains these values.
(b) Show that f is periodic, of period 2.
(c) Sketch the graph of f .
x
6. (Canada) Compute the number of real solutions of the equation sin x D 10
.
7. Find the maximum
value
attained
by
the
function
f
W
Œ1;
1
!
R,
given
by
p
f .x/ D 3x C 4 5 x2 .
8. (New Zealand) Let ˛ be a given irrational number. Prove that the function f W
R ! R, defined by
f .x/ D cos x C cos.˛x/;
for every x 2 R, is not periodic.
9. Find all integer values of n for which the function f W R ! R, given for x 2 R
by
f .x/ D cos .nx/ sin
5x
;
n
is periodic of period 3.
10. (Canada) Prove that the function f W R ! R given by f .x/ D sin.x2 / is not
periodic.
Chapter 7
More on Real Numbers
This chapter proceeds with the study of real numbers by presenting the notion of
convergence for (infinite) sequences and series of real numbers. Among other applications, we shall introduce one of the two most important numbers of Mathematics,1
the number e. We shall also present a famous result of Kronecker on dense subsets of
the real line, which will find several interesting applications, here and in the coming
chapters.
7.1 Supremum and Infimum
We begin this chapter by examining the completeness of R from another viewpoint,
for which we need to introduce some preliminary concepts.
A nonempty subset X R is bounded from above if there exists a real number
M such that
X .1; M:
In this case, we also say that M is an upper bound for X. Similarly, a nonempty set
X R is bounded from below if there exists a real number m such that
X Œm; C1/:
In this case, we say that m is a lower bound for X. Finally, a nonempty set X R
is bounded if X is simultaneously bounded from above and from below.
1
As the reader probably suspects, the other one is the number , which is defined as the numerical
value of the area of a circle of radius 1—see [4], for instance.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_7
201
202
7 More on Real Numbers
Yet in another way, a nonempty set X R is bounded (resp. bounded from
above, bounded from below) if there exists a positive real number a such that
a x a .resp. x a; x a/; 8 x 2 X:
On the other hand, a nonempty set X R which is not bounded (resp. not bounded
from above, not bounded from below) is said to be unbounded (resp. unbounded
from above, unbounded from below). In this case, given any positive real number
a, one can always find an element x 2 X such that
x … Œa; a .resp. x > a; x < a/:
Examples 7.1
(a) The set X D f1; 12 ; 13 ; 14 ; : : :g is bounded. Indeed, 0 is a lower bound and 1 is an
upper bound for X.
(b) Bounded intervals (cf. Definition 1.10 and subsequent discussion) are bounded
sets, in the sense of the above discussion.
Example 7.2 If a nonempty set X R is bounded from above, then the subset Y of
R given by Y D fxI x 2 Xg is bounded from below, and conversely. Indeed, a real
number a is an upper bound for X if and only if a is a lower bound for Y.
In spite of its apparent obviousness, we shall list our next example as an axiom
of the natural number system, known as its archimedian property, after the Greek
mathematician Archimedes of Syracuse.2 Some important consequences of it are
collected in Problems 2, 3 and 4, page 206.
Axiom 7.3 The set N of natural numbers is unbounded from above.
Continuing with the development of the theory, fix a bounded from above
nonempty set X R. If M 2 R is an upper bound for X, then X .1; M.
Nevertheless, it may happen that there exists M 0 < M which still is an upper
bound for X, i.e., is such that X .1; M 0 . Indeed, the condition X .1; M
doesn’t guarantee that, for M 0 < M, we have X \ .M 0 ; M ¤ ;; if it happens that
X \ .M 0 ; M D ;, then we will have X .1; M 0 , and M 0 will be another upper
bound for X, which is less than M.
On the other hand, given x 2 X, it is obvious that no M 0 < x is an upper bound
for X, since x 2 X n .1; M 0 , i.e., X 6 .1; M 0 . To put in another way, the set
of upper bounds of a nonempty, bounded from above set X R is bounded from
below.
2
Archimedes, who lived in the III century b.C., was the greatest scientist of his time. Among many
important contributions to Mathematics and Physics, his seminal ideas on how to compute the
area under a parabolic segment anticipated, in some 2000 years, the development of the Integral
Calculus by Newton and Leibniz.
7.1 Supremum and Infimum
203
As we shall see later (cf. Problem 9, page 242), the completeness of R, as
postulated in Sect. 1.3, is a consequence of the following statement, which sharpens
the discussion of the two previous paragraphs and will be taken as an axiom.
Axiom 7.4 If a nonempty set X R is bounded from above, then there exists a real
number M satisfying the two following properties:
(a) M is an upper bound for X.
(b) If M 0 < M, then M 0 is not an upper bound for X.
In the hypotheses and notations of the previous axiom, a moment’s thought shows
that there exists at most one real number M satisfying the properties there stated.
Indeed, if two distinct real numbers M1 and M2 did the job, suppose, without loss
of generality, that M1 < M2 . Then, on the one hand, item (a) would guarantee that
M1 and M2 are upper bounds for X; on the other, since M1 < M2 , item (b) would
guarantee that M1 is not an upper bound for X.
Yet in the hypotheses and notations of the axiom, the discussion of the last
paragraph allows us to say that M is the least upper bound or supremum of X.
We denote M D sup.X/ or M D lub.X/.
Examples 7.5
(a) The set X D f1; 12 ; 13 ; 14 ; : : :g has 1 as an upper bound. On the other hand, since
1 2 X, no real number less that 1 can be an upper bound of X, so that sup.X/ D
1.
(b) If X D .1; 2/ (an open interval), then 2, but no real number less than 2, is an
upper bound for X. Indeed, if 1 < a < 2, then the number 1Ca
is also greater
2
1Ca
1Ca
than 1 and less than 2, so that 2 2 X. Now, since a < 2 2 X, the number
a cannot be an upper bound for X. Therefore, sup.X/ D 2, and observe that
2 … X.
If ; ¤ Y R is bounded below, then one can prove, as an easy consequence of
Theorem 7.4 (cf. Problem 6), that Y admits a greatest lower bound m, which is
also said to be an infimum of Y. As in the case of nonempty, bounded above sets,
one can easily prove that a nonempty, bounded below set Y has a unique greatest
lower bound m; hence, we denote m D inf.Y/ or m D glb.Y/.
For future use, the coming results collect some useful properties of the notions
of sup and inf.
Proposition 7.6 Let X R be a nonempty, bounded above set, and M D sup.X/.
If n 2 N, then there exists x 2 X such that
M
1
< x M:
n
Proof Since M is the least upper bound of X and M 1n < M, the real number M 1n
is no longer an upper bound of X. Hence, there exists x 2 X such that x > M 1n .
However, since X .1; M, we must also have x M.
t
u
204
7 More on Real Numbers
Problem 10 states an analogous result for the greatest lower bound of nonempty,
bounded below sets.
Proposition 7.7 Let X; Y R be nonempty sets. If x y for all x 2 X and y 2 Y,
then X is bounded above, Y is bounded below and
sup.X/ inf.Y/:
Proof Fix y 2 Y arbitrarily. Since x y for all x 2 X, the number y is an upper
bound for X. Hence, X is bounded above and, letting M D sup.X/ (the least upper
bound of X), we have M y.
Since y 2 Y was arbitrarily chosen, the reasoning of the previous paragraph
shows that, for all y 2 Y, we have M y. Therefore, M is a lower bound for Y, so
that Y is bounded below and, letting m D inf Y (the greatest lower bound of Y), we
have M m.
t
u
The next result gives a sufficient condition for equality to happen in the previous
proposition.
Proposition 7.8 Let X; Y R be nonempty sets, such that X is bounded above, Y
is bounded below and sup X inf Y. If, for every n 2 N, there exist xn 2 X and
yn 2 Y satisfying yn xn < 1n , then sup.X/ D inf.Y/.
Proof Let M D sup.X/, m D inf.Y/ and suppose that M < m. Since x M < m y for all x 2 X and y 2 Y, we would have y x m M, for all x 2 X and y 2 Y.
1
On the other hand, by choosing a natural number n > mM
(which is possible,
thanks to Axiom 7.3), our hypotheses would guarantee the existence of real numbers
xn 2 X and yn 2 Y such that
yn xn <
1
< m M;
n
which is a contradiction!
Finally, since the assumption that sup.X/ < inf.Y/ leads us to a contradiction,
the only possibility is that sup.X/ D inf.Y/.
t
u
We finish this section by presenting a nontrivial (and important) application of
the concept of supremum of a bounded above nonempty set. To set the stage, we
recall that, in the previous chapters, we several times relied upon the existence of
square roots of positive real numbers. The next result establishes existence in a more
general setting.
Theorem 7.9 Given a positive real number x and a natural number n > 1, there
exists a unique positive real number y such that yn D x.
Proof Let’s consider just the case x > 1, referring the reader to Problem 1 for the
case 0 < x < 1 and observing that the case x D 1 is trivial. If
X D fa 2 RI a 0 and an < xg;
7.1 Supremum and Infimum
205
then X is nonempty, since 0 2 X. Also, X is bounded above, for, if ˛ x C 1,
then Problem 6, page 176, together with item (b) of Corollary 1.3, guarantees that
˛ n .x C 1/n > x C 1 > x; therefore, ˛ … X and, thus, X .1; x C 1/.
Being nonempty and bounded above, X has a least upper bound, say y. Since
1n D 1 < x, we have 1 2 X and, thus, y 1. We shall show that yn D x in the
following way:
(i) If yn < x, we shall obtain a positive real number z such that yn < zn < x.
Then, the first inequality will give y < z, whereas the second (with the aid of
the result of Problem 6, page 176) z 2 X. Therefore, we will get y < z 2 X,
thus contradicting the fact that y is an upper bound for X.
(ii) If yn > x, we shall obtain a positive real number z for which x < zn < yn . Then,
taking an arbitrary a 2 X, it will follow from an < x < zn < yn (again by the
result of Problem 6, page 176) that a < z < y, so that z will be an upper bound
for X which is less than y. This will contradict the fact that y is the least upper
bound for X.
Once we have proved that yn < x and yn > x both lead to contradictions, the only
possibility left will be yn D x.
For the proof of (i), suppose yn < x. If z D y C 1k , with k 2 N, then, once more
by the result of Problem 6, page 176, we have yn < zn . We assert that zn < x for all
sufficiently large k. Indeed, the binomial formula (cf. Theorem 4.20) gives
!
n
X
1 n
n 1 nj
n
n
z D yC
Dy C
y I
j kj
k
jD1
however, since
n
j
< 2n ,
1
kj
<
1
k
and ynj < yn for 1 j n, it follows that
z <y C
n
n
n
X
2n
jD1
k
yn D yn C
n.2y/n
I
k
n
n
therefore, we have zn < x provided yn C n.2y/
< x, i.e., k > n.2y/
.
k
xyn
1
n
For the proof of (ii), suppose y > x. Letting z D y k , with k 2 N, then, as
in the previous paragraph anterior, we have yn > zn . We claim that zn > x for all
sufficiently large k. To this end, arguing as above we have
!
n
n
X
1
n
1 nj
zn D y D yn C
.1/j
y
j kj
k
jD1
!
n
X
n.2y/n
n 1 nj
n
n
I
>y y
>
y
j kj
k
jD1
hence, zn > x whenever yn n.2y/n
k
> x, i.e., k >
n.2y/n
.
yn x
t
u
For another, conceptually simpler, proof of the above theorem, we refer the reader
to Example 8.36.
206
7 More on Real Numbers
Problems: Section 7.1
1. * Yet in respect to roots of positive real numbers, do the following items:
(a) If 0 < x < 1 and n > 1 is natural, show that the real number
1
x
1
p
n
1=x
(whose
existence is guaranteed by Theorem 7.9, since > 1) is the n-th root of x.
p
p
n a
p p p
(b) Given a; b > 0 and m; n > 1 naturals, show that n ab D n a n b, n ab D p
n
b
p
p
p
m n
mn
and
aD
a.
p
p
(c) Given 0 < a < b and n > 1 natural, show that n ap
< n b.p
(d) Given
ap> 0 and m > n > 1 naturals, show that m a > n a if a > 1, and
p
m
a < n a if 0 < a < 1.
p
(e) Let x < 0 be a real number and n 2 N be odd. If y D n x, show that
yn D x (the real number y is also called the n-th root of x) and extend the
properties of items (b)–(d) to this case.
2. * Use the archimedian property of the set of natural numbers to prove the
following items:
(a) If a 2 R is such that 0 a < 1n for every n 2 N, then a D 0.
(b) If a; b; c 2 R, with a > 0, then there exists n 2 N for which an C b > c.
3. * Let a and b be given rationals, with a < b. Prove that:
ba
p < b.
(a) a < aCb
2 < b and a < a C
2
(b) The interval .a; b/ contains infinitely many rational numbers and infinitely
many irrational numbers.
4. * The purpose of this problem is to generalize the result of the previous one,
showing that between any two given real numbers there is always a rational
number and an irrational number (thanks to these properties, we say that Q and
R n Q are dense in R). To this end, let a and b be given real numbers, with
a < b.
(a) Show that it suffices to consider the case a 0.
p
(b) Prove that there exists n 2 N such that 0 < 1n < b a and 0 < n2 < b a.
(c) Letting a 0 e n 2 N be chosen as
in (b),
show
that one of the numbers
p
p
p
2 2 2 3 2
1 2 3
;
;
;
:
:
:
and
one
of
the
numbers
;
;
;
:
: : belong to the interval
n n n
n
n
n
.a; b/.
5. * A rational number r 2 Œ0; 1 is said to be dyadic if there exist k; n 2 Z such
that 0 n 2k and r D 2nk . Prove that the set of dyadic rationals is dense in
Œ0; 1, i.e., that for every a 2 Œ0; 1 and > 0, there exists a dyadic rational in
the interval .a ; a C /.
6. * If Y R is nonempty and bounded below, prove that it has a greatest lower
bound.
7. Let X D fx 2 QI 0 < x < 1g and Y D fy 2 R n QI 0 < y < 1g. Prove that
inf.X/ D inf.Y/ D 0 and sup.X/ D sup.Y/ D 1:
7.1 Supremum and Infimum
207
p
p
8. If X D fj a bjI a; b 2 N and a ¤ bg, compute inf.X/.
9. Let
C1 D .0; 1/
C2 D .0; 1/ n .1=3; 2=3/ D .0; 1=3 [ Œ2=3; 1/
C3 D ..0; 1=3 n .1=9; 2=9// [ .Œ2=3; 1/ n .7=9; 8=9/
D .0; 1=9 [ Œ2=9; 1=3 [ Œ2=3; 7=9 [ Œ8=9; 1/:
More generally, for each n 2 N, obtain CnC1 from Cn , by
S erasing the open
middle third of each of the intervals that form Cn . If C D n1 Cn , show that
inf.C/ D 0 and sup.C/ D 1.
10. * Let Y R be a nonempty, bounded below set, with m D inf Y. If n 2 N,
prove that there exists y 2 Y such that
1
my<mC :
n
11. Let X; Y R be nonempty sets, such that X is bounded above, Y is bounded
below and sup.X/ D inf.Y/ D ˛. If ˛ … X [ Y, prove that there exist elements
xn 2 X and yn 2 Y for which yn xn < 1n , for every n 2 N.
12. * Let X R be a nonempty, bounded above set. Given c 2 R, let cX D fcxI x 2
Xg. Prove that:
(a) If c > 0, then sup.cX/ D c sup.X/.
(b) If c < 0, then cX is bounded below and inf.cX/ D c sup.X/.
Then, if X is bounded below, establish properties analogous to the ones listed
above, relating inf X to the sup or the inf of cX, according to whether c < 0 or
c > 0.
13. * Let X; Y R be nonempty sets and X C Y D fx C yI x 2 X and y 2 Yg.
(a) If X and Y are bounded above, prove that X C Y is bounded above and
sup.X C Y/ D sup.X/ C sup.Y/.
(b) If X and Y are bounded below, prove that X C Y is bounded below and
inf.X C Y/ D inf.X/ C inf.Y/.
14. * Let X; Y Œ0; C1/ be nonempty, bounded above sets. If XY D fxyI x 2
X and y 2 Yg, prove that XY is bounded above and such that sup.XY/ D
sup.X/ sup.Y/ and inf.XY/ D inf.X/ inf.Y/.
15. (Hungary) Let .Rn /n1 be a sequence of pairwise distinct rectangles in the
cartesian plane, each of which having all vertices with integer coordinates and
two sides along the axes. Prove that one can find two of them such that one
contains the other.
16. (IMO) Let f ; g W R ! R be functions satisfying, for all x; y 2 R, the relation
f .x C y/ C f .x y/ D 2f .x/g.y/:
If f is not identically zero and jf .x/j 1 for every x 2 R, prove that jg.x/j 1,
for every x 2 R.
208
7 More on Real Numbers
7.2 Limits of Sequences
Given a sequence .an /n1 of real numbers, we are interested in recognizing whether
or not their terms are approaching a certain real number l, as n increases. For
instance, if an D 1n , it is reasonable to say that the numbers an become closer and
closer to 0 as n increases, since the result of the division of 1 by n is increasingly
smaller as n increases. This naive point of view is formalized as follows.
Definition 7.10 A sequence .an /n1 of real numbers converges to a real number l
if, given an error > 0 for the value of l, there exists an index n0 2 N such that
jan lj < for every n > n0 .
Alternatively, if the sequence .an /n1 converges to l, we say that it is convergent
and that l is a limit of the sequence, which we denote by writing
n
an ! l or
lim an D l:
n!C1
Finally, a sequence which is not convergent is said to be divergent.
In general, if we diminish the error > 0 and the condition “jan lj < for
every n > n0 ” is to continue holding, then the natural number n0 in the definition
of convergent sequence tends to increase. In other words, in general n0 depends
on > 0. Anyhow, what is important to assure the convergence of the sequence
.an /n1 is that, for an arbitrarily given error > 0, we are capable of finding n0 2 N
such that
n > n0 ) jan lj < :
For the reader to get used to the important concept of convergent sequence, we
collect below some elementary examples of convergent and divergent sequences.
Examples 7.11
n
(a) If an D 1n , then an ! 0: indeed, for a given > 0, we have jan 0j < provided n > 1 ; thus, once we have chosen n0 2 N such that n0 > 1 , we will
have jan 0j < whenever n > n0 .
(b) If an D .1/n , then .an /n1 is divergent: indeed, since the terms of the sequence
are alternately equal to 1 and 1, it is impossible for them to (collectively)
become closer to a single real number l (formalize this intuition).
n
n
(c) If an D 1 C .1/
, then an ! 1: this is so because jan 1j D 1n , so that
n
jan 1j < for n > 1 .
(d) If .an /n1 is a constant sequence, with an D c for every n 1, then an ! c.
n
Example 7.12 If an D qn , with 0 < jqj < 1, then an ! 0.
1
1
> 1, we can write jqj
D 1 C ˛, with ˛ > 0. Therefore, taking the
Proof Since jqj
first two terms in the binomial expansion formula, we get
1
D .1 C ˛/n 1 C n˛
jqjn
7.2 Limits of Sequences
209
and, hence,
jan 0j D jqjn 1
:
1 C n˛
Thus, if we wish that jan 0j < , it suffices to impose
n > ˛1 1 1 .
1
1Cn˛
< or, equivalently,
t
u
p
p
Example 7.13 The sequence .an /n1 , given for n 1 by an D n C 1 n,
converges to 0.
Proof Note that an D
n0 >
1
,
42
p
1
p .
nC1C n
Thus, given > 0 and taking n0 2 N such that
we have
n > n0 )
p
p
p
p
p
1
n C 1 C n > n0 C 1 C n0 > 2 n0 > :
Therefore,
1
n > n0 ) jan 0j D p
p < :
nC1C n
t
u
The notion of convergent sequence doesn’t make it clear whether the correspondent limit is unique. Yet in another way, in principle it could happen that a certain
sequence converges to more than one limit. The coming result shows that this is not
so.
Proposition 7.14 If the sequence .an /n1 converges, then its limit is unique.
Proof Let l1 and l2 be distinct real numbers, and suppose that the given sequence
simultaneously converges to l1 and l2 . Toking D 12 jl1 l2 j > 0, the definition of
convergence guarantees the existence of n1 ; n2 2 N such that
n > n1 ) jan l1 j < and n > n2 ) jan l2 j < :
Therefore, triangle inequality gives
n > maxfn1 ; n2 g ) jl1 l2 j jan l1 j C jan l2 j < 2 D jl1 l2 j;
which is an absurd.
t
u
Thanks to the previous result, from now on we speak of the limit of a convergent
sequence. In this respect, the next proposition collects two basic, albeit very
important, properties of limits of convergent sequences. In order to state it properly,
we define a subsequence of a sequence .an /n1 as the restriction of the given
sequence to an infinite subset N1 D fn1 < n2 < n3 < g of N; in this case,
we denote it by .ank /k2N . Since the function j 7! nj from N1 to N is a bijection,
every subsequence can actually be seen as a sequence.
210
7 More on Real Numbers
Proposition 7.15 Let .an /n1 be a convergent sequence, with limn!C1 an D l.
Then:
(a) If an a (resp. an a), for every n 1, then l a (resp. l a).
(b) Every subsequence .ank /k1 of .an /n1 also converges to l.
Proof
(a) Suppose that an a for every n 1, and let’s show that l a (the other case
is completely analogous). By contradiction, if l < a, take D a l > 0. The
definition of convergence guarantees the existence of an index n0 2 N such that
n > n0 ) jan lj < ; in particular, given n > n0 , we have
an < l C D l C .a l/ D a;
which is an absurd.
n
(b) Let > 0 be given. Since an ! l, there exists a natural number n0 such that
jan lj < for n > n0 . Since n1 < n2 < n3 < , there exists an index ni such
that nj > n0 for j i; hence, for all such j, we have janj lj < , which is the
k
same as saying that ank ! l.
t
u
In words, item (b) of the previous proposition says that, if the terms of a certain
sequence come closer and closer to l as their indices increase, then the same is true
for the terms of every subsequence of the given sequence. Item (b) of the previous
proposition also has the following immediate corollary, which gives us a sufficient
(and quite useful) condition for the divergence of a sequence. sequência.
Corollary 7.16 If two subsequences of a given sequence converge to distinct limits,
then the original sequence is divergent.
Up to now, except for some very simple examples we haven’t seen how one could
find out the limit of a convergent sequence. In order to remedy this situation, we
need to understand how to perform simple arithmetic operations with convergent
sequences. We turn to this next, starting with an auxiliary result which is important
in its own.
We say that a sequence .an /n1 is bounded (resp. bounded from above,
bounded from below) if the set fa1 ; a2 ; : : :g is bounded (resp. bounded from above,
bounded from below), in the sense of the previous section.
Lemma 7.17 Every convergent sequence is bounded.
Proof If .an /n1 is a convergent sequence with limit l, then there exists n0 2 N such
that
n > n0 ) jan lj < 1:
This, together with the triangle inequality, gives
n > n0 ) jan j jan lj C jlj < 1 C jlj:
7.2 Limits of Sequences
211
Finally, letting L D maxf1 C jaj; ja1 j; ja2 j; : : : ; jan0 1 jg, we get jan j < L for every
n 2 N, so that the sequence is bounded.
t
u
Proposition 7.18 Let .an /n1 and .bn /n1 be convergent sequences, and c be any
real number.
(a)
(b)
(c)
(d)
If an
If an
If an
If an
n
n
! a, then can ! ca.
n
n
n
n
! a and bn ! b, then an ˙ bn ! a ˙ b and an bn ! ab.
n
n
! 0 and .bn /n1 is bounded, then an bn ! 0.
n
n
n
! a and bn ! b, with b; bn ¤ 0 for every n 1, then abnn ! ab .
Proof
(a) If c D 0, then can D ca D 0, and there is nothing to do. Suppose, then, that
n
c ¤ 0, and let > 0 be given. Since an ! a, there exists n0 2 N such that
n > n0 ) jan aj < jcj
. Hence,
n > n0 ) jcan caj D jcjjan aj < jcj D :
jcj
n
(b) For the first part, let’s prove that an Cbn ! aCb (to prove that an bn ! ab
n
n
is completely analogous). Given > 0, the convergences an ! a and bn ! b
assure the existence of n1 ; n2 2 N such that
n > n1 ) jan aj <
and n > n2 ) jbn bj < :
2
2
Therefore, taking n > maxfn1 ; n2 g, we get
j.an C bn / .a C b/j jan aj C jbn bj <
C D :
2
2
For the second part, let L > 0 be such that jbn j < L for every n 2 N. Given
> 0, take n0 2 N for which
n > n0 ) jan aj <
and jbn bj <
:
2L
2jaj C 1
Then
jan bn abj D jan bn abn C abn abj jan ajjbn j C jajjbn bj
<
L C jaj < C D :
2L
2jaj C 1
2
2
(c) Let L > 0 be such that jbn j < L for every n 1. Given > 0, let’s take n0 2 N
such that
n > n0 ) jan j <
:
L
212
7 More on Real Numbers
Then,
n > n0 ) jan bn 0j D jan jjbn j <
L D :
L
(d) By the second part of item (b), it suffices to show that
start by observing that
1
bn
n
! 1b . To this end,
ˇ
ˇ
ˇ1
ˇ
ˇ 1 ˇ D 1 jbn bj 1 jbn bj ;
ˇb
bˇ
jbj
jbn j
jbj jbj jbn bj
n
where we used the triangle inequality in the last passage above. Now, given
> 0, choose n0 2 N such that
n > n0 ) jbn bj <
jbj
; :
2 2
Then, for n > n0 , we have
ˇ
ˇ
ˇ1
ˇ
ˇ 1 ˇ 1 jbn bj D 2jbn bj < :
ˇb
b ˇ jbj jbj jbj=2
n
Example 7.19 Let a be a positive real number. If .an /n1
n
every n 1, then an ! 1.
t
u
p
n
is given by an D a for
Proof If a > 1, then an > 1. Write an D 1 C bn , so that bn > 0. Since
!
n
a D ann D .1 C bn /n 1 C
bn D 1 C nbn ;
1
we get 0 < bn <
n
a1
n .
Hence, the squeezing principle (cf. Problem 6) guarantees
n
7.18 gives an D 1 C bn ! 1.
that bn ! 0, and item (b) of Proposition
q
If 0 < a < 1, let a0n D
1
an
D
n
1
a,
n
so that a0n ! 1 by the first part. Then, item
n
(d) of Proposition 7.18 gives that an ! 1.
Example 7.20 The sequence .an /n1 , given by an D
to 1.
p
n
t
u
n for every n 1, converges
Proof As in the previous example, write an D 1 C bn for n 2. Since bn > 0,
we have
!
!
n
n 2 n.n 1/ 2
n
n
n D an D .1 C bn / 1 C
bn ;
bn C
b >
2
1
2 n
7.2 Limits of Sequences
213
so that
0 < b2n <
2
:
n1
n
Hence, once more from the squeezing principle, the above inequality gives bn ! 0
n
and, thus, an D 1 C bn ! 1.
u
t
For what comes next, recall that a sequence .an /n1 of real numbers is just a
function f W N ! R, for which we write an D f .n/. Hence, it is natural to say that
.an /n1 is monotonic increasing (resp. decreasing, nondecreasing, nonincreasing)
provided an < anC1 (resp. an > anC1 , an anC1 , an anC1 ), for every n 1.
The most important result on limits of sequences is the theorem below, which is
known in the mathematical literature as Bolzano-Weierstrass theorem.3
Theorem 7.21 (Bolzano-Weierstrass ) Every monotonic bounded sequence is
convergent.
Proof Suppose that .an /n1 is a nondecreasing bounded sequence (the other cases
can be dealt with similarly), i.e., that
a1 a2 a3 < M;
for some M > 0. Then, M is an upper bound for the set A D fa1 ; a2 ; a3 ; : : :g, so that
n
A has a sup, say sup A D l. We claim that an ! l. Indeed, let > 0 be given; since
l is no longer an upper bound for A, some element of it is greater than l ,
say, an0 > l . Therefore, since an0 an0 C1 an0 C2 , we conclude that
an > l for every n n0 . Thus, for n n0 , we have
l < an l < l C ;
as we wished to show.
t
u
The previous theorem, together with the definition of convergence, assures that if
a bounded sequence is monotonic from a certain term on, then it will be convergent.
We explore this comment by revisiting the last two examples.
p
Example 7.22 Let a be a positive real number. If .an /n1 is given by an D n a for
n
every n 1, then an ! 1.
Proof Assume a > 1 (the case 0 < a < 1 can be dealt with as in the Example 7.19).
Then, a1 > a2 > a3 > > 1, and the Bolzano-Weierstrass theorem guarantees
the existence of l D limn!C1 an 1. Item (a) of Proposition 7.15 gives l 1, and item (b) guarantees that every subsequence of .an /n1 also converges to l.
3
After Bernard Bolzano and Karl Weierstrass, German mathematicians of the XIX century.
214
7 More on Real Numbers
k
Therefore, ak.kC1/ ! l. Now, since ak.kC1/ D
Proposition 7.18 that
ak.kC1/ D
p
k a
p
kC1
a
, it follows from item (d) of
p
k
a k l
p ! D 1:
kC1
l
a
Therefore, l D 1.
p
n
t
u
Example 7.23 The sequence .an /n1 , given by an D n for every n 1, converges
to 1.
p p
p
3
Proof The initial terms
ofpthe sequence
are 2,
3, 4 4, . . . , and it is easy to
p
p
p
p
directly show that 2 < 3 3 and 3 3 > 4 4 > 5 5. Since 2n n2 for n 4 (by
induction, for instance), we get a2 an > 1 for n 4, so that the sequence is
bounded; hence, if we show that it is indeed decreasing from the third term on, its
convergence will follow from Bolzano-Weierstrass theorem, with limit l 1.
For what is left to do, for an integer n > 2 we have
p
n
n>
p
1 n
n C 1 , nnC1 > .n C 1/n , n > 1 C
:
n
nC1
Let’s prove the last inequality above. For n D 3,
n it’s immediate to check it
numerically; for n > 3, it suffices to show that 1 C 1n < 3. To this end, notice that
!
!
!
1 n
n 1
n 1
n 1
C
D1C
CC
1C
n
2 n2
1 n
n nn
and
!
1
nŠ
1 n.n 1/ : : : .n k C 1/
1
n 1
k1 :
D
D
<
k nk
kŠ.n k/Šnk
kŠ
nk
kŠ
2
Therefore,
!
!
!
1 n
n 1
n 1
n 1
C
1C
D1C
CC
n
2 n2
1 n
n nn
<1C1C
D3
1
1
1
C 2 C C n1
2
2
2
1
< 3:
2n1
In order to finish, we need to show that l D 1. To this end, first observe
that
the
pp
p
p
p
2k
2k
2k
k
subsequence a2k D 2k also converges to l. On the other hand, 2k D 2
k,
7.2 Limits of Sequences
215
pp k p
p
k
2k
with 2 ! 1. Now, it follows from Problem 2 that k k ! l. Therefore, by
applying item (b) of Proposition 7.18, we get
l D lim
k!C1
p
2k
2k D lim
k!C1
p
2k
2 lim
k!C1
q
p
p
k
k D l;
t
u
Sometimes, we have to show that a given sequence has at least a convergent
subsequence (even if, as a whole, the sequence does not converge). In this sense,
Theorem 7.25 below, also due to Weierstrass, provides a sufficient condition for the
existence of such a subsequence. Before we state and prove it, we need to discuss an
important auxiliary result, known as the lemma of nested intervals. In what follows
if I D Œa; b, with a; b 2 R, we shall let jIj D b a.
Lemma 7.24 For n 2 N, let In D Œan ; bn . If I1
I2T
I3
: : : and
limn!C1 jIn j D 0, then there exists a unique l 2 R such that n1 In D flg.
Proof First of all, note that the intersection of the intervals In , if not empty, has a
single element;Tindeed, if there existed reals a < b in such an intersection, we would
have Œa; b n1 In ; in particular, Œa; b In and, hence, jIn j b a for every
n 2 N, thus contradicting the fact that limn!C1 jIn j D 0.
Secondly, the inclusions I1
I2
I3
: : : give a1 a2 a3 b3 b2 b1 , and the Bolzano-Weierstrass
theorem assures the existence
T
of l D limn!C1 an . We claim that l 2 n1 In .
Since l D supfan I n 2 Ng, it follows that an l for every n 2 N. On the
other hand, for a fixed m 2 N, we have an bm for every n 2 N, and item (a)
of Proposition 7.15 gives l D limn!C1 an bm . However, since m was chosen
arbitrarily, we have l bm for every m 2 N. It thus follows that l 2 Œam ; bm D Im
for every m 2 N, as we wished to show.
t
u
We are now in position to prove Weierstrass theorem.
Theorem 7.25 (Weierstrass ) Every bounded sequence admits a convergent subsequence.
Proof Let .an /n1 be a given bounded sequence, and I0 D Œa; b be a closed and
bounded
containing
all of its terms. One (possibly both) of the intervals
aCb interval
a; 2 and aCb
;
b
,
call
it
I
1 , also contains infinitely many terms of the sequence.
2
Do the same with I1 , obtaining a closed and bounded interval I2 I1 such that jI2 j D
1
2 jI1 j and I2 contains infinitely many terms of the sequence .an /n1 . Proceeding
inductively, we construct a nested sequence I1
I2
I3
of closed and
bounded intervals, such that jIkC1 j D 12 jIk j and Ik contains infinitely many terms of
the sequence .an /n1 , for every
T k 1. Therefore, by the lemma of nested intervals,
there exists c 2 R such that k1 Ik D fcg.
Now, choose n1 2 N such that an1 2 I1 ; then, after having chosen nj 2 N such
that anj 2 Ij , choose njC1 2 N such that njC1 > nj and anjC1 2 IjC1 (this is possible
216
7 More on Real Numbers
by the way the Ij ’s were defined). This way, we inductively construct a subsequence
.ank /k1 of .an /n1 , such that ank 2 Ik for every k 1. Since jIk j D 21k jI0 j and
ank ; c 2 Ik , we conclude that jank cj k
1
jI j,
2k 0
for every k 1; since
1
jI j
2k 0
we conclude that ank ! c.
k
! 0,
t
u
The concept of convergent sequence gives a precise meaning to the geometric
intuition that the terms of the given sequence come closer and closer to a certain
real number (the limit of the sequence), as long as their indices increase. However,
it is also reasonable to expect that, if the terms of a given sequence come close
together, then the sequence should also converge. This is indeed the case and, in
order to establish it, we start with the following
Definition 7.26 A sequence .an /n1 is said to be a Cauchy sequence if, for every
> 0, there exists n0 2 N such that
m; n > n0 ) jam an j < :
The fundamental result concerning Cauchy sequences is the content of the
following
Theorem 7.27 A sequence .an /n1 is convergent if and only if is Cauchy.
Proof Let .an /n1 be a convergent sequence, with limit l. Given > 0, the definition
of convergence guarantees the existence of n0 2 N such that
:
2
n > n0 ) jan lj <
Hence, given naturals m; n > n0 , the triangle inequality gives
jam an j jam lj C jan lj <
C D ;
2
2
and the sequence is Cauchy.
Conversely, let .an /n1 be a Cauchy sequence. Then, there exists n0 2 N such
that jam an j < 1 for m; n > n0 . In particular, jam an0 C1 j < 1 for every m > n0 ,
and the sequence has all of its terms contained in the set
fa1 ; a2 ; : : : ; an0 g [ .an0 C1 1; an0 C1 C 1/;
so that it is bounded. Hence, by the theorem of Bolzano-Weierstrass, the sequence
k
.an /n1 has a convergent subsequence, say, ank ! l. Let us prove that, actually,
n
an ! l.
Given > 0, there exists N0 2 N such that
nk > N0 ) jank lj <
:
2
7.2 Limits of Sequences
217
On the other hand, since the sequence is Cauchy, there exists N1 2 N such that
m; n > N1 ) jam an j <
:
2
Letting M D maxfN1 ; N2 g and fixing nk > M, we have, for n > M and with the aid
of the triangle inequality,
jan lj jan ank j C jank lj <
C D ;
2
2
t
u
as we wished to show.
The coming example collects an interesting application of the above result.
Example 7.28 Let .an /n1 be a sequence of real numbers such that
janC2 anC1 j cjanC1 an j
for every n 2 N, where 0 < c < 1 is a real constant. Show that this sequence is
convergent.
Proof By Theorem 7.27, it suffices to show that .an /n1 is a Cauchy sequence. To
this end, iterating the inequality in the statement we get, for every k 2 N,
jakC1 ak j ck1 ja2 a1 j:
Let n and p be given natural numbers. The above inequality, together with the
triangle inequality, gives
X
nCp1
janCp an j jakC1 ak j kDn
D
<
c
X
nCp1
nCp c
1c
n1
ck1 ja2 a1 j
kDn
ja2 a1 j
1
ja2 a1 jcn1 :
1c
We now note that, by Example 7.12, the last expression above tends to 0 when
1
n ! C1. Therefore, given > 0, there exists n0 2 N such that 1c
ja2 a1 jcn1 <
for every n > n0 . Hence, for n > n0 and p 2 N, we get janCp an j < , so that
.an /n1 is, indeed, a Cauchy sequence.
t
u
218
7 More on Real Numbers
Problems: Section 7.2
1. * Let .an /n1 and .bn /n1 be convergent sequences of real numbers, with
limn!C1 an D a and limn!C1 bn D b. Generalize item (a) of Proposition 7.15, showing that if an bn for every n 1, then a b.
2. * Let .an /n1 be a sequence of positive real numbers converging to a > 0. Show
n p
p
that an ! a.
n
3. * Given a 2 R such that jaj > 1, show that anŠ ! 0 when n ! C1.
4. Generalize the result of Example 7.12, showing that, given k 2 N and a 2 R,
n
k
with jaj > 1, we have ann ! 0.
5. Let .an /n1 and .bn /n1 be sequences of real numbers and, for each n 2 N, let
tn 2 Œ0; 1 be given. Denote by .cn /n1 the sequence defined by
cn D .1 tn /an C tn bn ;
n
n
for every n 2 N. If an ; bn ! l, show that cn ! l.
6. * Prove the squeezing theorem: let .an /n1 , .bn /n1 and .cn /n1 be sequences
n
of real numbers such that an bn cn , for every n 2 N. If an ; cn ! l, for
n
some l 2 R, show that bn ! l.
7. Compute the following limits:
(a) limn!C1
p
n n
.
n2 C1
p
(b) limn!C1 . n2 C an C b n/, with a; b 2 R.
p
(c) limn!C1 n 1 C qn , where 0 < q < 1 is a real number.
p
(d) limn!C1 n an C bn , with a and b positive reals such that a > b.
8. * This problem extends the concept of limit of sequences to consider infinite
limits. We say that a sequence .an /n1 of real numbers converges to C1 (resp.
1) if, given M > 0, there exists n0 2 N such that n > n0 ) an > M (resp.
an < M). In this case, we denote limn!C1 an D C1 (resp. limn!C1 an D
n
n
1), or simply an ! C1 (resp. an ! 1). With respect to this concept,
and given sequences .an /n1 and .bn /n1 of real numbers, do the following
items:
n
n
(a) If an ! ˙1 and .bn /n1 is bounded, then an C bn ! ˙1.
n
(b) If an ! ˙1 and bn c > 0 (resp. bn c < 0) for every n 1, then
n
n
an bn ! ˙1 (resp. an bn ! 1).
n
(c) If bn ! C1 and there exists c > 0 such that an cbn (resp. an cbn )
n
for every n 1, then an ! ˙1.
9. Let q be a real number and .an /n1 be the sequence defined by an D qn . If
n
q > 1, show that an ! C1. If q < 1, show that .an /n1 does not converge
to either C1 or 1.
7.2 Limits of Sequences
219
n
10. * Given positive reals a and q, with q < 1, show that an q2 ! 0.
11. (IMO shortlist) Let .an /n1 be a sequence of real numbers such that, for every
m; n 2 N, we have
n
jam an j 2mn
:
C n2
m2
Show that the sequence is constant.
p
p
p
12. Let .an /n1 be the sequence defined
by a1 D 1, a2 D 1 C 1, a3 D 1 C 2,
r
q
q
p
p
p
p
a4 D 1 C 1 C 2, a5 D 1 C 1 C 1 C 2, . . . . Show that .an /n1
is convergent and compute its limit.
13. (Austrian-Polish) Let .an /n1 be a sequence of positive reals, such that
akC2 D
p
p
akC1 C ak ;
for every k 1. Prove that the sequence converges and compute its limit.
14. Let n > 1 be a given integer and t0 ; t1 ; : : : ; tn be given real numbers, such that
t0 C t1 C C tn D 0. Prove that the sequence .ak /k1 , defined by
p
p
p
ak D t0 k C t1 k C 1 C C tn k C n
converges to 0.
p
15. (Romania) The sequence .xn /n1 is such that xnC1 C 2 xn 2, for every
n 1. Find all possible values of x1986 .
16. (Leningrad) Let .an /n1 be a sequence of real numbers such that
jam C an amCn j 1
;
mCn
for all m; n 2 N. Prove that the sequence is an AP.
17. (Bulgaria) For each n 2 N, let
an D
nC1
2nC1
22
2n
21
C
CC
:
1
2
n
Prove that the sequence .an /n2 is decreasing and convergent, and compute its
limit.
18. (Romania) Let k be a fixed natural number and .an /n1 be the sequence defined
by
r
an D
with exactly n square roots.
q
kC
k CC
p
k;
220
7 More on Real Numbers
(a) Show that .an /n1 is convergent.
(b) Show that, if k is odd, then the limit of the sequence is an irrational number.
(c) Find all natural values of k for which sequence converges to an integer.
19. For each positive real a, let the sequence .an /n1 be defined by a1 D 1 and
akC1
1
a
ak C
;
D
2
ak
p
for every integer k 1. Prove that the sequence converges to a.
20. (TT) The set of natural numbers is partitioned into m disjoint, infinite and
nonconstant arithmetic progressions, of common ratios d1 , d2 , . . . , dm . Prove
that
1
1
1
C
CC
D 1:
d1
d2
dm
21. (OIMU) Let c and ˛ be positive real constants4 and Q be a square in the plane.
Prove that there doesn’t exist a surjection f W Œ0; 1 ! Q for which
d.f .x/; f .y// cjx yj˛C1=2
for all 0 x; y 1, where we let d.A; B/ D AB denote the euclidean distance
between the points A and B in the plane.
p
22. (Turkey) Let .an /n1 be a sequence of integers such that 0 < anC1 an < an ,
for every natural n. Given real numbers x and y, with 0 x < y 1, prove that
there exist natural numbers m and n such that
x<
am
< y:
an
23. (IMO shortlist) Let .an /n1 be a sequence of positive reals. Show that
p
n
1 C an > an1 2
for infinitely many values of n.
4
Powers of a positive basis with real exponents will be defined in Sect. 10.7. For the time being,
you may assume that ˛ is a positive rational, if you will.
7.3 Kronecker’s Lemma
221
7.3 Kronecker’s Lemma
In this section, we apply some of the ideas exposed so far in this chapter to study
the important concept of dense set, as well as to present an interesting geometric
application of it. We start by recalling the following definition.
Definition 7.29 Given an interval I, we say that a subset X of I is dense (in I) if,
for every a 2 I and > 0, it happens that
X \ .a ; a C / ¤ ;:
Intuitively, the density of X in I means that X is spreaded all over I. Problem 4,
page 206, shows that both Q and RnQ are dense in R, whereas Problem 5, page 206,
shows that the set of dyadic rationals, i.e., rational numbers of the form 2nk , where
n; k 2 ZC are such that 0 n 2k , is dense in Œ0; 1.
A quite useful result on the density (in R) of certain of its subsets is the content of
Theorem 7.31 and Corollary 7.32, which are collectively known in the mathematical
literature as Kronecker’s lemma.5 The proofs we present, albeit not being the
simplest ones, have the advantage of deriving from a circle of ideas which are
interesting in themselves. First of all, we need yet another definition.
Definition 7.30 A nonempty subset G of R is said to be an additive subgroup of
R if, for all x; y 2 G, we have x y 2 G.
Evidently, f0g, Z, Q and R itself are additive subgroups of R. For a less obvious
example, given real numbers x1 ; : : : ; xk , it is immediate to verify (see Problem 1)
that the set
Gx1 ;:::;xk D fa1 x1 C C ak xk I a1 ; : : : ; ak 2 Zg
(7.1)
is also an additive subgroup of R.
Now, let G be an arbitrary additive subgroup of R and take x 2 G. By the above
definition, we have 0 D x x 2 G. Thus, for x; y 2 G, we also have y D 0 y 2 G
and, hence, x C y D x .y/ 2 G; therefore, G is closed under the operation of
addition.
Hence, if ˛ 2 G, then 2˛ D ˛ C ˛ 2 G; moreover, if k˛ 2 G, for some k 2 N,
then .k C 1/˛ D k˛ C ˛ 2 G, so that m˛ 2 G, for every m 2 N. Since 0˛ D 0 2 G
and .k/˛ D k˛ 2 G for every k 2 N, it follows that
G˛ D fm˛I m 2 Zg G:
The coming result collects two central facts on additive subgroups of R.
5
After Leopold Kronecker, German mathematician of the XIX century.
(7.2)
222
7 More on Real Numbers
Theorem 7.31 (Kronecker) Let G ¤ f0g be an additive subgroup of R, and let
GC D G \ RC .
(a) If inf.GC / D 0, then G is dense in R.
(b) If inf.GC / D ˛ > 0, then ˛ 2 G and G D G˛ .
Proof
(a) Suppose inf.GC / D 0, let a 2 R and > 0 be given. We have to show that
G \ .a ; a C / ¤ ;. Since x 2 G , x 2 G, it suffices to analyse the case
a 0. If a < 0, we have 0 2 G \ .a ; a C / and there is nothing to do.
Suppose, then, that a 0.
The hypothesis inf.GC / D 0 guarantees the existence of x 2 GC such that
x < 2. Letting m be the greatest nonnegative integer such that mx a , we
claim that .m C 1/x 2 G \ .a ; a C /. Indeed, if .m C 1/x a C , we
would have
mx a < a C .m C 1/x;
so that
x D .m C 1/x mx .a C / .a / D 2;
thus contradicting the choice of x. Hence, .m C 1/x 2 .a ; a C / \ G.
(b) Suppose inf.GC / D ˛ > 0. We initially claim that ˛ 2 GC . By the sake
of arriving at a contradiction, suppose that ˛ … GC . Then, the definition of
infimum of a set would assure the existence of elements ˇ; 2 GC such that
˛ < ˇ < < 2˛. However, since G is an additive subgroup of R, it would
follow from here that ˇ 2 GC , with
0 < ˇ < 2˛ ˛ D ˛:
This contradicts the fact that ˛ D inf.GC /,
˘ thus showing that
x ˘˛ 2 GC .
x
Now, take any x 2 GC and let q D ˛ and r D x ˛ ˛ , so that q 2 ZC ,
0 r < ˛ and x D q˛ C r. If r > 0, then the fact that G is an additive subgroup
of R would imply r D x q˛ 2 GC , with 0 < r < ˛. Since this contradicts the
fact that ˛ D inf.GC /, we conclude that r D 0 and, hence, x D q˛ 2 G˛ .
Therefore,
GC fn˛I n 2 Ng
and, since the opposite inclusion was already established in (7.2), we actually
have GC D fn˛I n 2 Ng. Finally, since G D GC [ f0g [ G , where G D
fxI x 2 GC g, it is immediate to see that
G D fm˛I m 2 Zg D G˛ :
t
u
In the coming corollary, we stick to the notation set forth in (7.1).
7.3 Kronecker’s Lemma
223
Corollary 7.32 (Kronecker) If ˛ is an irrational number, then the additive
subgroup G1;˛ D fm C n˛I m; n 2 Zg of R is dense in R.
Proof By the sake of simplicity of notation, let G D G1;˛ . By the previous theorem,
in order to prove that G is dense in R, it suffices to prove that inf.GC / D 0.
If this was not the case, then, once more from Kronecker’s theorem, there would
exist a positive real number ˇ such that inf.GC / D ˇ > 0 and G D Gˇ . Since both
˛ and 1 C ˛ belong to G, there would exist distinct, nonzero integers m and n for
which
˛ D nˇ and 1 C ˛ D mˇ:
Now, since ˛ is irrational, the first equality above would give n ¤ 0 and
ˇ D ˛n … Q. On the other hand, we would also have
.m n/ˇ D .1 C ˛/ ˛ D 1;
1
2 Q.
so that ˇ D mn
We have then reached a contradiction, which came from the supposition that G
is not dense in R. Therefore, G is indeed dense in R, as we wished to show.
t
u
Our next corollary refines the conclusion of the previous one.
Corollary 7.33 If ˛ is an irrational number, then the following sets are dense
in R:
(a) A D fm C n˛I m; n 2 Z and m < 0 < ng.
(b) B D fm C n˛I m; n 2 Z and n < 0 < mg.
Proof Without any loss of generality, we can suppose that ˛ > 0. Let’s prove item
(a), the proof of item (b) being totally analogous. Given a 2 R and > 0, we want
to establish the existence of x 2 A such that a < x < a C . Suppose that
a 0 (the remaining cases are entirely analogous), and let ı D minf˛; 2g > 0.
We first claim that there exist m; n 2 Z such that m < 0 < n and m C n˛ 2
A \ .0; ı/. By contradiction, suppose that
m C n˛ 2 A \ .0; ı/ ) n 0:
Choose (by the former corollary) x0 D m0 C n0 ˛ 2 A \ .0; ı/, with the greatest
possible n0 0. Since Kronecker’s lemma guarantees that A \ .0; x0/ is infinite, we
can take x1 D m1 C n1 ˛ 2 A \ .0; x0 /, with n1 < n0 . Then,
x0 x1 D .m0 m1 / C .n0 n1 /˛ 2 A \ .0; x0 / A \ .0; ı/;
which is a contradiction, for, n0 n1 > 0. Hence, we can choose mCn˛ 2 A\.0; ı/,
with n > 0. This being said, if it were m 0, we would have m C n˛ ˛ ı,
which is another contradiction. Therefore, m < 0 and, thus, A \ .0; ı/ ¤ ;.
224
7 More on Real Numbers
Now, take x 2 A \ .0; ı/ and consider all numbers of the form kx, with k 2 ZC .
Letting k0 be the greatest nonnegative integer for which k0 x a , we claim that
.k0 C 1/x 2 .a ; a C /. Indeed, if it were .k0 C 1/x a C , we would have
k0 x a < a C .k0 C 1/x
and, hence,
ı > x D .k0 C 1/x k0 x .a C / .a / D 2:
This contradicts the choice of ı.
t
u
The discussion of the following example uses a few simple facts on plane
Euclidean Geometry, for which we refer the reader to [4].
Example 7.34 (Brazil) Let … be an euclidean plane and f W … ! … be a function
such that
d.P; Q/ D 1 ) d.f .P/; f .Q// D 1;
for all P; Q 2 …. Prove that f is an isometry of …, i.e., prove that, for all P; Q 2 …,
one has d.P; Q/ D d.f .P/; f .Q//, where d.X; Y/ D XY stands for the euclidean
distance between the points X and Y.
Proof For P 2 …, we let f .P/ be systematically denoted by P0 , so that d.P; Q/ D
PQ andpd.f .P/; f .Q// D P0 Q0 . Firstly, let’s show that f must preserve segments of
length 3.
Claim 1
PQ D
p
p
3 ) P0 Q0 D 3:
p
Indeed, given points P and Q in the plane for which PQ D 3, let’s construct
points R and S such that both QRS and PRS are equilateral triangles whose side
lengths are equal to 1 (cf. Fig. 7.1).
Fig. 7.1 pPQ D
P0 Q0 D 3
p
3)
Q
U
T
S
R
V
P
7.3 Kronecker’s Lemma
225
Fig. 7.2 PQ D 2 )
P0 Q0 D 2
S
P
T
R
Q
Let us counterclockwise rotate rhombus PRQS with center at P, until we get a
rhombus PTUV such that QU D 1.
Observe that the images P0 , R0 and S0 of P, R and S form an equilateral triangle of
side lengths equal to 1. Since Q0 R0 D Q0 S0 D 1, it p
follows that Q0 D P0 or P0 R0 Q0 S0
is a rhombus congruent to PRQS (so that P0 Q0 D 3). In order to discard the first
possibility, it suffices to note that, if it were P0 D Q0 , then T 0 , U 0 and V 0 would all be
points on a circle centered at P0 D Q0 , while being vertices of an equilateral triangle
of side lengths equal to 1, which is an absurd.
Claim 2 For every positive integer n, we have
PQ D n ) P0 Q0 D n:
It suffices to establish the case n D 2, the general case being totally analogous.
Let P and Q be such that PQ D 2, and let R be the midpoint of PQ, such that
PR D RQ D 1 (cf. Fig. 7.2). Let’s consider points S and T such that PRS, RST
and QRT are equilateral triangles of side lengths equal to 1, all situated on one of
!
the half-planes determined by line PQ. Using claim 1 twice, it’s immediate that
P0 Q0 D 2.
Analogously, we can prove that
p
p
PQ D n 3 ) P0 Q0 D n 3:
Claim 3 P0 Q0 PQ, for all points P and Q in the plane.
In order to prove this claim, letpPQ D l, such that l is neither a natural number
nor a real number of the form n 3, for some n 2 N. By Corollary 7.33 (see,
also, Problem 3), we can take sequences .mk /k1 e .nk /k1 of integers satisfying
the following conditions:
i. mk < 0 < nk , for every
p k 1;
ii. limk!C1 .mk C nk 3/ Dpl;
iii. maxf0; l 1g < mk C nk 3 < l, for every k 1.
p
3. To this
Let’s first show that
l, mk and nk p
pthere exists a triangle of side lengths
p
end, since mk C nk 3p< l, we have l C .mk / > nk 3; also, l C
.m
C
n
3/ > 0,
k
k
p
which implies l C nk 3 > mk ; finally, from l 1 < mk C nk 3 we get
p
p
nk 3 C .mk / > nk 3 C mk C 1 > l:
226
7 More on Real Numbers
R
R
√
nk 3
f
−mk
−mk
√
nk 3
Q
l
P
l
Q
P
Fig. 7.3 P0 Q0 PQ
Fig. 7.4 PQ D l )
P0 Q0 D l
X
P
Q
!
Therefore,
ptriangle inequality assures the existence of a point R 2 … n PQ such that
PR D nk 3 and RQ D mk ; since we already have PQ D l, there is nothing left
to do.
0 Q0 P0 R0 or,
It follows from what we did above (cf. Fig. 7.3) that P0 Q0 C Rp
0
0
0
0
0
0
0
0
which is the same, P Q P R R Q . However, since P R D nk 3 and R0 Q0 D
p
p
k
mk , we get P0 Q0 nk 3 C mk . On the other hand, since nk 3 C mk ! l, it
comes that P0 Q0 l D PQ.
Let’s consider again points P and Q in the plane, with PQ D l. Tesselate the
plane with equilateral triangles of side lengths all equal to 1, such that one of these
!
triangles has one of its vertices at P and one of its sides on line PQ (cf. Fig. 7.4).
By what we did above, the images by f of the vertices of such a triangulation form
the vertices of an analogous triangulation. On the other hand, if X is an arbitrary
vertex of the original triangulation (see Fig. 7.4 once more), then X 0 Q0 XQ, for
every point Q of the plane. Geometrically, this means that Q0 doesn’t belong to the
interior of the disk centered at X 0 and having radius XQ. However, since this is true
for every vertex of the triangulation, we must necessarily have P0 Q0 D l.
t
u
Apart from the following problems, other interesting applications of Kronecker’s
lemma will appear in the context of continuity in Problem 14, page 264, as well
as in Sect. 10.8 (cf. Examples 10.59 and 10.60), when we have at our disposal the
concepts and elementary properties of logarithms.
7.3 Kronecker’s Lemma
227
Problems: Section 7.3
1. * Given real numbers x1 , . . . , xk , verify that the set
Gx1 ;:::;xk D fa1 x1 C C ak xk I a1 ; : : : ; ak 2 Zg
is indeed an additive subgroup of R.
For the next problem, the reader might want to recall the definitions of integer
part and fractional part of a real number, given in Problems 9 and 10, page 152.
2. The purpose of this problem is to give a direct proof of Corollary 7.32. To this
end, given ˛ 2 R n Q, a 2 R and > 0, start by choosing p 2 N such that
1
p < 2 and do the following items:
(a) Show that at least two of the numbers
f˛g, f2˛g, . . . , f.p C 1/˛g belong to
a single interval of the form pk ; kC1
, for some integer k satisfying 0 k <
p
p.
(b) Use the result of (a) to show that there exist m0 ; n0 2 Z such that 0 <
m0 C n0 ˛ < 1p .
(c) Use the result of (b) to show that there exists r 2 Z such that r.m0 C n0 ˛/ 2
.a ; a C /.
3. * Given ˛; l 2 R, with ˛ irrational, show that there exist sequences .mk /k1 and
.nk /k1 of integers satisfying the following conditions:
(a) mk < 0 < nk , for every k 1;
(b) limk!C1 .mk C nk ˛/ D l.
For the next problem, we assume from the reader some familiarity with the
basics of plane analytic geometry and vector algebra in the plane. We refer to
Chaps. 6 and 8 of [4] for the necessary background.
4. A subset X of an euclidean plane … is said to be dense in provided X intersects
every disk of …. Now, let O be a fixed point in …. A nonempty subset X of … is
said to be an additive subgroup of … with respect to O provided the following
! ! !
condition is satisfied: for every A; B 2 X, if OA OB D OC, then C 2 X.
(a) If A1 ; : : : ; An 2 … and
XA1 ;:::;An D
n
nX
o
!
mk OAk I mk 2 Z; 8 1 k n ;
kD1
show that XA1 ;:::;An is an additive subgroup of … with respect to O, which is
not dense in it.
!
(b) Choose A and B in … such that O … AB . If ˛ 2 R n Q and
!
!
YA;B D fmOA C .n C p˛/OBI m; n; p 2 Zg;
228
7 More on Real Numbers
show that YA;B is an additive subgroup of … with respect to O, which is not
dense in it.
(c) Give an example of an additive subgroup X of … with respect to O, different
from … itself and dense in it.
(d) Let X be an additive subgroup of … with respect to O. Do the following
items:
i. Given a line r through O, look at it as a real line, with O representing 0.
Prove that X \ r is either the empty set or an additive subgroup of r.
ii. If there exist distinct lines r and s through O such that X \ r; X \ s ¤ ;,
prove that X is dense in ….
iii. If X is not dense in … and is not contained in a single line, prove that
there exists a direction d in … such that X is contained in the union of a
family of equally spaced lines, all parallel to d.
Although the next problem does not really use Kronecker’s lemma, this is the
best place to put it.
5. Let ˛ and ˇ be positive irrationals such that ˛1 C ˇ1 D 1. Our purpose is to show
that the sets
fbk˛cI k 2 Ng and fbkˇcI k 2 Ng
form a partition of the natural numbers. To this end, do the following items:
(a) Show that if ˛ > 1 is irrational
n o and n 2 N, then n is a term of the sequence
.bk˛c/k1 if and only if ˛n > 1 ˛1 .
n o n o
(b) Given n 2 N, show that ˛n C ˇn D 1.
n o
n o
(c) Conclude that either ˛n > 1 ˛1 or ˇn > 1 ˇ1 , but not both.
(d) Finish the proof.
In the notations above, one says that .ak /k1 and .bk /k1 , given by ak D bk˛c
and bk D bkˇc, are the Beatty sequences6 corresponding to ˛ and ˇ.
7.4 Series of Real Numbers
Let .an /n1 be a sequence of real numbers. By the series
C1
X
an ;
nD1
6
After Samuel Beatty, Canadian mathematician of the XX century.
7.4 Series of Real Numbers
229
P
or simply n1 an , we mean the sequence .sn /n1 , where sn D a1 Ca2P
C Can for
n 1. The real number sn is called the n-th partial sum of the series n1 an , and
we say that such a series converges to s 2 R if the sequence .sn /n1 of its partial
sums converges to s. In this case, we say that s is the sum of the series and write
X
an D s:
(7.3)
n1
P
In other words, whenever we write k1 ak D s, we will be saying that the terms
of the sequence of finite sums sn D a1 C a2 C C an come closer and closer to
the real number s, as long as n ! C1. It is in this sense that equality (7.3) must be
thought of, as a limit.
We shall sometimes have a sequence P
.an /n0 of reals, in which case the
corresponding series will be denoted by
n0 an . We leave to the reader the
(immediate) task of adapting the former and coming discussions to such a situation.
Our main interest in this section is to find out efficient criteria to decide whether
a given series converges or not. If it doesn’t converge, we shall say that it is a
divergent series. Let’s see two simple examples of divergent series.
P
P
Example 7.35 The series k1 k and k1 .1/k diverge.
Proof The first series diverges, for its n-th partial sum is sn D 1 C 2 C C n D
n.nC1/
2 , so that .sn /n1 is a divergent sequence. In the second case, the n-th partial
sum sn of the given series is such that sn D 0 if n is even and sn D 1 if n is odd, so
that .sn /n1 is also a divergent sequence.
t
u
P
Given a series n1 an , we refer to a generic term an as the general term of the
series. The following proposition gives a necessary condition on the general term of
a series, if it is to converge.
P
n
Proposition 7.36 If the series k1 ak converges, then ak ! 0.
Proof Given > 0, we want to
Pprove that there exists n0 2 N such that n > n0 )
jan j < . To this end, let l D k1 ak . By the definition of convergence for series,
there exists n0 2 N such that
n n0 ) j.a1 C a2 C C an / lj <
:
2
Therefore, it follows from the triangle inequality that, for n > n0 , we have
jan j j.a1 C a2 C C an / lj C jl .a1 C a2 C C an1 /j
C D :
2
2
t
u
230
7 More on Real Numbers
The converse to the above proposition is not true, namely, there are divergent
P
n
series k1 ak for which ak ! 0. The classical example is that of the harmonic
P
1
series, i.e., the series
k1 k , whose divergence is established in the coming
example and will find further use in these notes.
Example 7.37 Given n 2 N, let m be the only natural number such that 2m n <
2mC1 . Then,
n
X
1
kD1
k
m
C 1:
2
(7.4)
In particular, the harmonic series diverges.
Proof Note that, for every integer k > 1,
1
1
1
1
1
1
1
C k1
CC k > k C k CC k D :
C1
2 C2
2
2
2 ƒ‚
2…
2
„
2k1
2k1 times
Hence,
2m
1 X1
1 X1
1C C
1C C
2
j
2
j
jD3
jD3
n
D1C
m 1
1 X
1
C
C
C
2 kD2 2k1 C 1
2k
m
1 X1
D1C :
>1C C
2 kD2 2
2
m
t
u
comes next, we shall show that, for every rational r > 1, the series
P In what
1
converges.
To this end, we need to examine the convergence of a geometric
r
k1 k
series, i.e., of a series of the form
X
qk1 ;
k1
for a certain nonzero real number q. In this sense, we have the following important
result.
P
Proposition 7.38 Given q 2 R n f0g, the geometric series k1 qk1 converges if
and only if 0 < jqj < 1. Moreover, if this is so, then
X
k1
qk1 D
1
:
1q
7.4 Series of Real Numbers
231
Proof If jqj 1, the geometric series diverges, since its general term qk1 doesn’t
converge to 0. Suppose, then, that 0 < jqj < 1, and let sn D 1 C q C C qn1 be
the n-th partial sum of the series. By the formula for the sum of the terms of a finite
GP, we have
sn D
1
qn
1 qn
D
:
1q
1q 1q
Hence, in order to show that the series converges to
n
1
,
1q
it suffices to show that
q ! 0. But this was done in Example 7.12.
n
We can now discuss the promised example.
Example 7.39 If r > 1 is rational, then the series
t
u
P
1
k1 kr
converges.7
Proof By the Bolzano-Weierstrass
P theorem, it suffices to show that the sequence
.sn /n1 of the partial sums sn D nkD1 k1r is bounded. To this end, given n 2 N, take
m 2 N such that 2m > n. Then,
1
1
1
1
C r CC
CC m
sn 1 C
2r
3
.2m1 /r
.2 1/r
1
1
1
C 4 r C C 2m1 .m1/r
2r
4
2
1
1
1
< 1 C r1 C r1 C C .m1/.r1/
2
4
2
X 1
<
:
2.r1/k
k0
< 1C2
However, since r > 1, we have 0 <
proposition that
X
1
k0
2.r1/k
Therefore, we conclude that 0 < sn <
.sn /n1 is, indeed, bounded.
1
2r1
D
< 1, and it follows from the previous
2r1
:
2r1 1
2r1
2r1 1
for every n 2 N, so that the sequence
t
u
P
Remark 7.40 For the sake of curiosity, we inform the reader that k1 k1r D .r/,
where
W .1; C1/ ! R stands for the famous Riemann’s zeta function.8
Had powers kr , with r > 0 real, been defined (this will be done inP
Sect. 10.7), the reasoning
presented in the proof would work equally well to show that the series k1 k1r converges.
8
After Bernhard Riemann, German mathematician of the XIX century. For more on Riemann, see
the footnote at page 354.
7
232
7 More on Real Numbers
2
An elementary computation of .2/ D 6 will be hinted to at Problem 12, page 470.
We also refer to Chap. 9 of [5]. Note, however, that for an odd natural number
m > 1 the computation of the exact numerical value of .m/ is an open problem in
Mathematics.
We finish our initial list of examples of series with an additional application of
Bolzano-Weierstrass theorem to the convergence of series. In the coming example,
we introduce one of the most important constants of Mathematics, the number e,
which will play a major role in Sect. 10.7.
P
Example 7.41 The series k0 kŠ1 converges to an irrational number e, such that
2 < e < 3. In symbols,
eD
X 1
:
kŠ
k0
(7.5)
Proof Let .sn /n0 be the sequence of the partial sums of the given series, i.e.,
sn D 1 C
1
1
1
C CC :
1Š
2Š
nŠ
For this sequence, we clearly have 1 D s0 < s1 < s2 < ; on the other hand, since
kŠ > 2k1 for every integer k > 2, we have, for an integer n 4,
sn D
n
n
X
1
1 X 1
8 X 1
35
1
<1C1C C C
;
< C
D
k1
k1
kŠ
2
6
2
3
2
12
kD0
kD4
k4
where we used the formula for the sum of a geometric series in the last equality
above. Hence, the sequence .sn /n0 is increasing and bounded, thus, convergent.
Now, item (b) of Proposition 7.15, together with s2 D 2 and sn < 35
for every
12
integer n 4, gives 2 < e < 3 (for an approximation of e with five correct decimal
places, see Problem 8).
We shall now show that e is irrational.9 To this end, observe initially that, for
natural numbers 1 < n < m, we have
m
X 1
X
1
sm sn D
<
kŠ knC1 kŠ
kDnC1
1
1
1
1C
C
C
D
.n C 1/Š
nC2
.n C 2/.n C 3/
X
1
nC2
1
1
<
;
D
k
.n C 1/Š k0 .n C 2/
.n C 1/Š n C 1
(7.6)
9
For the reader’s knowledge, we observe that the number e is, indeed, transcendent, i.e., it is not
the root of a polynomial with rational coefficients. A proof of this fact is beyond the scope of these
notes, and can be found in [11].
7.4 Series of Real Numbers
233
where, once more, we used the formula for the sum of a geometric series in the last
equality above.
Therefore, yet for natural numbers 1 < n < m, it follows from the above
computations that
sm D sn C
m
X
nC2
1
1
< sn C
;
kŠ
.n C 1/Š n C 1
kDnC1
and item (a) of Proposition 7.15 gives
e D lim sm sn C
m!C1
nC2
1
:
.n C 1/Š n C 1
Thus,
sn < e sn C
nC2
1
:
.n C 1/Š n C 1
Multiplying this last inequality by .n 1/Š and noticing that sn D sn1 C
conclude that
.n 1/Šsn1 C
1
nŠ ,
we
nC2
1
1
< .n 1/Še .n 1/Šsn1 C C
;
n
n
n.n C 1/2
Writing tn D nŠsn 2 N and observing that
nC2
1
1
2
1
C
D C
C
n
n.n C 1/2
n
.n C 1/2
n.n C 1/2
1
1
2
21
C 2C
< 1;
D
2
3
4
34
48
for every integer n > 2, we finally arrive at the estimate
tn1 < .n 1/Še < tn1 C 1;
which is valid for every integer n > 2.
Now, suppose that e D pq , with p; q 2 N. Making n D q C 1 > 2 in the above
inequalities, we would get
tq < .q 1/Šp < tq C 1;
with tq 2 N. This is a contradiction.
t
u
As an alternative to (7.5), we have the following result, which will be quite useful
in Sect. 10.7.
234
7 More on Real Numbers
n
Theorem 7.42 e D limn!C1 1 C 1n .
n
Proof Let an D 1 C 1n . Arguing as in the proof of Example 7.23, we get
!
n
X
n 1
an D 2 C
k nk
kD2
D 2C
n
X
1 n.n 1/.n 2/ : : : .n k C 1/
kŠ
nk
kD2
D 2C
n
X
1
2
k1
1
1
1
::: 1 kŠ
n
n
n
kD2
<2C
n
n
X 1
X
X
1
1
D
<
D e:
kŠ
kŠ
kŠ
kD2
kD0
k0
The above computations also give
n
X
1
2
k1
1
1
1
::: 1 an D 2 C
kŠ
n
n
n
kD2
<2C
nC1
X
1
2
k1
1
1
1
::: 1 D anC1 :
kŠ
nC1
nC1
nC1
kD2
Therefore, .an /n1 is monotone increasing and bounded above, and hence there
exists l D limn!C1 an . In particular, since an < e for every n 2 N, item (b) of
Proposition 7.15 gives l e.
Also from the computations above, given natural numbers n > m 2 we can
write
an D 2 C
n
X
1
2
k1
1
1
1
::: 1 kŠ
n
n
n
kD2
>2C
m
X
1
2
k1
1
1
1
::: 1 kŠ
n
n
n
kD2
m
X
1
2
m1
1
1
1
::: 1 >2C
kŠ
n
n
n
kD2
!
m
X
1
2
m1
1
> 1
1
::: 1 2C
:
n
n
n
kŠ
kD2
7.4 Series of Real Numbers
235
Therefore, it follows from Problem 1, page 218, we have
!
m
X
2
m1
1
1
1
::: 1 2C
l D lim an lim 1 n!C1
n!C1
n
n
n
kŠ
kD2
D2C
m
m
X
X
1
1
D
:
kŠ
kŠ
kD2
kD0
Pm 1
However, since m 2 N was arbitrarily chosen, we conclude that l kD0 kŠ
for every m 2 N. Then, letting m ! C1 and invoking again item (b) of
Proposition 7.15, we finally obtain
m
X 1
X
1
D
D e:
m!C1
kŠ
kŠ
kD0
k0
l lim
t
u
Back to the general development of the theory, the next result is the analogue, for
series, of Proposition 7.18, and teaches us how to operate with convergent series.
P
P
Proposition 7.43 If k1 ak and k1 bk are convergent series and c is a real
number, then:
P
P
P
(a) Pk1 cak converges and k1P
cak D c k1 ak . P
P
(b)
k1 .ak C bk / converges and
k1 .ak C bk / D
k1 ak C
k1 bk .
Proof
P
(a) If sn denotes the n-th
Ppartial sum of the series k1 ak , then the n-th partial
sum of the series
P k1 cak equals csn . Hence, according to item (a) of
Proposition 7.18, k1 cak converges, and
X
cak D lim csn D c lim sn D c
k1
n!C1
n!C1
X
ak :
k1
P
P
(b) If sn and tn are the n-th partial sums of the series
k1 ak and
k1 bk ,
P
respectively, then the n-th partial sum of the series k1 .ak Cbk / equals sn Ctn .
Therefore, item (b) of Proposition 7.18 assures the convergence of this last
series, with
X
X
X
.ak C bk / D lim .sn C tn / D lim sn C lim tn D
ak C
bk :
k1
n!C1
n!C1
n!C1
k1
k1
t
u
236
7 More on Real Numbers
A quick analysis of the arguments presented in Examples 7.39 and 7.41 gives the
following more general result, known as the comparison test for the convergence
of series.
Proposition 7.44 Let .ak /k1 and .bk /k1 be sequences
of positive real numbers,
P
such that
a
b
for
every
k
1.
If
the
series
b
converges,
then so does the
k
k1 k
P k
series k1 ak . Moreover,
X
k1
ak X
bk :
k1
Pn
Pn
Proof Letting sn D
kD1 ak and tn D
kD1 bk , it follows from 0 < ak bk
that 0 < sn tn , for every n 2 N. Since the sequence .tn /n1 converges, it is
bounded. Hence, the sequence .sn /n1 is monotonic and bounded, thus convergent,
by Bolzano-Weierstrass theorem. To what is left to prove, it suffices to make n !
C1 in the inequality sn tn and apply the result of Problem 1, page 218.
t
u
Example
P 7.45 Is there
P a sequence .ak /k1 of positive real numbers such that both
series k1 ak and k1 k21ak converge?
Solution Suppose there is such a series. Then, item (b) of Proposition 7.43, together
with the inequality between the arithmetic and geometric means, would give us
X s
X 1
X
X2
1
1
:
ak C 2
ak C
D
2 ak 2 D
2
k ak
k ak
k ak
k
k1
k1
k1
k1
k1
X
Therefore, by the comparison test for series, the harmonic series would be convergent, which is an absurd.
t
u
The coming example uses the theory of series to give a proof of the uncountability of R.
Example 7.46 Problem 23, 172, shows that the family F of infinite subsets of N is
uncountable. On the other hand, if A D fm1 < m2 < m3 < g is such a P
set, then
the comparison test, together with the convergence of the geometric series j1 21j ,
P
guarantees the convergence of the series k1 2m1k .
Let B D fn1 < n2 < n3 < g be another infinite subset of N. If we show that
X 1
X 1
¤
;
2 mk
2nk
k1
k1
P
then the correspondence A 7! k1 2m1k defines an injection from F into R, and this
guarantees that R is uncountable. (Otherwise, by composing such a function with a
bijection from R to N, we would get an injection from F to N, thus contradicting
the uncountability of F .)
7.4 Series of Real Numbers
237
What is left to do is quite similar to the proof of Example 4.12. Indeed, suppose
we had
X 1
X 1
D
:
2 mk
2nk
k1
k1
Then,
X 1
X 1
X 1
1
1
<
D
D n 1 ;
m
m
n
j
1
k
k
2
2
2
2
21
k1
k1
jn
1
so that m1 n1 . By reversing the roles of the two series, we analogously conclude
that m1 n1 , so that m1 D n1 . Thus,
X 1
X 1
D
;
2 mk
2nk
k2
k2
and a similar reasoning gives m2 D n2 . Finally, by continuing this way, we get
mk D nk for every k 1, so that A D B.
P
Back to the development of the theory, for a series
k1 ak with infinitely
many positive and negative terms the results obtained so far say nothing about its
convergence. We remedy this situation from now on, starting from the following
P
Definition
P 7.47 A series k1 ak is said to be absolutely convergent provided the
series k1 jak j converges.
The usefulness of the concept of absolutely convergent series stems from the
coming proposition, as well as the subsequent example.
Proposition 7.48 Every absolutely convergent series is convergent.
P
Proof Let k1 ak be an absolutely convergent series and, for each integer n 1,
let sn D a1 CP
a2 C C anP
and tn D ja1 j C ja2 j C C jan j be the n-th partial sums
of the series k1 ak and k1 jak j. Given integers m > n 1, we have
jsm sn j D janC1 C anC2 C C am j
janC1 j C janC2 j C C jam j
D tm tn :
Since .tn /n1 converges, it is a Cauchy sequence; hence, given > 0, there exists
n0 2 N such that m > n > n0 ) jtm tn j < . With these and n0 , it follows from
the above inequality that
m > n > n0 ) jsm sn j tm tn < ;
238
7 More on Real Numbers
and .sn /n1 is also a Cauchy sequence. Therefore, Theorem 7.27 guarantees the
convergence of the sequence .sn /n1 , as we wished to show.
u
t
The converse to the previous proposition is not valid, namely, there are convergent series which are not absolutely convergent. The classical example is a direct
application of the coming result, which is known in the mathematical literature as
the Leibniz criterion10 for the convergence of alternate series.
nonincreasing sequence of positive reals
Proposition 7.49 (Leibniz) If .an /n1 is a P
such that an ! 0, then the alternate series k1 .1/k1 ak converges.
Proof For each n 2 N, let sn D a1 Ca2 C Can . Condition a1 a2 a3 > 0
easily gives
s1 s3 s5 s6 s4 s2 :
(7.7)
On the other hand, for each m 2 N we have
js2m1 s2m j D a2m ! 0;
which clearly guarantees, in conjunction with (7.7), that .sn /n1 is a Cauchy
sequence. Therefore, .sn /n1 is convergent, as desired.
t
u
P
k1
Example 7.50 The alternate series k1 .1/k converges, by a simple application
of the former proposition (we shall compute its value in Problem 8, page 484).
Nevertheless, the series formed by the absolute values of its terms (the harmonic
series) diverges.
We now discuss quite a useful criterion for the convergence of series of nonzero
real numbers. It is based on the asymptotic behavior11 of the quotient of neighboring
terms of the series, and is known as the ratio test.
7.51 Let .an /n1 be a sequence of nonzero real numbers, such that
ˇProposition
ˇ
P
ˇ anC1 ˇ
ˇ an ˇ ! l. If l < 1, then the series k1 ak is absolutely convergent; if l > 1, then
P
the series k1 ak is divergent.
P
Proof Let’s prove that the series k1 ak is absolutely convergent if l < 1 (the
proof of its divergence in case l > 1 is completely analogous).
10
After Gottfried Wilhelm Leibniz, German mathematician and philosopher of the XVII century.
Together with Sir Isaac Newton, Leibniz is considered to be one of the creators of the Differential
and Integral Calculus. Up to this day, we still use some of the notations invented by Leibniz more
than 300 years ago.
11
In this context, this expression refers to the behavior of some expression that depends on n 2 N,
when n ! C1.
7.4 Series of Real Numbers
239
Letting l < 1, we can take a real number q such that l < q < 1. The convergence
n
! l assures the existence of n0 2 N such that
janC1 j
jan j
n n0 )
janC1 j
q:
jan j
Hence, for n n0 , we have
jan j D jan0 j
n1
Y
kDn0
jakC1 j
jan0 jqnn0 :
jak j
P
Thus, P
for n n0 , the terms of the series k1 jak j are majorized by those of the
series k1 jan0 jqnn0 , which converges,P
by Propositions 7.43 and 7.38. Therefore,
it follows from
the
comparison
test
that
k1 jak j converges, which is the same as
P
saying that k1 ak is absolutely convergent.
t
u
In the
P notations of the former proposition, we observe that, if l D 1, then the
series k1 ak may converge or diverge. Indeed, for an D 1n we have
albeit the series
while the series
following
P
k
akC1
k
D
! 1;
ak
kC1
1
k1 k
P
diverges; on the other hand, for an D
1
k1 k2
1
n2
we have
k2
akC1
k
D
! 1;
2
ak
.k C 1/
converges. On the positive side of things, we present the
Example 7.52
number m and a real number q > 1, explain whether
P Given a natural
m
the series k1 .1/k1 kqk converges or diverges.
m
Solution Letting an D .1/n1 nqn
ˇ
ˇ
m
ˇ anC1 ˇ
ˇ
ˇ D .n C 1/
ˇ a ˇ
qnC1
n
for n 1, we have
qn
nC1 m 1 n 1
m D
! < 1:
n
n
q
q
Therefore, by the ration test, the given series is absolutely convergent, hence,
convergent.
t
u
We close this section by discussing the product of two absolutely convergent
series.
P
P
Theorem 7.53 Let i1 ai and j1 bj be absolutely convergent series. If
ck D
X
iCjDk
ai bj D
k1
X
iD1
ai bki
(7.8)
240
7 More on Real Numbers
for k 1, then
P
k1 ck
is absolutely convergent and such that
X X
X
ck D
ai
bj :
k1
i1
j1
Proof It suffices to show that, given > 0, there exists n0 2 N such that, for n > n0 ,
we have
2n
ˇX
ˇ 2n1
X X ˇ
X X ˇ
ˇ
ˇ
ˇX
ˇ
ck ai
bj ˇ < and ˇ
ck ai
bj ˇ < :
ˇ
kD1
i1
j1
kD1
i1
j1
Let’s guarantee the existence of n0 2 N for which the first inequality above is true
(the analysis of the validity of the second inequality is entirely analogous).
Given n 2 N, it follows from triangle inequality that
2n
2n X
n
n
ˇX
ˇ
X
X
X X ˇ ˇX
ˇ
ˇ ˇ
ˇ
ck ai
bj ˇ ˇ
ai bj ai
bj ˇ
ˇ
kD1
i1
j1
kD1
iCjDk
n
ˇ X
ˇ
Cˇ
ai
iD1
n
X
iD1
bj jD1
jD1
X
ai
i1
n
X
ˇ
ˇ
bj ˇ
jD1
n
ˇ X X
X X ˇ
ˇ
ˇ
Cˇ
ai
bj ai
bj ˇ:
i1
jD1
i1
j1
Let A, B and C denote the first, second and third summands of the right hand side
above, respectively, so that
n
ˇX ˇ ˇX
ˇ
ˇX ˇ ˇX ˇ
ˇ
ˇ ˇ
ˇ
ˇ
ˇ ˇ
ˇ
BDˇ
ai ˇ ˇ
bj ˇ and C D ˇ
ai ˇ ˇ
bj ˇ:
i>n
The sequence
jD1
j>n
i1
P
n
jD1
bj
, being convergent, is bounded; therefore, there
ˇ P n1 ˇ
ˇ
ˇ
exists M > 0 such that ˇ njD1 bj ˇ < M, for every n 1. On the other hand, since
P
P
the series i1 ai and j1 bj converge, we have
X
i>n
ai D
X
i1
ai n
X
iD1
n
ai !
X
i1
ai X
ai D 0
i1
P
n
and, analogously, j>nPbj ! 0. In order to estimate C, we can suppose, without
loss of generality, that i1 ai ¤ 0. Then, we can choose n1 ; n2 2 N such that
ˇX ˇ
ˇX ˇ
ˇ
ˇ
ˇ
ˇ
ˇI
and n > n2 ) ˇ
ai ˇ <
bj ˇ < ˇP
n > n1 ) ˇ
ˇ
ˇ
3M
3
i1 ai
i>n
j>n
7.4 Series of Real Numbers
241
hence, for n > maxfn1 ; n2 g, we have
ˇX ˇ
ˇ
ˇ
ˇD :
M D
and C < ˇ
ai ˇ ˇP
ˇ
ˇ
3M
3
3
3
a
i1 i
i1
B<
In what concerns A, notice firstly that
2n X
n
n
ˇ
ˇ ˇ X
ˇX
X
X
ˇ
ˇ ˇ
ˇ
ai bj ai
bj ˇ D ˇ
ai bj ˇ
A Dˇ
iCjDk
kD1
iD1
jD1
maxfi;jg>n
iCj2n
X
2n X
X
jai bj j jai jjbj j C
iDnC1 jn
maxfi;jg>n
2n X
X
jai jjbj j
jDnC1 in
iCj2n
2n
X
jai j
X
iDnC1
jbj j C
2n
X
j1
jbj j
X
jDnC1
jai j :
i1
P
P
To estimate A, we can suppose that i1 jai j ¤ 0 and j1 jbj j ¤ 0. Now,
Pn
P
P
since the series i1 jai j and j1 jbj j converge, the sequences
iD1 jai j n1 and
P
n
are Cauchy; therefore, there exist n3 ; n4 2 N such that
jD1 jbj j
n1
n > n3 )
2n
X
jai j <
iDnC1
6
P
j1
jbj j
and n > n4 )
2n
X
jbj j <
jDnC1
6
P
i1
jai j
:
Then, for n > maxfn3 ; n4 g, we get
A
6
P
j1 jbj j
X
j1
jbj j C
6
P
i1 jai j
X
i1
jai j D
:
3
Finally, by letting n0 D maxfn1 ; n2 ; n3 ; n4 g and taking n > n0 , all of the previous
estimates are valid, so that
2n
ˇX
X X ˇ
ˇ
ˇ
ck ai
bj ˇ A C B C C < 3 D :
ˇ
3
kD1
i1
j1
t
u
242
7 More on Real Numbers
Problems: Section 7.4
1. Let .an /n1 be a sequence of positive real numbers defined by a1 D 12 and
P
anC1 D a2n C an , for every n 2 N. Prove that k1 ak 1C1 converges and show
that
X
k1
1
D 2:
ak C 1
2. Sequence
.an /n1 is a nonconstant AP of nonzero real numbers. Prove that
P
1
k1 ak akC1 converges and compute its sum.
P
converges and
3. Given a real number a > 1, prove that the series k1 2k1
ak
compute its sum.
P
1
4. Decide whether the series k1 p p
converges or diverges.
kC
k2 1
P
5. Prove that the series k>1000 p 3 1 2 converges.
k 1000k
6. Let .an /n1 be an infinite, nonconstant AP of positive terms. Prove that:
P
1
diverges.
(a)
Pk1 a1k
(b)
k1 a k converges.
2
7. (NMC) Let A be a finite set of naturals, all of the form 2a 3b 5c , for some
nonnegative integers a, b and c. Prove that
X1
x2A
x
< 4:
8. * The purpose of this problem is to show that e Š 2:71828, with five correct
decimal places. To this end, do the following items:
(a) For every integer n > 10, show that
1
1
1
1
1
1
1
C
CC
<
1C
C 2 C C n10 :
10Š
11Š
nŠ
10Š
11
11
11
(b) Use P
item (a), together with the fact that 10Š > 2 106 , to show that 0 <
1
6
e 10
kD1 kŠ < 10 .
(c) Conclude from (b) that 2:71828 approximates e with five correct decimal
places.
9. * Given a sequence .a1 ; a2 ; a3 ; : : :/ of digits, prove that there exists a single
x 2 R such that, for a given n 2 N, we have
0x
a
1
10
C
a2
1
ak
C C k n;
102
10
10
7.4 Series of Real Numbers
10.
11.
12.
13.
14.
243
for every natural number k n. In such a case (and as the reader is certainly
used to), we write x D 0:a1 a2 a3 : : : and say that 0:a1 a2 a3 : : : is the decimal
representation of x.
Show that every real number x 2 .0; 1/ admits a unique decimal expansion of
the form x D 0:a1 a2 a3 : : :, with an ¤ 0 for infinitely many values of n. Then,
use this fact to construct a surjective function f W Œ0; 1
P! Œ0; 1 Œ0; 1.
Let .an /n1 be a sequence of real numbers such that k1 a2k converges. Prove
P
that, for every rational ˛ > 12 , k1 ka˛k also converges.
P
Let .an /n1 be a sequence
positive real numbers, such that the series k1 ak
P ofp
converges. Prove that k1 ak akC1 also converges.
Let .Fn /n1 be the Fibonacci sequence, i.e., the sequence defined by F1 D 1,
F2 D 1 and FkC2 D FkC1 C Fk , for every integer k 1. Show that the series
P
1
k1 Fk converges.
p
Let .an /n1 be a sequence of positive reals such that n an ! l.
P
(a) If l < 1, show that the series Pk1 ak converges.
(b) If l > 1, show that the series k1 ak diverges. P
(c) If l D 1, give examples showing that the series k1 ak may converge or
diverge.
TheP
convergence criterion given by the case l < 1 is known
as the root test.
P
15. Let k1 ak be an absolutely convergent series, with k1 ak D 0. Show that
X
n1
a2
an1
a1
C
CC 2
2
2
.n 1/
.n 2/
1
converges and compute the corresponding sum.
P
.1/k
1
16. Prove
Pthat e D k0 kŠ .
17. Let k1 ak be an absolutely convergent series and ' W N ! N be a bijection.
P
P
(a) If an 0 for every n 2 N, prove that k1 ak D k1 a'.k/ .
C
(b) Write an D aC
n an , where
P an CD maxfa
P n ; 0g and an D minfan ; 0g, for
every n 2 N. Prove that P
k1 ak and Pk1 ak are both convergent.
(c) Conclude that, in general, k1 ak D k1 a'.k/.
In view of item (c) above, we say that an absolutely convergent series is
commutatively convergent.
The convergence criterion for series stated in the next problem is due to N.
Abel, and is known in the mathematical literature as Abel’s convergence test
or Abel’s convergence criterion.
18. Let .an /n1 and .bn /n1 be two sequences of real numbers satisfying the
following conditions:
(a) The sequence .sn /n1 , defined by sn D a1 C C an for every n 2 N, is
bounded.
244
7 More on Real Numbers
n
(b) b1 b2 b3 > 0 and bn ! 0.
P
Prove that the series k1 ak bk converges.
19. Show that Abel’s criterion implies Leibniz criterion.
20. Do the following items:
(a) Given a; h 2 R, with h ¤ 2l for every l 2 Z, show that
k
X
sin a C
sin.a C jh/ D
.k1/h
2
sin .kC1/h
2
sin h2
jD0
(b) Use Abel’s criterion to show that
P
k1
sin k
k
:
converges.
21. In a cartesian coordinate system centered at O, let .An /n1 be the sequence of
points such that A1 D .1; 0/ and:
(i) Triangle OAn AnC1 is rectangle at An and such that An AnC1 D 1.
(ii) Triangles OAnC1 AnC2 and OAn AnC1 have disjoint interiors, for every n 1.
!
Prove that, when n ! C1, half-line OAn revolves infinitely many times
around O.
22. Let Tn be a right triangle whose side lengths are 4n2 , 4n4 1 and 4n4 C 1, with
n 2 N. Let ˛n be the measure, in radians, of its internal angle opposite to the
side of length 4n2 . Show that
X
k1
˛k D
:
2
Chapter 8
Continuous Functions
In this chapter we formalize the concept of continuous function, intuitively thought
of as that of a function whose graph is a curve without interruptions. As a result of
our discussion, we shall present sufficient criteria for a function to be continuous
and, among other important results, shall show that every continuous function
possesses the intermediate value property. Also, several interesting examples are
scattered throughout the chapter.
8.1 The Concept of Continuity
Let’s initially consider the function f W R ! R given by f .x/ D fxg (the fractional
part function), whose graph is sketched in Fig. 8.1. After a quick look at it, we
would certainly feel comfortable in saying that such a graph is discontinuous, for,
it presents several (actually, infinitely many) jumps. Also, note that this apparently
doesn’t happen with the graph of the function g W R ! R given by g.x/ D x2 , which
should surely be called continuous.
It does emerge the question of how to find a reasonable criterion to identify
the existence or absence of jumps in graphs, thus discerning between the two
possibilities above. In order to develop some intuition on how to do it, let’s restrict
the domain of the function x 7! fxg to the interval Π34 ; 32 . If we denote this new
function still by f , then we easily conclude that
Im.f / D 0;
1
3
[
;1 I
2
4
in particular, there exist values of y in the closed interval bounded by f . 34 / and f . 32 /,
for example y D 58 , such that no x 2 Π34 ; 32 satisfies f .x/ D y.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_8
245
246
8 Continuous Functions
y
...
−4
...
−3
−2
−1
O
1
2
3
x
Fig. 8.1 Graph of x 7! fxg
On the other hand, it is easy to verify that this doesn’t happen with the function g
of the first paragraph. More precisely, let’s fix an interval Œa; b R and let g.Œa; b/
denote the image of the restriction of g to Œa; b, i.e.,
g.Œa; b/ D fg.x/ 2 RI x 2 Œa; bgI
then, by setting c D maxfa; bg, it’s easy to show that
8 2 2
< a ; b ;
g.Œa; b/ D
0; c2 ;
: 2 2
b ;a ;
if 0 a < b
if a < 0 < b :
if a < b 0
In particular, for every y in the closed interval bounded by g.a/ and g.b/, there exists
p
x 2 Œa; b (x D ˙ y, according to the case at hand), such that g.x/ D y.
We are now in position to generalize the previous discussion with the following
definition. To this end, in all that follows, unless we explicitly state otherwise, I
denotes an interval and X a union of intervals of the real line.
Definition 8.1 A function f W X ! R has the intermediate value property (cf.
Fig. 8.2) if, for every interval Œa; b X and every y0 situated in the closed interval
bounded by f .a/ and f .b/, there exists x0 2 Œa; b such that y D f .x/.
The previous discussion assures that the function f .x/ D fxg, x 2 R, does not
have the intermediate value property, while the function g.x/ D x2 , x 2 R, has
such a property. Therefore, it compels us to say that, if a function f W X ! R has
the intermediate value property, then its graph should be continuous, i.e., with no
jumps. Nevertheless, the reader can easily verify that the function f W Œ0; 2 ! R,
given by
f .x/ D
x;
if 0 x 1
;
x 12 ; if 1 < x 2
8.1 The Concept of Continuity
247
y
Gf
f (a)
y = y0
a
x0
x
b
f (b)
Fig. 8.2 The intermediate value property
Gf
y
P0
a
x0
b
x
2δ
Fig. 8.3 Continuity from the correct viewpoint
does satisfy the intermediate value property, albeit its graph presents a jump at
x D 1. Thus, the intermediate value property is not the correct way to formulate
the concept of continuous function.
In order to adequately define the concept of continuous function, let’s analyse the
whole situation from another point of view. Let f W .a; b/ ! R be a given function,
x0 2 .a; b/ be a fixed real number and P0 .x0 ; f .x0 // 2 Gf . Also, let x 2 .a; b/ n fx0 g
and P.x; f .x// 2 Gf . For the graph of f to be called continuous at x0 , our geometric
intuition says that, the closer x is of x0 , the closer P should be of P0 .
More precisely, this closedness means (cf. Fig. 8.3) that, for a fixed, arbitrarily
chosen error r > 0 for the position of the point P0 (i.e., given an arbitrary disk
D.P0 I r/, centered at P0 and with radius r), we should have P 2 D whenever the
abscissa x of P approximates x0 within a sufficiently small error, say less than a
certain ı > 0 (one reads delta). In symbols, given an arbitrary r > 0, there must
exist ı > 0 such that
jx x0 j < ı ) PP0 < r:
(8.1)
248
8 Continuous Functions
y
Gf
P0
a
2
x0
b
x
2δ
Fig. 8.4 Elaborating the correct definition of continuous function
Since every disk in the cartesian plane contains arbitrarily small rectangles with
sides parallel to the axis, it is easy to show (cf. Problem 4) that the validity of the
condition (8.1) is equivalent to the following alternative geometric description (cf.
Fig. 8.4): given an arbitrary error > 0 (one reads epsilon) for the value of f at x0 ,
i.e., given an arbitrary horizontal strip
R
.f .x0 / ; f .x0 / C /
of the cartesian plane, symmetric with respect to the point P0 .x0 ; f .x0 // (the gray
strip in Fig. 8.4)), there must exist another error ı > 0 such that, for every x 2 X
satisfying jx x0 j < ı, the point P.x; f .x// belongs to the gray strip. This being said,
we can finally state the formal definition of continuity of a function (at a point).
Definition 8.2 A function f W X ! R is continuous at the point x0 2 X if the
following condition is satisfied: given > 0, there exists ı > 0 such that
x 2 X; jx x0 j < ı ) jf .x/ f .x0 /j < :
(8.2)
The function f is said to be continuous if it is continuous at every point x0 2 X.
Example 8.3 Every constant function is continuous.
Proof Let c be a given real number and f W R ! R be the function constant and
equal to c. To show that f is continuous at x0 2 R, the definition of continuity asks
that, given > 0, we find ı > 0 such that, for x 2 R, the validity of inequality
jx x0 j < ı implies that of jf .x/ f .x0 /j < . However, since jf .x/ f .x0 /j D
jc cj D 0, inequality jf .x/ f .x0 /j < is always true, independently of any
restriction on jxx0 j. Yet in another way, taking ı equal to any positive real number,
we will always have that jx x0 j < ı ) jf .x/ f .x0 /j < , for, the inequality
jf .x/ f .x0 /j < cannot be false.
t
u
8.1 The Concept of Continuity
249
In order to establish the continuity of other, less simple functions, we need to
elaborate a little more the former definition, and we do this next.
Suppose we want to establish the continuity of f W X ! R at x0 2 X. According
to the given definition, we should assume that an error > 0 for the value of f .x0 /
is given and, then, be capable of finding another error ı > 0 for x0 that turns true
implication (8.2). In general, the following strategy is a good one:
(i) Starting from x 2 X subjected to an error jx x0 j < ı, we estimate by excess
the error jf .x/ f .x0 /j in terms of ı, obtaining an inequality of the type jf .x/ f .x0 /j < E.ı/, where E represents a certain function of ı.
(ii) Then, we impose that such an error E.ı/ doesn’t surpass the desired error ,
thus finding the appropriate values of ı. Usually, this second step reduces to
solving, for ı > 0, inequality E.ı/ .
Once we execute the two steps above, if ı > 0 satisfies E.ı/ , we will clearly
have that
x 2 X and jx x0 j < ı ) jf .x/ f .x0 /j < E.ı/ ;
as we wished to show.
As a last remark, the coming examples will make it clear that, along the execution
of items (i) and (ii), the function E will be generally implied.
Example 8.4 The modular function (cf. Example 6.60) is continuous.
Proof Given x0 2 R, if jx x0 j < ı, then triangle inequality gives
jf .x/ f .x0 /j D jjxj jx0 jj jx x0 j < ı:
Hence, if ı , it will follow from the above that jf .x/ f .x0 /j < whenever
jx x0 j < ı.
t
u
Example 8.5 Let a and b be real numbers, with a ¤ 0. If f W R ! R is given by
f .x/ D ax C b, then f is continuous.
Proof Again, for a given x0 2 R, se jx x0 j < ı, we have
jf .x/ f .x0 /j D j.ax C b/ .ax0 C b/j D jajjx x0 j < jajı:
),
jaj
we have jf .x/ f .x0 /j < whenever
t
u
p
Example 8.6 The square root function f W Œ0; C1/ ! R, such that f .x/ D x for
x 0, is continuous.
p
p
p
Proof If x0 D 0 and 0 x < ı, then jf .x/ f .x0 /j D x < ı. Since ı ,
ı 2 , for 0 < ı 2 we have jf .x/ f .x0 /j < if jx 0j < ı.
Suppose, now, that x0 > 0. Then, for jx x0 j < ı, we have
Hence, if jajı (or, equivalently, ı <
jx x0 j < ı.
p
p
jx x0 j
1
ı
jf .x/ f .x0 /j D j x x0 j D p
p p jx x0 j < p :
x0
x0
x C x0
250
8 Continuous Functions
Fig. 8.5 j sin xj jxj
B
P
x
A
A
O
Q
P
B
p
p
Note that pıx0 , ı < x0 . Therefore, taking 0 < ı x0 , we conclude
that jx x0 j < ı ) jf .x/ f .x0 /j < , as wished.
t
u
In what comes next, we establish the continuity of the sine and cosine functions.
To this end, we shall need the following auxiliary result.
Lemma 8.7 For every x 2 R, we have j sin xj jxj.
Proof Firstly, let’s show that sin x x whenever 0 x 2 . This inequality is
obvious for x D 0; for 0 < x 2 , mark (cf. Fig. 8.5) point P in the first quadrant of
_
the unit circle such that `.AP/ D x.
!
Letting P0 be the symmetric of P with respect to AA0 and Q the intersection of
_
_
P0 P and A0 A, we have P0 P D 2 QP and `. P0 P/ D 2`. AP/. Since the length of every
arc of a circle is greater than that of the corresponding chord, we get
_
_
2 sin x D PP0 < `.PP0 / D 2`.AP/ D 2x:
Now, since sin.x/ D sin x, it’s immediate that j sin xj jxj for jxj Finally, for jxj > 2 , we have
j sin xj 1 <
.
2
< jxj:
2
t
u
Example 8.8 Sine and cosine functions are continuous.
Proof Fix x0 2 R. If jx x0 j < ı, it follows from the sum-to-product identities,
together with the previous lemma, that
8.1 The Concept of Continuity
251
ˇ x x
0
ˇ
j cos x cos x0 j D 2 ˇsin
2
ˇ x x
ˇ
0
2 ˇsin
2
ˇ ˇˇ x C x ˇˇ
0 ˇ
ˇˇ
ˇ ˇsin
ˇ
2
ˇ
ˇx x ˇ
ˇ
ˇ
0ˇ
ˇ 2ˇ
ˇ
2
D jx x0 j < ı:
Hence, if ı , then j cos x cos x0 j < whenever jx x0 j < ı, as wished.
Finally, the reasoning for the sine function is completely analogous.
t
u
Later (cf. Proposition 8.19), we shall prove that if I is an interval and f ; g W I ! R
are continuous at x0 2 I, then the functions f ˙ g; f g W I ! R are also continuous
at x0 . On the other hand, we shall show in Proposition 8.21 that, if g doesn’t vanish
in I, then the function gf W I ! R is also continuous at x0 . For the time being, we
assume these facts without proof, using them to present some additional examples
of continuous functions.
Example 8.9 Given a 2 R and k 2 N, the function x 7! axk (from R to itself) can
be seen as the product of a constant function (equal to a) and k copies of the identity
function, x 7! x. Therefore, Examples 8.3 and 8.5, together with the discussion at
the last paragraph (extended to the product of k continuous functions), assure that
the function x 7! axk is also continuous.
Applying once more the discussion at the paragraph that precedes this example
(this time extended to the sum of a finite number of continuous functions), we
conclude that every polynomial function, i.e., every function f W R ! R of the
form
f .x/ D an xn C an1 xn1 C C a1 x C a0 ;
for some n 2 N and a0 ; a1 ; : : : ; an 2 R, is continuous too.
Example 8.10 Given a polynomial function f W R ! R (see the preceding
example), we say that x0 2 R is a root of f if f .x0 / D 0. It is possible to prove
(cf. Chap. 15 of [5], for instance), that every polynomial function has a finite
(possibly zero) number of real roots. Hence, if f and g are polynomial functions
and Rg D fx1 < x2 < < xk g is the set of real roots of g, the complement
R n Rg D .1; x1 / [ .x1 ; x2 / [ : : : [ .xk1 ; xk / [ .xk ; C1/
is a finite union of open intervals. This way, we get the (well defined) rational
function
f
W R n Rg ! R:
g
The previous example, together with the discussion at the paragraph that precedes
it, assures the continuity of rational functions, wherever they are defined.
252
8 Continuous Functions
Example 8.11 Function f W R ! R, given by
f .x/ D
x4 sin x .x2 C 1/ cos x
;
x8 C 2 C cos x
is continuous. Indeed, f is the quotient of the functions g; h W R ! R, such that
g.x/ D x4 sin x .x2 C 1/ cos x and h.x/ D x8 C 2 C cos x. The previous examples
and discussions guarantee the continuity of g and h, and function h doesn’t vanish,
for x8 0 and 2 C cos x > 0.
We close this section by examining the continuity of a composition. In this sense,
the coming result is known as the chain rule for continuous functions.
Proposition 8.12 If X; Y R are unions of intervals and f W X ! Y, g W Y ! R
are continuous functions, then g ı f W X ! R is also continuous.
Proof Let x0 2 X and y0 D f .x0 /. Given > 0, the continuity of g guarantees the
existence of ı > 0 such that
y 2 Y; jy y0 j < ı ) jg.y/ g.y0 /j < :
(8.3)
On the other hand, the continuity of f assures the existence of ı 0 > 0 such that
x 2 X; jx x0 j < ı 0 ) j f .x/ f .x0 / j < ı:
„ƒ‚… „ƒ‚…
y
y0
Therefore, it follows from the above relations (with y D f .x/ in (8.3)) that
x 2 X; jx x0 j < ı 0 ) jf .x/ f .x0 /j < ı ) jg.f .x// g.f .x0 //j < ;
thus establishing the continuity of g ı f at x0 .
t
u
The two coming examples show typical applications of the chain rule for
continuous functions.
Example 8.13 Function f W R ! R, given by f .x/ D sin.x2 /, is continuous. Indeed,
if g; h W R ! R are given by g.x/ D sin x and h.x/ D x2 , then g and h are continuous
and f D g ı h; hence, the chain rule assures the continuity of f .
Example 8.14 Function f W R ! R, given by
s
f .x/ D
x4
x2 j sin.x2 /j C 1
;
p
C x2 C 2 C j sin xj
is continuous. Indeed, since f is the composition of the square root function (which
x2 j sin.x2 /jC1
p
is continuous) with the function g W R ! R given by g.x/ D x4 Cx
2 C2C j sin xj ,
the chain rule assures that it suffices to establish the continuity of g. Now, g is the
8.1 The Concept of Continuity
253
p
quotient of the functions x 7! x2 j sin.x2 /j C 1 and x 7! x4 C x2 C 2 C j sin xj,
which are continuous by the previous examples, together with the chain rule. Finally,
observe that both the numerator and denominator of g.x/ are positive.
Let us finish this section with a result on the set of points of discontinuity of
monotonic functions.
Proposition 8.15 If I R be an interval and f W I ! R is an nondecreasing
function. then the set of points of discontinuity of f is finite or countably infinite.
Proof Let
D D fx0 2 II x0 is not an endpoint of I and f is discontinuous at x0 g:
For each x0 2 D, let mx0 D supff .x/I x 2 I and x < x0 g and Mx0 D infff .x/I x 2
I and x > x0 g. Problem 13 guarantees that mx0 < M
x0 . Now, Problem 4, page 206
guarantees that we can choose rx0 2 Q \ mx0 ; Mx0 . On the other hand, if x0 < y0
are both in D, the fact that f is nondecreasing gives
x C y
0
0
my 0 ;
Mx0 f
2
so that the intervals mx0 ; Mx0 and my0 ; My0 are disjoint.
Hence, the correspondence x0 7! rx0 defines an injective function from D into Q,
and it follows from Problem 21, page 172, that D is finite or countably infinite. u
t
In the notations of the statement of the previous proposition, we call the reader’s
attention to the fact that Problem 14 will guarantee that every countably infinite
subset of I is the set of points of discontinuity of a nonincreasing function.
Problems: Section 8.1
1. Prove that:
(a) The function f W R n f0g ! R, given by f .x/ D 1x , has the intermediate
value property and is continuous, but the function f W R ! R given by
1
; if x ¤ 0
f .x/ D x
0; if x D 0
doesn’t have such property.
(b) The function f W R ! R, given by
f .x/ D
x2 C 1; if x 0
x;
if x < 0
doesn’t have the intermediate value property.
254
8 Continuous Functions
2. Let I R be an interval and f W I ! R be a continuous function. For a fixed
x0 2 I, let g W I ! R be the function given by
g.x/ D
c; if x D x0
:
f .x/; if x ¤ x0
Show that g is continuous if and only if c D f .0/, i.e., if and only if g D f .
3. In each of the following items, find out whether there exists a real value of c
that turns the function f W R ! R into a continuous one. Justify your answers.
8
< 3x 2; if x < 0
(a) f .x/ D c; if x D 0
.
:
2; if x > 0
x cos x; if x ¤ 0
(b) f .x/ D
.
c; if x D 0
8
< 3x 2; if x < 0
(c) f .x/ D c; if x D 0
.
: 2
x ; if x > 0
1
; if x ¤ 0
.
(d) f .x/ D x
c; if x D 0
4. * Prove that every function continuous in the sense of relation (8.1) is
continuous in the sense of Definition 8.2, and vice-versa.
5. * Let D D R n f=2 C kI k 2 Zg. Use the discussion at the paragraph that
precedes Example 8.9 to establish the continuity of the tangent function,
tan W D ! R
:
x 7! tan x
6. If f W X ! R is a continuous function, explain why the function jf j W X ! R is
also continuous.
7. Use the results of this section to establish
the continuity of the function f W
p
xC1
Œ1; C1/ ! R, given by f .x/ D cosx3 C2
.
8. Let n >p1 be a natural number and f W Œ0; C1/ ! R be the n-th root function,
f .x/ D n x for x 0.
(a) Show that f is continuous at x0 D 0.
(b) If x0 > 0 and x 0, show that
jx x0 j
jf .x/ f .x0 /j D p
p
p
p
n
n
n
n1
n2
. x/
C . x/
x0 C C . n x0 /n1
1
< p
jx x0 j:
. n x0 /n1
(c) Use the result of (b) to conclude that f is continuous at x0 .
9. Use the results of this
qsection to establish the continuity of the function f W R !
R, given by f .x/ D
3
x4 2x3 C1
.
x2 C1
8.1 The Concept of Continuity
255
10. Justify the continuity of the function f W R ! R, given by
f .x/ D
0; if x D 0
:
x sin 1x ; if x ¤ 0
11. * The Dirichlet function1 is the function f W Œ0; 1 ! R such that
f .x/ D
0 if x … Q
:
1 if x 2 Q
Prove that f is discontinuous at every x 2 Œ0; 1.
12. Let f W Œ0; 1 ! R be the function given by
f .x/ D
0; if x … Q
:
1
; if x D mn with m 2 ZC ; n 2 N and gcd.m; n/ D 1
n
Prove that f is discontinuous at every rational number and continuous at every
irrational number2 of the interval Œ0; 1.
13. Let I R be an interval and f W I ! R be a nondecreasing function. If x0 2 I
is not an endpoint of I, then
supff .x/I x 2 I and x < x0 g infff .x/I x 2 I and x > x0 g;
with equality if and only if f is continuous at x0 .
14. Let D D fx1 ; x2 ; x3 ; : : :g be an arbitrary countably infinite set of reals. Let f W
R ! R be given by
f .x/ D
X 1
;
2n
n2NI
xn <x
where the sum is interpreted as being equal to 0 if fn 2 NI xn < xg D ;. Prove
that:
(a) f is well defined and nondecreasing.
(b) The set of points of discontinuity of f is D.
1
After Gustav Lejeune Dirichlet, German mathematician of the XIX century. Dirichlet made several important contributions to Mathematics, notably to Analysis and Number Theory. Even though
the most famous of these is perhaps Dirichlet’s theorem for primes in arithmetic progressions
(about which we shall have more to say on [5]), Dirichlet made very important contributions to
the theory of Fourier Series, which we will tangentially touch upon in Chap. 11.
2
It is possible to prove that, given an interval I, there doesn’t exist a function f W I ! R continuous
at (every point of) I \ Q and discontinuous at I \ .R n Q/. A proof can be found in [1].
256
8 Continuous Functions
8.2 Sequential Continuity
In this section we relate the concepts of limit of a convergent sequence and
continuity of a function. The main result is the coming one, which will be used
several times in all that follows and assures that continuous functions are exactly
those that preserve convergent sequences.
In all that follows, I denotes an interval of the real line.
Theorem 8.16 A function f W I ! R is continuous if and only if the following
condition is satisfied: for every a 2 I and every sequence .an /n1 of elements of I,
we have
lim an D a ) lim f .an / D f .a/:
n!C1
n!C1
Proof Let’s first assume that f is continuous. Then, given a 2 I and > 0, there
exists ı > 0 such that
x 2 I; jx aj < ı ) jf .x/ f .a/j < :
Now, let .an /n1 be a sequence of elements of I converging to a. Then, there exists
n0 2 N such that
n > n0 ) jan aj < ı:
Taking together the two conditions above, we get
n > n0 ) jan aj < ı ) jf .an / f .a/j < ;
which is the same as saying that limn!C1 f .an / D f .a/.
Conversely, suppose that f is not continuous at a 2 I. Then, the definition of
continuity guarantees the existence of > 0 such that, for every ı > 0, we have
jf .x/ f .a/j for some xı 2 I satisfying jxı aj < ı. In particular, taking ı D 1n ,
with n 2 N, and writing an D x1=n , we get
jan aj <
1
n
and jf .an / f .a/j :
Then, the above conditions assure that the sequence .an /n1 thus constructed
converges to a, whereas the sequence .f .an //n1 doesn’t converge to f .a/.
t
u
Let us illustrate the importance of the previous result with two examples, the
first of which revisits Example 7.19 and shows that the idea used in the proof of
Example 7.23 is quite natural.
8.2 Sequential Continuity
257
Example 8.17 If a > 0 and an D
p
n
n
a, then an ! 1.
Proof We make the proof in the case a 1, the proof in the case 0 < a < 1 being
completely analogous. If a 1, then an anC1 1, for every n 2 N. Hence, the
sequence .an /n1 is monotonic and bounded, thus convergent.
Setting l D limn!C1 an , it follows from item (a) of Proposition 7.15 that
l 1. On the other hand, the continuity of the square root function (according
to Example 8.6) and the previous result guarantee that
p
n
n p
n p
an ! l ) an ! l ) a2n ! l;
p
where in the last implication we used the fact that an D a2n .
Since every subsequence of a convergent sequence converges to the same limit
n
of the whole sequence, we havepa2n ! l. Therefore, the uniqueness of limits of
convergent sequences gives l D l, so that l D 1.
t
u
Example 8.18 (Sweden) Find all continuous functions f W R ! R such that f .x/ C
f .x2 / D 0, for every x 2 R.
Solution If f is any function satisfying the stated conditions, we shall show that f
is identically zero.
Making x D 0 and x D 1, we get f .0/ D 0 and f .1/ D 0. Now, for x > 0 and
applying the stated condition several times, we get
p
p
p
f .x/ D f . x/ D f . 4 x/ D f . 8 x/ D :
p
n
In particular, jf . 2 x/j D jf .x/j for every n 2 N.
p
n
n
Letting an D 2 x, we already know that an ! 1. On the other hand, since jf j
is also continuous, Theorem 8.16 guarantees that
n
jf .an /j ! jf .1/j D 0:
n
However, since jf .an /j D jf .x/j for every natural n, we also have jf .an /j ! jf .x/j.
Therefore, the uniqueness of the limit of a convergent sequence gives jf .x/j D 0.
Finally, for x < 0 we have x2 > 0. Hence, by what we’ve done above, f .x2 / D 0,
so that f .x/ D f .x/ C f .x2 / D 0.
t
u
Theorem 8.16 allows us to easily prove the claims made at the paragraph that
precedes Example 8.9. We start by establishing the continuity of the sum and
product of two continuous functions.
Proposition 8.19 If f ; g W I ! R are continuous at x0 2 I, then f ˙ g; f g W I ! R
are also continuous at x0 .
Proof Let .an /n1 be a sequence in I, such that limn!C1 an D x0 . According to
Theorem 8.16, the continuity of f and g at x0 gives
lim f .an / D f .x0 / and
n!C1
lim g.an / D g.x0 /:
n!C1
258
8 Continuous Functions
It thus follows from item (b) of Proposition 7.18 that
lim .f ˙ g/.an / D lim .f .an / ˙ g.an //
n!C1
n!C1
D lim f .an / ˙ lim g.an /
n!C1
n!C1
D f .x0 / ˙ g.x0 / D .f ˙ g/.x0 /:
Hence, once more from Theorem 8.16, we conclude that f ˙ g are continuous at x0 .
The reasoning for f g uses item (c) of Proposition 7.18 and is totally similar to
what we did above, so that it will be left to the reader.
t
u
Before we look at the continuity of the quotient of two continuous functions,
we need an auxiliary result, known as the sign-preserving lemma for continuous
functions and which is important in itself.
Lemma 8.20 Let I R be an interval and f W I ! R be a continuous function. If
x0 2 I is such that f .x0 / > 0 (resp. f .x0 / < 0), then there exists ı > 0 such that
x 2 I; jx x0 j < ı ) f .x/ >
f .x0 /
f .x0 /
.resp. f .x/ < /:
2
2
In particular, f is still positive (resp. negative) in I \ .x0 ı; x0 C ı/.
Proof Let’s do the proof in the case f .x0 / > 0, the proof in the other case being
totally analogous.
The definition of continuity guarantees that, for D f .x20 / > 0, there exists ı > 0
such that
x 2 I; jx x0 j < ı ) jf .x/ f .x0 /j < D
f .x0 /
:
2
On the other hand, this inequality implies
f .x/ f .x0 / > or, which is the same, f .x/ >
f .x0 /
,
2
f .x0 /
2
for every x 2 I \ .x0 ı; x0 C ı/.
t
u
If g W I ! R is continuous at x0 2 I and such that g.x0 / ¤ 0, the sign-preserving
lemma assures the existence of ı > 0 such that g doesn’t vanish in the open interval
J D I \ .x0 ı; x0 C ı/. Hence, by restricting g to J, if necessary, we can suppose,
without loss of generality, that g doesn’t vanish in I.
Proposition 8.21 Let f ; g W I ! R be continuous at x0 2 I. If g doesn’t vanish in I,
then gf W I ! R is continuous at x0 .
Proof As in the proof of the previous proposition, if .an /n1 is a sequence in I, such
that limn!C1 an D x0 , then
lim f .an / D f .x0 / and
n!C1
lim g.an / D g.x0 /:
n!C1
8.2 Sequential Continuity
259
Now, item (e) of Proposition 7.18 gives
limn!C1 f .an /
f .x0 /
f
f
f .an /
.an / D lim
D
D
D .x0 /:
n!C1 g
n!C1 g.an /
limn!C1 g.an /
g.x0 /
g
lim
Invoking Theorem 8.16 once more, we conclude that
f
g
is continuous at x0 .
t
u
In what comes next, we introduce a stronger notion of continuity, and show that
it is equivalent to the usual one for functions defined in intervals I D Œa; b R.
Definition 8.22 A function f W I ! R is uniformly continuous if the following
condition is satisfied: for every > 0, there exists ı > 0 such that
x; y 2 I; jx yj < ı ) jf .x/ f .y/j < :
(8.4)
In words, the difference between the definitions of continuous and uniformly
continuous functions is that, for a uniformly continuous function and given > 0,
the number ı > 0 whose existence is assured by the definition depends only on
the chosen error > 0, being the same for all x; y 2 I; on the other hand, in the
definition of continuity of a function at a point x0 of its domain, such a ı depends
on the error > 0 as well as on x0 .
It is clear from the former definition that every uniformly continuous function
is, in particular, continuous: it suffices to make y D x0 , where x0 2 I is arbitrarily
chosen. On the other hand, Problems 4 and 6 bring examples of continuous functions
which are not uniformly continuous.
We now pause to give an example of a uniformly continuous function which will
be useful for later purposes (cf. Example 11.22). To this end, for x 2 R we let d.x/
denote the distance of x to the nearest integer, so that
d.x/ D
fxg; if 0 fxg 12
:
1 fxg; if 12 fxg < 1
Example 8.23 In ˇthe above notations,
let a > 0 be given. If x; y 2 R are such that
ˇ
jx yj < 1a , then ˇd.ax/ d.ay/ˇ ajx yj. In particular, x 7! d.ax/ is uniformly
continuous.
Proof If jx yj < 1a , then jax ayj < 1. In this case, the integers closest to ax
and ay, if not equal, are consecutive. Assume this is so, and let n and n C 1 be the
integers closest to ax and ay, respectively. Then,
n ax < n C
1
< ay < n C 1;
2
260
8 Continuous Functions
so that
ˇ
ˇ
ˇ
ˇ ˇ
ˇ
ˇd.ax/ d.ay/ˇ D ˇ.n C 1 ay/ .ax n/ˇ D ˇ2n C 1 ay axˇ
ˇ
ˇ
1
1
ˇ
ˇ
D ˇ n C ay C n C ax ˇ
2
2
ˇ
ˇ ˇ
ˇ
1
1
ˇ
ˇ ˇ
ˇ
ˇn C ayˇ C ˇn C axˇ
2
2
D jay axj D ajx yj:
For the last part, given > 0 take ı D minf a1 ; a g > 0. Then,
jx yj < ı ) jx yj <
ˇ
ˇ
1
) ˇd.ax/ d.ay/ˇ ajx yj < a D :
a
a
t
u
The coming result proves that, for a function defined on a closed and bounded
interval, the concepts of continuity and uniform continuity coincide.
Theorem 8.24 Every continuous function f W Œa; b ! R is uniformly continuous.
Proof By contradiction, suppose that f is continuous but (8.4) is not valid. Then,
there exists > 0 such that, for every ı > 0, we can find x; y 2 Œa; b satisfying
jx yj < ı and jf .x/ f .y/j :
In particular, choosing ı D 1n , with n 2 N, we conclude that there exist elements
xn ; yn 2 Œa; b such that
jxn yn j <
1
n
and jf .xn / f .yn /j :
(8.5)
Since .xn /n1 is a sequence in Œa; b, Weierstrass Theorem 7.25 guarantees the
existence of a subsequence .xnk /k1 of .xn /n1 , converging to some x0 2 Œa; b.
Then, it follows from triangle inequality that
jynk x0 j jynk xnk j C jxnk x0 j <
1
k
C jxnk x0 j ! 0:
nk
Hence, .ynk /k1 also converges to x0 , and Theorem 8.16, together with the continuity
of x 7! jxj, gives
lim jf .xnk / f .ynk /j D jf .x0 / f .x0 /j D 0:
k!C1
8.2 Sequential Continuity
261
On the other hand, as a particular case of (8.5), we have jf .xnk / f .ynk /j ,
and such an inequality obviously contradicts the former limit.
u
t
The previous theorem admits the following important consequence.
Corollary 8.25 Every continuous function f W Œa; b ! R is bounded.
Proof Since such an f is uniformly continuous, given D 1 there exists ı > 0
such that
x; y 2 Œa; b; jx yj < ı ) jf .x/ f .y/j < 1:
(8.6)
Now, choose real numbers a D x0 < x1 < < xk D b satisfying xi xi1 < ı
for 1 i k, and let
M D maxfjf .x1 /j; : : : ; jf .xk /jg:
(8.7)
For x 2 Œa; b, there exists 1 j k such that x 2 Œxj1 ; xj ; in particular, jx xj1 j jxj xj1 j < ı. Then, the triangle inequality, together with (8.6) and (8.7),
gives
jf .x/j jf .x/ f .xj /j C jf .xj /j < 1 C jf .xj /j 1 C M:
Finally, since x 2 Œa; b as in the last paragraph was arbitrarily chosen, we get
jf .x/j M C 1 for every x 2 Œa; b, so that f is bounded.
u
t
The last property of continuous functions we shall study in this section is also
due to Weierstrass, and states that every continuous function f W Œa; b ! R assumes
extreme values in the interval Œa; b.
Theorem 8.26 (Weierstrass) If f W Œa; b ! R is continuous, then there exist
xm ; xM 2 Œa; b such that
f .xm / D minff .x/I x 2 Œa; bg and f .xM / D maxff .x/I x 2 Œa; bg:
Proof Let’s prove the existence of xm , that of xM being totally analogous. Since f is
bounded, it makes sense to let
m D infff .x/I x 2 Œa; bg:
The definition of greatest lower bound assures (with the aid of Problem 10,
page 207) the existence of a sequence .xn /n1 in Œa; b such that f .xn / ! m.
Since every bounded sequence possesses a convergent subsequence, we can take
a subsequence .xnk /k1 of .xn /n1 , converging to a real number x0 2 Œa; b. Then,
Theorem 8.16 gives
lim f .xnk / D f .x0 /:
k!C1
262
8 Continuous Functions
On the other hand, since the sequence .f .xnk //k1 is a subsequence of .f .xn //n1
(which, in turn, converges to m), we conclude from Proposition 7.15, that .f .xnk //k1
also converges to m. Therefore, the uniqueness of the limit of sequences gives
f .x0 / D m.
t
u
Problems: Section 8.2
1. Find all continuous3 functions f W R ! R such that, for all x; y 2 R, we have
f .x C y/ D f .x/ C f .y/.
2. (Romania) Let f W R ! R be a surjective function satisfying the following
property: for every divergent sequence .an /n1 , the sequence .f .an //n1 is also
divergent. Prove that f is bijective and that f 1 is continuous.
3. * Given an interval I R, we say that a function f W I ! R is lipschitzian4 if
there exists a constant c > 0 (called the Lipschitz constant of f ) such that
jf .x/ f .y/j cjx yj;
for all x; y 2 I. For example, we saw in Example 8.8 that
j sin x sin yj jx yj
4.
5.
6.
7.
and j cos x cos yj jx yj
for all x; y 2 R, so that both sine and cosine functions are lipschitzian, with
Lipschitz constant equal to 1. Prove that every lipschitzian function f W I ! R
is uniformly continuous.
For an integer n > 1, show that the function f W R ! R, given by f .x/ D xn , is
not uniformly continuous.
Let I R be an interval and f W I ! R be a continuous function. Suppose
that there exist > 0 and sequences .an /n1 and .bn /n1 in I, such that jan n
bn j ! 0, but jf .an / f .bn /j for every n 1. Show that f is not uniformly
continuous.
Show that the function f W .0; C1/ ! R, given by f .x/ D sin 1x , is not
uniformly continuous.
The purpose of this problem is to present another proof of Theorem 8.26. To
this end, recall that Corollary 8.25 assures the boundedness of a continuous
function f W Œa; b ! R, so that there exist
m D infff .x/I x 2 Œa; bg and M D supff .x/I x 2 Œa; bg:
3
It can be shown that the hypothesis of continuity is essential here. More precisely, there exists
infinitely many functions f W R ! R such that f .x C y/ D f .x/ C f .y/ for all x; y 2 R but f is not
linear. A proof of this fact is beyond the scope of these notes and can be found in [13].
4
After Rudolf Lipschitz, German mathematician of the XIX century.
8.2 Sequential Continuity
263
Now, do the following items:
1
, for
(a) If m … Im.f / and g W Œa; b ! R is the function given by g.x/ D f .x/m
every x 2 Œa; b, show that there exist c > 0 such that g.x/ c, for every
x 2 Œa; b.
(b) Yet under the hypothesis of item (a), conclude that f .x/ m C 1c for every
x 2 Œa; b and arrive at a contradiction.
(c) Argue in an analogous way to show that M 2 Im.f /.
8. Given real numbers a < b, give an example of a continuous function f W
.a; b/ ! R, which is unbounded from above and from below.
9. If f W .a; b/ ! R is a continuous function and P.x0 ; y0 / is a point not belonging
to the graph of f , we define the distance from P to Gf , denoted d.PI Gf /, by
d.PI Gf / D inff A0 PI A0 2 Gf g:
If jy0 f .x0 /j < minfjx0 aj; jx0 bjg, prove that there exists A 2 Gf such that
d.PI Gf / D AP.
10. * Let f W R ! R be the polynomial function given by
f .x/ D an xn C an1 xn1 C C a1 x C a0
for x 2 R, where a0 ; a1 ; : : : ; an 2 R and an ¤ 0. If n is even and an > 0, prove
that:
(a)
f .x/a0
xn
f .x/a0
xn
an jan1 j
ja1 j
jxj jxjn1 if x
1 Pn1
jD1 jaj j if jxj 1.
jxj
¤ 0.
an (b)
(c) There exists A > 0 such that f .x/ > a0 for jxj > A.
(d) There exists x0 2 ŒA; A such that f .x0 / D minff .x/I x 2 Rg.
11. * The purpose of this problem is to prove the famous Banach fixed point
theorem5 in the real line. To this end, let 0 < c < 1 and f W R ! R be a
function such that
jf .x/ f .y/j cjx yj; 8 x; y 2 R:
(8.8)
Do the following items:
(a) Choose an arbitrary x0 2 R and let .xn /n1 be such that xk D f .xk1 /, for
every k 2 N. Prove that jxkC1 xk j cjxk xk1 j for every k 2 N.
(b) Use the result of Example 7.28 to conclude that .xn /n1 is convergent.
(c) If ˛ D limn!C1 xn , use the continuity of f (which is guaranteed by
Problem 3) to show that ˛ is the unique fixed point of f .
5
After Stefan Banach, Polish mathematician of the XX century.
264
8 Continuous Functions
12. Give an example of a continuous function f W R ! R, without fixed points and
such that jf .x/ f .y/j < jx yj, for all x; y 2 R.
13. * Let f W Œ0; 1 ! R be a continuous function such that f .r/ 0 for every
dyadic rational (cf. Problem 5, page 206) r 2 Œ0; 1. Prove that f .x/ 0, for
every x 2 Œ0; 1.
14. (Berkeley) Let ˛ be a given irrational number. Find all continuous functions
f W R ! R such that
f .x/ D f .x C 1/ D f .x C ˛/;
for every x 2 R.
8.3 The Intermediate Value Theorem
In this section, we show that a continuous function defined on an interval does
satisfy the intermediate value property. Then, we present several interesting applications of this fact. We start by analysing the following special case, known as
Bolzano’s theorem.
Theorem 8.27 (Bolzano) Let f W Œa; b ! R be a continuous function. If
f .a/f .b/ < 0, then there exists c 2 .a; b/ such that f .c/ D 0.
Proof Suppose, without loss of generality, that f .a/ < 0 < f .b/, and let
A D fx 2 Œa; bI f is negative on Œa; xg:
Since a 2 A by hypothesis, we have A ¤ ;. On the other hand, A is bounded (for,
A Œa; b), so that we can take c D sup A. We shall show that f .c/ D 0.
We initially assert that c > a. Indeed, since f .a/ < 0, Lemma 8.20 guarantees
the existence of 0 < ı < b a such that f .x/ < 0 for x 2 Œa; a C ı/; therefore,
c a C ı.
Now, suppose f .c/ < 0. Then, c < b (for, f .b/ > 0) and, once more from the
sign-preserving lemma, there would exist 0 < ı < b c such that f is negative in
.c ı; c C ı/. However, since c D sup A, we could take d 2 .c ı; c/ \ A, so that
f < 0 in Œa; d. Hence, we would have f < 0 in Œa; d [ .c ı; c C ı D Œa; c C ı/,
which is in contradiction to the fact that c D sup A.
Finally, suppose f .c/ > 0. Then (by invoking the sign-preserving lemma yet
another time), there would exist ı > 0 such that f is positive in .c ı; c C ı/ \ Œa; b;
in particular, A \ .c ı; c D ;, and we would have sup A c ı, which is also a
contradiction.
Therefore, the only left possibility is that f .c/ D 0.
t
u
Example 8.28 If f W R ! R is a polynomial function of the form
8.3 The Intermediate Value Theorem
265
f .x/ D an xn C an1 xn1 C C a1 x C a0 ;
with a0 ; a1 ; : : : ; an 2 R, an ¤ 0 and n odd, then the image of f is R. In particular, f
has at least one real root.
Proof Given d 2 R, let g.x/ D f .x/ d. Then, g W R ! R is a polynomial function
satisfying the same hypotheses as f and, for c 2 R, we have f .c/ D d if and only
if g.c/ D 0. This reasoning reduces the example to proving the existence of c 2 R
such that f .c/ D 0. To this end, we shall use Bolzano’s theorem.
Without loss of generality, let an > 0. For x ¤ 0, several applications of the
triangle inequality give
f .x/
a1
an1
a0
C C n1 C n
D an C
n
x
x
x
x
ˇa
a1
a0 ˇˇ
ˇ n1
C C n1 C n ˇ
an ˇ
x
x
x
ˇ a ˇ ˇa ˇ
ˇa ˇ
ˇ 1 ˇ ˇ 0ˇ
ˇ n1 ˇ
an ˇ
ˇ ˇ n1 ˇ ˇ n ˇ
x
x
x
ja1 j
jan1 j
ja0 j
n1 n :
D an jxj
jxj
jxj
If jxj 1, then jxj jxj2 jxjn , so that
1 X
f .x/
an jaj jI
n
x
jxj jD0
n1
1
an
Pn1
jD0 jaj j.
The argument at the previous paragraph shows that f x.x/
n > 0 at x D ˙A, whenever
n
o
1 Pn1
> max 1; an jD0 jaj j . However, since n is odd, it follows that f .A/ < 0 <
in turn, this last expression is positive for jxj >
A
f .A/, and Bolzano’s theorem assures the existence of c 2 ŒA; A such that f .c/ D 0.
t
u
Our next result is a refinement of Bolzano’s theorem, which shows that every
continuous function defined on an interval satisfies the intermediate value property.
For this reason, this result is known as the intermediate value theorem (we
abbreviate IVT).
Theorem 8.29 (IVT) Let f ; g W Œa; b ! R be continuous functions. If f .a/ < g.a/
and f .b/ > g.b/ (or vice-versa), then there exists c 2 .a; b/ such that f .c/ D g.c/.
In particular, if a real number d belongs to the interval with endpoints f .a/ and f .b/,
then there exists c 2 Œa; b such that f .c/ D d.
266
8 Continuous Functions
Proof For the first claim, note that h D f g is continuous and such that h.a/h.b/ D
.f .a/g.a//.f .b/g.b// < 0. Hence, Bolzano’s theorem gives c 2 .a; b/ for which
h.c/ D 0, i.e., such that f .c/ D g.c/.
The particular case of the second claim is obtained by taking g to be the constant
function, equal to d.
t
u
As a first application of the IVT, let us show that the image of a continuous
function defined on an interval is also an interval.
Corollary 8.30 If I R is an interval and f W I ! R is a continuous function, then
the image of f is also an interval. If I D Œa; b, then there exist real numbers c d
such that Im.f / D Œc; d.
Proof Let’s first consider the case I D Œa; b. By Weierstrass Theorem 8.26, there
exist xm ; xM 2 Œa; b such that f attains its minimum and maximum values at xm and
xM , respectively. Letting f .xm / D c and f .xM / D d, we have Im.f / Œc; d. On the
other hand, for a fixed y 2 Œc; d, the IVT gives a real x, belonging to the interval of
endpoints xm and xM and such that f .x/ D y. In particular, Im.f / Œc;S
d.
For the general case, note that every interval I can be written as I D n1 Œan ; bn ,
with Œa1 ; b1 Œa2 ; b2 : : : (for instance, if I D .a; b/, we could take an D a C 1n
1
and bn D b 1n , provided n > 2.ba/
—the remaining cases can be dealt with in
analogous ways). Now, letting Œcn ; dn be the image of the interval
S Œan ; bn by f ,
it’s easy to show that Œc1 ; d1 Œc2 ; d2 : : : and, hence, that S
n1 Œcn ; dn is an
interval J. On the other hand, it’s also easy to see that, since I D n1 Œan ; bn , we
have
[
Im.f / D
Œcn ; dn D J:
n1
t
u
In what follows, we discuss some other interesting applications of the IVT.
Example 8.31 Let f W Œ0; 1 ! Œ0; 1 be a continuous function. Prove that there
exists a real number 0 c 1 such that f .c/ D c (i.e., prove that f has at least one
fixed point).
Proof If f .0/ D 0 or f .1/ D 1, there’s nothing left to do; otherwise, suppose f .0/ >
0 and f .1/ < 1. Letting g W Œ0; 1 ! R be the function given by g.x/ D x, we then
have f .a/ > g.a/ and f .b/ < g.b/. Hence, the IVT gives 0 < c < 1 satisfying
f .c/ D g.c/ or, which is the same, f .c/ D c.
t
u
Example 8.32 (Bulgaria) Let m; n 1 be given integers. For x > 0, compute the
number of real solutions of the equation
p
p
p
1
1
1
1
C 2 C 3 C C m D x C x C 3 x C C n x:
x
x
x
x
Solution We claim that the given equation has just one real solution. To this end,
note first that, for x > 0 and a natural k, the function x 7! x1k is decreasing, whereas
8.3 The Intermediate Value Theorem
267
p
the function x 7! k x is increasing. Now, since a finite sum of increasing functions
on the same domain is increasing and a finite sum of decreasing functions on the
same domain is decreasing (prove these facts!), we conclude that the functions f ; g W
.0; C1/ ! R given by
f .x/ D
p
p
p
1
1
1
1
C 2 C 3 C C m and g.x/ D x C x C 3 x C C n x
x
x
x
x
are decreasing and increasing, respectively. On the other hand, since the solutions
of the given equation correspond to the positive values of x for which f .x/ D g.x/,
Problem 1, page 183, guarantees that such equation has at most one real solution.
In order to show that a real solution indeed exists, let’s use the IVT, initially
observing that both f and g are continuous functions. We then consider two cases
separately:
• for x > 1, we have
p
p
p
1
1
1
1
and x > x > > n1 x > n x;
< m1 < < 2 <
xm
x
x
x
p
p
so that f .x/ < mx and g.x/ > n n x. In particular, f .x/ < g.x/ if mx < n n x, i.e., if
o
n
n
n
x > mn nC1 . Hence, f .x/ < g.x/ for x > max 1; mn nC1 .
• for 0 < x < 1, we have
p
p
p
1
1
1
1
and x < x < < n1 x < n x;
> m1 > > 2 >
m
x
x
x
x
p
p
m
n
so that f .x/ > x and g.x/ < n x. In particular, f .x/ > g.x/ if mx > n n x, i.e., if
n n o
n
x < mn nC1 . Hence, f .x/ > g.x/ for 0 < x < min 1; mn nC1 .
Finally, the previous discussion assures that we can take real numbers 0 < a <
1 < b satisfying f .a/ > g.a/ and f .b/ < g.b/, and the IVT provides c 2 .a; b/ such
that f .c/ D g.c/.
t
u
Example 8.33 (Romania) There exists a continuous function f W R ! R such that
f .x/ 2 Q , f .x C 1/ … Q‹
Solution Suppose such an f exists, and let g W R ! R be given by g.x/ D
f .x C 1/ f .x/. Then, g is continuous (by the chain rule for continuous functions)
and, by the stated conditions, transforms every real number into an irrational
number. However, since every nondegenerate interval contains irrational numbers
(see Problem 4, page 206), the only way of not contradicting the IVT is that g is
constant. Thus, there exists an irrational number ˛ such that f .x C 1/ f .x/ D ˛ for
every x 2 R. Therefore,
f .x C 2/ f .x/ D f .x C 2/ f .x C 1/ C f .x C 1/ f .x/ D 2˛;
also for every x 2 R.
(8.9)
268
8 Continuous Functions
We now assert that there exists x0 2 R such that f .x0 / 2 Q. Indeed, take any
real number a; if f .a/ is not rational, it follows from the hypotheses that f .a C 1/ is
rational, and it suffices to take x0 D a or x0 D a C 1.
Finally, for x0 as in the former paragraph, the hypotheses on f guarantee that
f .x0 / 2 Q ) f .x0 C 1/ … Q ) f .x0 C 2/ 2 Q:
Hence, f .x0 C 2/ f .x0 / 2 Q, which contradicts (8.9) and finishes the proof.
t
u
For the coming example, recall that a function f W R ! R is even if f .x/ D f .x/,
for every real x.
Example 8.34 Let f W R ! R be a continuous function such that f .f .x// D x2 C 1,
for every x 2 R. Prove that f is even.
Proof First of all, note that for every x 2 R we have
f .f .x// D x2 C 1 ) f .f .f .x/// D f .x2 C 1/
) f .x/2 C 1 D f .x2 C 1/
) f .x/2 C 1 D f .x/2 C 1
) f .x/ D ˙f .x/
Now, if f .˛/ D f .ˇ/ D 0, then
˛ 2 C 1 D f .f .˛// D f .0/ D f .f .ˇ// D ˇ 2 C 1;
so that ˛ D ˙ˇ and f has at most two zeros. Now, there are three possibilities:
• f .x/ ¤ 0, for every real x: by the IVT, f has a constant sign in R. However, since
f .x/ D ˙f .x/ for every real x, it follows that f .x/ D f .x/ for every real x, and
f is even.
• f has a single zero, say at x D ˛: since f .˛/ D ˙f .˛/ D 0, we must have
˛ D ˛, so that ˛ D 0. On the other hand,
f .f .0// D 02 C 1 D 1 ) f .0/ ¤ 0
which is a contradiction.
• f has exactly two zeros, at x D ˛ and x D ˛, for some ˛ > 0: again by the IVT,
f has a constant sign in .˛; ˛/. Let g.x/ D f .x/ x. If f > 0 in .˛; ˛/, then
g.0/ D f .0/ 0 > 0
and g.˛/ D f .˛/ ˛ D ˛ < 0;
so that there exists 0 < c < ˛ for which g.c/ D 0. If f < 0 in .˛; ˛/, then
g.0/ D f .0/ 0 < 0
and g.˛/ D f .˛/ .˛/ D ˛ > 0;
8.3 The Intermediate Value Theorem
269
so that there exists ˛ < c < 0 for which g.c/ D 0. In any case, f admits a fixed
point c. However,
f .c/ D c ) c D f .f .c// D c2 C 1 ) c2 c C 1 D 0;
t
u
which is impossible.
We close this section by using the IVT to prove the continuity of the inverse of a
continuous function defined on an interval.
Theorem 8.35 Let I R be an interval and f W I ! R be a continuous
function. Then, f is injective if and only if f is increasing or decreasing. Moreover,
in this case:
(a) The image J of f is an interval of the same type (i.e., open, half-open or closed)
as I.
(b) f 1 W J ! I is continuous.
Proof If f is not injective, then f is neither increasing nor decreasing. Conversely,
if f is neither increasing nor decreasing, then there exist a < b < c in I such that
f .a/ f .b/ f .c/ or f .a/ f .b/ f .c/. Suppose that f .a/ f .b/ f .c/ (the
other case is entirely analogous) and choose d 2 R such that
maxff .a/; f .c/g d f .b/:
The IVT assures the existence of x0 2 .a; b/ and x1 2 .b; c/ (therefore, x0 ; x1 2 I)
such that f .x0 / D d and f .x1 / D d. In particular, f .x0 / D f .x1 / and f is not
injective.
(a) Corollary 8.30 showed that J is an interval. Since f is injective, we can suppose
that f is increasing (again, the case of a decreasing f is analogous). Let I D
.a; b/, with a; b 2 R (the remaining cases can also be dealt with in analogous
ways).
If Im.f / D .c; d or .1; d, with c; d 2 R, take x0 2 .a; b/ such that
f .x0 / D d. Since f is increasing, for x 2 .x0 ; b/ we have f .x/ > f .x0 / D d,
thus contradicting the fact that f .x0 / 2 Im.f /. Analogously, we show that
Im.f / ¤ Œc; d/,Œc; C1/, Œc; d, with c; d 2 R. Therefore, Im.f / is also an open
interval.
(b) Let’s look at the case of an increasing f , with I D .a; b/ and Im.f / D .c; d/
(once more, the analysis of the other ones is quite similar). We start by observing
(according to Problem 8, page 176), that f 1 is also increasing. Now, for a fixed
y0 2 .c; d/, let x0 D f 1 .y0 /. Given > 0, we want ı > 0 such that
y 2 .c; d/; jy y0 j < ı ) jf 1 .y/ f 1 .y0 /j < :
(8.10)
270
8 Continuous Functions
To this end, let ı0 D minfy0 c; d y0 g and take 0 < ı < ı0 (so that the
condition jy y0 j < ı suffices to guarantee that y 2 .c; d/). Letting x D f 1 .y/,
we have y D f .x/ and can rewrite (8.10) in the following way: we want 0 < ı <
ı0 such that
jf .x/ f .x0 /j < ı ) jx x0 j < :
Notice that we can pick > 0 so small that x0 ˙ 2 .a; b/ (otherwise, change the
originally given > 0 by a smaller one, so that this additional condition is satisfied).
Recalling that f is increasing, take
0 < ı < minfı0 ; f .x0 C / f .x0 /; f .x0 / f .x0 /g:
Then,
f .x/ f .x0 / < ı ) f .x/ f .x0 / < f .x0 C / f .x0 /
) f .x/ < f .x0 C / ) x < x0 C and, analogously,
f .x/ f .x0 / > ı ) x > x0 :
In any case, the choice of ı gives
jf .x/ f .x0 /j < ı ) ı < f .x/ f .x0 / < ı
) < x x0 < ) jx x0 j < ;
t
u
as we wished to show.
As a first application of the previous result, we shall now give an alternative proof
for Theorem 7.9 and Problem 8, page 254.
Example 8.36 Given n 2 N, the n-th power function is f W R ! R given by
f .x/ D xn . If n is even, write also f to denote its restriction to Œ0 C 1/. Since
f is continuous, increasing and f .k/ D kn > k for every k 2 N, the IVT gives
Im.f / D Œ0; C1/. Hence, for every y 0 there exists a single x 0 such that
p
xn D y, and we denote by x D n y. The previous theorem now guarantees the
continuity of the function
f 1 W Œ0; C1/ ! Œ0; C1/
:
p
n y
y
7!
8.3 The Intermediate Value Theorem
271
Analogously, if n is odd, then f is continuous, increasing, Im.f / D R and (letting
p
n y be as above)
f 1 W R ! R
p
y 7! n y
is continuous. In both cases, f 1 is called the n-th root function.
The next example establishes the continuity of the inverse trigonometric functions, introduced in Problem 1, page 198.
Example 8.37 The restriction of the sine function to the interval Π2 ; 2 , which
we shall also denote by sin W Œ 2 ; 2 ! Œ1; 1, is a continuous and increasing
bijection. The arc-sine function is its inverse arcsin W Œ1; 1 ! Œ 2 ; 2 , so that, for
x 2 Œ1; 1 and y 2 Œ 2 ; 2 ,
arcsin x D y , sin y D x:
Thanks to Theorem 8.35, the arc-sine function os increasing and continuous.
Similarly, the inverse of cos W Œ0; ! Œ1; 1 is the arc-cosine function arccos W
Œ1; 1 ! Œ0; , which is continuous (again by Theorem 8.35) and decreasing.
Note also that, for x 2 Œ1; 1 and y 2 Œ0; , we have
arccos x D y , cos y D x:
Finally, the arc-tangent function arctan W R ! 2 ; 2 is the inverse of the
function,
so that it is also continuous
restriction tan W 2 ; 2 ! R of the tangent
and increasing. Also, for x 2 R and y 2 2 ; 2 , we have
arctan x D y , tan y D x:
Problems: Section 8.3
1. The continuous function f W .1; 1/ [ .1; C1/ ! R, given by
f .x/ D
x; if x 2 .1; 1/
;
x2 C 1; if x 2 .1; C1/
satisfies f .0/ < 32 < f .2/ but there doesn’t exist any c in the domain of f such
that f .c/ D 32 . Why does this example do not contradict the IVT?
2. Compute the number of real solutions of each of the following equations:
(a) jxj C 1 D x4 .
272
8 Continuous Functions
(b) cos x D x2 .
(c) sin x D 4x .
3. Let f W R ! R be given by f .x/ D x3 sin x, for every x 2 R. Show that its graph
intersects every non vertical real line an infinite number of times.
4. Let n > 1 be an integer. Write down an explicit expression for a continuous
function f W Œ0; 1 ! Œ0; 1 with exactly n fixed points.
5. * Let I be an interval and X be a nonempty subset of I having the following
properties:
(i) For every x0 2 X, there exists ı > 0 such that I \ .x0 ı; x0 C ı/ X.
(ii) If .an /n1 is a sequence of points of X and l 2 I is such that an ! l, then
l 2 X.
Show that X D I.
6. (Bulgaria) Let n > 1 be an integer and a1 , a2 , . . . , an be given positive real
numbers. Prove that the equation
p
p
p
1 C a1 x C 1 C a2 x C C 1 C an x D nx
has exactly one positive real solution.
7. Let x1 , x2 , . . . , xn be real numbers in the interval Œ0; 1. Prove that there exists
x 2 Œ0; 1 such that
jx x1 j C jx x2 j C C jx xn j D
n
:
2
8. (Leningrad) Let f W R ! R be a continuous function satisfying, for every
x 2 R, the relation f .x/f .x C 2/ C f .x C 1/ D 0. If f .0/ 0, show that there
exist infinitely many values of x for which f .x/ D 0.
9. (Leningrad) Let f W R ! R be a continuous function such that f .x/f .f .x// D 1,
for every real x. If f .1000/ D 999, compute f .500/.
10. (Australia) Find all real values of a for which, for every continuous function
f W Œ0; 1 ! R satisfying the condition f .0/ D f .1/, there exists x0 2 Œ0; 1 a
such that f .x0 / D f .x0 C a/.
11. Let f W R ! R be such that f .x C 1/f .f .x/ C 1/ C 1 D 0, for every real x. Prove
that f is not continuous.
12. Let f W Œ0; 1 ! Œ0; 1 be a continuous function, such that .f ı f /.x/ D x for
every x 2 Œ0; 1.
(a) Show that f is an increasing or decreasing bijection.
(b) If f is increasing, show that f .x/ D x for every x 2 Œ0; 1.
(c) Show that there are infinitely many possibilities for f , if it is decreasing.
13. (Leningrad) The continuous functions f ; g W Œ0; 1 ! Œ0; 1 are such that f ı
g D g ı f . If f is nondecreasing, prove that there exists 0 a 1 such that
f .a/ D g.a/ D a.
8.3 The Intermediate Value Theorem
273
14. (Crux) Let f W R ! R be a continuous function which assumes positive
and negative values. Given a natural number k > 2, prove that there exists a
nonconstant AP .a1 ; a2 ; : : : ; ak / such that
f .a1 / C f .a2 / C C f .ak / D 0:
15. (TT) Prove that, for each natural number n, the graph of any continuous
increasing function f W Œ0; 1 ! Œ0; 1 can be covered by n closed rectangles,
each of which having area n12 and sides parallel to the coordinate axis.
16. (Leningrad) Let f W R ! R be a continuous function such that, for every real x,
we have f .x C f .x// D f .x/. Prove that f is constant.
17. (Belarus) Find all functions f ; g; h W R ! R such that, for every x; y 2 R, we
have
f .x C y3 / C g.x3 C y/ D h.xy/:
Chapter 9
Limits and Derivatives
In this chapter we study derivatives of functions and some of their applications.
Along the way, we show how to use derivatives to solve the problem of finding
the monotonicity (resp. concavity) intervals of a differentiable (resp. twice differentiable) function, as well as to build an accurate sketch of the graph of such functions.
In turn, the analysis of such problems will motivate several interesting applications
of the concept of derivative to problems of maxima and minima.
9.1 Some Heuristics I
As a motivation for what is to come, in this section we heuristically discuss two
problems which gave birth to the notion of derivative of a function.
Let’s start by trying to define the tangent line to the graph of a function f W
.a; b/ ! R at a point A.x0 ; f .x0 // on it, so that x0 2 .a; b/. Taking x1 2 .a; b/ n fx0 g
!
and letting B1 .x1 ; f .x1 //, we say that AB1 is a secant to the graph of f and passing
!
!
through A. In Fig. 9.1, we consider secants AB1 and AB2 to the graph of f .
A little geometric intuition makes it plausible to guess that a generic secant line
!
AB should come closer and closer to the tangent line to the graph of f at A, as long
as B approaches A along the graph (or, which is the same, as long as x approaches
x0 in .a; b/).
If x1 2 .a; b/ n fx0 g and B.x1 ; f .x1 //, elementary analytic geometry (cf. Chap. 6
!
of [4], for instance) shows that AB has equation
y f .x0 / D
f .x1 / f .x0 /
.x x0 /I
x 1 x0
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_9
(9.1)
275
276
9 Limits and Derivatives
!
Fig. 9.1 Secants AB1 and
!
AB2
B2
y
to the graph of f
f
a
B1
A
x0
x2 x1
x
b
on the other hand, if we can define the line tangent to the graph of f at A, and it is
not vertical, then its equation must be of the form
y f .x0 / D m.x x0 /;
(9.2)
for some m 2 R.
Comparing (9.1) and (9.2), and taking into account the discussion at the
.x0 /
previous paragraph, we are led to the conclusion that the quotients f .xx11/f
should
x0
approximate m better and better, provided x1 approximates x0 better and better. Thus,
we conclude that the line tangent to the graph of f at A is that of Eq. (9.2), where m
.x0 /
is the limit value of the quotients f .xx11/f
, when x1 comes closer and closer to x0 .
x0
.x0 /
when x1 approaches x0 is the derivative
This limit value of the quotients f .xx11/f
x0
df
.x0 /.
of the function f at x0 , which we denote f 0 .x0 / or, in classical notation, dx
Writing x in place of x1 , we summarize the previous discussion by “defining”
df
f .x/ f .x0 /
.x0 / D lim
;
x!x0
dx
x x0
where the notation lim means that the closer x is to x0 , the closer
to
df
.x /.
dx 0
x!x0
f .x/f .x0 /
xx0
is
Hence, for small values of jx x0 j, we expect to have
f .x/ f .x0 /
df
.x0 / Š
:
dx
x x0
Classically,
ˇx
ˇxthe differences f .x/ f .x0 / and x x0 were respectively denoted by
f ˇx0 and xˇx0 , and were called the variations of f and of the independent variable
ˇx
in the interval Œx0 ; x. Accordingly, f ˇ is the rate of change of f (with respect to
x x0
the independent variable) in the interval Œx0 ; x. Therefore, for small jxx0 j, we have
f ˇˇx
df
.x0 / Š
ˇ :
dx
x x0
9.2 Limits of Functions
277
Now, suppose we are in a situation where x represents time and f represents some
quantity that evolves with time. Writing t in place of x and t0 in place of x0 , it follows
from the above that
df
f ˇˇt
.t0 / D lim
ˇ ;
t!t0 t t0
dt
so that dfdt .t0 / can naturally be called the instantaneous rate of change of f with
respect to t, at instant t0 .
As a particular situation of that of the previous paragraph, if f .t/ measures
the displacement of a point object along a real line at instant t, then dfdt .t0 / is the
instantaneous velocity of the point object at instant t0 , and is the starting point of the
discussion of Kinematics at most Physics books.
The circle of ideas we developed up to this point, and much more, bears the
name of Differential Calculus and is considered to be the creation of two of the
most brilliant minds mankind ever produced, Newton and Leibniz.1
The rest of this chapter is devoted to a thorough development of the concepts,
results and applications of the Differential Calculus.
9.2 Limits of Functions
Let be given a continuous function f W .a; b/ ! R and x0 2 .a; b/. Recalling our
intuitive discussion so far, we conclude that if there exists a reasonable notion of
tangent line to the graph of f at the point A.x0 ; f .x0 //, and if such a line is not
vertical, then its equation must have the form (9.2), where
m D lim
x!x0
f .x/ f .x0 /
:
x x0
(9.3)
Here, the expression at the right hand side above intuitively represents the limit value
.x0 /
of the quotients f .x/f
as x goes to (i.e., approaches) x0 , provided such a “limit”
xx0
exists in a proper sense (to be precised in a while).
Since our discussion rests upon rather naive grounds, we start to fix this now by
setting some preliminaries and giving a rigorous definition of the notion of limit of
a function at a point.
Definition 9.1 Given a 2 R, a neighborhood of a is an interval of the form .a r; a C r/, where r is a positive real number. In this case, we say that r is the radius
of the neighborhood .a r; a C r/ and that each x 2 .a r; a C r/ approximates a
with error less than r.
1
Gottfried Wilhem Leibniz, German mathematician and philosopher of the XVII century. Together
with Sir Isaac Newton, Leibniz is considered to be one of the creators of the Differential and
Integral Calculus. Some of the notations used in Calculus up to this day go back to Leibniz, having
survived the ruthless test of time.
278
9 Limits and Derivatives
Let I R be an interval, let x0 2 I and f W I n fx0 g ! R be a given function. For
a given real number L, our first task is to formulate a precise definition for the claim
that f .x/ can be taken as close to L as we wish, provided we take x 2 I sufficiently
close to (but different from) x0 .
A little reflection allows us to conclude that a reasonable formulation of this
concept is obtained if we ask that, to each given neighborhood of L, there exists a
neighborhood of x0 that is applied to the former one by f .
Since a neighborhood of x0 is completely determined by its radius r, and
x 2 .x0 r; x0 C r/ n fx0 g , 0 < jx x0 j < r;
we can summarize the previous discussion in the following definition (see Fig. 9.2).
Definition 9.2 Let I R be an interval, let x0 2 I and f W I n fx0 g ! R be a given
function. We say that f has limit L when x goes to x0 , and write
lim f .x/ D L;
(9.4)
x!x0
if, to each given real number > 0, there corresponds a real number ı > 0 such that
x 2 I; 0 < jx x0 j < ı ) jf .x/ Lj < :
(9.5)
In words, (9.5) happens if, to each arbitrarily given error > 0 for L, there
corresponds an error ı > 0 for x0 such that, if x 2 I n fx0 g approximates x0 in I with
error less that ı, then f .x/ approximates L with error less than .
Geometrically, we want that, to each x 2 I sufficiently close to (but different
from) x0 , the point of the graph of f with abscissa x belongs to the gray strip
of Fig. 9.2.
y
2
L
a
x0
2δ
Fig. 9.2 Limit of a function at a point
b
Gf
x
9.2 Limits of Functions
279
At this point (and as it happened in our study of continuity), it is worth stressing
that, to justify the validity of a specific limit by applying the definition given above
is the same as to play the following cat and mouse game: to each arbitrarily given
error > 0 for the limit candidate L, we ought to be able to find an error ı > 0 for
x0 (which, in general, will depend both on as on x0 itself) so that the validity of
the condition 0 < jx x0 j < ı for some x 2 I implies jf .x/ Lj < .
Having gone through the study of continuity, the reader has surely acquired some
familiarity with the dynamics of finding which ı > 0 are adequate to some given
> 0. In spite of this, and as a remembrance, we shall carefully work out two
examples below. In all that follows, and whenever there is no danger of confusion,
we shall frequently omit explicit references to the domain or codomain of the
functions f involved, thus focusing on the expressions that define f .x/ in terms of
the independent variable x. Whenever this is so, we shall implicitly assume that the
domain of f is the maximal one (cf. Sect. 6.1), whereas its codomain is the set R of
real numbers.
Examples 9.3
(a) limx!2 .2x C 7/ D 3: let > 0 be given. Departing from x 2 R such that
0 < jx 2j < ı, we have
j.2x C 7/ 3j D j 2x C 4j D 2jx 2j < 2ı:
Therefore, choosing ı > 0 so that 2ı , we conclude that
x 2 R; 0 < jx 2j < ı ) j.2x C 7/ 3j < 2ı ;
as wished.
(b) limx!3 x2 D 9: again, let > 0 be given. Starting with x 2 R such that 0 <
jx 3j < ı, we have
jx2 9j D jx 3jjx C 3j < ıjx 3 C 6j
ı.jx 3j C 6/ < ı.ı C 6/;
where we used the triangle inequality in the next to last passage above. Hence,
if it is possible to choose ı > 0 so that ı.ı C 6/ , we will have
x 2 R; 0 < jx 3j < ı ) jx2 9j < ı.ı C 6/ ;
as wished. It now suffices to show that it is possible to choose such a ı > 0,
and to this end we just have to solve the inequality ı.ı C 6/ . By doing so,
we get
0<ı
p
C 9 3:
Looking back at the previous examples, we conclude that if we have a function
f W I n fx0 g ! R and want to prove that limx!x0 f .x/ D L, then, as in Sect. 8.1, the
280
9 Limits and Derivatives
key to find out which ı > 0 works for a given > 0 is to reason in the following
way: starting with an x 2 I such that 0 < jx x0 j < ı, we estimate from above the
error jf .x/Lj in terms of ı, thus getting an inequality of the form jf .x/Lj < E.ı/.
Here, E represents a certain function of ı; in item (a) we got E.ı/ D 2ı, whereas
in item (b) we found E.ı/ D ı.ı C 6/. Then, we impose that such an error E.ı/
does not surpass , thus finding the appropriate values of ı (usually, this second step
reduces to solving, for ı > 0, the inequality E.ı/ ). Finally, if ı > 0 satisfies
E.ı/ , then we clearly have that
x 2 I; 0 < jx x0 j < ı ) jf .x/ Lj < E.ı/ ;
as wished.
Notice that the only difference between the discussion above and that of Sect. 8.1
lies in the fact that, here, we have to start by assuming jx x0 j > 0, i.e., x ¤ x0 . On
the other hand, and as the coming proposition explains, such a difference was not
relevant to the previous examples because both functions x 7! 2x C 7 and x 7! x2
are continuous.
Proposition 9.4 Let I R be an interval and f W I ! R be a given function. For
x0 2 I, we have f continuous at x0 if and only if
lim f .x/ D f .x0 /:
(9.6)
x!x0
Proof First of all, let f be continuous at x0 . Definition 8.2 guarantees that, given
> 0, there exists ı > 0 such that
x 2 I; jx x0 j < ı ) jf .x/ f .x0 /j < :
In particular, if x 2 I and 0 < jx x0 j < ı, we still have jf .x/ f .x0 /j < , so
that (9.5) is satisfied. Therefore, limx!x0 f .x/ D f .x0 /.
Conversely, let limx!x0 f .x/ D f .x0 /. According to the definition of limit, given
> 0, there exists ı > 0 such that
x 2 I; 0 < jx x0 j < ı ) jf .x/ f .x0 /j < :
However, since the condition jf .x/ f .x0 /j < is trivially satisfied for x D x0 , we
can certainly write the implication above as
x 2 I; jx x0 j < ı ) jf .x/ f .x0 /j < :
t
u
Hence, f is continuous at x0 .
x3 3x2 C2
Example 9.5 Given f W R n f˙1g ! R such that f .x/ D x2 1 for every x ¤ ˙1,
we want to compute limx!1 f .x/. To this end, we initially observe that 1 is a root
of both the numerator and the denominator of the defining expression for f .x/, with
9.2 Limits of Functions
281
x3 3x2 C 2 D .x 1/.x2 2x 2/ and x2 1 D .x 1/.x C 1/. Therefore, on
R n f˙1g we have
f .x/ D
.x 1/.x2 2x 2/
x2 2x 2
D
:
.x 1/.x C 1/
xC1
Now, since the function g W R n f1g ! R given by g.x/ D
the previous proposition gives
x2 2x2
xC1
is continuous,
3
lim f .x/ D lim g.x/ D g.1/ D :
x!1
x!1
2
A rather important fact on limits of functions is that if limx!x0 f .x/ D L, with
L ¤ 0, then there exists a neighborhood of x0 such that f has the same sign of L
in all points of its domain which belong to this neighborhood (with the possible
exception of x0 itself). This is essentially the content of the coming result, which
extends Lemma 8.20 to limits of functions and for this reason is also known as the
sign-preserving lemma.
Lemma 9.6 Let I R be an interval, x0 2 I and f W I n fx0 g ! R be a given
function. If limx!x0 f .x/ D L, with L ¤ 0, then there exists ı > 0 such that
x2I
and 0 < jx x0 j < ı )
L=2 < f .x/ < 3L=2;
if L > 0
:
3L=2 < f .x/ < L=2; if L < 0
Proof Suppose L > 0 (the other case is completely analogous). By the definition of
limit, given D L2 > 0 there exists ı > 0 such that
x 2 I; 0 < jx x0 j < ı ) jf .x/ Lj <
Hence, for each such x we have L2 f .x/ L <
.
f .x/ < 3L
2
L
2,
L
:
2
which is the same as
L
2
<
t
u
We now collect some arithmetic properties on limits that pretty much simplify
their computation. For item (c) of the coming proposition, it is worth observing
the following: given an interval I R, x0 2 I and g W I n fx0 g ! R such that
limx!x0 g.x/ D M ¤ 0, Lemma 9.6 guarantees the existence of r > 0 such that g
doesn’t vanish on Jnfx0 g, where J D I\.x0 r; x0 Cr/. Hence, when considering the
1
function 1g such that 1g .x/ D g.x/
, we will always implicitly assume that its domain
is restricted to J n fx0 g.
Proposition 9.7 Let I R be an interval, let x0 2 I and f ; g W I n fx0 g ! R be
given functions. If limx!x0 f .x/ D L and limx!x0 g.x/ D M, then:
282
9 Limits and Derivatives
(a) limx!x0 .f ˙ g/.x/ D L ˙ M.
(b) limx!x0 .f g/ D L M.
(c) limx!x0
f
g
.x/ D
L
M,
provided M ¤ 0.
Proof In all cases below, we let > 0 be given.
(a) Let’s do the proof for f C g, the proof for f g being completely analogous. As
was previously hinted, our strategy will be to try to estimate j.f Cg/.x/.LCM/j
by excess, in terms of jf .x/ Lj and jg.x/ Mj. This is easily accomplished
with the aid of the triangle inequality, which gives
j.f C g/.x/ .L C M/j D j.f .x/ L/ C .g.x/ M/j
jf .x/ Lj C jg.x/ Mj:
Thus, in order that j.f C g/.x/ .L C M/j < for x 2 I close to (but different
from) x0 , it suffices for us to have jf .x/ Lj < 2 and jg.x/ Mj < 2 .
Since 2 > 0 and limx!x0 f .x/ D L, limx!x0 g.x/ D M, the definition of limit
guarantees the existence of ı1 ; ı2 > 0 such that
x 2 I; 0 < jx x0 j < ı1 ) jf .x/ Lj <
2
x 2 I; 0 < jx x0 j < ı2 ) jg.x/ Mj <
:
2
and
Therefore, letting ı D minfı1 ; ı2 g > 0 and taking x 2 I such that 0 < jx x0 j <
ı, we get
j.f C g/.x/ .L C M/j jf .x/ Lj C jg.x/ Mj
C D :
2
2
(b) Here again, we estimate j.fg/.x/ LMj by excess in terms of jf .x/ Lj and
jg.x/ Mj. To this end, it follows from triangle inequality that
j.fg/.x/ LMj D jf .x/.g.x/ M/ C .f .x/ L/Mj
jf .x/jjg.x/ Mj C jf .x/ LjjMj
.jf .x/ Lj C jLj/jg.x/ Mj C jf .x/ LjjMj
D jf .x/ Ljjg.x/ Mj C jLjjg.x/ Mj C jMjjf .x/ Lj:
Therefore, in order that j.fg/.x/LMj < for x 2 I close to (but different from)
x0 , it suffices for us to have jf .x/ Ljjg.x/ Mj, jLjjg.x/ Mj and jMjjf .x/ Lj
all less than 3 . In turn, these inequalities hold provided we have
9.2 Limits of Functions
283
r
jf .x/ Lj; jg.x/ Mj <
jg.x/ Mj <
3.jLj C 1/
;
3
and jf .x/ Lj <
:
3.jMj C 1/
r
In short, it suffices to have
jf .x/ Lj < min
;
3 3.jMj C 1/
and
r
jg.x/ Mj < min
;
:
3 3.jLj C 1/
o
o
np
np
>
0
and
Now, letting 1 D min
;
D
min
;
2
3 3.jMjC1/
3 3.jLjC1/ >
0, the definition of limit guarantees the existence of errors ı1 ; ı2 > 0 such that
jf .x/ Lj < 1 for every x 2 I such that 0 < jx x0 j < ı1 , and jg.x/ Mj < 2
for every x 2 I such that 0 < jx x0 j < ı2 . Hence, setting ı D minfı1 ; ı2 g > 0,
we conclude that x 2 I and 0 < jx x0 j < ı imply both jf .x/ Lj < 1 and
jg.x/ Mj < 2 , as needed.
(c) The sign-preserving lemma gives ı0 > 0 such that jg.x/j > jMj
for all x 2 I
2
satisfying 0 < jx x0 j < ı0 . Sticking to the notations set forth in the paragraph
that immediately precedes this proposition, we look at gf as the function gf W
J n fx0 g ! R, where J D I \ .x0 ı0 ; x0 C ı0 /.
Since
f
g
D f 1g , by item (b) it suffices to show that limx!x0
jMj
2
1
g.x/
D
1
M.
To this
for x 2 J n fx0 g, we obtain (for such
end, and taking into account that jg.x/j >
an x)
ˇ
ˇ
ˇ 1
1 ˇˇ
jg.x/ Mj
2
ˇ
ˇ g.x/ M ˇ D jg.x/jjMj M 2 jg.x/ Mj:
ˇ
ˇ
2
ˇ
ˇ 1
M1 ˇ < , we need to have jg.x/ Mj < M2 . To fulfill
Therefore, in order that ˇ g.x/
the last condition above, we choose (by invoking once more the definition of limit)
ı1 > 0 for which
x 2 J; 0 < jx x0 j < ı1 ) jg.x/ Mj <
M2 :
2
Finally, letting ı D minfı0 ; ı1 g > 0 and taking x 2 J such that 0 < jx x0 j < ı,
M2 we have jg.x/j < jMj
t
u
2 and jg.x/ Mj < 2 , as needed.
284
9 Limits and Derivatives
An easy induction allows us to extend the formulas of items (a) and (b) of the
previous proposition to a finite number of functions. More specifically, if I is an
interval, x0 2 I and f1 ; : : : ; fn W I n fx0 g ! R are such that limx!x0 fj .x/ D Lj for
1 j n, then
lim .f1 ˙ f2 ˙ ˙ fn /.x/ D L1 ˙ L2 ˙ ˙ Ln
(9.7)
lim .f1 f2 : : : fn /.x/ D L1 L2 : : : Ln :
(9.8)
x!x0
and
x!x0
In particular, if k 2 N and f W I n fx0 g ! R is such that limx!x0 f .x/ D L, then
lim f .x/k D Lk :
x!x0
From now on, we assume the validity of these remarks without further comments.
Our next result is usually referred to as the squeezing theorem. As we shall see
right after its proof, it is quite useful for the actual computation of limits.
Proposition 9.8 Let I be an interval, x0 2 I and f ; g; h W I n fx0 g ! R be such
that g.x/ belongs to the interval of ends f .x/ and h.x/ for every x 2 I n fx0 g. If
limx!x0 f .x/ D limx!x0 h.x/ D L, then limx!x0 g.x/ exists and is also equal to L.
Proof Given > 0, we want to find ı > 0 for which the conditions x 2 I and
0 < jx x0 j < ı imply jg.x/ Lj < . To this end, if f .x/ g.x/ h.x/, then
f .x/ L g.x/ L h.x/ L and is easy to conclude that
jg.x/ Lj maxfjf .x/ Lj; jh.x/ Ljg:
If h.x/ g.x/ f .x/, one reaches the inequality above in essentially the same way.
Now, the definition of limit gives ı1 ; ı2 > 0 such that
x 2 I; 0 < jx x0 j < ı1 ) jf .x/ Lj < and
x 2 I; 0 < jx x0 j < ı2 ) jh.x/ Lj < :
Therefore, letting ı D minfı1 ; ı2 g > 0 and taking x 2 I such that 0 < jx x0 j < ı,
we get jf .x/ Lj < and jh.x/ Lj < , so that
jg.x/ Lj maxfjf .x/ Lj; jh.x/ Ljg < :
t
u
Concerning the computation of limits, the coming corollary is quite a useful tool.
For its statement, recall that a function f W I ! R is bounded if there exists M > 0
such that
9.2 Limits of Functions
285
jf .x/j M; 8 x 2 I:
Corollary 9.9 Let I be an interval, x0 2 I and f ; g W I n fx0 g ! R be such that f
is bounded and limx!x0 g.x/ D 0. Then, limx!x0 f .x/g.x/ D 0, even if limx!x0 f .x/
doesn’t exist.
Proof If jf .x/j M for every x 2 I n fx0 g, then 0 jf .x/g.x/j Mjg.x/j for every
x 2 I n fx0 g. Since limx!x0 jg.x/j D 0 (check this!), it follows from the squeezing
principle that limx!x0 jf .x/g.x/j D 0. Therefore, limx!x0 f .x/g.x/ D 0.
t
u
Example 9.10 If f W R n f0g ! R is given by f .x/ D sin 1x , then jf .x/j 1 for
every x 2 R n f0g. Since limx!0 x D 0, it follows from the previous proposition
that limx!0 x sin 1x D 0. Note that this result agrees with Problem 10, page 255,
and (9.6).
The squeezing theorem also allows us to compute the fundamental trigonometric limit, which will reveal itself to be of crucial importance in the next section.
Lemma 9.11 limx!0
sin x
x
D 1.
Proof Since we wish to compute a limit, we can restrict ourselves to the interval
_
_
jxj < 2 . Suppose first that x > 0. Letting `.AB/ D x be the length of the arc AB
(cf. Fig. 9.3), we have
_
sin x D BD < AB < `.AB/ D x;
so that sinx x < 1. On the other hand, it is well known (cf. Chap. 5 of [5], for instance)
x
that the area of the circular sector AOB equals 2
D 2x , whereas the area of the
Fig. 9.3 The fundamental
trigonometric limit
C
B
O
D
A
286
9 Limits and Derivatives
triangle AOC equals 12 AC; hence, we have
_
x D `.AB/ < AC D tan x;
so that cos x < sinx x .
Combining the two inequalities deduced above, we obtain
cos x <
sin x
<1
x
(9.9)
for 0 < x < 2 . However, since both x 7! cos x and x 7! sinx x are even functions,
these inequalities still hold true for 2 < x < 0. Then, it follows from (9.9),
together with the squeezing theorem and the continuity of the cosine function, that
limx!0 sinx x exists and equals 1.
t
u
Remark 9.12 The attentive reader may object the former proof under the following
argument: the very definitions of the sine and cosine functions, as given in
elementary school, rely upon the fact that a circle of radius 1 has length 2; in
turn, as we shall see in Sect. 10.10, a thorough computation of length of a circle
involves the use of the derivative of the sine function, so that we are in presence of
a circular reasoning. Although this is indeed the case, we observe that we will fix
this situation in Sect. 11.5, when we shall construct the sine and cosine functions
without invoking any concepts or results of Euclidean Geometry. In particular, at
that future time we will show, by purely analytical means, that limx!0 sinx x D 1 and
sin0 x D cos x. For the time being, we suggest that the reader takes the proof of the
previous lemma simply as a heuristic argument for justifying the validity of both
these results.
Let be given an open interval I, a point x0 2 I and a function f W I n fx0 g ! R.
In a way entirely analogous to what we have done so far, we can define for f the
concept of one-sided or lateral limits at x0 . More precisely:
(i) If I \ .1; x0 / ¤ ;, we say that L is the left-handed limit of f as x goes to x0 ,
and write limx!x0 f .x/ D L, if, for each given > 0, there exists ı > 0 such
that
x 2 I; x0 ı < x < x0 ) jf .x/ Lj < :
(9.10)
(ii) If I \ .x0 ; C1/ ¤ ;, we say that L is the right-handed limit of f as x goes to
x0 , and write limx!x0 C f .x/ D L, if, for each given > 0, there exists ı > 0
such that
x 2 I; x0 < x < x0 C ı ) jf .x/ Lj < :
(9.11)
The idea behind the notation x ! x0 (resp. x ! x0 C) should be clear to the
reader: we write x ! x0 (resp. x ! x0 C) in left-handed (resp. right-handed)
9.2 Limits of Functions
287
y
2
L
a
x0
b
δ
x
Gf
Fig. 9.4 The right-handed limit limx!x0 C f .x/
limits, for x approaches x0 from the left (resp. from the right), i.e., through values
smaller (resp. larger) than x0 itself. In yet another way, when x ! x0 (resp. x !
x0 C), we have x D x0 .something positive/ (resp. x D x0 C .something positive/).
Figure 9.4 gives a geometric interpretation for the notion of right-handed limit.
Note that, for 0 < jx x0 j < ı, the point .x; f .x// only belongs to the gray strip if
x0 < x < x0 C ı. We left to the reader the task of drawing a figure that provides the
analogous geometric interpretation for left-handed limits.
Yet with respect to one-sided limits, it is immediate to state and prove for them
results analogous to those of Proposition 9.7 (cf. Problem 4). We also stress that,
for a function f W .a; x0 / ! R, the notions of limite and left-handed limit of
f at x0 coincide (for, there is simply no way of approaching x0 from the right,
while belonging to the domain of f ). Thus, in such a case we shall generally write
limx!x0 f .x/ D L instead of limx!x0 f .x/ D L. Obviously, for a given function
f W .x0 ; b/ ! R, analogous remarks are true for limits and right-handed limits at x0 .
In this respect, see also Problem 5.
We close this section by defining the notions of infinite limits and limits at
infinity, and sketching some elementary (though useful) results on them.
Definition 9.13 Let be given an interval I, a point x0 2 I and a function f W I n
fx0 g ! R. We write
lim f .x/ D C1
x!x0
if, for any given M > 0, there exists ı > 0 such that
x 2 I; 0 < jx x0 j < ı ) f .x/ > M:
We leave to the reader the task of stating definitions analogous to the one above
for the concepts of limx!x0 f .x/ D 1 and limx!x0 ˙ f .x/ D ˙1.
288
9 Limits and Derivatives
The next result establishes two simple, yet useful, arithmetic rules for infinite
limits. Since its proof resembles that of Proposition 9.7, we also leave it as an
exercise for the reader.
Proposition 9.14 Let be given an interval I, a point x0 2 I and functions f ; g W
I n fx0 g ! R.
(a) If lim f .x/ D L > 0 and lim g.x/ D ˙1, then lim f .x/g.x/ D ˙1.
x!x0
x!x0
x!x0
x!x0
x!x0
x!x0
(b) If lim f .x/ D L < 0 and lim g.x/ D ˙1, then lim f .x/g.x/ D 1.
The prototype of infinite limits are those collected in the coming example, whose
proof is also an easy exercise for the reader.
Example 9.15
1
(a) If f W .x0 ; C1/ ! R is given by f .x/ D xx
, then limx!x0 f .x/ D C1.
0
1
(b) If f W .1; x0 / ! R is given by f .x/ D xx0 , then limx!x0 f .x/ D 1.
Let’s now turn our attention to limits at infinity.
Definition 9.16 Given f W .a; C1/ ! R and L 2 R, we write limx!C1 f .x/ D L
if, for a given > 0, there exists A > a such that
x > A ) jf .x/ Lj < :
As before, we leave to the reader the task of stating definitions analogous to
the above one for limx!1 f .x/ D L and limx!˙1 f .x/ D ˙1. On the other
hand, note that Propositions 9.7, 9.8 and 9.14 remain true if we change I n fx0 g by
.a; C1/ (resp. .1; b/) and x0 by C1 (resp. 1); moreover, the proofs of the
corresponding results in these cases are essentially the same.
Let us see an interesting (and useful) example of application of Proposition 9.14.
Example 9.17 Let n 2 N and f W R ! R be the polynomial function given by
f .x/ D an xn C an1 xn1 C C a1 x C a0 ;
with a0 ; a1 ; : : : ; an 2 R and an > 0. Prove that:
(a) If n is even, then limjxj!C1 f .x/ D C1.
(b) If n is odd, then limx!˙1 f .x/ D ˙1.
Proof Since limjxj!C1 xn D C1 if n iseven, limx!˙1 xn D ˙1 if n is odd and
a1
a0
limjxj!C1 an C an1
x C C xn1 C xn D an (verify these claims!), it suffices to
write
a1
an1
a0
f .x/ D xn an C
C C n1 C n
x
x
x
and, then, apply Proposition 9.14 (with ˙1 in place of x0 ).
t
u
9.2 Limits of Functions
289
One of the most important applications of infinite limits and limits at infinity is in
identifying horizontal and vertical asymptotes of (the graph of) a function, according
to the coming
Definition 9.18 If I is an interval, x0 2 I and f W I n fx0 g ! R is a function such
that limx!x0 f .x/ D ˙1, then we say that line x D x0 is a vertical asymptote of
(the graph of) f . Analogously, if f W .a; C1/ ! R (resp. f W .1; b/ ! R) is such
that limx!C1 f .x/ D L (resp. limx!1 f .x/ D L), then we say that line y D L is a
horizontal asymptote of (the graph of) f .
Examples 9.19
(a) Figure 6.17 sketches the graph of the tangent function in the interval . 2 ; 2 /.
In view of the limits limx! 2 sin x D 1 and limx! 2 cos x D 0, it follows from
Proposition 9.14 that limx! 2 tan x D C1. Analogously, limx! 2 tan x D
1, so that lines x D ˙ 2 are vertical asymptotes of this graph.
2x
(b) Figure 9.5 sketches the graph of f W R n f1g ! R, given by f .x/ D xC1
.
This graph can be obtained from that of the inverse proportionality function (cf.
Fig. 6.14) with the aid of Problem 10, page 194. Indeed, since
2x
2x C 2 2
2
D
D2
;
xC1
xC1
xC1
(9.12)
y
(0, 2)
(−1, 0)
O
Fig. 9.5 Graph of x 7!
2x
xC1
x
290
9 Limits and Derivatives
we can sketch the graph of f in the following way: firstly, we sketch the graph
of x 7! 1x ; then, we translate it one unit to the left, thus getting the graph of
1
x 7! xC1
; thirdly, we stretch this second graph in the vertical direction by factor
2
2, obtaining the graph of x 7! xC1
; we then reflect the result along the horizontal
2
axis, arriving at the graph of x 7! xC1
; we end by translating this last graph
2x
two units above, thus finally obtaining the graph of x 7! xC1
.
1
Since limx!1˙ 2x D 2 and limx!1˙ xC1
D ˙1, it follows from
2x
Proposition 9.14 that limx!1˙ xC1 D 1, so that line x D 1 is indeed a
vertical asymptote of the graph. On the other hand, it easily follows from (9.12)
2x
that limx!˙1 xC1
D 2, so that line y D 2 is a horizontal asymptote of the graph.
Finally, note that these results are in perfect accordance with the geometric intuition
given by the sketch of the graph.
Problems: Section 9.2
1. Let I be an interval and x0 2 I. Given a function f W I n fx0 g ! R, prove that if
limx!x0 f .x/ exists, then it is unique.
2. * Let I be an interval, x0 2 I and f ; g W I n fx0 g ! R be such that there exist
limx!x0 f .x/ D L and limx!x0 g.x/ D M. If f .x/ g.x/ for every x 2 I n fx0 g,
prove that L M.
3. * Establish the generalizations of items (a) and (b) of Proposition 9.7, as
discussed right after the proof of it.
4. * Extend Proposition 9.7 to one-sided limits.
5. * Let f W .a; b/ ! R be a given function, and x0 2 .a; b/. Prove that limx!x0 f .x/
exists if and only if the one-sided limits limx!x0 C f .x/ and limx!x0 f .x/ exist
and are equal. In this case, show also that limx!x0 f .x/ equals this common
value.
6. Prove Proposition 9.14 and its analogue for limits at infinity.
7. * Let 1 a < b C1 and f W .a; b/ ! R be a continuous and increasing
(resp. decreasing) function. If limx!aC f .x/ D L and limx!b f .x/ D M, with
1 L; M C1, prove that Im .f / D .L; M/.
8. Compute the following limits:
x sin x
sin 2 .x1/
(a) limx!0 1cos
(d) limx!1 x1
.
x.
x
(b) limx!0 1cos
.
cos. x
/
2
x
(e) limx!1 x12 .
sin.2x/
p
3
(c) limx!0 sin.3x/
(f) limx!0 1 x2cos x .
9. If f ; g W .a; C1/ ! R are such that f is bounded and limx!C1 g.x/ D 0,
prove that limx!C1 f .x/g.x/ D 0. Then, use this fact to compute the two limits
below:
(a) limx!C1
(b) limx!C1
sin x
.
x
sin x
.
xCcos x
9.2 Limits of Functions
291
10. * Let I be an interval, x0 2 I and f W I n fx0 g ! R be such that limx!x0 f .x/ D L.
If .an /n1 is a sequence in I n fx0 g satisfying limn!C1 an D x0 , prove that
limn!C1 f .an / D L. Then, use this fact to conclude that limx!0 sin 1x does not
exist.
11. * Given f W .A; C1/ ! R, we say that line y D axCb is an oblique asymptote
(of the graph) of f provided limx!C1 .f .x/ .ax C b// D 0. If this is so, prove
that
a D lim
x!C1
f .x/
x
and b D lim .f .x/ ax/:
x!C1
Then, elaborate the concept of oblique asymptote y D axCb for f W .1; A/ !
R and show how to compute a and b in such a case.
12. In each of the following items, find out (cf. previous problem) the horizontal,
vertical and oblique asymptotes of the given function:
(a) f W R n f0g ! R such that f .x/ D x C 1x for every real x ¤ 0.
p
(b) f W R n .a; a/ ! R such that f .x/ D ba x2 a2 for jxj a. (Notice—
cf. Chap. 6 of [4], for instance—that the graph of f is the portion of the
2
2
hyperbola ax2 by2 D 1 situated in the upper semiplane of the cartesian
plane.)
(c) f W .0; C1/ ! R such that f .x/ D x2 sin 1x for x > 0.
q
p
p
p
13. Compute limx!C1
xC xC x x .
14. Let f W .0; C1/ ! .0; C1/ be an increasing function. If limx!C1
prove that limx!C1
f .nx/
f .x/
f .2x/
f .x/
D 1,
D 1 for every n 2 N.
p
15. Let .an /r
n1 be the sequence defined by a1 D 2 2 and, for an integer n > 1,
q
p
p
n
an D 2 2 2 C 2 C C 2, with n square root signs. In this respect,
do the following items:
q
p
p
(a) If bn D 2 C 2 C C 2 for n > 1, with n1 square root signs, show
that b2n D 2 C bn1 and conclude that bn < 2 for every n > 1.
(b) Write bn D 2 cos n , with 0 < n < 2 , to conclude that n D 12 n1 for
every n > 2.
(c) Show that an D 2nC1 sin 2n D 2nC1 sin 21n . Then, use this fact to show that
limn!C1 an D .
16. (IMO) Find all functions f W .0; C1/ ! .0; C1/ satisfying the following
conditions:
(a) f .xf .y// D yf .x/ for all x; y > 0.
(b) limx!C1 f .x/ D 0.
292
9 Limits and Derivatives
9.3 Basic Properties of Derivatives
With the concept of limit at our disposal, we return to the discussion that lead
to (9.3).
Definition 9.20 Let I R be an open interval and f W I ! R be a given function.
For a fixed x0 2 I, we say that f is differentiable at x0 if the limit
lim
x!x0
f .x/ f .x0 /
x x0
does exist. In this case, this limit is called the derivative of f at x0 , and is
denoted f 0 .x0 /.
In the settings of the previous definition, we observe that the existence of f 0 .x0 /
is equivalent to the existence of the limit
lim
h!0
f .x0 C h/ f .x0 /
:
h
Indeed, on the one hand, letting x0 C h D x we have h D x x0 , and obviously
h ! 0 , x ! x0 . On the other hand, when we write f .x0 C h/ we are tacitly
assuming that h is so small that x0 C h 2 I; however, since we are computing a limit
and I is open, such a supposition doesn’t impose any restriction upon the definition
of derivative.
In short, f being differentiable at x0 2 I, we have
f 0 .x0 / D lim
x!x0
f .x/ f .x0 /
f .x0 C h/ f .x0 /
:
D lim
h!0
x x0
h
(9.13)
From now on, whenever convenient we refer to any one of the fractions above as the
Newton’s quotient of f at x0 .
The coming examples compute derivatives of some simple functions.
Example 9.21
(a) If f W R ! R is a constant function, then f is differentiable and f 0 .x0 / D 0 for
every x0 2 R.
(b) If n 2 N and f W R ! R is such that f .x/ D xn for every x 2 R, then f is
differentiable, with f 0 .x0 / D nxn1
for every x0 2 R.
0
(c) If n 2 Z is negative and f W R n f0g ! R is given by f .x/ D xn , then f is
differentiable and f 0 .x0 / D nx0n1 for every x0 2 R n f0g.
Proof (a) Let’s apply the first equality in (9.13): letting f .x/ D c for every x 2 R,
we get
f 0 .x0 / D lim
x!x0
f .x/ f .x0 /
cc
D lim
D lim 0 D 0:
x!x
x!x0
x x0
0 x x0
9.3 Basic Properties of Derivatives
293
(b) Let’s apply the second equality in (9.13): by Newton’s binomial formula, we
get
!
!
n
1
1 X n nk k
f .x0 C h/ f .x0 /
n
n
n
x h x0
D ..x0 C h/ x0 / D
h
h
h kD0 k 0
!
!
n
n
X
X
n nk k1
n nk k1
n1
D
x h
x h :
D nx0 C
k 0
k 0
kD1
kD2
Then, successively applying (9.7) and (9.6), we obtain
!
!
n
X
n
f 0 .x0 / D lim nx0n1 C
x0nk hk1
h!0
k
kD2
!
!
n
X
n nk k1
n1
D nx0 C
x0 h
D nx0n1 :
lim
h!0
k
kD2
(c) Let n D m, with m 2 N. Then,
f .x/ f .x0 /
1
D
x x0
x x0
1
1
xm xm
0
1
D m m
x x0
xm x m
0
x x0
:
Hence, Proposition 9.7 and (9.6) furnish
m
1
x m xm
x xm
0
0
D lim m m lim
x!x0
x!x0 x x
x!x0
x x0
x x0
0
m
1
x xm
0
:
D 2m lim
x0 x!x0 x x0
f 0 .x0 / D lim
1
xm xm
0
Now, applying the result of item (b) (with m in place of n), we get
f 0 .x0 / D 1
mxm1
D mx0m1 D nx0n1 :
0
x2m
0
t
u
Example 9.22 The sine and cosine functions are differentiable, with sin0 x0 D cos x0
and cos0 x0 D sin x0 for every x0 2 R.
Proof Let’s do the proof for the sine function, the proof for the cosine function
being completely analogous.
294
9 Limits and Derivatives
It follows from the product formulas of Trigonometry that
sin h2
h
sin.x0 C h/ sin x0
D h cos x0 C
:
h
2
2
Since the cosine function is continuous, Proposition 9.7, (9.6) and the fundamental
trigonometric limit (cf. Lemma 9.11) give
sin0 x0 D lim
h!0
D lim
h!0
sin h2
h
2
sin h2
h
2
h
cos x0 C
2
h
lim cos x0 C
h!0
2
sin h
cos x0 D cos x0 :
h!0 h
D lim
t
u
Remarks 9.23 Before we proceed with the development of the theory, let’s pause to
make two useful remarks.
1. Since derivatives are limits, given a function f W Œa; b/ ! R we can consider the
right-handed derivative of f at a, i.e., the one-sided limit
fC0 .a/ D lim
x!aC
f .x/ f .a/
;
xa
provided it exists. Accordingly, for f W .a; b ! R we can consider the lefthanded derivative at b, i.e., the limit
f0 .b/ D lim
x!b
f .x/ f .b/
xb
(also as before, provided such a limit exists). Moreover, whenever we say that
a function f W Œa; b ! R is differentiable, we shall implicitly assume that
its derivatives at x D a and x D b are one-sided ones, and shall write simply
f 0 .a/ and f 0 .b/ (instead of fC0 .a/ and f0 .b/) to denote them. From now on, such
conventions will be in force, without further comments.
2. As in Sect. 9.1, we shall sometimes denote the derivative of a function f at a point
df
x0 by using Leibniz’s classical notation dx
.x0 /, so that
f .x/ f .x0 /
df
.x0 / D lim
:
x!x
dx
0
x x0
The definition of derivative, together with the heuristic discussion of Sect. 9.1,
allows us to present another important
9.3 Basic Properties of Derivatives
Fig. 9.6 Tangent to the graph
of x 7! 1x
295
y
P
A
O
x
Q
Definition 9.24 Let I be an interval and f W I ! R be a function differentiable at
x0 2 I. The tangent line to the graph of f at the point .x0 ; f .x0 // is the line that
passes through this point and has slope f 0 .x0 /.
If f W I ! R is differentiable at x0 , it follows from the definition above and
elementary analytic geometry (cf. Chap. 6 of [4], for instance) that the tangent line
to its graph at .x0 ; f .x0 // has equation
y f .x0 / D f 0 .x0 /.x x0 /:
(9.14)
Example 9.25 Let H be the portion of the curve xy D 1 contained in the first
quadrant of a cartesian plane,2 and P and Q be points along the axes, such that
!
line PQ is tangent to H at the point A (cf. Fig. 9.6). Prove that:
(a) AP D AQ.
(b) If O is the origin of the cartesian plane, then the area of triangle POQ doesn’t
depend on the position of A along H.
Proof It is clear that H coincides with the graph of f W .0;C1/ ! R, given by
f .x/ D 1x . If x0 > 0 and A is the point .x0 ; f .x0 // D x0 ; x10 , then the line tangent
to Gf at A has equation y x10 D f 0 .x0 /.x x0 /. Example 9.21 gives f 0 .x0 / D x12 ,
so that the equation of the tangent line can be written as y which is the same, x20 y C x D 2x0 .
1
x0
0
D x12 .x x0 / or,
0
Making x D 0 and then y D 0 in such an equation, we get P 0; x20
Q.2x0 ; 0/, or vice-versa. In any case, items (a) and (b) follow immediately:
and
2
Such a portion is a branch of an equilateral hyperbola (cf. Chap. 6 of [4], for instance), but we
shall not use this fact.
296
9 Limits and Derivatives
(a) It’s a well know fact (again, see Chap. 6 of [4], for instance) that the midpoint of
PQ has abscissa and ordinate respectively equal to 12 x20 D x10 and 12 2x0 D x0 .
Hence, such a point coincides with A.
(b) Since POQ is a right triangle, we have
A.POQ/ D
1
1 2
OP OQ D 2x0 D 2:
2
2 x0
t
u
Continuing with the analysis of the concept of derivative, let’s show that
differentiability is stronger than continuity.
Proposition 9.26 If a function f W I ! R is differentiable at x0 2 I, then f is
continuous at x0 .
Proof Note first that, for x 2 I n fx0 g, we have
f .x/ D f .x0 / C
f .x/ f .x0 /
.x x0 /:
x x0
On the other hand, since we are assuming the existence of limx!x0
f 0 .x0 /, it follows from Proposition 9.7 that
f .x/f .x0 /
xx0
D
f .x/ f .x0 /
f .x0 / C
.x x0 /
x!x0
x x0
f .x/ f .x0 /
lim .x x0 /
D f .x0 / C lim
x!x0
x!x0
x x0
lim f .x/ D lim
x!x0
D f .x0 / C f 0 .x0 / 0 D f .x0 /:
Therefore, Proposition 9.4 assures the continuity of f at x0 .
t
u
The coming example shows that there exist continuous functions which are not
differentiable in an arbitrary finite set of points of their domains.3
Example
P9.27 Given distinct a1 ; a2 ; : : : ; an 2 R, the function f W R ! R such that
f .x/ D njD1 jxaj j is continuous, though not differentiable at any of a1 , a2 , . . . , an .
Proof The given function is obviously continuous. For what is left to do note that,
for a fixed 1 k n, we have
X
f .x/ D jx ak j C
jx aj j;
1jn
j¤k
3
Later, in Chap. 11, we shall see an example of a continuous function f W R ! R that is not
differentiable at any point.
9.3 Basic Properties of Derivatives
297
and in an appropriate neighborhood of ak each summand jx aj j, with j ¤ k, equals
x aj or aj x. Therefore, it suffices to show that x 7! jx ak j is not differentiable
at ak .
For the sake of notation, write f .x/ D jx aj, with a 2 R. Then, for x ¤ a, we
have
f .x/ f .a/
jx aj
1; if x > a
D
D
:
1; if x < a
xa
xa
.a/
.a/
Hence, limx!aC f .x/f
D 1 and limx!a f .x/f
D 1, so that Problem 5,
xa
xa
f .x/f .a/
page 290, assures that limx!a xa does not exist.
u
t
If I R is an interval and f W I ! R is differentiable at each x0 2 I, we shall
simply say that f is differentiable function. In this case, the derivative function
f 0 W I ! R of f associates to x 2 I the derivative f 0 .x/ of f at x.
If I and J are intervals and f W I ! J is a differentiable bijection, the last
proposition guarantees that f is continuous. Moreover, Theorem 8.35 assures that
f 1 W J ! I is also continuous. Our next result explains when f 1 is differentiable.
Theorem 9.28 Let I and J be intervals and f W I ! J be a differentiable bijection.
For x0 2 I and y0 D f .x0 / 2 J, we have f 1 W J ! I differentiable at y0 if and only
if f 0 .x0 / ¤ 0. Moreover, if this is so, then
.f 1 /0 .y0 / D
1
f 0 .x
0/
:
(9.15)
Proof For h sufficiently close to (but different from) 0, let x D x0 C h, l D f .x0 C
h/ f .x0 / and y D y0 C l. Then, y D f .x0 / C l D f .x0 C h/, so that
h D .x0 C h/ x0 D f 1 .y0 C l/ f 1 .y0 /:
Hence, the continuity of f and f 1 guarantee that h ! 0 , l ! 0. Also, notice that
1
l h
f .x0 C h/ f .x0 /
f .y0 C l/ f 1 .y0 /
D D 1:
h
l
h l
Now, suppose that f 1 is differentiable at y0 . Making h ! 0 (which, as we have
just seen, is equivalent to making l ! 0), it follows from item (b) of Proposition 9.7
and the computations above that
1
f .y0 C l/ f 1 .y0 /
f .x0 C h/ f .x0 /
1 D lim
h!0
h
l
1
f .x0 C h/ f .x0 /
f .y0 C l/ f 1 .y0 /
lim
D lim
h!0
l!0
h
l
D f 0 .x0 /.f 1 /0 .y0 /:
Therefore, f 0 .x0 / ¤ 0 and .f 1 /0 .y0 / D
1
.
f 0 .x0 /
298
9 Limits and Derivatives
Conversely, suppose that f 0 .x0 / ¤ 0. Since
1
f 1 .y0 C l/ f 1 .y0 /
D
;
l
.f .x0 C h/ f .x0 //=h
item (c) of Proposition 9.7, together with the fact that l ! 0 , h ! 0, furnishes
lim
l!0
f 1 .y0 C l/ f 1 .y0 /
l
1
l!0 .f .x0 C h/ f .x0 //=h
1
1
D 0
:
D lim
h!0 .f .x0 C h/ f .x0 //=h
f .x0 /
D lim
Therefore, f 1 is differentiable at y0 .
t
u
Example 9.29 Let n > 1 be
p an integer and f W Œ0; C1/ ! Œ0; C1/ be the n-th
root function, i.e., f .x/ D n x D x1=n . Then, f is not differentiable at x D 0 but is
differentiable at every x > 0, with f 0 .x/ D 1n x1=n1 .
Proof Recall that f D g1 , where g W Œ0; C1/ ! Œ0; C1/ is such that g.y/ D yn .
Hence, it follows from the previous result that f is differentiable at x D g.y/ if
and only if g0 .y/ ¤ 0. However, since g0 .y/ D nyn1 , we have f differentiable at
x D g.y/ if and only if y ¤ 0por, which is the same, x ¤ 0. In this last case, recalling
that x D g.y/ D yn , y D n x, (9.15) gives
f 0 .x/ D
1
g0 .y/
D
1
1
1
D p
D x1=n1 :
n
n1
n1
ny
n
n. x/
t
u
For an important generalization of the previous example, see Problem 4,
page 310.
We close this section by presenting the concept of higher order derivatives. To
this end, let be given an interval I and a function f W I ! R.
We say that f is twice differentiable at x0 2 I if there exists a neighborhood
.x0 r; x0 C r/ of x0 such that f is differentiable in I \ .x0 r; x0 C r/ and f 0 W
I \ .x0 r; x0 C r/ ! R is differentiable at x0 2 I. In this case, we also say that
.f 0 /0 .x0 / is the second derivative of f at x0 , and write f 00 .x0 / D .f 0 /0 .x0 /. If f is
twice differentiable at each x0 2 I, we shall simply say that f is twice differentiable
in I, and let f 00 W I ! R denote its second derivative function.
More generally, let k 2 N and suppose we have already defined what is meant
by the k-th derivative f .k/ .x0 / of f at x0 2 I. If f is k times differentiable in
I (i.e., if f .k/ .x0 / exists for every x0 2 I), we let f .k/ W I ! R denote the k-th
9.3 Basic Properties of Derivatives
299
derivative4 function of f . If f is k times differentiable in a neighborhood .x0 r; x0 C
r/ of x0 and f .k/ W I \ .x0 r; x0 C r/ ! R is differentiable at x0 , then we say that f
is k C 1 times differentiable at x0 , and write
f .kC1/ .x0 / D .f .k/ /0 .x0 /
to denote its .k C 1/-th derivative at x0 .
If f W I ! R is k times differentiable in I and f .k/ W I ! R is continuous, we
shall say that f is k times continuously differentiable (or simply continuously
differentiable, if k D 1) in I. Finally, if f is k times differentiable in I for every
k 2 N, then we say that f is infinitely differentiable in I.
Example 9.30 If n 2 Z n f0g and f W R ! R is given by f .x/ D xn , it easily follows
from item (b) of Example 9.21 and Problem 1 that f is infinitely differentiable in
R n f0g (in R, if n 0), with
f .k/ .x/ D n.n 1/ : : : .n k C 1/xnk
for every k 2 N. In particular, if n 0 and k > n, then f .k/ vanishes identically.
Example 9.31 Functions sin; cos W R ! R are infinitely differentiable, with
sin.4k/ D sin; sin.4kC1/ D cos; sin.4kC2/ D sin; sin.4kC3/ D cos
and
cos.4k/ D cos; cos.4kC1/ D sin; cos.4kC2/ D cos; cos.4kC3/ D sin;
for every k 2 N.
Problems: Section 9.3
1. * Let I R be an interval and f W I ! R be differentiable at x0 2 I. For a fixed
c 2 R, prove that:
(a) Function g W I ! R given by g.x/ D cf .x/ is differentiable at x0 , with
g0 .x0 / D cf 0 .x0 /.
(b) Function h W I ! R given by h.x/ D f .x/ C c is differentiable at x0 , with
h0 .x0 / D f 0 .x0 /.
Since f .k/ is also used to denote the composition of f W I ! I with itself, k times, we will rely on
the context to clear any danger of confusion.
4
300
9 Limits and Derivatives
The next problem establishes a simple particular case of the chain rule (cf.
Theorem 9.19).
2. * Let
at x0 2 .a; b/. For c > 0 (resp. c < 0), let
f W .a; b/ ! R be differentiable
g W ac ; bc ! R (resp. g W bc ; ac ! R) be the function given by g.x/ D f .cx/.
Show that g is differentiable at xc0 , with g0 xc0 D cf 0 .x0 /.
3. * Let I be an interval and k 2 N. If f ; g W I ! R are k times differentiable in
x0 2 I, prove that f C g and fg are k times differentiable at x0 .
4. * In Example 8.37, we defined arcsin W Œ1; 1 ! Œ 2 ; 2 and arccos W
Œ1; 1 ! Œ0; as the inverses of sin W Œ 2 ; 2 ! Œ1; 1 and cos W Œ0; !
Œ1; 1, respectively. Use Theorem 9.28 to prove that the restrictions of arcsin
and arccos to the open interval .1; 1/ are differentiable, with
1
arcsin0 x D p
1 x2
and
1
arccos0 x D p
;
1 x2
para todo x 2 .1; 1/.
5. Compute the indicated derivatives (in each case, assume that the domain of the
corresponding function is the maximal one):
(a) f 0 .1/; f .x/ D
1
2.
4x
p
x
.
p5
(e)
(f)
(g)
(h)
(b) f 0 .2/; f .x/ D
(c) f 0 .2/; f .x/ D p3x.
(d) f 0 .1/; f .x/ D 3 2x.
f 0 ./;
f .x/ D sin.3x/.
f 0 2 ; f .x/ D 4 cos 3x .
f 0 . 14 /; f .x/ D arcsin.2x/.
f 0 . 16 /; f .x/ D 2 arccos.3x/.
6. Let f ; g W .a; b/ ! R be differentiable functions whose graphs pass through a
given point A. We say that these graphs are tangent to each other at A if their
tangent lines at A coincide. In this respect, do the following items:
(a) Show that, for m; n 2 N, the graphs of f .x/ D xn and g.x/ D xm are tangent
to each other at .0; 0/, but are tangent at .1; 1/ only if m D n.
(b) Let f W .a; b/ ! .a; b/ be a differentiable bijection such that f 1 is
also differentiable. If the graphs of f and f 1 are tangent to each other at
A.x0 ; y0 /, compute the possible values of f 0 .x0 /f 0 .y0 /.
7. Let I R be an interval and f W I ! R be continuous at x0 2 I. If the function
x 7! xf .x/ (x 2 I) is differentiable at x0 , prove that f is also differentiable at x0 .
8. * Let I be an open interval, x0 2 I and f W I ! R be differentiable at x0 , with
f 0 .x0 / D L. Given sequences .an /n1 and .bn /n1 in I, with an < x0 < bn for
every n 1, show that
lim
n!C1
f .bn / f .an /
D L:
bn an
9.3 Basic Properties of Derivatives
301
9. Given a differentiable function f W R ! R, we say that a real number x0 is a root
of f provided f .x0 / D 0. Newton’s method5 for the computation of numerical
approximations of the roots of f guarantees that, under certain conditions and
for an appropriate choice of ˛ 2 R, the sequence .an /n1 defined by a1 D ˛ and
anC1 D an ff0.a.ann// for n 1 (of course assuming f 0 .an / ¤ 0 for every n 1)
converges to a root of f . In this respect, do the following items:
(a) Show that anC1 is the point at which the tangent line to the graph of f at
.an ; f .an // intersects the horizontal axis.
(b) Given a > 0, explain how Newton’s method make it natural the definition
of the sequence .an /n1 of Problem 19, page 220.
p
10. Given R > 0, let f W .R; R/ ! R be given by f .x/ D R2 x2 . Prove that:
(a) The graph of f is the upper semicircle of radius R, centered at the origin of
the cartesian plane and with diameter along the horizontal axis.
(b) For x0 2 .R; R/, we have f 0 .x0 / D p x20 2 .
R x0
(c) Given x0 2 .R; R/ and A.x0 ; f .x0 //, the tangent line to the graph of f ,
obtained according to Definition 9.24, coincides with the straight line that
passes through A and is perpendicular to the radius OA.
11. Let I be an open interval and f W I ! R be a differentiable function. Given
a; b 2 R, suppose that there exists a straight line r, tangent to the graph of f and
passing through the point .a; b/. If .x0 ; y0 / is the point at which r is tangent to
the graph of f , show that
y0 D f .x0 /
:
y0 b D f 0 .x0 /.x0 a/
Then, let r and r0 be the straight lines that pass through the point .1; 1/ and
2
are tangent to the parabola y D x4 ; use the system above to find the points at
0
which r e r touch the parabola.
For the next problem, the reader may find it helpful to read the proof of
Theorem 6.62 again, as well as to review the elementary facts on conics (cf.
Chap. 6 of [4], for instance).
12. Let P be the parabola with focus F and directrix d. If P is a point on d and
!
!
A; B 2 P are such that AP and BP are tangent to P, show that F 2 AB.
For the next problem, the reader may find it convenient to have at his/her
disposal an extension of the definition of limx!x0 f .x/ to the case of a function
f W X ! R, where X R is a nonempty set and x0 2 R is the limit of a
sequence of distinct elements of X (for instance—and this will
p essentially be
the case of our interest –, we can have X D Q and x0 D 2). Fortunately,
such an extension is formally identical to the one we have already met: we say
5
We shall have more to say on Newton’s method in Problem 7, page 455.
302
9 Limits and Derivatives
that limx!x0 f .x/ exists and is equal to L if, for every > 0, there exists ı > 0
(depending on ) such that
x 2 X and 0 < jx x0 j < ı ) jf .x/ Lj < :
13. (OIMU) Let f W .0; C1/ ! R be given by
(
f .x/ D
0; if x … Q
:
1
; if x D pq with p; q 2 NI gcd.p; q/ D 1
q3
If k 2 N is not a perfect square, show that f is differentiable at x D
p
k.
9.4 Computing Derivatives
This section is devoted to the presentation of the usual differentiation rules, i.e.,
formulas that show how to compute derivatives of certain functions constructed in
terms of two given differentiable functions.
For item (c) of the coming proposition, we recall that if g W I ! R is
differentiable at x0 2 I, then g is continuous at x0 ; therefore, if g.x0 / ¤ 0, then
there exists an interval J I, containing x0 and such that g ¤ 0 in J. Thus, we think
of gf as defined in such a J.
Proposition 9.32 If f ; g W I ! R are differentiable at x0 2 I, then:
(a) f ˙ g is differentiable at x0 , with .f ˙ g/0 .x0 / D f 0 .x0 / ˙ g0 .x0 /.
(b) fg is differentiable at x0 , with .fg/0 .x0 / D f 0 .x0 /g.x0 / C f .x0 /g0 .x0 /.
(c) If g.x0 / ¤ 0, then gf is differentiable at x0 , with
0
f 0 .x0 /g.x0 / f .x0 /g0 .x0 /
f
.x0 / D
:
g
g.x0 /2
Proof
(a) Let us prove that f C g is differentiable at x0 , with .f C g/0 .x0 / D f 0 .x0 / C g0 .x0 /
(the claims relative to f g can be proved in quite analogous ways). Since
.f C g/.x/ .f C g/.x0 /
f .x/ f .x0 /
g.x/ g.x0 /
D
C
x x0
x x0
x x0
for x 2 I n fx0 g, the arithmetic properties on limits of functions guarantee that,
if f 0 .x0 / and g0 .x0 / exist, then .f C g/0 .x0 / also does and
9.4 Computing Derivatives
303
.f C g/.x/ .f C g/.x0 /
x x0
g.x/ g.x0 /
f .x/ f .x0 /
D lim
C
x!x0
x x0
x x0
.f C g/0 .x0 / D lim
x!x0
D lim
x!x0
f .x/ f .x0 /
g.x/ g.x0 /
C lim
x!x0
x x0
x x0
D f 0 .x0 / C g0 .x0 /:
(b) Firstly, note that
.fg/.x/ .fg/.x0 /
D
x x0
g.x/ g.x0 /
f .x/ f .x0 /
g.x/ C
f .x0 /:
x x0
x x0
(9.16)
Now, since g is differentiable at x0 , it follows from Proposition 9.26 that g is
continuous at x0 , so that
lim g.x/ D g.x0 /:
x!x0
Therefore, letting x ! x0 in (9.16) and applying the arithmetic properties on
limits of functions, we get
lim
x!x0
.fg/.x/ .fg/.x0 /
f .x/ f .x0 /
D lim
lim g.x/
x!x0
x!x0
x x0
x x0
C f .x0 / lim
x!x0
g.x/ g.x0 /
x x0
D f 0 .x0 /g.x0 / C f .x0 /g0 .x0 /:
(c) Let us firstconsider the case in which f is constant and equal to 1, and let us
0
g0 .x0 /
show that 1g .x0 / D g.x
2 . Since g.x/ ¤ 0 for x 2 J I, we have for
0/
x 2 J n fx0 g that
1
x x0
1
1
g.x/ g.x0 /
g.x/ g.x0 /
D
x x0
1
:
g.x/g.x0 /
From the equality above and invoking the differentiability (and continuity) of g
at x0 , we get
lim
x!x0
1
x x0
1
1
g.x/ g.x0 /
D g0 .x0 / 1
:
g.x0 /2
304
9 Limits and Derivatives
For the general case, we use the particular case above, together with the result
of (b):
0
0
1 0
f
1
1
.x0 / C f .x0 /
.x0 / D f .x0 / D f 0 .x0 /
.x0 /
g
g
g
g
D
g0 .x0 /
f 0 .x0 /g.x0 / f .x0 /g0 .x0 /
f 0 .x0 /
f .x0 / D
:
g.x0 /
g.x0 /2
g.x0 /2
t
u
Example 9.33 Let f W R ! R be the polynomial function given by
f .x/ D an xn C an1 xn1 C C a1 x C a0 ;
with a0 ; a1 ; : : : ; an 2 R and an ¤ 0. Applying Example 9.21 and item (a) of the
previous proposition several times, we conclude that f is differentiable, with
f 0 .x/ D nan xn1 C .n 1/an1 xn2 C C 2a2 x C a1
for every x 2 R. Then, f 0 W R ! R is also polynomial, and an easy inductive
argument shows that f is indeed infinitely differentiable, with f .nC1/ vanishing
identically.
For the next example, recall from Trigonometry that the secant function sec W
R n f 2 C kI k 2 Zg ! R is given by sec x D cos1 x , for every real x in its domain.
Example 9.34 If D D f 2 C kI k 2 Zg, then the tangent function tan W R n D ! R
is differentiable, with tan0 x D sec2 x for every x 2 R n D. Indeed, item (c) of the
previous proposition, together with the formulae deduced in Example 9.22, furnish
tan0 x D
cos2 x C sin2 x
1
sin0 x cos x sin x cos0 x
D
D
D sec2 x:
2
2
cos x
cos x
cos2 x
We now show that the arc-tangent function (cf. Example 8.37) is differentiable
and compute its derivative.
Example 9.35 Function arctan W R ! 2 ; 2 is differentiable, with
arctan0 y D
for every y 2 R.
1
1 C y2
Proof First of all, observe that tan0 x D sec2 x ¤ 0 for every x 2 2 ; 2 . Hence,
it follows from Theorem 9.28 that arctan is differentiable. Moreover, if y 2 R and
x D arctan y, then y D tan x and, thanks to (9.15) and the result of the former
example, we get
9.4 Computing Derivatives
305
arctan0 y D
1
1
1
1
D
D
D
:
tan0 x
sec2 x
1 C tan2 x
1 C y2
t
u
The most important of all differentiation rules is the chain rule. Loosely
speaking, it establishes the differentiability of the composite of two differentiable
functions, and shows how to compute such a derivative.
More precisely, let I and J be intervals and g W I ! J and f W J ! I be given
functions, with g differentiable at x0 2 I and f differentiable at y0 D g.x0 / 2 J.
Suppose that, in some neighborhood I \ .x0 r; x0 C r/ of x0 , equation g.x/ D y0
has x0 as its only root. Then, for x 2 I \ .x0 r; x0 C r/, we can safely write
f .g.x// f .g.x0 // g.x/ g.x0 /
.f ı g/.x/ .f ı g/.x0 /
D
:
x x0
g.x/ g.x0 /
x x0
Since g is continuous at x0 , we get limx!x0 g.x/ D g.x0 /, so that
lim
x!x0
.f ı g/.x/ .f ı g/.x0 /
f .g.x// f .g.x0 //
g.x/ g.x0 /
lim
D lim
x!x0
x!x0
x x0
g.x/ g.x0 /
x x0
D lim
y!y0
f .y/ f .y0 / 0
g .x0 / D f 0 .y0 /g0 .x0 /
y y0
D f 0 .g.x0 //g0 .x0 /:
As we shall see in Theorem 9.19, this formula continues to hold in the general
case, albeit the proof is much more involving, for, g.x/ D y0 can have infinitely
many solutions in any neighborhood of x0 (for instance, think of g as the function f
of Problem 11, and solve the equation g.x/ D 0). Hence, we need to develop some
preliminaries.
Let I R be an open interval and f W I ! R be differentiable at x0 2 I. Set
I x0 D fh 2 RI x0 C h 2 Ig;
so that I x0 is an open interval containing 0. Define r W I x0 ! R by letting
r.h/ D f .x0 C h/ f .x0 / f 0 .x0 /h:
Since f is continuous at x0 , we have
lim r.h/ D lim .f .x0 C h/ f .x0 / f 0 .x0 /h/ D 0 D r.0/;
h!0
h!0
so that r is continuous at 0. Also, the definition of f 0 .x0 / gives
r.h/
f .x0 C h/ f .x0 /
0
D lim
f .x0 / D 0:
lim
h!0 h
h!0
h
306
9 Limits and Derivatives
(This last computation, together with r.0/ D 0, actually shows that r is differentiable
at 0 with r0 .0/ D 0. Nevertheless, we won’t use this fact here.)
Conversely, we have the following auxiliary result.
Lemma 9.36 Let I R be an open interval, f W I ! R be a given function and
x0 2 I. If there exists a real number L such that the function r W I x0 ! R, given
by r.h/ D f .x0 C h/ f .x0 / Lh, satisfies the condition limh!0 r.h/
D 0, then f is
h
differentiable at x0 , with f 0 .x0 / D L.
Proof Since
r.h/
h
D
f .x0 Ch/f .x0 /
h
lim
h!0
L, the condition limh!0
r.h/
h
f .x0 C h/ f .x0 /
D L:
h
D 0 is equivalent to
t
u
Let I be an open interval and f W I ! R be a function continuous at x0 2 I.
For every L 2 R, the graph of the affine function h 7! f .x0 / C Lh passes
through .x0 ; f .x0 //. This being said, it is frequently useful to rephrase the previous
discussion by saying that f is differentiable at x0 if and only if f admits a best affine
approximation in a neighborhood of x0 , in the sense of Lemma 9.36.
In other words, letting f W I ! R be differentiable at x0 2 I, we have
f .x0 C h/ D f .x0 / C f 0 .x0 /h C r.h/;
(9.17)
with r W I x0 ! R such that limh!0 r.h/
h D 0. On the other hand, if h 7! f .x0 / C Lh
is an affine approximation of f in a neighborhood of x0 , such that setting r.h/ D
f .x0 C h/ f .x0 / Lh we have limh!0 r.h/
h D 0, then f is differentiable at x0 and
L D f 0 .x0 /.
We shall sometimes refer to (9.17), valid for all h 2 I x0 , as Taylor’s formula
of order 1 for f in a neighborhood of x0 . Notice that its real content is captured by
the fact that limh!0 r.h/
h D 0 if f is differentiable at x0 .
For what comes next, it’s more useful for us to write (9.17) as
f .x0 C h/ D f .x0 / C f 0 .x0 /h C R.h/h;
(9.18)
with R W I x0 ! R given by
R.h/ D
r.h/
h ;
if h ¤ 0
:
0; if h D 0
Notice that the continuity of r implies that of R at every h 2 I x0 different from 0,
whereas condition limh!0 r.h/
D 0 assures that R is continuous also at 0.
h
Conversely, if f can be written as
f .x0 C h/ D f .x0 / C Lh C R.h/h
9.4 Computing Derivatives
307
in I x0 , with R being continuous in I x0 and such that R.0/ D 0, then a slight
modification of the proof of Lemma 9.36 gives that f is differentiable at x0 , with
f 0 .x0 / D L.
We can finally state and prove the chain rule in the general case.
Theorem 9.37 (Chain Rule) Let I and J be open intervals and g W I ! J and
f W J ! R given functions. If g is differentiable at x0 2 I and f is differentiable at
g.x0 / 2 J, then f ı g W I ! R is differentiable at x0 , with
.f ı g/0 .x0 / D f 0 .g.x0 //g0 .x0 /:
(9.19)
Proof Let y0 D g.x0 /. According to (9.18), the differentiability of f at y0 allows us
to write
f .y0 C t/ D f .y0 / C f 0 .y0 /t C s.t/t
(9.20)
for every t 2 J y0 , with s W J y0 ! R continuous and such that s.0/ D 0.
Analogously, the differentiability of g at x0 furnishes, for h 2 I x0 ,
g.x0 C h/ D g.x0 / C g0 .x0 /h C r.h/h;
(9.21)
with r W I x0 ! R continuous and such that r.0/ D 0.
Now, for h 2 I x0 , it follows from (9.21) that
.f ı g/.x0 C h/ .f ı g/.x0 / D f .g.x0 C h// f .g.x0 //
D f .g.x0 / C g0 .x0 /h C r.h/h/ f .y0 /
D f .y0 C t.h// f .y0 /;
where t.h/ D g0 .x0 /h C r.h/h. Hence, for h 2 I x0 , it follows from this and (9.20)
that
.f ı g/.x0 C h/ .f ı g/.x0 / D f .y0 C t.h// f .y0 /
D f 0 .y0 /t.h/ C s.t.h//t.h/
D f 0 .y0 /.g0 .x0 /h C r.h/h/ C s.t.h//t.h/
D f 0 .g.x0 //g0 .x0 /h C f 0 .y0 /r.h/h C s.t.h//t.h/
D f 0 .g.x0 //g0 .x0 /h C f 0 .y0 /r.h/h
C s.t.h//.g0 .x0 / C r.h//h
D .f ı g/.x0 / D f 0 .g.x0 //g0 .x0 /h C R.h/h;
where R.h/ D f 0 .y0 /r.h/ C s.t.h//.g0 .x0 / C r.h// for h 2 I x0 .
308
9 Limits and Derivatives
Finally, the discussion at the next to last paragraph before the statement of
the chain rule guarantees that, in order to show that f ı g is differentiable at x0
and that (9.19) is true, it suffices to prove that R is continuous at I x0 , with
R.0/ D 0. But this is straightforward from the chain rule for continuous functions
(cf. Proposition 8.12), together with the fact that r, s and t are continuous, with
r.0/ D s.0/ D t.0/ D 0.
t
u
Example 9.38 The chain rule allows us to compute the derivative of the cosine
function from that of the sine function. Indeed, since cos x D .sin ıg/.x/, where
g W R ! R is given by g.x/ D 2 x, it follows from the chain rule that
cos0 x D .sin ıg/0 .x/ D sin0 g.x/ g0 .x/ D cos g.x/ .1/
x D sin x:
D cos
2
p
Example 9.39 Let R > 0 and f W .R; R/ ! R be given by f .x/ D Rp2 x2 .
Define g W .0; C1/ ! R and h W .R; R/ ! .0; C1/ by letting g.x/ D x and
1
h.x/ D R2 x2 , so that f D g ı h. Since g0 .y/ D 2p
for y > 0, the chain rule gives,
y
for x 2 .R; R/,
1
x
f 0 .x/ D .g ı h/0 .x/ D g0 .h.x//h0 .x/ D p
:
.2x/ D p
2
2 h.x/
R x2
A faster way of using the chain rule to compute the derivative of f at x 2 .R; R/
is this: first write f .x/2 Cx2 D R2 , then differentiate both sides with respect to x (and
x
with the aid of the chain rule) to get 2f .x/f 0 .x/ C 2x D 0; hence, f 0 .x/ D f .x/
D
x
p 2 2 .
R x
Example 9.40 We can use the chain rule, together with the formula for differentiation of a product, to establish the differentiation formula for a quotient. To this
end, let f ; g W I ! R be differentiable in I, with g.x/ ¤ 0 for every x 2 I. Letting
W R n f0g ! R denote the inverse proportionality function, so that .x/ D x1 for
every x ¤ 0, it follows from item (c) of Example 9.21 that is differentiable, with
f .x/
0
.x/ D x2 for every x 2 R n f0g. Now, notice that g.x/
D f .x/ . ı g/.x/ for
every x 2 I, so that the formula for differentiation of a product and the chain rule
give
0
f
.x/ D f 0 .x/ . ı g/.x/ C f .x/ . ı g/0 .x/
g
D f 0 .x/ 1
C f .x/ g.x/
0
.g.x//g0 .x/
D
f 0 .x/
C f .x/.g.x/2 /g0 .x/
g.x/
D
f 0 .x/g.x/ f .x/g0 .x/
:
g.x/2
9.4 Computing Derivatives
309
The coming example is considerably more sophisticated than the previous ones.
Example 9.41 Prove that f W R ! R, given by f .x/ D sin.x2 /, is not periodic.
Proof First of all, note that if f W R ! R is continuously differentiable and periodic,
then its derivative f 0 W R ! R, besides being continuous, is also periodic. Indeed,
if f .x/ D f .x C p/ for some real p > 0 and every x 2 R, then the chain rule gives
f 0 .x/ D f 0 .x C p/ for every x. Therefore,
0
/;
Im.f 0 / D Im.fjŒ0;p
and the continuity of f 0 , together with Corollary 8.30, guarantees that f 0 is a bounded
function.
Now, letting f be the given function and g W R ! R be given by g.x/ D x2 , we
have f .x/ D .sinıg/.x/, so that the chain rule furnishes
f 0 .x/ D sin0 g.x/ g0 .x/ D 2x cos.x2 /:
Hence, if f .x/ D sin.x2 / were periodic, the discussion at the preceding paragraph
would guarantee that f 0 would be bounded. Since this is obviously false, we have
reached a contradiction.
t
u
The following corollary of the chain rule will be of interest later.
Corollary 9.42 Let I and J be open intervals and g W I ! J and f W J ! R given
functions. If g is n times differentiable at x0 2 I and f is n times differentiable at
g.x0 / 2 J, then f ı g W I ! R is n times differentiable at x0 .
Proof We make induction on n, relying on the chain rule for the case n D 1. Assume
the corollary to be true when n D k, and let f and g be k C 1 times differentiable
at x0 .
In a neighborhood of x0 , we have .f ı g/0 D .f 0 ı g/g0 , with f 0 and g0 being k
times differentiable at that point. The induction hypothesis guarantees that f 0 ı g is k
times differentiable at x0 , from where Problem 3, page 300 (applied to f 0 ı g and g0 )
assures that .f ı g/0 D .f 0 ı g/g0 is also k times differentiable at x0 . Therefore, f ı g
is k C 1 times differentiable at x0 .
t
u
Problems: Section 9.4
1. Compute the derivatives of the given functions, assuming in each case that the
function and its derivative are defined in their maximal domains:
(a) f .x/ D x3p 2x2 C 7x C 1.
(b) f .x/ D x 3 x C x4 .
(c) f .x/ D 1x C x12
(d) f .x/ D cos x C tan x.
310
9 Limits and Derivatives
(e) f .x/ D x3psin x.
(f) f .x/ D 7 x.x2 C 4/.
(g) f .x/ D
(h) f .x/ D
x2 3x
.
2x3 C1
sin x
.
1Ccos x
2. Let I be an interval and f W Ip! .0; C1/ be differentiable at x0 2 I. If g W
I ! R is given by g.x/ D n f .x/, show that g is differentiable at x0 , with
1
g0 .x0 / D 1n f .x0 / n 1 f 0 .x0 /.
3. Compute the derivatives of the given functions, assuming in each case that the
function and its derivative are defined in their maximal domains:
(a)
(b)
(c)
(d)
f .x/ D .cos x/2 .
2
f .x/ D cos.x
/.
p
p
f .x/ D p1 x 1 x.
f .x/ D 1 x cos.x2 /.
(e) f .x/ D arcsin.1 x2 /.
2/
(f) f .x/ D x arccos.x
.
x2 C1
p
tg x
(g) f .x/ D 1Cx3 .
(h) f .x/ D sin.sin.cos x//.
4. Let r be a nonzero rational number and f W .0; C1/ ! R be given by f .x/ D xr .
Prove that f is differentiable, with f 0 .x/ D rxr1 for every x > 0.
For the next problem we recall a few facts on roots of polynomial functions.
Given a polynomial function f with real coefficients and degree n > 1, and a
real root ˛ of f , the division algorithm for polynomials (cf. Chap. 14 of [5])
guarantees the existence of an integer 1 k n and a polynomial g of degree
n k such that
f .x/ D .x ˛/k g.x/;
(9.22)
with g.˛/ ¤ 0. In this case, we say that k is the multiplicity of ˛ as a root of f .
5. Let f W R ! R be a nonconstant polynomial function. Show that:
(a) If ˛ is a root of f and f .x/ D .x ˛/k g.x/, with g.˛/ ¤ 0, then, for every
x 2 R such that f .x/ ¤ 0, we have
f 0 .x/
k
g0 .x/
D
C
:
f .x/
x˛
g.x/
(b) If f .x/ D a.x ˛1 /.x ˛2 / : : : .x ˛n /, with a; ˛1 ; : : : ; ˛n 2 R and a ¤ 0,
then, for x ¤ ˛1 ; : : : ; ˛n , we have
X 1
f 0 .x/
D
:
f .x/
x ˛j
jD1
n
6. If f W R n f3g ! R is given by f .x/ D
7. Let f W Œ0; C1/ ! R be given by
s
r
f .x/ D
xC
2xC1
x3 ,
xCC
with ten square root signs. Compute f 0 .0/.
compute f .5/ .0/.
q
xC
p
x C 1;
9.4 Computing Derivatives
311
pk .x/
8. For k 2 N, show that arctan.k/ .x/ D .1Cx
2 /k , where pk is a polynomial of degree
k 1 and leading coefficient .1/k1 kŠ.
9. Find the common tangents to the parabolas of equations y D x2 and y D 2 C
.x 3/2 .
For the next problem, the reader may find it helpful to review the elementary
material on conics, at the level of Chap. 6 of [4], for instance.
2
2
10. Let E be the ellipse of equation ax2 C by2 D 1 and H the hyperbola of equation
x2
a02
2
by02 D 1. We say that E and H are confocal if they have the same foci. In
this respect, show that:
(a) a2 b2 D a02 C b2 .
(b) E and H intersect in four distinct points; moreover, if .x0 ; y0 / is one of them,
then
1
1
1
1
1
1
1
1
2
2
x0
C 02 2 D 2 C 02 and y0
C 02 2 D 02 2 :
2
02
2
02
a b
a b
b
b
a b
a b
a
a
(c) If P is one of the intersection points of item (b), then the straight lines
tangent to E and H at P are perpendicular.
11. Do the following items:
(a) If f W R ! R is given by
f .x/ D
x2 sin 1x ; if x ¤ 0
;
0; if x D 0
show that it is differentiable at every real number and that f 0 is not
continuous at 0.
(b) For real numbers x1 < x2 < : : : < xn , give an example of a differentiable
function g W R ! R such that g0 is discontinuous at x1 , x2 , . . . , xn .
p
12. Prove that f W .0; C1/ ! R, given by f .x/ D sin x, is not periodic.
13. (Putnam) Let a1 , a2 , . . . , an be given reals and f W R ! R be such that f .x/ D
a1 sin x C a2 sin.2x/ C C an sin.nx/ for every x 2 R. If jf .x/j j sin xj for
every x 2 R, prove that ja1 C 2a2 C C nan j 1.
14. (OBMU) Given nonzero real numbers a1 ; a2 ; : : : ; an , show that the function f W
R ! R, such that
f .x/ D
n
X
jD1
for every x 2 R, has period 2.
aj cos.jx/
312
9 Limits and Derivatives
15. * Prove the following version of l’Hôpital’s rule6 : let f ; g W Œa; b/ ! R be two
functions differentiable at a and such that f .a/ D g.a/ D 0. If g ¤ 0 in .a; b/
and g0 .a/ ¤ 0, show that
f 0 .a/
f .x/
D 0 :
x!a g.x/
g .a/
lim
9.5 Rôlle’s Theorem and Applications
Let f W Œa; b ! R be continuous in Œa; b, differentiable in .a; b/ and such that
f .a/ D f .b/ D 0. In addition, suppose that f is nonzero in at least one point of the
interval .a; b/. Arguing heuristically, translate the horizontal axis parallel to itself,
until it becomes tangent to the graph of f at a point .c; f .c// (cf. Fig. 9.7). Then, on
the one hand, such a tangent line must have slope f 0 .c/; on the other, being parallel
to the horizontal axis, its slope must be equal to 0, so that f 0 .c/ D 0.
The first result of this section, which is known in the literature as Rôlle’s
theorem,7 puts the discussion of the previous paragraph in solid grounds.
Lemma 9.43 (Rôlle) Let f W Œa; b ! R be continuous in Œa; b and differentiable
in .a; b/. If f .a/ D f .b/ D 0, then there exists c 2 .a; b/ such that f 0 .c/ D 0.
Proof Let c; d 2 Œa; b be the points in which f attains its minimum and maximum
values, respectively. If c; d 2 fa; bg, then for every x 2 Œa; b we have 0 D f .c/ f .x/ f .d/ D 0. Therefore, f vanishes identically in Œa; b and there is nothing left
to do.
Else, suppose that c 2 .a; b/ (the case of d 2 .a; b/ can be dealt with
analogously). Since f .x/ f .c/ for every x 2 Œa; b, we conclude that
Fig. 9.7 Rôlle’s theorem
y
f
a
c
b
x
6
After Guillaume F. Antoine, Marquis de l’Hôpital, French mathematician of the XVII century. A
more refined version of l’Hôpital’s rule will be presented at Proposition 9.50.
7
After Michel Rôlle, French mathematician of the XVII century.
9.5 Rôlle’s Theorem and Applications
x 2 .c; b )
313
f .x/ f .c/
f .x/ f .c/
0 and x 2 Œa; c/ )
0:
xc
xc
Hence, it follows from the results of Problems 2 and 5, page 290, that
f 0 .c/ D lim
x!cC
f .x/ f .c/
f .x/ f .c/
0 and f 0 .c/ D lim
0:
x!c
xc
xc
Thus, f 0 .c/ D 0.
t
u
The next example shows a typical application of Rôlle’s theorem.
Example 9.44 Given a; b; c 2 R, show that the equation 4ax3 C 3bx2 C 2cx D
a C b C c has at least one real root between 0 and 1.
Proof First of all, note that a standard application of the IVT is not conclusive,
for, if f .x/ D 4ax3 C 3bx2 C 2cx .a C b C c/, then f .0/ D .a C b C c/ and
f .1/ D 3a C 2b C c, so that f .0/ and f .1/ can have equal signs (for instance, for
a D 1 and b D c D 23 ).
On the other hand, if f .x/ D ax4 Cbx3 Ccx2 .aCbCc/x, then f .0/ D f .1/ D 0.
Hence, Rôlle’s theorem assures the existence of x0 2 .0; 1/ such that f 0 .x0 / D 0.
Now, note that this is the same as saying that x0 is a root of the given equation. u
t
Let’s (heuristically) analyse a situation more general than that of Rôlle’s theorem.
More precisely, let’s consider again a function f W Œa; b ! R, continuous in Œa; b
and differentiable in .a; b/, but such that f .a/ and f .b/ can assume any values
whatsoever. The slope of the secant line to the graph of f passing through .a; f .a//
.a/
and .b; f .b// equals f .b/f
ba . We translate it parallel to itself until it becomes tangent
to the graph at a point .c; f .c//, with c 2 .a; b/ (cf. Fig. 9.8). Then, on the one hand,
the slope of this tangent line equals f 0 .c/; on the other, since it is parallel to the
original secant line, its slope equals that of the secant, so that
f 0 .c/ D
f .b/ f .a/
:
ba
This result is the content of the mean value theorem (MVT, shortly) of
Lagrange, and it is the content of the coming result.
Theorem 9.45 (Lagrange) If f W Œa; b ! R is continuous in Œa; b and
differentiable in .a; b/, then there exists c 2 .a; b/ such that
f .b/ f .a/
D f 0 .c/:
ba
Proof Let g W Œa; b ! R be given by
g.x/ D f .x/ x a
bx
f .a/ C
f .b/ ;
ba
ba
314
9 Limits and Derivatives
Fig. 9.8 Lagrange’s mean
value theorem
y
f (b)
f (a)
f (c)
a
c
b
x
for every x 2 Œa; b. (Note that the function of x between parentheses is precisely the
one that defines the secant to the graph of f passing through .a; f .a// and .b; f .b//.)
Clearly, g is continuous in Œa; b and differentiable in .a; b/, with g.a/ D g.b/ D 0.
Hence, by Rôlle’s theorem, there exists c 2 .a; b/ such that g0 .c/ D 0. Now, an easy
computation gives
0
0
g .x/ D f .x/ f .b/
f .a/
C
;
ba
ba
so that
0 D g0 .c/ D f 0 .c/ Example 9.46 For x; y 2
2 ; 2
f .b/ f .a/
:
ba
, show that j tan x tan yj jx yj.
t
u
Proof If x D y, there is nothing to do. Otherwise, it follows from the MVT of
Lagrange the existence of a real c between x and y, such that
tan x tan y D .tan c/.x y/ D
1
.x y/:
cos2 c
However, since j cos cj 1, we get
j tan x tan yj D
1
jx yj jx yj:
cos2 c
t
u
The following corollary sets an important consequence of Lagrange’s MVT.
Corollary 9.47 Let I R be an interval and f ; g W I ! R be two differentiable
functions. If f 0 D g0 , then f g is constant.
9.5 Rôlle’s Theorem and Applications
315
Proof Since f 0 D g0 if and only if .f g/0 D 0, it suffices to show that f 0 D 0
implies f constant.
Let f 0 D 0 and fix a < b in I. Since f is differentiable in I, it is certainly
continuous in Œa; b and differentiable in .a; b/. Therefore, Lagrange’s MVT gives
.a/
c 2 .a; b/ such that f .b/f
D f 0 .c/ D 0. Hence, f .b/ D f .a/, and since a and b
ba
were arbitrarily chosen in I, it follows that f is constant.
t
u
Example 9.48 If f ; g W R ! R are differentiable functions such that f 0 .x/ D g.x/
and g0 .x/ D f .x/ for every x 2 R, then f .x/ D f .0/ cos x C g.0/ sin x and g.x/ D
g.0/ cos x f .0/ sin x, for every x 2 R.
Proof Letting h.x/ D f .x/ sin x C g.x/ cos x and l.x/ D f .x/ cos x g.x/ sin x, we
have h.0/ D g.0/, l.0/ D f .0/ and
h0 .x/ D f 0 .x/ sin x C f .x/ cos x C g0 .x/ cos x g.x/ sin x
D .f 0 .x/ g.x// sin x C .f .x/ C g0 .x// cos x D 0:
Analogously, l0 .x/ D 0 for every x 2 R, so that the previous corollary guarantees
that h and l are constant functions. Hence,
f .x/ sin x C g.x/ cos x D g.0/
:
f .x/ cos x g.x/ sin x D f .0/
The above inequalities can be seen as a linear system of two equations in the
unknowns f .x/ and g.x/. We can easily solve it to get f .x/ D f .0/ cos x C g.0/ sin x
and g.x/ D g.0/ cos x f .0/ sin x.
t
u
In what comes next, we discuss an important generalization of Lagrange’s MVT
to two given functions, known as Cauchy’s mean value theorem. Note that (9.23)
reduces to the MVT of Lagrange if g.x/ D x.
Theorem 9.49 (Cauchy) If f ; g W Œa; b ! R are continuous in Œa; b and
differentiable in .a; b/, then there exists c 2 .a; b/ such that
.f .b/ f .a//g0 .c/ D .g.b/ g.a//f 0 .c/:
(9.23)
Proof Let h W Œa; b ! R be given by
h.x/ D .f .b/ f .a//g.x/ .g.b/ g.a//f .x/;
for every x 2 Œa; b. It is immediate that h is continuous in Œa; b and differentiable
in .a; b/, so that, by Lagrange’s MVT, there exists c 2 .a; b/ satisfying
h0 .c/ D
h.b/ h.a/
:
ba
316
9 Limits and Derivatives
Now, since h.a/ D h.b/ D f .b/g.a/ f .a/g.b/, we have h0 .c/ D 0 and it suffices
to observe that
h0 .c/ D .f .b/ f .a//g0 .c/ .g.b/ g.a//f 0 .c/:
t
u
In Calculus, it is often the case we have continuous functions f ; g W Œa; b/ ! R
such that f .a/ D g.a/ D 0, and we need to evaluate (if it exists) the limit
f .x/
:
x!a g.x/
lim
In general, we refer to such a quotient as an indeterminacy of the form 00 as x D a.
Problem 15, page 311, taught us how to deal with it when f and g are differentiable
at a, with g0 .a/ ¤ 0. Cauchy’s MVT allows us to get a more refined version of that
result, which is also known in the literature as l’Hôpital’s rule.
Proposition 9.50 (l’Hôpital’s rule for 00 ) Let f ; g W .a; b/ ! R be differentiable
functions such that limx!a f .x/ D limx!a g.x/ D 0. If g; g0 ¤ 0 in .a; b/, then
f 0 .x/
f .x/
D L ) lim
D L:
x!a g0 .x/
x!a g.x/
lim
Proof Start by continuously extending f and g to Œa; b/, setting f .a/ D g.a/ D 0.
For x 2 .a; b/, it follows from Cauchy’s MVT that
f .x/ f .a/
f 0 .c.x//
f .x/
D
D 0
;
g.x/
g.x/ g.a/
g .c.x//
(9.24)
for some c.x/ 2 .a; x/; in particular, we clearly have limx!a c.x/ D a.
0
Now, if limx!a gf 0.x/
.x/ D L, a slight modification in the hint given to Problem 10,
page 291, guarantees that limx!a
that
f .x/
limx!a g.x/
f 0 .c.x//
g0 .c.x//
D L. Hence, it readily follows from (9.24)
D L.
t
u
p x . (Note that
Example 9.51 We use l’Hôpital’s rule for 00 to compute limx!0 1cos
x
p
x is not differentiable at x D 0, so that we cannot usep the result of
x 7!
Problem 15, page 311). Indeed, both f .x/ D 1 cos x and g.x/ D x (for x > 0)
satisfy the hypotheses of the previous proposition and are such that
p
f 0 .x/
sin x
D lim
p D lim 2 x sin x D 0
x!0 g0 .x/
x!0 1=.2 x/
x!0
lim
Therefore, limx!0
f .x/
g.x/
D 0.
9.5 Rôlle’s Theorem and Applications
317
Problems: Section 9.5
1. Let f W R ! R be a twice differentiable function, such that f 00 .x/ C f .x/ D 0
for every x 2 R. Prove that
f .x/ D f .0/ cos x C f 0 .0/ sin x;
for every x 2 R.
2. Let > 0 and f W R ! R be twice differentiable and such that f 00.x/ C f .x/ D
0 for every x 2 R. Prove that
p
p
f 0 .0/
f .x/ D f .0/ cos. x/ C p sin. x/;
for every x 2 R.
3. * Let I be an interval and f W I ! R be a twice differentiable function, such
that f 00 is constant. Prove that f either vanishes identically or is a polynomial
function of degree at most 2. More precisely, if x0 2 I, show that
f .x/ D f .x0 / C f 0 .x0 /.x x0 / C
f 00 .x0 /
.x x0 /2 :
2
4. (OIMU) Find all functions f W R ! R such that .f .x/ f .y//2 jx yj3 , for
every x; y 2 R.
5. Let I R be an interval and c 2 I. Let f W I ! R be a function continuous in I
and differentiable in I n fcg. If limx!c f 0 .x/ D l, prove that f is differentiable in
c, with f 0 .c/ D l.
an
6. (Putnam) Let a0 ; a1 ; : : : ; an be real numbers such that a10 C a21 C C nC1
D 0.
n
Show that the polynomial function a0 C a1 x C C an x has at least one real
root.
7. Show that the equation x2 D x sin x C cos x has exactly two real roots.
8. Redo Problem 8, page 300, this time under the assumption that f W I ! R is
continuously differentiable in I.
9. In each item below, compute the given limit:
1cos x
.
2
xp
1 3 cos x
limx!0
.
x2
p
cos x 3 cos x
.
2
x
sin2 .x/
limx!1 .x3 C1/2 .
p
(a) limx!0
(c) limx!0
(b)
(d)
10. Let f ; g W .0; 2 / ! R be given by f .x/ D x3=2 sin 1x and g.x/ D x1=3 cos x C 1.
f .x/
Compute, if it exists, limx!0 g.x/
.
11. If f W .a; b/ ! R is twice differentiable, show that, for every x0 2 .a; b/,
f .x0 C h/ C f .x0 h/ 2f .x0 /
D f 00 .x0 /:
h!0
h2
lim
318
9 Limits and Derivatives
12. Let f W Œa; b/ ! R be twice continuously differentiable in Œa; b/, with f .a/ D 0
and f 00.a/ ¤ 0.
(a) Show that there exists c 2 .a; b/ such that f .x/ .x a/f 0 .x/ ¤ 0, for every
x 2 .a; c/.
.xa/f .x/
0
00
(b) Compute limx!a f .x/.xa/f
0 .x/ in terms of f .a/, f .a/ and f .a/ (item (a)
assures that the denominator doesn’t vanish in the interval .a; c/).
13. * Prove the following version of l’Hôpital’s rule for indeterminacies 00 at C1:
let a > 0 and f ; g W .a; C1/ ! R be differentiable functions such that g; g0 ¤ 0
in .a; C1/. If limx!C1 f .x/ D limx!C1 g.x/ D 0, then
f 0 .x/
f .x/
D L ) lim
D L:
x!C1 g0 .x/
x!C1 g.x/
lim
14. Let f W R ! R be a differentiable function such that limx!C1 f .x/ and
limx!C1 xf 0 .x/ exist and are finite. Compute the possible values of this second
limit.
15. Prove the following version of l’Hôpital’s rule for indeterminacies ˙1
˙1 at a 2
R: let f ; g W .a; b/ ! R be differentiable functions such that g; g0 ¤ 0 in .a; b/
and limx!a g.x/ D C1 (or 1). Then,
f 0 .x/
f .x/
D L ) lim
D L:
0
x!a g .x/
x!a g.x/
lim
16. The differentiable function f W R ! R is such that jf 0 .x/j c < 1 for every real
x, where c is a positive real constant. Show that there exists only one x0 2 R
such that f .x0 / D x0 .
17. * Let I R be an open interval, x0 2 I and f W I ! R a continuous function,
which is differentiable in I n fx0 g. If there exists L D limx!x0 f 0 .x/, prove that f
is differentiable at x0 , with f 0 .x0 / D L.
18. (Romania) Prove that there does not exist a differentiable function f W
.0; C1/ ! .0; C1/ such that f .x/2 f .x C y/.f .x/ C y/ for every
x; y 2 .0; C1/.
19. (OBMU) Find all differentiable functions f W R ! R such that f .0/ D 0 and
jf 0 .x/j jf .x/j for every x 2 R.
9.6 The First Variation of a Function
We saw in Sect. 8.2 that every continuous function f W Œa; b ! R attains
extreme (i.e., maximum and minimum) values in Œa; b. In this section, we shall
discuss procedures that, among other things, allow us to effectively compute the
corresponding extreme points of such an f , provided it is differentiable in .a; b/.
9.6 The First Variation of a Function
319
In all that follows, I denotes an interval, whose interior is the interval obtained
by excluding its endpoints, if they exist. Thus, if I D Œa; b, Œa; b/, .a; b or .a; b/,
then the interior of I is the open interval .a; b/; analogously, if I D Œa; C1/ or
.a; C1/, its interior is .a; C1/, and if I D .1; b or .1; b/, its interior is
.1; b/.
Broadening the situation of the first paragraph above, our purpose in this section
is to solve the problem of finding extreme points and values of functions f W I ! R
which are continuous in I and differentiable in the interior of I. To this end, we begin
with the following
Definition 9.52 Given a function f W I ! R, we say that x0 2 I is a local maximum
(resp. local minimum) point for f if there exists ı > 0 such that f .x0 / f .x/ (resp.
f .x0 / f .x/), for every x 2 I \ .x0 ı; x0 C ı/.
In Fig. 9.9, points x0 , x00 and x000 are of local minimum for the function
f W .a; b/ ! R whose graph is depicted. Observe that, in a neighborhood of each
of these points, the values of f are no less than those it assumes at x0 , x00 and x000
(as indicated by the three dashed horizontal line segments). Also, x00 is the only
global minimum point of f (i.e., a point in which f attains the minimum possible
value in .a; b/), whereas x0 and x000 are not global minimum points. Finally, notice
that f has also two maximum local points (identify them!), albeit none of which is
a global maximum point.
Generically, local maximum or minimum points of f W I ! R are called its
extreme points, or simply its extrema. If I is an open interval and f W I ! R is
differentiable, we shall show below that its extrema (if any) can be found among the
points where f 0 vanishes. Since these vanishing points will play quite an important
role in subsequent discussions, it is worth to start by naming them.
Definition 9.53 Let I R be an open interval and f W I ! R a differentiable
function. We say that x0 2 I is a critical point of f if f 0 .x0 / D 0.
y
Gf
x0
a
x0
Fig. 9.9 Local minimum points of f W I ! R
x0
b
x
320
9 Limits and Derivatives
The notion of critical point has an obvious geometric interpretation: letting I be
an open interval, it follows from (9.14) that the critical points of a differentiable
function f W I ! R are precisely those points in which the tangent lines to its graph
are horizontal.
We can now state and prove the following fundamental result, which is known
as the first derivative test for extreme points. Although its proof is quite similar to
that of Rôlle’s theorem, we repeat it here for the sake of completeness.
Proposition 9.54 If I R is an open interval and f W I ! R is a differentiable
function, then every extreme point of f is also critical.
Proof Assume that x0 2 I is a local minimum point of f (the proof in the other case
is completely analogous). Take ı > 0 such that .x0 ı; x0 C ı/ I (since I is open,
such a ı always exists) and
0 < jx x0 j < ı ) f .x/ f .x0 / 0:
For x0 < x < x0 C ı we have
f .x/f .x0 /
xx0
f 0 .x0 / D lim
x!x0 C
0, so that
f .x/ f .x0 /
0:
x x0
An analogous reasoning with x0 ı < x < x0 gives us
f .x/ f .x0 /
0:
x!x0 x x0
f 0 .x0 / D lim
Therefore, f 0 .x0 / D 0:
t
u
As a corollary to the first derivative test, we have the following search criterion
for extreme points.
Corollary 9.55 Let f W I ! R be continuous in I and differentiable in the interior
of I. If f attains a global (maximum or minimum) value in I, then the corresponding
extreme point is either an endpoint of I or a critical point of f .
Proof Suppose that f attains a global minimum value in I (the case of a global
maximum value can be treated analogously). Let x0 2 I be such that f .x0 / D
minff .x/I x 2 Ig. If x0 is an endpoint of I, there is nothing left to do. Otherwise,
x0 belongs to the interior of I; however, since the interior de I is an open interval,
the first derivative test assures that x0 is a critical point of f .
t
u
The coming example makes it clear that the above corollary solves the problem
of finding the global extreme values of a continuous function f W Œa; b ! R,
differentiable in .a; b/, as long as we know how to find the roots of the equation
f 0 .x/ D 0.
9.6 The First Variation of a Function
321
Example 9.56 Find the maximum value of f W Œ0; 1 ! R, given by f .x/ D x x4 .
Solution Since f is continuous in Œ0; 1, we know from Theorem 8.26 that f assumes
a maximum value in Œ0; 1. By the previous corollary, the corresponding maximum
point is 0, 1 or a critical point of f . Now, since f 0 .x/ D 1 4x3 , we conclude that
1
1
3
p
is the only critical point of f , with f . p
3
3 / D p
3 . Finally, since f .0/ D f .1/ D
4
0<
3
p
3
4
4
, it follows that the maximum value of f is
4
3
p
3
4
.
t
u
Our next result, a direct consequence of Lagrange’s MVT, teaches us how to get
the monotonicity intervals of a differentiable function. From now on, we shall refer
to it as the study of the first variation of a differentiable function.8
Proposition 9.57 If f W I ! R is continuous in I and differentiable in the interior
of I, then:
(a) f 0 0 in the interior of I if and only if f is nondecreasing in I.
(b) If f 0 > 0 in the interior of I, then f is increasing in I, but the converse is not
necessarily true.
(c) f 0 0 in the interior of I if and only if f is nonincreasing in I.
(d) If f 0 < 0 in the interior of I, then f is decreasing in I, but the converse is not
necessarily true.
Proof Let’s prove only items (a) and (b), the proofs of items (c) and (d) being
entirely analogous.
(a) Suppose first that f is nondecreasing in I. For a given x0 in the interior of I,
take ı > 0 such that .x0 ı; x0 C ı/ I. If x0 < x < x0 C ı, then x 2 I and
.x0 /
f .x/ f .x0 / 0. Hence, f .x/f
0 and, making x ! x0 C, we get
xx0
f 0 .x0 / D lim
x!x0 C
f .x/ f .x0 /
0:
x x0
Conversely, suppose f 0 0 in the interior of I. If a; b 2 I are such that
a < b, the continuity of f in I guarantees its continuity in Œa; b. On the other
hand, since .a; b/ is contained in the interior of I, we have f differentiable in
.a; b/, so that Lagrange’s MVT guarantees the existence of c 2 .a; b/ for which
f .b/ f .a/
D f 0 .c/ 0:
ba
Therefore, f .b/ f .a/, and f is nondecreasing in I.
(b) Assume that f 0 > 0 in the interior of I, and let a < b be two points of I. Again
by Lagrange’s MVT, there exists c 2 .a; b/ for which
8
This terminology alludes to the key role of the first derivative in this result, as well as to the fact
that, classically, the first derivative of a (differentiable) function was called its first variation.
322
9 Limits and Derivatives
f .b/ f .a/
D f 0 .c/ > 0:
ba
Hence, f .b/ > f .a/ and f is increasing in I.
To see that the converse is not generally true, let f W R ! R be given by
f .x/ D x3 . Then, f is increasing in the whole real line, albeit f 0 .0/ D 0.
u
t
The coming example shows how to use Proposition 9.57 to approach problems
involving maxima and minima of differentiable functions.
2
C3
Example 9.58 Let f W Œ0; C1/ ! R be given by f .x/ D xxC1
. Show that f attains
a global minimum in Œ0; C1/ and compute this minimum value.
Solution If f is to attain a global minimum in Œ0; C1/, then Corollary 9.55
guarantees that the corresponding extreme point is either 0 or a critical point of
f . Computing f 0 , we get
f 0 .x/ D
2x.x C 1/ .x2 C 3/ 1
x2 C 2x 3
D
;
.x C 1/2
.x C 1/2
so that f 0 .x/ D 0 if and only if x D 1 (recall that we must have x 0).
On the other hand, since x2 C 2x 3 D .x C 3/.x 1/, we conclude that
0
f is negative in the interval .0; 1/ and positive in the interval .1; C1/. Hence,
Proposition 9.57 says that f decreases in Œ0; 1 and increases in Œ1; C1/, so that it
really attains a global minimum, at x D 1. Therefore, the minimum value of f is
f .1/ D 2.
t
u
We now apply the first variation of functions to give a third proof of the inequality
between the arithmetic and geometric means.
Example 9.59 Given an integer n > 1 and positive reals a1 ; a2 ; : : : ; an we have
a1 C a2 C C an p
n a1 a2 : : : an ;
n
with equality if and only if a1 D a2 D D an .
Proof Let’s make induction on n > 1, relying in the discussion that led to (5.2) for
the case n D 2. Given an integer k > 2, suppose we have established the inequality
for k 1 arbitrary positive real numbers, with equality if and only if they are all
equal.
Given k positive reals a1 ; : : : ; ak1 ; ak , let f W .0; C1/ ! R be defined for
x > 0 by
p
f .x/ D a1 C C ak1 C x k k a1 : : : ak1 x:
We shall show that f .x/ 0 for every x > 0, with equality if and only if x D a1
and a1 D D ak1 ; in particular, it will follow that f .ak / 0, with equality if and
only if a1 D D ak1 D ak , as wished. To this end, note that f is differentiable,
with
9.6 The First Variation of a Function
323
f 0 .x/ D 1 p
1
k
a1 : : : ak1 x k 1 :
p
Hence, letting x0 D k1 a1 : : : ak1 , we have f 0 < 0 in .0; x0 /, f 0 .x0 / D 0 and f 0 > 0
in .x0 ; C1/, so that Proposition 9.57 assures f to decrease in .0; x0 and increase in
Œx0 ; C1/. Therefore, f attains a global minimum (only) at x D x0 .
Finally, a straightforward substitution gives
p
f .x0 / D a1 C C ak1 .k 1/ k1 a1 : : : ak1 ;
(9.25)
so that f .x0 / 0 by induction hypothesis. Thus,
f .x/ f .x0 / 0
for every x > 0, with equality if and only if x D x0 and f .x0 / D 0. It now suffices
to observe that, thanks to (9.25) and the induction hypothesis, we have f .x0 / D 0 if
and only if a1 D : : : D ak1 .
t
u
We shall sometimes use the first variation in the form of the following
Corollary 9.60 Let I be an open interval, f W I ! R be twice differentiable and
x0 2 I be a critical point of f . If f 00 > 0 (resp. f 00 < 0) in I, then x0 is the only global
minimum (resp. maximum) point of f . In particular, x0 is the only critical point of f .
Proof Suppose f 00 > 0 in I (the case f 00 < 0 in I is totally analogous). Since f 00 D
.f 0 /0 , item (b) of Proposition 9.57 (applied to f 0 , in place of f ) guarantees that f 0
increases in I. However, since f 0 .x0 / D 0, it follows that f 0 .x/ < 0 for x < x0 and
f 0 .x/ > 0 for x > x0 (in particular, x0 is the only critical point of f ). Then, items
(b) and (d) of Proposition 9.57 (this time applied to f ) guarantee that f decreases in
I\.1; x0 and increases in I\Œx0 ; C1/. Therefore, x0 is the only global minimum
point of f .
t
u
With the aid of the previous corollary, we can given an interesting application of
the first variation to Euclidean Geometry.
Example 9.61 We are given an angle †AOB such that Ab
OB < 90ı (cf. Fig. 9.10),
!
!
as well as a point P inside it. Show how to choose points X 2 OA and Y 2 OB such
that X, P and Y are collinear and XY has the minimum possible length.
Fig. 9.10 Line segment XY
of minimum length
A
X
P
O
Y
B
324
9 Limits and Derivatives
Proof In what follows, we shall rely on [4] for the necessary background in
Euclidean Geometry.
Taking angles in radians, let Ab
OP D ˛, Bb
OP D ˇ and Ob
XY D , so that ˛
!
and ˇ are known and is variable (i.e., depends on the position of X along OA).
Since the sum of the internal angles of triangle OXY equals radians, we have
0 < < ˛ ˇ and Ob
YX D ˛ ˇ .
Applying the Sine Law (cf. Sect. 7.3 of [4]) to triangles XOP and YOP, we get
PX
OP
D
sin ˛
sin
and
PY
OP
D
sin ˇ
;
sin.˛ C ˇ C /
so that
sin ˇ
sin ˛
C
:
XY D PX C PY D OP
sin
sin.˛ C ˇ C /
Hence, letting f W .0; ˛ ˇ/ ! R be given by
f. / D
sin ˛
sin ˇ
C
;
sin
sin.˛ C ˇ C /
it suffices to show that there exists a single 2 .0; ˛ ˇ/ in which f attains its
global minimum.
The first derivative test assures that, if f has a global minimum point, then it must
be a critical one. On the other hand, an easy computation furnishes
f 0. / D sin ˛ cos
sin2
sin ˇ cos.˛ C ˇ C /
:
sin2 .˛ C ˇ C /
Substituting D ˛ ˇ, we easily get
lim f 0 . / D 1
!0
and
lim f 0 . / D C1:
!
Therefore, the IVT (applied to f 0 , which is clearly continuous on ) guarantees the
existence of 0 2 .0; / such that f 0 . 0 / D 0. Now,
f 00 . / D
sin ˇ.1 C cos2 .˛ C ˇ C //
sin ˛.1 C cos2 /
C
> 0;
sin3
sin3 .˛ C ˇ C /
since ˛; ˇ; ; ˛ C ˇ C 2 .0; /. Hence, Corollary 9.60 says that
global minimum point of f .
0
is the only
t
u
9.6 The First Variation of a Function
325
We finish this section by applying Proposition 9.57 to prove the following result,
known as Darboux’s theorem.9
Theorem 9.62 (Darboux) If I R is an interval and f W I ! R is a differentiable
function, then the function f 0 W I ! R, even if discontinuous, satisfies the mean
value property.
Proof Let a < b in I and d be a real number between f .a/ and f .b/. We wish to
establish the existence of c 2 Œa; b such that f .c/ D d. To this end, we consider two
cases separately:
(i) f 0 .a/ < 0 < f 0 .b/ (or vice-versa): if f is not injective in Œa; b, then there exist
˛ < ˇ in Œa; b such that f .˛/ D f .ˇ/. By Lagrange’s MVT, there exists c 2
.˛/
.˛; ˇ/ (hence, c 2 .a; b/) such that f 0 .c/ D f .ˇ/f
D 0. If f is injective in
ˇ˛
Œa; b, it follows from Theorem 8.35 that f is monotonic in Œa; b. Therefore, the
previous proposition gives f 0 0 in Œa; b or f 0 0 in Œa; b, which is not the
case.
(ii) f 0 .a/ < d < f 0 .b/ (or vice-versa): let g W I ! R be given by g.x/ D f .x/ dx,
so that g is differentiable, with g0 .x/ D f 0 .x/ d for every x 2 I. Hence,
g0 .a/ < 0 < g0 .b/ or vice-versa, and item (i) assures the existence of c 2 .a; b/
such that g0 .c/ D 0. This is the same as f 0 .c/ D d.
t
u
Problems: Section 9.6
1. Prove Proposition 6.25 applying the results of this section.
2. In each of the following items, find the maximum and minimum values (if these
exist) of the given functions:
p
x
(a) f W Œ0; C1/ ! R given by f .x/ D x2 C16 .
(b) f W R ! R given by f .x/ D ax2xCb , where a and b are given positive real
numbers.
(c) f W .0; C1/ ! R given by f .x/ D x2 C ax , where a is a given positive real
number.
2
(d) f W .0; C1/ ! R given by f .x/ D x3xCa , where a is a given positive real
number.
3. Let f W R ! R be given by f .x/ D x3 C ax2 C bx C c, where a; b; c 2 R are real
constants. Discuss the first variation of f in terms of a, b and c.
3
2
4. * Show that x sin x x x3Š and cos x 1 x2 , for every x 0.
5. In a square ABCD of diagonals AC and BD, the sides have length 2. Mark the
midpoint P of side AB and, then, a point Q 2 AD and a point R 2 CD such that
9
After Jean-Gaston Darboux, French mathematician of the XIX and XX centuries.
326
9 Limits and Derivatives
Pb
QR D 90ı . Find the position of Q along side AD such that the area of triangle
PQR is as large as possible.
6. Let f W .a; b/ ! R be a differentiable function and P.x0 ; y0 / be a point not
belonging to the graph of f . If there exists A 2 Gf such that
AP D minf A0 PI A0 2 Gf g;
!
we say that AP is the distance from P to Gf . In such a case, prove that AP ?r,
where r denotes the tangent line to Gf at A.
7. Let f W .a; b/ ! R and g W .c; d/ ! R be differentiable functions whose graphs
do not intersect each other. If there exist points A.˛; f .˛// and B.ˇ; f .ˇ// such
that
AB D minf A0 B0 I A0 2 Gf and B0 2 Gg g;
(9.26)
we say that AB is the distance between the graphs of f and g. In such a case,
Show that f 0 .˛/ D g0 .ˇ/.
Given differentiable functions f W .a; b/ ! R and g W .c; d/ ! R whose
graphs do not intersect, the result of the previous problem assures that the
problem of finding points A 2 Gf and B 2 Gg satisfying (9.26) is equivalent
to that of minimizing F W .a; b/ .c; d/ ! R given by
F.x; y/ D .x y/2 C .f .x/ g.y//2 ;
subject to the constraint f 0 .x/ D g0 .y/. Nevertheless, the pair of functions f ; g W
.0; C1/ ! R given by f .x/ D 1x and g.x/ D 0 shows that it is not always
the case that such a pair of points A 2 Gf and B 2 Gg actually exist. The next
problem applies the above discussion to a case in the positive side of things.
8. Compute the distance between the graphs of functions f ; g W R ! R such that
f .x/ D x2 e g.x/ D 1 .x 3/2 , admitting the (geometrically plausible) fact
that such a distance is attained by a certain pair of points A 2 Gf and B 2 Gg .
9. In Fig. 9.11, we have BC D 2 BH. Compute the largest possible value of the
measure of angle Bb
AC.
10. ABCD is an isosceles trapezoid of bases AD and BC, with AD > BC, and legs
AB and CD. If BC D a and AB D CD D b, compute the measure of the angles
Bb
AD D Ab
DC for the area of ABCD to be as large as possible.
Fig. 9.11 Maximizing the
measure of angle Bb
AC
A
H
B
C
9.6 The First Variation of a Function
327
A
B
a
b
r
P
A
B
Fig. 9.12 Minimizing AP C BP
Fig. 9.13 Minimizing
AP BP
A
B
a
b
l
A
B
r
11. We are given an angular region †AOB, such that Ab
OB < 90ı (cf. Fig. 9.10),
!
!
and a point P inside it. Choose points X 2 OA and Y 2 OB in such a way that
X, P are Y collinear. Show that PX PY is minimum if and only if OXY is an
isosceles triangle, with basis XY.
12. Let a line r and points A and B be given, as shown in Fig. 9.12. The distances
from A and B to r are respectively equal to a and b, and c is the length of the
line segment joining the feet of the perpendiculars dropped from A and B to r.
Find the position of a point P along r such that the sum AP C BP is as small as
possible.
!
!
0
0
0
0
0 0
13. In
p Fig. 9.13, we have AA ; BB ?r, AA D a, BB D b and A B D l. If l 2
2
2.a C b /, show that there exists a single point P 2 r such that AP BP
attains its minimum possible value. Show also that P 2 A0 B0 .
nC1 m n
14. Show that mC1
n , for every m; n 2 N.
nC1
15. For each real number k > 1, compute the minimum
possiblepvalue of x C y,
p
where x and y are real numbers satisfying .x C 1 C x2 /.y C 1 C y2 / D k.
2
1
16. (Romania) Compute the minimum possible value of x C y C xCy
C 2xy
, for
distinct and positive real numbers x and y.
A
B
B
A
17. (BMO) In a triangle ABC, let sin23 b
cos48 b
D sin23 b
cos48 b
. Compute the
2
2
2
2
AC
ratio BC .
18. (OIMU) Let f W Œ0; 1 ! Œ0; 1 be continuous in Œ0; 1, differentiable in .0; 1/
and such that f .0/ D 0 and f .1/ D 1. Prove that there exist distinct a; b 2 Œ0; 1
such that f 0 .a/f 0 .b/ D 1.
19. (Leningrad) Let f W R ! R be a polynomial function such that
f .x/ f 0 .x/ f 00 .x/ C f 000 .x/ 0
for every x 2 R. Prove that f .x/ 0 for every x 2 R.
328
9 Limits and Derivatives
9.7 The Second Variation of a Function
Along this section we continue to study how the behavior of the derivatives of a
function influence the shape of its graph. As before, in all that follows I R denotes
an interval.
Definition 9.63 A function f W I ! R is said to be convex if, for all a; b 2 I and
t 2 Œ0; 1, we have
f ..1 t/a C tb// .1 t/f .a/ C tf .b/:
As t varies from 0 to 1, basic Analytic Geometry guarantees that the points of the
form .1 t/.a; f .a// C t.b; f .b// trace out the line segment secant with endpoints
.a; f .a// and .b; f .b//. Hence, f W I ! R is convex if and only if, for all a < b in
I, the portion of the graph of f between lines x D a and x D b does not intersect
the open half-plane situated above the straight line passing through .a; f .a// and
.b; f .b//. Yet in another way, letting
RC .f / D f.x; y/ 2 R2 I x 2 I and y f .x/g;
f is convex if and only if RC .f / is a convex region of the plane.
The next definition refines the notion of convexity for functions (pay particular
attention to the range of values of t, when compared to those in the former
definition).
Definition 9.64 A function f W I ! R is strictly convex if, for all distinct a; b 2 I
and every t 2 .0; 1/, we have
f ..1 t/a C tb// < .1 t/f .a/ C tf .b/:
Analogously to the above, we say that f W I ! R is (strictly) concave if f is
(strictly) convex. Hence, f is concave if and only if
f ..1 t/a C tb// .1 t/f .a/ C tf .b/
for all a; b 2 I and t 2 Œ0; 1. Accordingly, f W I ! R is strictly concave if and only
if f ..1 t/a C tb// > .1 t/f .a/ C tf .b/ for all distinct a; b 2 I and every t 2 .0; 1/.
Properties of (strictly) concave functions are easily derived from those of
(strictly) convex ones, just changing f by f when necessary. For this reason, in
all that follows we shall restrict ourselves to the analysis of convex and strictly
convex functions, leaving to the reader the task of adapting it to concave and strictly
concave ones.
9.7 The Second Variation of a Function
329
Definition 9.63 guarantees that every strictly convex function is convex. Nevertheless, the converse is not true in general, as the coming example shows.
Example 9.65 Every affine function is convex, albeit not strictly convex. Indeed, if
f W R ! R is given by f .x/ D Ax C B, with A; B 2 R, it is immediate to verify that
f ..1 t/a C tb/ D .1 t/f .a/ C tf .b/, for all a; b 2 R and every t 2 Œ0; 1.
In Corollary 9.70 we will show that, if I is an open interval, then a twice
differentiable f W I ! R is convex if and only if f 00 .x/ 0 for every x 2 I. We
will also show that, if f 00 .x/ > 0 for every x 2 I, then f is strictly convex. By
assuming the truth of these results for the time being, we now list some examples
of convex and concave functions (note that when we sketched the graphs of these
functions we implicitly assumed that they were indeed convex or concave, according
to the case at hand).
Example 9.66
(a) The restriction of the inverse proportionality function to the set of positive reals
is strictly convex. Indeed, letting f .x/ D 1x for x > 0, we have f 00 .x/ D x23 > 0.
(b) Letting n > 1 be an integer, the function f W .0; C1/ ! R given by f .x/ D xn
is also strictly convex, for, f 00 .x/ D n.n 1/xn2 > 0.
(c) The sine function is strictly concave in the interval .0; /, since sin00 x D
sin x < 0 in .0; /.
(d) Since tan00 x D 2 tan x sec x > 0 in 0; 2 , the tangent function is strictly convex
in this interval.
Remark 9.67 If f W I ! R is convex, then, letting t D 12 in Definition 9.63, we
f .a/Cf .b/
get f . aCb
for all a; b 2 I. Conversely, if f is continuous and such that
2 / 2
f .a/Cf .b/
aCb
f. 2 / for
all a; b 2 I, it is possible to show (see Problem 8) that f
2
is convex. By the same token, it is also possible to show (see Problem 9) that a
.b/
continuous f W I ! R is strictly convex if f . aCb
/ < f .a/Cf
for all distinct a; b 2 I.
2
2
In turn, these remarks allow us to use the elementary inequalities we had studied to
establish the (strictly) convex or concave character of several common functions. In
this respect, take a look at Problem 11.
Back to the development of the theory, let f W I ! R be a convex function. Then,
given a < b in I and 0 < t < 1, we have
f ..1 t/a C tb/ .1 t/f .a/ C tf .b/:
Letting x D .1 t/a C tb, we have t D
inequality above can be written as
f .x/ xa
ba
and 1 t D
xa
bx
f .a/ C
f .b/:
ba
ba
bx
.
ba
(9.27)
Therefore, the last
330
9 Limits and Derivatives
f (b)
(1 − t)f (a) + tf (b)
f (a)
f ((1 − t)a + tb)
a
(1 − t)a + tb
b
Fig. 9.14 Graph of a convex function
In turn, this gives
1
f .x/ f .a/
xa
xa
and, analogously,
f .b/f .x/
bx
xa
f .b/ f .a/
bx
f .a/ C
f .b/ f .a/ D
ba
ba
ba
f .b/f .a/
ba .
In short, letting f be convex in I, we have
f .b/ f .a/
f .b/ f .x/
f .x/ f .a/
xa
ba
bx
(9.28)
for all a < x < b in I.
Conversely, if
f .x/ f .a/
f .b/ f .x/
;
xa
bx
(9.29)
for all a < x < b in I, then .b x/.f .x/ f .a// .x a/.f .b/ f .x// or, which is
the same,
.b a/f .x/ .b x/f .a/ C .x a/f .b/:
Substituting x D .1 t/a C tb in this inequality, we immediately recover (9.27), so
that f is convex.
Notice that the quotients in (9.28) are the slopes of the secants to the graph of f
and passing through the pairs of points .a; f .a// and .x; f .x//, .a; f .a// and .b; f .b//,
.x; f .x// and .b; f .b//, respectively. On the other hand, such slopes are the tangents
of the trigonometric angles measured from the horizontal axis to those secants. Since
the tangent function increases in each of the intervals .0; 2 / and . 2 ; /, the reader
can easily use the comments above to grasp the “turning behavior” of those secants,
as we walk along the horizontal axis in the positive direction (see Fig. 9.14).
9.7 The Second Variation of a Function
331
In what follows, we use the previous discussion to show that a convex function,
defined in an open interval, is continuous.
Proposition 9.68 If I is an open interval and f W I ! R is convex, then f is
continuous.
Proof Fix x0 2 I and a < x0 < b also in I (this is possible due to the openness
of I). For x 2 .x0 ; b/, the first inequality in (9.28), with x0 in place of a, gives
f .x/f .x0 /
.x0 /
f .b/f
xx0
bx0 ; in turn, the second inequality in (9.28), with x0 in place of x
and x in place of b, gives
we get
f .x/f .x0 /
xx0
f .x0 /f .a/
x0 a .
By combining these two inequalities,
f .x0 / f .a/
f .b/ f .x0 /
.x x0 / f .x/ f .x0 / .x x0 /:
x0 a
b x0
Letting x ! x0 C, it follows from these inequalities, together with the squeezing
theorem (cf. Proposition 9.8) that
lim f .x/ D f .x0 /:
x!x0 C
Arguing in an analogous way, we conclude that limx!x0 f .x/ D f .x0 /. Therefore,
Problem 5, page 290, guarantees that limx!x0 f .x/ D f .x0 /, and Proposition 9.4
shows that f is continuous at x0 .
t
u
As we have previously commented, we shall now show in a moment that the
(strictly) convex character of a twice differentiable function is intimately related to
the sign of its second derivative. This, in turn, will be a straightforward consequence
of the coming result, in which we assume that the function under consideration is
merely differentiable.
Theorem 9.69 If I is an open interval and f W I ! R is differentiable, then:
(a) f is convex in I if and only if f 0 is nondecreasing in I.
(b) f is strictly convex in I if and only if f 0 is increasing in I.
Proof Firstly, let f be convex in I. Given a < b in I and x 2 .a; b/, it follows
from (9.28) that
f .b/ f .a/
f .b/ f .x/
f .x/ f .a/
:
xa
ba
bx
332
9 Limits and Derivatives
However, since f is differentiable in I, letting x ! aC in the first inequality and
x ! b in the second one, we obtain
f 0 .a/ f .b/ f .a/
f 0 .b/:
ba
Hence, f 0 is nondecreasing in I.
If f is strictly convex, we refine the above reasoning bay taking c 2 .a; b/ and
x 2 .a; c/, y 2 .c; b/. Then, (9.28) and the strict convexity of f give
f .x/ f .a/
f .c/ f .a/
f .b/ f .c/
f .b/ f .y/
<
<
<
:
xa
ca
bc
by
Now, letting x ! aC and y ! b, we get
f 0 .a/ f .b/ f .c/
f .c/ f .a/
<
f 0 .b/;
ca
bc
so that f is increasing in I.
Conversely, suppose that f 0 is nondecreasing in I. To show that f is convex, we
previously saw that it suffices to prove that
f .x/ f .a/
f .b/ f .x/
xa
bx
for all a < x < b in I. Since I is open and f is differentiable in I, we have f
continuous in Œa; b and differentiable in .a; b/. Hence, Lagrange’s MVT assures
.a/
the existence of ˛ 2 .a; x/ and ˇ 2 .x; b/ such that f .x/f
D f 0 .˛/ and
xa
f .b/f .x/
D f 0 .ˇ/. However, since ˛ < ˇ and f 0 is assumed to be nondecreasing,
bx
we have f 0 .˛/ f 0 .ˇ/. Thus,
f .b/ f .x/
f .x/ f .a/
D f 0 .˛/ f 0 .ˇ/ D
;
xa
bx
as we wished to show.
In case f 0 is increasing in I, the above reasoning gives
f .b/ f .x/
f .x/ f .a/
D f 0 .˛/ < f 0 .ˇ/ D
:
xa
bx
Hence, we get a strict inequality in (9.29) for all a < x < b in I, and f is strictly
convex.
t
u
Corollary 9.70 Let I be an open interval and f W I ! R be twice differentiable.
(a) f is convex in I if and only if f 00 0 in I.
(b) If f 00 > 0 in I, then f is strictly convex in I.
9.7 The Second Variation of a Function
333
Proof Item (a) follows immediately from item (a) of the previous result, together
with item (a) of Proposition 9.57 (applied to f 0 ).
In what concerns (b), if f 00 > 0 in I, then item (b) of Proposition 9.57 (applied
once more to f 0 ) guarantees that f 0 increases in I. Therefore, item (b) of the previous
theorem assures that f is strictly convex.
t
u
Let us see a simple application of this result.
p p
x
Example 9.71 Let f W R n f 2; 2g ! R be given by f .x/ D 2x
2 . Find the
intervals of the real line in which f is strictly convex or strictly concave.
Solution Computing the first and second derivatives of f , we get f 0 .x/ D
and f 00 .x/ D
2x
.
.2x2 /3 .6Cx2 /
2Cx2
.2x2 /2
Hence,
p
8
8
x > 0 and 2 x2 > 0
0<x< 2
<
<
x
f 00 .x/ > 0 ,
>0,
,
:
or
or p
:
:
2 x2
x < 0 and 2 x2 < 0
x< 2
p
We readily
p conclude that f is strictly convex in each of thepintervals .1;
p 2/
and .0; 2/, and strictly concave in each of the intervals . 2; 0/ and . 2; C1/.
t
u
The former corollary is usually referred to as the study of the second variation
of a twice differentiable function. We now wish to look at the points in which the
graph of f changes from being convex to being concave, or vice-versa. Before that,
it is useful to have the following definition at our disposal.
Definition 9.72 Let I be an open interval and f W I ! R be a continuous function.
A point x0 2 I is said to be an inflection point of f if there exists ı > 0 such that f
is convex in I \ .x0 ı; x0 / and concave in I \ .x0 ; x0 C ı/, or vice-versa.
The coming corollary gives a necessary condition to be satisfied by the inflection
points of twice differentiable functions.
Corollary 9.73 Let I be an open interval and f W I ! R be a twice differentiable
function. If x0 2 I is an inflection point of f , then f 00 .x0 / D 0.
Proof Suppose f is convex in .x0 ı; x0 / and concave in .x0 ; x0 C ı/, with ı > 0 so
small that .x0 ı; x0 Cı/ I (the other case can be dealt with in an analogous way).
Then, Corollary 9.70 gives f 00 .x/ 0 in .x0 ı; x0 / and f 00 .x/ 0 in .x0 ; x0 C ı/.
Suppose f 00 .x0 / < 0, and fix an arbitrary a 2 .x0 ı; x0 /, so that f 00 .x0 / <
00
0/
0 f 00 .a/. Since f 00 .x0 / < f .x
< f 00 .a/, Darboux’s Theorem 9.62 assures the
2
00
0/
< 0. However, since b 2 .a; x0 / )
existence of b 2 .a; x0 / such that f 00 .b/ D f .x
2
00
b 2 .x0 ı; x0 /, we should also have f .b/ 0, which is a contradiction.
Analogously, we also reach a contradiction by supposing that f 00 .x0 / > 0. Hence,
the only possibility is that f 00 .x0 / D 0.
t
u
334
9 Limits and Derivatives
We now state and prove Jensen’s inequality,10 a result that gives greater
flexibility to the applications of convex functions to problems of maxima and
minima.
Theorem 9.74 (Jensen) Let I R be an open interval and f W I ! R be a convex
function. If x1 ; x2 ; : : : ; xn 2 I and t1 ; t2 ; : : : ; tn 2 .0; 1/, with t1 C t2 C C tn D 1,
then t1 x1 C t2 x2 C C tn xn 2 I and
f .t1 x1 C t2 x2 C C tn xn / t1 f .x1 / C t2 f .x2 / C C tn f .xn /:
(9.30)
Moreover, if f is strictly convex, then equality happens if and only if x1 D x2 D
D xn .
Proof Suppose f is strictly convex (the case of a merely convex f is totally
analogous) and let’s prove Jensen’s inequality by induction on n > 1. Case n D 2
follows from the definition of strict convexity for f , since condition t1 C t2 D 1 is
equivalent to t1 D t and t2 D 1 t.
Suppose that, for a certain n > 1 and all x1 ; x2 ; : : : ; xn 2 I and t1 ; t2 ; : : : ; tn 2
.0; 1/, with t1 C t2 C C tn D 1, we have t1 x1 C t2 x2 C C tn xn 2 I and
f .t1 x1 C t2 x2 C C tn xn / t1 f .x1 / C t2 f .x2 / C C tn f .xn /;
with equality if and only if x1 D x2 D D xn . Let’s consider x1 , x2 , . . . , xn ,
xnC1 2 I and t1 ; t2 ; : : : ; tn ; tnC1 2 .0; 1/ such that t1 C t2 C C tn C tnC1 D 1.
Define
yD
with sj D
tj
1tnC1
t1 x1 C t2 x2 C C tn xn
D s1 x1 C s2 x2 C C sn xn ;
1 tnC1
for 1 j n. Then, s1 ; s2 ; : : : ; sn > 0 and
n
X
jD1
sj D
n
X
1
1
tj D
.1 tnC1 / D 1;
1 tnC1 jD1
1 tnC1
so that sj 2 .0; 1/ for 1 j n. Hence, induction hypothesis gives y 2 I and, thus,
t1 x1 C t2 x2 C C tn xn C tnC1 xnC1 D .1 tnC1 /y C tnC1 xnC1 2 I:
Now, by invoking the strict convexity of f we obtain
f .t1 x1 C t2 x2 C C tnC1 xnC1 / D f ..1 tnC1 /y C tnC1 xnC1 /
.1 tnC1 /f .y/ C tnC1 f .xnC1 /;
10
After Johan Jensen, Danish engineer and mathematician of the XIX and XX centuries.
9.7 The Second Variation of a Function
335
with equality if and only if y D xnC1 . On the other hand, also by induction
hypothesis we have
f .y/ D f .s1 x1 C s2 x2 C C sn xn / s1 f .x1 / C s2 f .x2 / C C sn f .xn /
D
1
.t1 f .x1 / C t2 f .x2 / C C tn f .xn //;
1 tnC1
with equality if and only if x1 D x2 D D xn .
Taking the two inequalities above together, we conclude that
f .t1 x1 C t2 x2 C C tnC1 xnC1 / .1 tnC1 /f .y/ C tnC1 f .xnC1 /
.t1 f .x1 / C C tn f .xn // C tnC1 f .xnC1 /;
with equality happening if and only if y D xnC1 and x1 D x2 D D xn . Finally,
it’s immediate to verify that these conditions are equivalent to x1 D x2 D D
xn D xnC1 .
t
u
Most often, we shall apply Jensen’s inequality in the form of the coming
corollary, which is stated for the reader’s convenience.
Corollary 9.75 (Jensen) If I R is an open interval and f W I ! R is a convex
function, then, for x1 ; x2 ; : : : ; xn 2 I, we have
f
x 1 C x2 C C xn
n
f .x1 / C f .x2 / C C f .xn /
:
n
(9.31)
Moreover, if f is strictly convex, then equality happens if and only if x1 D x2 D
D xn .
Proof Let t1 D t2 D D tn D
1
n
t
u
in the previous result.
We leave to the reader the task of stating Jensen’s inequality in the case of a
(strictly) concave function f . Presently, we concentrate ourselves in discussing some
applications of it.
Example 9.76 Given n > 1 positive reals a1 , a2 , . . . , an , prove that
1
1
1
.a1 C a2 C C an /
C
CC
a1
a2
an
n2 ;
with equality if and only if a1 D a2 D D an .
Proof Example 9.66 showed that the inverse proportionality function f W
.0; C1/ ! R, given by f .x/ D 1x , is strictly convex. Hence, it follows from (9.31)
that
336
9 Limits and Derivatives
n
Df
a1 C a2 C C an
a1 C a2 C C an
n
f .a1 / C f .a2 / C C f .an /
n
1=a1 C 1=a2 C C 1=an
D
n
with equality if and only if a1 D a2 D D an . Now, it suffices to observe that
inequality
n
1=a1 C 1=a2 C C 1=an
a1 C a2 C C an
n
t
u
is precisely what we wished to get.
Example 9.77 (BMO) Let n > 1 and a1 ; : : : ; an be positive reals with sum equal to
1. For each 1 i n, let bi D a1 C C ai1 C aiC1 C C an . Prove that
a2
an
n
a1
;
C
CC
1 C b1
1 C b2
1 C bn
2n 1
with equality if and only if a1 D a2 D D an D 1n .
Proof Substituting bi D 1 ai for 1 i n, it suffices to prove that
a2
an
n
a1
:
C
CC
2 a1
2 a2
2 an
2n 1
To this end, we claim first that f W .1; 2/ ! R, given by f .x/ D
4
convex. Indeed, one readily computes f 00 .x/ D .2x/
3 > 0.
Hence, by applying Jensen’s inequality we get
n
X
iD1
1X
ai
n iD1
n
f .ai / nf
!
D nf
x
,
2x
is strictly
2
1
D
;
n
2n 1
with equality if and only if a1 D a2 D D an , i.e., if and only if a1 D a2 D D
an D 1n .
t
u
Our next example applies Jensen’s inequality to solve an interesting geometric
problem.
Example 9.78 Let be a semicircle of radius R and diameter A0 A1 . For each integer
n > 2, show that there exists a single convex n-gon A0 A1 A2 : : : An1 satisfying the
following conditions:
(a) A2 ; : : : ; An1 2 .
(b) The area of A0 A1 A2 : : : An1 is as large as possible.
9.7 The Second Variation of a Function
337
Fig. 9.15 A convex n-gon
inscribed in a semicircle
A3
A2
An−1
A0
A1
O
Proof Figure 9.15 depicts a convex n-gon A0 A1 A2 : : : An1 inscribed in . Let
Ai b
OAiC1 D ˛i for 1 i n1, where O is the center of (and with the convention
that An D A0 ). Hence, ˛1 C ˛2 C C ˛n1 D . Applying the sine formula for the
area of a triangle (cf. Sect. 7.3 of [4]), we get
A.A0 A1 : : : An1 / D
n1
X
A.Ai OAiC1 / D
iD1
n1
X
1
iD1
2
R2 sin Ai b
OAiC1
1 2X
R
sin ˛i :
2 iD1
n1
D
Now, since the sine function is strictly concave in the interval Œ0; , it follows
from (9.31) that
n1
X
1 X
sin ˛i .n 1/ sin
˛i
n 1 iD1
iD1
n1
!
D .n 1/ sin
;
n1
. Therefore, there is indeed only
with equality if and only if ˛1 D D ˛n1 D n1
one convex n-gon satisfying the prescribed conditions.
t
u
Problems: Section 9.7
1. Let I; J R be intervals, and g W I ! J and f W J ! R be two strictly convex
functions. If f is increasing, prove that f ı g W I ! R is also strictly convex.
2. Let I; J R be intervals and f W I ! J be a continuous bijection. If f is strictly
convex, prove that f 1 W J ! I is either strictly convex or strictly concave in J.
3. * Prove that the sum of a finite number of strictly convex (resp. strictly concave)
functions on a common domain is also a strictly convex (resp. strictly concave).
x
4. Let f W .1; 1/ ! R be the function given by f .x/ D p1x
. Find the intervals
in which f is convex (resp. concave).
338
9 Limits and Derivatives
5. Let a; b; c 2 R and f W R ! R be given by f .x/ D x3 C ax2 C bx C c. Find
the inflection points of f , as well as the intervals along which it is convex or
concave.
6. Let f W .0; C1/ ! R be given by f .x/ D x2 sin 1x . Show that f has infinitely
many inflection points.
7. Let f W .0; C1/ ! R be a twice differentiable, nondecreasing and convex
function. If g W .0; C1/ ! R is given by g.x/ D xf .x/, show that g is also
convex.
.y/
8. Let f W I ! R be a continuous function such that f . xCy
/ f .x/Cf
for all
2
2
xCy
f .x/Cf .y/
x; y 2 I (resp. f . 2 / for all x; y 2 I). Prove that f is convex (resp.
2
concave).
f .a/Cf .b/
9. Let f W I ! R be continuous and such that f . aCb
(resp. f . aCb
2 / <
2
2 / >
f .a/Cf .b/
),
for
all
distinct
a;
b
2
I.
Prove
that
f
is
strictly
convex
(resp.
concave).
2
10. Let f W .0; C1/ ! R be a continuous, increasing (resp. nondecreasing) and
convex (resp. strictly convex) function. Prove that g W .0; C1/ ! R, given by
g.x/ D xf .x/, is also strictly convex.
11. Use the results of Problems 8 and 9 to establish the strictly concave or convex
character of each of the functions of Example 9.66.
12. Let I be an open interval and f W I ! R be a twice differentiable convex
function. Show that, for every x0 2 I, the graph of f is not below its tangent line
at x0 . More precisely, show that, for all x; x0 2 I, we have
f .x/ f .x0 / C f 0 .x0 /.x x0 /:
13. (Romania) Let I R be an interval and f W I ! R be strictly convex
and increasing. Prove that the sequence .f .n//n1 does not contain an infinite
arithmetic progression.
14. Let k > 1 be an integer and a1 ; a2 ; : : : ; an be positive reals. Prove that
ak1 C ak2 C C akn
n
a1 C a2 C C an
n
k
;
with equality if and only if a1 D a2 D D an .
15. Let be the circle of center O and radius R, n > 2 be a given integer and
A1 A2 : : : An be a convex n-gon inscribed in .
(a) Suppose that O does not belong to the interior of A1 A2 : : : An . Use a
geometric argument to show that there are convex n-gons inscribed in
and with area strictly greater than that of A1 A2 : : : An .
(b) Use Jensen’s inequality to show that, among all convex n-gons inscribed in
, the regular ones are those of maximal area.
16. Let be given a circle , with center O and radius r, and an integer n > 2. Among
all convex n-gons A1 A2 : : : An circumscribed to , prove that the regular ones
are those of minimum perimeter.
9.8 Sketching Graphs
339
17. Let ABC be an acute triangle of sides AB D c, AC D b and BC D a. If R is its
circumradius (i.e., the radius of the circle circumscribed to ABC), show that
p
a C b C c 3R 3;
with equality if and only if ABC is equilateral.
18. (Romania—adapted) Let ABC be an equilateral triangle of height h, let P be an
interior point and x, y and z be the distances from P to the sides of ABC.
(a) Prove that x C y C z D h.
(b) Find the least possible value of
hx
hCx
C
hy
hCy
C
hz
.
hCz
19. (Turkey) Let n > 1 be an integer and x1 ; x2 ; : : : ; xn be positive reals with sum
equal to 1. Prove that
n
X
iD1
xi
p
1 xi
r
n
n1
p
p
x1 C C xn
;
p
n1
with equality in any of the above inequalities if and only if x1 D D xn D 1n .
9.8 Sketching Graphs
Let be given an interval I and a continuous function f W I ! R, which is twice
differentiable in the interior of I. In principle, the theory developed so far allows us
to draw a reasonably accurate sketch of the graph of f . Indeed:
(i) The results collected in Problem 10, page 194, allows us to possibly reduce the
task of sketching the graph of f to doing so in an appropriate interval J I.
(ii) The first derivative test guarantees that the extreme points of f are either the
endpoints of I or the solutions of equation f 0 .x/ D 0.
(iii) The study of the first variation of f assures that the intervals along which f
increases (resp. decreases) are the solution sets of the inequality f 0 .x/ > 0
(resp. f 0 .x/ < 0).
(iv) The inflection points of f are found among the solutions of equation
f 00 .x/ D 0.
(v) The study of the second variation of f guarantees that the intervals along
which f is strictly convex (resp. strictly concave) are the solution sets of the
inequality f 00 .x/ > 0 (resp. f 00 .x/ < 0).
(vi) The tangent lines to the graph of f in its inflection points help us in sketching
the graph in a neighborhood of each of those points.
(vii) Definition 9.18 and Problem 11, page 291, teach us how to find (if any) the
asymptotes to the graph of f .
340
9 Limits and Derivatives
(viii) The computation of the limits limx!˙1 f .x/ (in case they make sense) allows
us to better understand the behavior of the graph of f for large values of jxj.
(ix) Plotting some other points along the graph can be quite helpful in sketching it.
In this section, we illustrate the application of the program just delineated to
sketching the graphs of some simple functions.
Example 9.79 Let D D Rnf 2 CkI k 2 Zg. Sketch the graph of the secant function
sec W D ! R.
Solution
(i) First of all, note that it suffices to sketch the graph of sec in . 2 ; 2 /[. 2 ; 3
2 /,
for, since the cosine function is periodic of period 2, so is the secant function.
On the other hand, since cos.x C / D cos x for every x 2 R, we have
sec.x C / D sec x for every x 2 D; hence, Problem 10, page 194,
guarantees that the portion of the graph of sec in the interval . 2 ; 3
2 / is
obtained from that in the interval . 2 ; 2 / by means of a reflection along
the horizontal axis, followed by a translation of units, parallel to that same
axis.
In view of the above, from now on we restrict the analysis of the graph of the
secant function to the interval . 2 ; 2 /.
(ii) and (iii) Since
sec0 x D .cos1 x/0 D .cos x/2 . sin x/ D
sin x
D tan x sec x;
cos2 x
(9.32)
we have sec0 x D 0 , sin x D 0 , x D 0. (Indeed, 0 < cos x 1 )
sec x 1, with equality if and only if cos x D 1, i.e., if and only if x D 0.
Hence, x D 0 is an extreme point—thus, a critical one—of
On the other
sec.)
0
hand, sec increases
,
sec
x
>
0
,
sin
x
>
0
,
x
2
0;
2 ; therefore, sec
decreases in 2 ; 0 .
(iv) and (v) Firstly, Example 9.34 and (9.32) furnish
sec00 x D tan0 x sec x C tan x sec0 x D sec3 x C tan2 x sec x
D sec x.sec2 x C tan2 x/:
Therefore, sec00 ¤ 0, and sec has no inflection points. Also, since sec is strictly
convex if and only if sec00 x > 0, which in turn happens if and only if cos x >
0, we conclude that sec is strictly convex in . 2 ; 2 /.
(vii) Since sec is periodic, it graph does not contain horizontal or oblique asymptotes. On the other hand, we easily get limx! 2 sec x D limx! 2 C sec x D
C1, so that the vertical lines x D ˙ 2 are asymptotes of the graph of sec.
(viii) Since sec is periodic and nonconstant, there does not exist limx!˙1 f .x/.
(ix) Computing cos x for x D 0; ˙ 12
, ˙ 6 , ˙ 4 , ˙ 3 and ˙ 11
, we plot points
12
p
p2
.0; 1/, ˙ 6 ; 3 , ˙ 4 ; 2 and ˙ 3 ; 2 on the graph of sec.
9.8 Sketching Graphs
341
y
y=1
− π2
π
2
0
π
3π
2
x
y = −1
Fig. 9.16 Graph of sec in . 2 ; 2 / [ . 2 ;
3
/
2
Gathering together the above information, we sketch the graph of sec in
the interval . 2 ; 2 / [ . 2 ; 3
/ as shown in Fig. 9.16.
2
t
u
Example 9.80 Sketch the graph of the polynomial function f W R ! R, given by
f .x/ D 2x3 C 4x2 C 2x 1.
Solution
(ii) and (iii) Since f 0 .x/ D 6x2 C 8x C 2, the critical points of f are x D 1 and
x D 13 . Thus, f 0 .x/ < 0 , x 2 .1; 13 /, and f decreases , f 0 .x/ <
0 , x 2 .1; 13 /. Therefore, f increases , x 2 .1; 1/ [ . 13 ; C1/.
It follows that x D 1 (resp. x D 13 ) is a local maximum (resp. minimum)
point of f .
(iv) and (v) Since f 00 .x/ D 12x C 8 and f is strictly convex (resp. concave) in an
interval I if and only if f 00 .x/ > 0 (resp. f 00 .x/ < 0) in I, we conclude that
f is strictly convex (resp. concave) in . 23 ; C1/ (resp. .1; 23 /). Finally,
x D 23 is the only inflection point of the graph of f .
2
(vi) Notice that f . 23 / D 31
27 , and the tangent line to the graph of f in x D 3
2
2
0
has slope f . 3 / D 3 . Letting .0; c/ be the point where such a tangent line
2
43
intersects the vertical axis, we have c.31=27/
0.2=3/ D 3 , so that c D 27 .
342
9 Limits and Derivatives
Fig. 9.17 Graph of the
polynomial function
f .x/ D 2x3 C 4x2 C 2x 1
y
5
4
1
2
1
2
−1 − 3 − 3
− 32
x
−1
35
− 27
− 74
31
− 27
43
− 27
(vii) Since f is defined in the whole real line, its graph does not possess vertical
2
asymptotes. On the other hand, since limx!˙1 f .x/
x D limx!˙1 .2x C 4x C
1
2 x / D C1, Problem 11, page 291, guarantees that the graph of f does not
have oblique asymptotes.
(viii) It follows from Example 9.17 that limx!˙1 f .x/ D ˙1.
(ix) Since f .0/ D 1, it follows from (viii) and the IVT that f has a positive root
˛. Computing f .x/ for x D 32 , 1, 13 , 0, 14 , 13 , 12 and 1, we plot points
1
7
1 1
1 5
. 32 ; 74 /, .1; 1/, . 13 ; 35
27 /, .0; 1/, . 4 ; 32 /, . 3 ; 27 /, . 2 ; 4 / and .1; 7/
/
on the graph of f . (Recall that, in item (vi), we had already get point . 23 ; 31
27
on the graph.) In particular, we conclude that ˛ 2 . 14 ; 13 /.
Finally, as in the previous example, we gather the above information to sketch
the graph of f in Fig. 9.17.
t
u
Example 9.81 Sketch the graph of f W R n f1g ! R, given by f .x/ D x2 C
1
.
xC1
Solution
3
2
1
2x C4x C2x1
, we have f 0 .x/ D 0 ,
(ii) and (iii) Since f 0 .x/ D 2x .xC1/
2 D
.xC1/2
3
2
2x C 4x C 2x 1 D 0. Hence, the only critical point of f is the only positive
root ˛ of the polynomial function of the previous example, so that ˛ 2 . 14 ; 13 /.
Then, f 0 .x/ > 0 , 2x3 C 4x2 C 2x 1 > 0 , x > ˛, and we conclude that
f increases in .˛; C1/. Note that ˛ is a minimum local point of f .
2
(iv) and (v) Since f 00 .x/ D 2 C .xC1/
3 , we have f strictly convex (resp. concave)
1
1
> 1 (resp. xC1
< 1) in I. Therefore, f is
in an interval I if and only if xC1
strictly convex in each of the intervals .1; 2/ and .1; C1/, and strictly
concave in .2; 1/. Moreover, its only inflection point is x D 2.
9.8 Sketching Graphs
343
(vi) Note that f .2/ D 3, and the tangent line to the graph of f at x D 2 has
slope f 0 .2/ D 5. Letting .0; c/ be the point in which such a tangent line
c3
intersects the vertical axis, we have 0.2/
D 5, so that c D 7.
(vii) Since x D 1 is the only point for which f .x/ is not defined, the vertical
line x D 1 is the only candidate for vertical asymptote of the graph. This
is indeed the case, for, limx!1 f .x/ D 1 and lim
x!1C f .x/ D C1.
f .x/
1
On the other hand, since limx!˙1 x D limx!˙1 x C x.xC1/
D ˙1,
Problem 11, page 291, assures that the graph has no oblique asymptotes.
(viii) It follows from Example 9.17 that limx!˙1 f .x/ D C1.
(ix) Computing f .x/ for x D 52 , 32 , 54 , 34 , 12 , 14 , 0, 14 , 12 , 1 and 2, we plot
3
7
5
39
3 73
1 9
1 67
points . 52 ; 67
12 /, . 2 ; 4 /, . 4 ; 16 /, . 4 ; 16 /, . 2 ; 4 /, . 4 ; 48 /, .0; 1/,
1
7
1 5
3
13
. 4 ; 32 /, . 2 ; 4 /, .1; 2 / and .2; 3 / on the graph of f . (Recall that, in item (vi),
we had already get point .2; 3/ on the graph.)
Once more, we collect all of the above to sketch the graph of f , as shown in
Fig. 9.18. Note that, for the purpose of a better qualitative picture, we adopted
different scales on the horizontal and vertical axes.
t
u
Fig. 9.18 Graph of
1
f .x/ D x2 C xC1
y
3
−2
−1
α
x
344
9 Limits and Derivatives
Problems: Section 9.8
1. Sketch the graph of the cosecant function, csc W R n fkI k 2 Zg ! R, such that
csc x D sin1 x .
2. In each of the items below, sketch the graph of the given function (defined in
its maximal domain). For each such function (and whenever pertinent), compute
explicitly or estimate the critical points, monotonicity intervals, inflection points,
tangent lines at inflection points, intervals of convexity or concavity, asymptotes
and behavior at infinity:
(a)
(b)
(c)
(d)
f .x/ D x4 C 2x2 2x C 1.
f .x/ D x C 1x .
f .x/ D x2 1C1 .
f .x/ D x3 1C1 .
(e)
(f)
(g)
(h)
f .x/ D
f .x/ D
f .x/ D
f .x/ D
C x2 1C1 .
x
.
x2 C1
x
.
2
x 3xC2
x
.
2
2
.x C1/.x 1/
x
2
3. * Let f W .a; b/ ! R be twice differentiable at x0 2 .a; b/, with f 00 .x0 / ¤ 0.
Since f is in particular differentiable at x0 , the tangent line r to the graph at
A.x0 ; f .x0 //, is not vertical. We define the osculating circle to the graph at A as
the circle tangent to r at A that better approximates the shape of the graph in
a neighborhood of x0 , in the following sense: taking c 2 .a; x0 / and d 2 .x0 ; b/
such that one of the arcs of situated in the strip defined by the vertical lines
x D c and x D d is the graph of a function g W .c; d/ ! R, we have g.x0 / D f .x0 /,
g0 .x0 / D f 0 .x0 / and g00 .x0 / D f 00 .x0 /. In this respect, do the following items:
(a) Let O.˛; ˇ/ be the center of . If f 0 .x0 / D 0, show that ˛ D x0 ; if f 0 .x0 / ¤ 0,
.x0 /
show that ˇf
D f 0 .x1 0 / .
˛x0
(b) p
Show that the equation of is .x ˛/2 C .y ˇ/2 D R2 , where R D
.x0 ˛/2 C .f .x0 / ˇ/2 is its radius.
(c) For x 2 .c; d/, substitute y D g.x/ in (b) and differentiate the relation thus
0 .x /2
x0 ˛
0
obtained to get g0 .x0 / D ˇg.x
D f 0 .x0 / and g00 .x0 / D 1Cf
ˇf
.x0 / .
0/
0
2
.x0 /
1Cf 0 .x0 /2
0
. From
(d) Conclude that ˇ D f .x0 / C 1Cf
f 00 .x0 / and ˛ D x0 f .x0 /
f 00 .x0 /
this, and with respect to the line r, show that O is located in the upper (resp.
lower) half-plane, provided f is strictly convex (resp. strictly concave) in a
neighborhood of x0 .
0
2 3=2
(e) Show that R D .1Cfjf 00.x.x00//j/ . (In particular, observe that the osculating circle
at A is uniquely determined by the conditions given at the statement of the
problem.)
u D .1; f 0 .x //, of origin A, is parallel to the
(f) Show that the geometric vector !
0
u a counterclockwise rotation with
line r. Show also that, upon applying to !
!
center A and angle 90ı , we get a vector of the form AO, with 2 R .
The radius R of is the curvature radius of the graph of f at x0 (or A); the
curvature of the graph at x0 is the real number
9.8 Sketching Graphs
345
Fig. 9.19 Osculating circle
to the graph of x 7! 1x at A
y
O
Γ
A
x
kD
R
D
f 00 .x0 /
;
.1 C f 0 .x0 /2 /3=2
2 f˙1g is the sign of . (Observe that the bigger R, the closer jkj
where D jj
is to 0. This fact reflects gives geometric intuition to the term curvature.)
p
(g) Let R > 0 and f W .R; R/ ! R be given by f .x/ D R2 x2 (so that the
graph of f is a semicircle of radius R). Show that the curvature radius of the
graph of f is constant and equal to R.
As an illustration, Fig. 9.19 brings the osculating circle to the graph of f .x/ D 1x ,
x > 0, at a point A.
4. The purpose of this problem is to sketch the graph of f W R ! R, given by
f .x/ D
0; se x D 0
:
x sin 1x ; se x ¤ 0
To this end, do the following items:
(a) Show that it suffices to consider the case x 0.
(b) Show that the set of critical points of f in .0; C1/ is the set of reals x such
that x D 1y , where y > 0 satisfies the equation sin y y cos y D 0.
(c) Show that the solutions y > 0 to this equation form a sequence .yn /n1 , such
that n < yn < n C 2 and the sequence .n C 2 yn /n1 decreases to 0
when n ! C1.
(d) If xn D y1n for n 1, show that x2k1 is a local minimum point and x2k is a
local maximum point for f , for every k 1.
(e) Show that jf .x1 /j > jf .x2 /j > jf .x3 /j > ! 0.
(f) f increases in Œx1 ; C1/, with limx!C1 f .x/ D 1.
1
1
(g) Mostre que f is strictly convex in . 2k
; .2k1/
/ and strictly concave in
1
1
. .2kC1/ ; 2k /, for every integer k 1.
(h) Sketch the graph of f .
Chapter 10
Riemann’s Integral
This chapter completes the task of establishing the fundamentals of Calculus, this
time studying the operation of integration on functions. As we shall see in the next
section, in its most simple form this reduces to the computation of areas under
the graphs of nonnegative continuous functions f W Œa; b ! R, suggesting that
geometric intuition will play a strong role throughout. Among other byproducts of
the coming discussions, we prove the irrationality of and study the properties
of two of the most ubiquitous functions of Mathematics, the exponential and
logarithmic functions with base e.
10.1 Some Heuristics II
Our story begins in the III century B.C., with the great Archimedes, who considered
(and solved) the problem of computing the area under an arc of parabola, by using
the method of exhaustion.
In modern notation and in a slightly more general situation, the heuristics behind
such a method is the following: given a nonnegative function f W Œa; b ! R (cf.
Fig. 10.1), one wants to compute the area of the region R of the cartesian plane,
situated under the graph of f and above the horizontal axis, so that
R D f.x; y/ 2 R2 I a x b and 0 y f .x/g:
To this end, one starts by dividing the interval Œa; b into k equal intervals, with
the aid of the points a D t0 < t1 < < tk D b such that ti ti1 D ba
k for
1 i k. Then (cf. Fig. 10.2), one considers the portion of R contained in the
vertical strip bounded by the straight lines x D ti1 and x D ti , and approximates its
area, from below, by that of the greatest rectangle contained therein and having the
Œti1 ; ti as one of its sides; accordingly, one approximates the area of that portion of
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_10
347
348
10 Riemann’s Integral
Fig. 10.1 The region R
under the graph of f
f
R
a
Fig. 10.2 Approximating the
area of R from above and
from below
b
x
b
x
f
mi
a
Mi
ti−1 ti
R, from above, by that of the smallest rectangle containing it and having the interval
Œti1 ; ti as one of its sides.
One now approximates the whole area of R from below and from above,
computing, in each case, the sums of the areas of the k rectangles of the previous
paragraph. Assuming that f attains its minimum and maximum values in each
interval Œti1 ; ti (which is always the case if f is continuous), and letting mi and
Mi respectively denote such values (so that mi and Mi are the lengths of the heights
of the rectangles one is considering), one obtains the sums
A.f I k/ WD
k
X
k
ba X
mi
k
iD1
(10.1)
k
ba X
Mi :
k
iD1
(10.2)
mi .ti ti1 / D
iD1
and
A.f I k/ WD
k
X
iD1
Mi .ti ti1 / D
Letting A.R/ denote the area of R, one then gets
A.f I k/ A.R/ A.f I k/;
and hopes that A.f I k/ and A.f I k/ do approximate A.R/ better and better as k !
C1.
In order to illustrate the difficulties involved, let’s consider the more particular
situation of the function f W Œ0; b ! R given by f .x/ D xn , where n is a given
natural number. As before, letting k 2 N and 0 D t0 < t1 < < tk D b be
the partition of Œ0; b such that ti ti1 D bk for 1 i k, we have ti D bik for
n
0 i k. Also, since x 7! xn is increasing, it follows that mi D f .ti1 / D ti1
and
10.1 Some Heuristics II
349
Mi D f .ti / D tin . Thus,
A.f I k/ D
nC1 X
k
k k
b X n
b X bi n
b
ti D D
in
k iD1
k iD1 k
k
iD1
(10.3)
and, analogously,
nC1 X
k
b
.i 1/n :
A.f I k/ D
k
iD1
(10.4)
It easily follows from these computations that
A.f I k/ A.f I k/ D
bnC1
I
k
(10.5)
hence, A.f I k/ A.f I k/ ! 0 as k ! C1. Therefore, at least in this case, we
conclude that A.f I k/ and A.f I k/ approximate A.R/ better and better, as the number
k of rectangles increases.
To compute the actual value of A.R/, we use the result of Problem 21, page 86,
which guarantees that
k
k
X
.k 1/nC1 X n
<
.i 1/n <
i:
nC1
iD1
iD1
Combining these inequalities with (10.3) and (10.4), we get
A.f I k/ <
1 nC1
bnC1
1
< A.f I k/:
nC1
k
(10.6)
nC1
nC1 However, since both A.R/ and bnC1 1 1k
belong to the closed interval
A.f I k/; A.f I k/ , relation (10.5) gives
ˇ
1 nC1 ˇˇ
bnC1
bnC1
ˇ
1
:
ˇ A.f I k/ A.f I k/ <
ˇA.R/
nC1
k
k
Making k ! C1, we then get
1 nC1
bnC1
bnC1
1
:
A.R/ D lim
D
k!C1 n C 1
k
nC1
350
Fig. 10.3 Approximating the
area under the graph of f
10 Riemann’s Integral
y
f
a
x0
x
b
x
Let’s now return (again in a modern setting) to the general problem of the
computation of the area of the region under the graph of a given nonnegative
function f W Œa; b ! R (cf. Fig. 10.3).
For each real number x 2 Œa; b, let’s denote by A.x/ the area of the region of
the plane situated under the graph of f , above the horizontal axis and between the
vertical lines of abscissas a and x (with the convention that A.a/ D 0). For a fixed
x0 2 Œa; b/ and taking x0 < x b, we have
A.x/ A.x0 / D A.Œx0 ; x/;
(10.7)
where A.Œx0 ; x/ denotes the area of the portion of the region just described, situated
between the vertical lines of abscissas x0 and x.
For small values of x x0 (when compared to the values of f along the interval
Œx0 ; x), it is reasonable to assume that a good approximation for the value of
A.Œx0 ; x/ is the area of the trapezoid with base lengths f .x0 / and f .x/ and having
the line segment Œx0 ; x as one of its legs (the gray trapezoid of Fig. 10.3). Since the
altitude of such a trapezoid equals x x0 , it follows from (10.7) that
f .x0 / C f .x/
.x x0 /
A.x/ A.x0 / D A.Œx0 ; x/ Š
2
for small values of x x0 . This way, we conclude that
f .x0 / C f .x/
A.x/ A.x0 /
Š
x x0
2
(10.8)
for small values of x x0 , and it is also reasonable to assume that, the closest x is to
x0 , the better is such an approximation.
Therefore, if f is a continuous function, (10.8) suggests that the area function
A W Œa; b ! R is differentiable, with
f .x / C f .x/
A.x/ A.x0 /
0
A0 .x0 / D lim
D f .x0 /;
D lim
(10.9)
x!x0
x!x0
x x0
2
In words, the heuristic reasoning above suggests that we can recover the
continuous function f W Œa; b ! R out of the corresponding area function
nC1
A W Œa; b ! R. In the case of f .x/ D xn , for instance, we got A.x/ D xnC1 , so
that A0 .x/ D xn .
10.2 The Concept of Integral
351
In modern language, given a nonnegative continuous function f W Œa; b ! R, we
say that the area function A W Œa; b ! R is an indefinite integral for f , and write
Z x
A.x/ D
f .t/dt:
(10.10)
a
Rx
Here, the symbol a recalls the fact that the area under the graph of f , from a to x,
was computed
with the aid of sums that approximated the desired area better and
R
better ( is a stylized capital S); in turn, the symbol f .t/dt recalls the fact that the
summands of those sums had the form mi .ti ti1 /, where mi denoted the smallest
value of f along the interval Œti1 ; ti (hence, mi D f .t/ for some t 2 Œti1 ; ti ) and ti ti1 D ti was the ith difference along the interval Œa; b (Leibniz systematically
used the Greek letter to denote finite differences, and such a notation survived to
this day).
Taking (10.9) and (10.10) together, we get
Z
d x
f .t/dt D f .x/;
(10.11)
dx a
and this is essentially the content of the Fundamental Theorem of Calculus,
which suggests that the operations of integration (i.e., of computing integrals) and
derivation are somewhat inverses of each another.
The rest of this chapter puts all of the above in solid grounds, developing several
interesting applications along the way.
10.2 The Concept of Integral
This section introduces the concept of integral for bounded functions f W Œa; b ! R
and establishes the integrability of two important classes of such functions, namely,
the continuous and the monotone ones. To this end, we begin by fixing some
notations.
A partition of an interval Œa; b is the choice of a finite subset
P D fa D x0 < x1 < x2 < < xk D bg
(10.12)
of Œa; b. Given such a partition and a bounded (but not necessarily nonnegative)
function f W Œa; b ! R, we shall systematically denote
mj D infff .x/I xj1 x xj g WD inf f
Œxj1 ;xj and
Mj D supff .x/I xj1 x xj g WD sup f :
Œxj1 ;xj Note that both mj and Mj are well defined, thanks to the boundedness of f .
352
10 Riemann’s Integral
The analogues of the lower and upper approximations for the area of the region
under the graph of f , discussed in the previous section, are the lower sum and the
upper sum of f with respect to P, which are respectively defined by
s.f I P/ D
k
X
mj .xj xj1 / and S.f I P/ D
jD1
k
X
Mj .xj xj1 /:
jD1
For a fixed partition P, we certainly have mj Mj for every j, so that s.f I P/ S.f I P/. Hence, also in accordance with the discussion undertaken in the previous
section, we would like to declare a bounded function f W Œa; b ! R as being
integrable if
supfs.f I P/I P is a partition of Œa; bg D inffS.f I P/I P is a partition of Œa; bg:
However, some care is needed, for, the inequality s.f I P/ S.f I P/ (which is valid
when we take the same partition P at both sides) does not necessarily imply a
relation between the supremum and infimum above. This is due to the fact that
there is no relation between the generic partition P that declares the set at the left
hand side above and the one that declares the set at the right hand side (the choice
of the same letter to represent them was just a matter of notational convenience).
Thus, before we proceed towards a coherent definition of integrable function and
integral, we ought to compare s.f I P/ and S.f I Q/, for two generic partitions P and
Q of Œa; b. We do that in the coming lemma.
Lemma 10.1 Let f W Œa; b ! R be a bounded function. Given partitions P and Q
of Œa; b such that P Q, we have
s.f I P/ s.f I Q/ and S.f I Q/ S.f I P/:
Proof Let P be as in (10.12). We initially consider the case in which Q D P [ fx0 g,
with x0 ¤ x0 ; x1 ; : : : ; xk , and let i 2 f1; 2; : : : ; kg be such that xi1 < x0 < xi . Then,
S.f I Q/ D
i1
X
Mj .xj xj1 / C . sup f /.x0 xi1 /
Œxi1 ;x0 jD1
C
0
sup f .xi x / C
Œx0 ;xi k
X
Mj .xj xj1 /:
jDiC1
Now, since Œxi1 ; x0 ; Œx0 ; xi Œxi1 ; xi , we get
sup f ; sup f sup f D Mi ;
Œxi1 ;x0 Œx0 ;xi Œxi1 ;xi (10.13)
10.2 The Concept of Integral
so that
sup f .x0 xi1 / C
Œxi1 ;x0 353
sup f .xi x0 / Mi .x0 xi1 / C Mi .xi x0 /
Œx0 ;xi D Mi .xi xi1 /:
Hence, it follows from (10.13) that
S.f I Q/ i1
X
Mj .xj xj1 / C Mi .xi xi1 / C
jD1
D
k
X
k
X
Mj .xj xj1 /
jDiC1
Mj .xj xj1 / D S.f I P/:
jD1
We now consider an arbitrary partition Q containing P. Since Q is a finite set,
we can pass from P to Q in a finite number of steps, by adjoining to P the points of
Q n P, one at a time. In doing so, we obtain partitions P D P1 P2 Pl D Q
such that, for 2 i l, we get Pi by adding a single point to Pi1 . Then, the first
part of the proof gives
S.f I P/ D S.f I P1 / S.f I P2 / S.f I Pl / D S.f I Q/:
Finally, the proof of the inequality involving lower sums is completely analogue
and will be left to the reader (see Problem 1).
t
u
As an immediate consequence of the previous lemma, given a bounded function
f W Œa; b ! R and (any) partitions P and Q of Œa; b, we have
s.f I P/ s.f I P [ Q/ S.f I P [ Q/ S.f I Q/:
(10.14)
In particular, given a partition Q of Œa; b, the above inequalities guarantee that
S.f I Q/ us an upper bound for the set fs.f I P/I P is a partition of Œa; bg of the lower
sums of f . Hence, Theorem 7.4 guarantees that such a set does have a supremum,
which will be denoted supP s.f I P/:
sup s.f I P/ D supfs.f I P/I P is a partition of Œa; bg:
P
On the other hand, since the supremum of a set bounded from above is the smallest
of its upper bounds (one of which is S.f I Q/), we conclude that
sup s.f I P/ S.f I Q/:
P
354
10 Riemann’s Integral
However, since the partition Q was arbitrarily chosen, we conclude that this last
inequality is valid for every partition Q of Œa; b. To put in another way, the real
number supP s.f I P/ is a lower bound for fS.f I Q/I Q is a partition of Œa; bg, the set
of the upper sums of f . Hence, it follows from Problem 6, page 206, that such a set
has an infimum, which will be denoted by infQ S.f I Q/:
inf S.f I Q/ D inffS.f I Q/I Q is a partition of Œa; bg:
Q
Finally, since the infimum of a set bounded from below is the largest of its lower
bounds (one of which is supP s.f I P/), it follows that
sup s.f I P/ inf S.f I Q/:
Q
P
(10.15)
We are finally in position to present the central definition of this chapter.
Definition 10.2 A bounded function f W Œa; b ! R is Riemann integrable1 if
sup s.f I P/ D inf S.f I P/:
P
P
As we have anticipated at the first paragraph of this section, continuous and
monotone functions with domain Œa; b are integrable. Nevertheless, with what we
have at our disposal at this point, it is more convenient to start by presenting the
classical example of a bounded nonintegrable function.
Example 10.3 Recall from Problem 11, page 255) that Dirichlet’s function is f W
Œ0; 1 ! R such that
f .x/ D
0 if x … Q
:
1 if x 2 Q
Fixed a partition P D f0 D x0 < x1 < x2 < < xk D 1g of Œ0; 1, Problem 4, page
206, assures that all of the intervals Œxj1 ; xj contain rational and irrational numbers.
Therefore, for 1 j k we have mj D 0 and Mj D 1, so that
s.f I P/ D
k
X
mj .xj xj1 / D 0
jD1
1
When we look at large to Riemann’s legacy to Geometry and Analysis, we easily come to the
conclusion that they largely surpass the adequate formalization of the concept of integral that bears
his name. (As the attentive reader has certainly noticed, this chapter is entitled Riemann’s Integral!)
Indeed, Semi-Riemannian Geometry gave Albert Einstein the adequate theoretical apparatus for the
development of General Relativity. Other of Riemann’s creations, known today as Riemann’s zeta
function, has shown to be an ubiquitous object in Number Theory.
10.2 The Concept of Integral
355
and
S.f I P/ D
k
X
Mj .xj xj1 / D
jD1
k
X
.xj xj1 / D 1:
jD1
Then,
sup s.f I P/ D 0 < 1 D inf S.f I P/;
P
P
and f is nonintegrable.
In order to discuss the integrability of functions in a more transparent way, it is
worth to recast the definition of integrable function as in the coming result, which is
known as Cauchy’s integrability criterion.
Theorem 10.4 (Cauchy) A bounded function f W Œa; b ! R is integrable if and
only if the following condition is satisfied: for every > 0, there exists a partition
P of Œa; b such that
S.f I P / s.f I P / < :
(10.16)
Proof Firstly, suppose that f is integrable, so that supP s.f I P/ D infP S.f I P/. Let
this common value be denoted by I, and let > 0 be given. Since I D supP s.f I P/
is the least upper bound for the set of lower sums of f and I 2 < I, there exists a
partition P1 of Œa; b such that I 2 < s.f I P1 /. Analogously, since I D infP S.f I P/
is the greatest lower bound for the set of upper sums of f and I C 2 > I, there exists
a partition P2 of Œa; b such that S.f I P2 / < I C 2 . Letting P D P1 [ P2 , it follows
from (10.14) that
S.f I P / s.f I P / S.f I P2 / s.f I P1 / < I C
I
D :
2
2
Conversely, assume that the stated condition is satisfied. Then, given > 0, and
taking a partition P as in the statement, it follows from (10.15) that
0 inf S.f I P/ sup s.f I P/ S.f I P / s.f I P / < :
P
P
However, since > 0 was arbitrarily chosen, it follows that infP S.f I P/ supP s.f I P/ D 0, as wished.
t
u
We can finally establish the integrability of monotone and continuous functions.
Theorem 10.5 Every monotone function f W Œa; b ! R is integrable.
Proof Suppose f to be nondecreasing, the case of a nonincreasing function being
totally analogous. Then, the image of contained in the interval Œf .a/; f .b/, so that it
356
10 Riemann’s Integral
is clearly bounded. Let > 0 be given. By Cauchy’s integrability criterion, in order
to establish the integrability of f it suffices to find a partition P D fa D x0 < x1 <
< xk D bg of Œa; b such that condition (10.16) is satisfied for P D P.
Given a partition P as above, f being nondecreasing implies
mj D infff .x/I xj1 x xj g D f .xj1 /
and, analogously, Mj D f .xj /. Hence,
S.f I P/ s.f I P/ D
k
X
Mj .xj xj1 / jD1
D
k
X
mj .xj xj1 /
jD1
k
X
.f .xj / f .xj1 //.xj xj1 /:
jD1
Letting ı D maxfxj xj1 I 1 j kg, the last equality above and (3.15) give
S.f I P/ s.f I P/ <
k
X
.f .xj / f .xj1 //ı D ı.f .b/ f .a//:
jD1
Therefore, it suffices to choose P such that ı.f .b/ f .a// < , which happens
whenever ı < f .b/f.a/C1 , for instance.
t
u
Theorem 10.6 Every continuous function f W Œa; b ! R is integrable.
Proof Corollary 8.25 guarantees that such an f is bounded. Hence, again by
Cauchy’s integrability criterion and given > 0, in order to establish the
integrability of f it suffices to find a partition P D fa D x0 < x1 < < xk D bg of
Œa; b such that (10.16) is satisfied for P D P.
As in the proof of the previous of the previous result,
S.f I P/ s.f I P/ D
k
X
jD1
D
k
X
Mj .xj xj1 / k
X
mj .xj xj1 /
jD1
(10.17)
.Mj mj /.xj xj1 /:
jD1
Now, Theorem 8.24 assures that f is uniformly continuous. Therefore, according
to Definition 8.22, given 0 > 0 there exists ı > 0 such that
x; y 2 Œa; b and jx yj < ı ) jf .x/ f .y/j < 0 :
10.2 The Concept of Integral
357
Thus, starting with a partition P such that xj xj1 < ı for 1 j k, we
conclude that
xj1 x; y xj ) jx yj < ı ) jf .x/ f .y/j < 0 :
However, since Mj D supff .x/I xj1 x xj g and mj D infff .y/I xj1 y xj g,
it easily follows from the above inequality that
xj xj1 < ı ) Mj mj 0 ; 8 1 j k:
(10.18)
Finally, letting P be a partition such that xj xj1 < ı for 1 j k, it follows
from (10.17), (10.18) and (3.15) that
S.f I P/ s.f I P/ D
k
X
.Mj mj /.xj xj1 /
jD1
k
X
0 .xj xj1 / D 0 .b a/:
jD1
Therefore, in order to get S.f I P/ s.f I P/ < it suffices to start with 0 D
.
2.ba/
t
u
As anticipated in the end of the previous section, if f W Œa; b ! R is an integrable
function, we let
Z b
f .x/dx
(10.19)
a
denote the integral of f on the interval Œa; b. Note also that it does not matter
Rb
whether we denote the integral of f on Œa; b as above or writing a f .t/dt. Indeed,
such a change in notation is equivalent to changing the notation for the independent
variable of f , which certainly does not alter the value of the integral.
In spite of what we have done so far, unfortunately we do not have yet a general
procedure for computing integrals of (integrable) specific functions f W Œa; b ! R.
We shall remedy this situation in Sect. 10.5. However, see Problems 2, 4, 5 and 10.
Problems: Section 10.2
1. * Complete the proof of Lemma 10.1 showing that, if f W Œa; b ! R is bounded
and P and Q are partitions of Œa; b such that P Q, then s.f I P/ s.f I Q/.
2. * If f W Œa; b ! R is a constant function, say f .x/ D c for every x 2 Œa; b,
show that
Z
b
f .x/dx D c.b a/:
a
358
10 Riemann’s Integral
3. Let f W Œa; b ! R be a monotone function and Pk D fa D x0 < x1 < <
xk D bg be a uniform partition of Œa; b, i.e., such that xj xj1 D ba
k for
1 j k. Show that
Z
b
f .x/dx D lim s.f I Pk / D lim S.f I Pk /:
a
k!C1
k!C1
4. * Given a nonnegative function f W Œa; b ! R, check that the approximations (10.1) and (10.2) for the area of the region R under the graph of f are
particular cases of lower and upper sums of f . Then, use this fact, (10.6) and the
result of the previous problem to show that, given n 2 N and b > 0, we have
Z
b
xn dx D
0
bnC1
:
nC1
Rbp
5. In order to compute a x dx, do the following items:
(a) Show that, for reals u v 0, one has
p
p
p
p
1
1 p
2 p
.u u v v/ .u v/. u v/ < . u C v/.u v/
3
3
2
and
p
p
1 p
2 p
. u C v/.u v/ < .u u v v/:
2
3
(b) If Pk D fa D x0 < x1 < < xk D bg is a uniform partition of Œa; b, use
the result of (a) to show that
p
p
p
p
p
p
2 p
.b a/. b a/
b a
.b b a a/ < s. xI Pk / C
3
3k
2k
and
p
p
p
2 p
b a
< .b b a a/:
2k
3
p
Rbp
p
(c) Conclude that a x dx D 23 .b b a a/.
p
R1 1
dx D 2. 2 1/. (We shall compute this integral in item
6. Assume that 0 pxC1
1
1
C p2Cn
C C p12n for
(d) of Problem 1, page 397.) If an D p1n p1Cn
p
n 2 N, show that .an /n1 converges to 2. 2 1/.
p
s. xI Pk / C
10.2 The Concept of Integral
359
R1 1
7. We shall compute 0 1Cx
2 dx D 4 in Example 10.42. Use this fact to show that,
1
1
1
if an D n n2 C1
2 C n2 C22 C C n2 Cn2 , then the sequence .an /n1 converges
to 4 .
8. Let f W Œa; b ! R be a bounded function and ˛ > 0 be given. If for each k 2 N
there exists a partition Pk of Œa; b satisfying S.f I Pk / s.f I Pk / < ˛k , show that
f is integrable, with
Z
b
f .x/dx D lim s.f I Pk / D lim S.f I Pk /:
k!C1
a
k!C1
For the coming problem, the reader might find it convenient to read again
the statement of Problem 3, page 262.
9. Let f W Œa; b ! R be a Lipschitz function, with Lipschitz constant c, and let
Pk D fa D x0 < x1 < < xk D bg be a uniform partition of interval Œa; b.
2
.
(a) Show that S.f I Pk / s.f I Pk / < c.ba/
k
(b) For 1 j k, choose a real number kj 2 Œxj1 ; xj and define the sum
P
†.f I Pk I k / by †.f I Pk I k / D kjD1 f .kj /.xj xj1 /. Prove that
Z
b
f .x/dx D lim †.f I Pk I k /:
a
k!C1
10. * Do the following items:
(a) Given a; h 2 R, with h ¤ 2l for every l 2 Z, prove that
k
X
sin.a C jh/ D
sin a C
.k1/h
2
sin .kC1/h
2
sin h2
jD0
and
k
X
cos.a C jh/ D
cos a C
.k1/h
2
sin h2
jD0
(b) Show that
Rb
a
sin x dx D cos a cos b and
sin .kC1/h
2
Rb
a
cos x dx D sin b sin a.
11. Let f W Œ0; 1 ! R be given by
f .x/ D
0; if x D 0 or x … Q
:
1
; if x D mn with m; n 2 N and gcd.m; n/ D 1
n
Prove that f is integrable.
360
10 Riemann’s Integral
10.3 Riemann’s Theorem and Some Remarks
In this section we state and prove an important theorem of Riemann on the
characterization of Riemann integrable functions, as well as make some other
important remarks on the theory of integration we are developing. In particular, we
rigorously define the concept of area for the region under the graph of a nonnegative
integrable function, which was informally discussed in the first section of the
chapter.
I. RIEMANN’S THEOREM. Usually, the presentation of the concept of integral in
Calculus courses makes use of the concept of Riemann sums. More precisely, given
a bounded function f W Œa; b ! R, one defines f as being (Riemann) integrable
if there exists a real number I (the Riemann integral of f ) satisfying the following
condition: given > 0, there exists ı > 0 such that, for every partition P D fa D
x0 < x1 < < xk D bg of Œa; b and every choice of points j 2 Œxj1 ; xj ,
1 j k, one has
k
ˇX
ˇ
ˇ
ˇ
f .j /.xj xj1 / I ˇ < :
maxfjxj xj1 jI 1 j kg < ı ) ˇ
(10.20)
jD1
In the above notations, we say that maxfjxj xj1 jI 1 j kg is the norm of
P
the partition P and that kjD1 f .j /.xj xj1 / is the Riemann sum of f with respect
to P and to the chosen intermediate points D .j /1jk . Letting
jPj D maxfjxj xj1 jI 1 j kg and †.f I PI / D
k
X
f .j /.xj xj1 /;
jD1
we summarize (10.20) by saying that I is the limit of the Riemann sums †.f I PI /,
when jPj ! 0 and for every choice D .j / of intermediate points relative to P.
Moreover, in this case we simply write
I D lim †.f I PI /:
jPj!0
The following result shows that the above definition of integral coincides with the
one we have adopted in the previous section. The proof of it can surely be omitted
on a first reading.
Theorem 10.7 (Riemann ) A bounded function f W Œa; b ! R is integrable (in
the sense of the previous section) if and only if the limit limjPj!0 †.f I PI / of its
Riemann sums exists, does not depending on the chosen intermediate points of .
Moreover, in this case one has
Z
b
a
f .x/dx D lim †.f I PI /:
jPj!0
10.3 Riemann’s Theorem and Some Remarks
361
Rb
Proof We first assume that f is integrable in Œa; b, with a f .x/dx D I.
Given > 0, we want to find ı > 0 such that, for every partition P of Œa; b, it
is the case that jPj < ı ) j†.f I PI / Ij < , for every choice of intermediate
points. To this end, start by choosing (by Cauchy’s integrability criterion) a partition
P0 D fa D x0 < x1 < : : : < xk D bg of Œa; b such that S.f I P0 / s.f I P0 / < 4 .
For a general partition P D fa D y0 < y1 < : : : < yl D bg of Œa; b, there are two
different kinds of intervals .yi1 ; yi /: those which contain at least one of the xj ’s and
those which do not.
If we choose 0 < ı < 12 jP0 j, then jPj < ı implies yi yi1 < 12 .xj xj1 / for all
1 i l, 1 j k. Therefore, each interval .yi1 ; yi / contains at most one of the
xj ’s and each interval .xj1 ; xi / contains at least three of the yi ’s. This way, we can
write
†.f I PI / D †0 .f I PI / C †00 .f I PI /;
where †0 .f I PI / (resp. †00 .f I PI /) denotes the collection of those summands of
†.f I PI / corresponding to indices i such that .yi1 ; yi / \ P0 ¤ ; (resp. .yi1 ; yi / \
P0 D ;).
Let D .1 ; : : : ; l / be an arbitrary choice of intermediate points for P, and
observe that
j†.f I PI / Ij j†0 .f I PI /j C j†00 .f I PI / Ij:
(10.21)
We now let M D supŒa;b jf j and estimate each of the summands at the right hand
side above.
(i) Estimating j†0 .f I PI /j: the restriction imposed on ı forces †0 .f I PI / to have
at most k1 summands (at most one for each of x1 , . . . , xk1 ). Since jf .i /j M
and yi yi1 < ı, we get
j†0 .f I PI /j D j†0 f .i /.yi yi1 /j †0 jf .i /j.yi yi1 / < Mı.k 1/:
(ii) Estimating j†00 .f I PI / Ij: we start by noticing that
j†00 .f I PI / Ij j†00 .f I PI / s.f I P0 /j C js.f I P0 / Ij
< j†00 .f I PI / s.f I P0 /j C ;
4
(10.22)
so we are left to estimating j†00 .f I PI / s.f I P0 /j. To this end, first recall that
if 1 i l is one of the indices that fall into †00 , then .yi1 ; yi / does not
contain any of the xj ’s, but there is exactly one index 1 j k such that
Œyi1 ; yi Œxj1 ; xj . Hence, we can write
†00 .f I PI / D †001 .f I PI / C C †00k .f I PI /;
362
10 Riemann’s Integral
where †00j .f I PI / stands for those summands of †00 .f I PI / corresponding to
indices 1 i l such that Œyi1 ; yi Œxj1 ; xj . Therefore, if mj D infŒxj1 ;xj f ,
then the triangle inequality gives
j†00 .f I PI / s.f I P0 /j k
X
j†00j .f I PI / mj .xj xj1 /j:
jD1
In order to deal with this last sum, fix 1 j k and let 1 i l be such that
Œyi1 ; yi Œxj1 ; xj . Recall that our choice of ı forces Œxj1 ; xj to contain at least
another point of P. Therefore, if rj i 1 and sj i are respectively the least and
greatest indices such that .yrj ; ysj / .xj1 ; xj /, then sj rj 2 and we have
yrj 1 < xj1 yrj < : : : < yi1 < yi < : : : < ysj xj < ysj C1 :
This way, and letting Sj D j†00j .f I PI / mj .xj xj1 /j and Mj D supŒxj1 ;xj f , we
estimate Sj by writing
sj
sj
ˇ X
X
ˇ
f .t /.yt yt1 / mj .yrj xj1 / mj .yt yt1 /
Sj ˇ
tDrj C1
tDrj C1
ˇ
ˇ
mj .xj ysj /ˇ
sj
X
jf .t / mj j.yt yt1 / C jmj j.yrj xj1 / C jmj j.xj ysj /
tDrj C1
sj
X
.Mj mj /.yt yt1 / C 2jmj jı
tDrj C1
.Mj mj /.ysj yrj 1 / C 2Mı
.Mj mj /.xj xj1 / C 2Mı;
where we used the facts that jf .t / mj j Mj mj , 0 < yrj xj1 ; xj ysj < ı and
ysj yrj 1 xj xj1 .
Hence, we get from (10.22) that
j†00 .f I PI / s.f I P0 /j k
X
jD1
Sj k
X
.Mj mj /.xj xj1 / C 2Mı
jD1
D S.f I P0 / s.f I P0 / C 2Mık <
C 2Mık:
4
10.3 Riemann’s Theorem and Some Remarks
363
Finally, gathering together (10.21), (10.22) and the estimates of items (i) and (ii),
we arrive at
j†.f I PI / Ij < C Mı.3k 1/:
2
Therefore, if we further ask that 0 < ı <
jPj < min
n1
2
jP0 j;
,
2M.3k1/
we get
o
) j†.f I PI / Ij < ;
2M.3k 1/
as we wished to show.
The proof of the converse is left to the reader as an exercise (see Problem 9). u
t
In spite of the fact that the introduction of the integral by means of Riemann
sums was the actual way the theory historically evolved, the approach by means of
lower and upper sums makes it technically much easier to deal with the material of
the next section. Nevertheless, as we shall see right now, the characterization of the
integral by Riemann’s theorem allows us to deal with the heuristic discussion of the
previous section (concerning area computations) in a mathematically precise way.
Given a bounded and nonnegative function f W Œa; b ! R, let R denote the
region under the graph of f (and above the horizontal axis and between the vertical
lines x D a and x D b):
R D f.x; y/ 2 R2 I a x b and 0 y f .x/g:
Let Pk D fa D x0 < x1 < < xk D bg be a uniform partition of Œa; b, i.e., such
that xj xj1 D ba
for 1 j k.
k
If f is continuous, then Theorem 8.26 guarantees that, for 1 j k, there exist
kj ; kj0 2 Œxj1 ; xj such that
f .kj / D min f and f .kj0 / D max f :
Œxj1 ;xj Œxj1 ;xj Hence, the lower and upper approximations A.f I k/ and A.f I k/ for the area of R
(respectively defined by (10.1) and (10.2), page 348)) respectively coincide with the
Riemann sums †.f I Pk I k / and †.f I Pk I k0 / (here, note that k D .kj / and k0 D
.kj0 /. However, since jPk j D 1k , it follows from Riemann’s theorem that
Z
b
f .x/dx D lim †.f I Pk I k / D lim A.f I k/
k!C1
a
k!C1
and, analogously,
Z
b
f .x/dx D lim A.f I k/:
a
Thus,
Rb
a
k!C1
f .x/dx is the only reasonable value for the area of R.
364
10 Riemann’s Integral
On the other hand, if f is merely integrable (albeit also nonnegative), then, except
for changing the minimum and maximum values of f along Œxj1 ; xj by its infimum
and supremum along such an interval, A.f I k/ and A.f I k/ also reduce to the lower
and upper sums s.f I Pk / and S.f I Pk /, respectively.
We sum up the above discussion in the following
Definition 10.8 Let f W Œa; b ! R be a nonnegative integrable function. We define
the area of the region R under its graph as
Z
b
A.R/ D
f .x/dx:
a
In the next section, the above definition will give heuristic arguments towards the
validity of some of the operational properties of Riemann’s integral.
II. LEBESGUE’S THEOREM AND INTEGRAL. In Theorem 10.6, we proved
that every continuous function f W Œa; b ! R is integrable. We also saw, in
Example 10.3, that the Dirichlet function (which, by Problem 11, page 255, is
discontinuous at every point of the interval Œ0; 1) is not integrable. On the other
hand, we shall prove in the next section (cf. Proposition 10.19) that a piecewise
continuous function f W Œa; b ! R (i.e., a function with a finite number of points
of discontinuity x0 2 Œa; b, such that for all of them the lateral limits limx!x0 ˙ f .x/
do exist) is still integrable. A more dramatic example is given by the function of
Problem 11, page 359, which is integrable but has an infinite number of points of
discontinuity.
The last paragraph suggests that there might be a more intimate relation between
the integrability of a bounded function f W Œa; b ! R and the size of the set Df of
its points of discontinuity. This is indeed the case, and a precise formulation of such
a connection requires the concept of a set of measure zero, which we now give.
Definition 10.9 We say that X R is a set of (Lebesgue2 ) measure zero, or a
null set if, given > 0, there exist open intervals I1 , I2 , I3 , . . . such that
X
[
Ij and
j1
X
jIj j < ;
j1
where jIj j stands for the length of Ij .
The following result links the concept of null set to that of Riemann integrable
functions.
Theorem 10.10 (Lebesgue) A bounded function f W Œa; b ! R is Riemann
integrable if and only if the set Df of its points of discontinuity has measure zero.
2
After Henri Lebesgue, French mathematician of the XX century.
10.3 Riemann’s Theorem and Some Remarks
365
We shall not prove Lebesgue’s theorem here, and refer the interested reader to
[1, 9] or [20]. The reason is that, apart from the Problem 4 (which will be obtained
by more elementary methods in the coming section) and Problem 5, page 468
(for which we present two proofs, one of which does not make use of Lebesgue’s
theorem), we will not use it anywhere else in the book. Nevertheless, it is important
to realize that it gives a unified explanation for the status, with respect to integration,
of all of the functions of the next to last paragraph. Indeed, since the set of points of
discontinuity of a continuous (resp. piecewise continuous) function is empty (resp.
finite), hence of measure zero, such functions are integrable. On the other hand, it
is possible to prove that Q is a null set, while Œ0; 1 is not (for the case of Q, see
Problem 2); this explains why the function of Problem 11, page 359 is integrable
(for, its set of points of discontinuity is Q \ Œ0; 1, which—being a subset of Q—is
certainly a null set), as well as why the Dirichlet function is not integrable (for, its
set of points of discontinuity is the whole interval Œ0; 1).
Let I denote the set of irrational numbers. It is possible to prove that the set
I \ Œa; b, of the irrational numbers of the closed interval Œa; b, is not a null set.
Hence, Lebesgue’s theorem assures that a function f W Œa; b ! R whose set of
points of discontinuity is I \ Œa; b is not integrable. However, as we stressed before,
such a function does not exist (a proof can be found in [1], for instance).
Lebesgue’s contributions to integration theory go far beyond Theorem 10.10. At
the dawn of the XX century, he introduced a much more refined and flexible concept
of integral in his doctor’s thesis. In order to get a glimpse of it, we shall first of all
need the following
Definition 10.11 The characteristic function of a set A R is the function XA W
Œa; b ! R such that
XA .x/ D
1; if x 2 A
:
0; if x … A
Now, let f W Œa; b ! Œ0; L be a nonnegative bounded function. In the context of
Riemann’s integral, we start by choosing a partition P D fa D x0 < x1 < <
xk D bg of the domain Œa; b of f . Then, we define the step functions fP; ; fP;C W
Œa; b ! R by letting
fP; D
k
X
mj XŒxj1 ;xj and fP;C D
jD1
k
X
Mj XŒxj1 ;xj ;
jD1
where mj D infŒxj1 ;xj f and Mj D supŒxj1 ;xj f . In the next section, we shall show
(cf. Example 10.20) that such functions are Riemann integrable, with
Z
b
fP; .x/dx D
a
k
X
jD1
Z
b
mj .xj xj1 / and
fP;C .x/dx D
a
k
X
jD1
Mj .xj xj1 /:
366
10 Riemann’s Integral
Thus, in accordance with the previous section, f is Riemann integrable if
Z b
Z b
sup
fP; .x/dx D inf
fP;C .x/dxI
P
P
a
a
moreover, in this case the integral of f coincides with this common value.
On the contrary, in the realm of Lebesgue’s integral, we start with a partition
Q D f0 D y0 < y1 < < yl D Lg of the codomain Œ0; L of f . Then, for
1 j k 1 we take the inverse image Aj of the interval Œyj1 ; yj / by f , which is
defined by
Aj D fx 2 Œa; bI f .x/ 2 Œyj1 ; yj /gI
we also consider the inverse image Ak of the interval Œyk1 ; yk , such that
Ak D fx 2 Œa; bI f .x/ 2 Œyk1 ; yk g:
If these sets are not too complicated (in a sense that doesn’t interest us at this
moment), one can show that it is possible to associate to Aj a nonnegative real
number m.Aj /, which we call the Lebesgue measure of Aj . Moreover, this is done
in such a way that if Aj is an interval then m.Aj / coincides with the length of Aj ; in
this case, we say that Aj is a (Lebesgue) measurable set.
Assuming that all of the Aj ’s are measurable, we consider the simple function
fQ D
k
X
yj1 XAj
jD1
and define its Lebesgue integral, denoted
Z
Œa;b
fQ D
R
Œa;b fQ ,
k
X
by
yj1 m.Aj /:
(10.23)
jD1
Finally, if for every partition Q of Œ0; L the resulting sets Aj are measurable,
then we say that f is a Rmeasurable function; if this is so, we define the Lebesgue
integral of f , denoted Œa;b f , by
Z
Z
f D sup
Œa;b
fQ I Q is a partition of Œ0; L :
(10.24)
Q
Although we have “defined” the Lebesgue integral just for nonnegative bounded
functions, it is possible to consider the notion of Lebesgue integral in the context
of bounded measurable (but not necessarily nonnegative) functions f W Œa; b !
R. (Actually, we could consider the even more general case of an unbounded
measurable function, but this will not play a role here).
10.3 Riemann’s Theorem and Some Remarks
367
It is possible to show that the Lebesgue integral is more comprehensive than
the Riemann integral, in two ways: on the one hand, every (bounded) Riemann
integrable function f W Œa; b ! R is also Lebesgue integrable and the values of
both integrals of f coincide; on the other, there exist bounded functions f W Œa; b !
R which are not Riemann integrable but are Lebesgue integrable (the Dirichlet
function is one such—see Problem 5).
Moreover, as we shall comment on Sect. 11.2 (cf. Remark 11.15), Lebesgue
integration also has at its disposal a number of convergence results whose proofs
are inaccessible in the context of Riemann integration, but which reveal themselves
to be central tools for the study of the deeper properties of sequences and series
of functions. Actually, such results are one of the main reasons behind the fact that
Lebesgue integration is more adequate than Riemann integration for many purposes
in Mathematics and its applications. To name one relevant example, it is completely
indispensable to the modern study of Partial Differential Equations and Differential
Geometry.
If this is so, the reader might ask why not to study the Lebesgue integral in
advance, relegating the Riemann integral to a museum. The reason is that, on the
one hand, in spite of its greater flexibility, an adequate presentation of the concepts
needed to develop Lebesgue’s notion of integral is considerably more complicated
than what we have done so far to the Riemann integral (in (10.24), the sets Aj may
be very complicated); on the other hand, and as we shall see later in this chapter,
Riemann’s integral suffices to the discussion of several interesting problems.
For the interested reader, we suggest the references [20] or [27] for rather
elementary introductions to the Lebesgue measure and integral.
Problems: Section 10.3
R1
1. In item (c) of Problem 1, page 397, we shall show that 0 x sin.x/dx D 1 .
k Pn k
Use this fact to show that, if an D kD1 n2 sin n , then .an /n1 converges
and limn!C1 an D 1.
2. * Show that Q is a null set.
S
3. Prove that if A1 ; A2 ; : : : R are null sets, then so is j1 Aj .
4. Use Lebesgue’s theorem to prove that if f ; g W Œa; b ! R are Riemann
integrable, then fg W Œa; b ! R is also Riemann integrable.
5. Show that Dirichlet’s
function f W Œ0; 1 ! R (cf. Example 10.3) is Lebesgue
R
integrable, with Œ0;1 f D 0.
6. If f W Œa; b ! Œ0; L is a monotone function, we have showed in Theorem 10.5
that f is Riemann integrable. Show that f is also Lebesgue integrable, with
Rb
R
a f .x/dx D Œa;b f .
368
10 Riemann’s Integral
7. (Berkeley) Let f W Œ0; C1/ ! Œ0; C1/ be a continuous and increasing
bijection. Show that
Z a
Z b
f .x/dx C
f 1 .x/dx ab;
0
0
for every positive reals a and b.
8. (Leningrad) Let f ; g W Œ0; 1 ! Œ0; 1 be continuous functions, with f
nondecreasing. Prove that
Z 1
Z 1
Z 1
.f ı g/.x/dx f .x/dx C
g.x/dx:
0
0
0
9. * Complete the proof of Riemann’s theorem.
10.4 Operating with Integrable Functions
This section is devoted to the derivation of some useful operational properties for
the Riemann integral. In spite of the fact that such properties will be extensively
used along the rest of the book, their proofs can be omitted in a first reading, with
essentially no loss of continuity.
Along all of this section, given bounded functions f ; g W Œa; b ! R and a
partition P D fa D x0 < x1 < < xk D bg of Œa; b, we shall denote the
infimums of f and g in Œxj1 ; xj by mj .f / and mj .g/, respectively; accordingly, Mj .f /
and Mj .g/ will denote the supremums of f and g in the same interval Œxj1 ; xj , also
respectively.
If f ; g W Œa; b ! R are nonnegative integrable functions, with f g, then the
regions Rf e Rg , respectively situated under the graphs of f and g, are such that
Rf Rg ; we therefore expect that A.Rf / A.Rg /. This is indeed the case, and
we usually refer to it by saying that the Riemann integral is monotonic. The general
case is as follows.
Proposition 10.12 If f ; g W Œa; b ! R are integrable functions such that f g,
Rb
Rb
then a f .x/dx a g.x/dx.
Proof Letting P D fa D x0 < x1 < < xk D bg denote a partition of Œa; b, it
follows at once from f g that Mj .f / Mj .g/ for 1 j k. Hence,
S.f I P/ D
k
X
Mj .f /.xj xj1 / jD1
However, since
Rb
a
k
X
Mj .g/.xj xj1 / D S.gI P/:
jD1
f .x/dx S.f I P/, we conclude that
Z b
f .x/dx S.gI P/;
a
10.4 Operating with Integrable Functions
for every partition P of Œa; b. Thus,
upper sums S.gI P/, so that
Z
b
369
Rb
a
f .x/dx is a lower bound for the set of the
Z
b
f .x/dx inf S.gI P/ D
P
a
g.x/dx:
a
t
u
For what comes next, we consider again a nonnegative integrable function f W
Œa; b ! R and a positive real number c. Problem 10, page 194, guarantees that
the region Rcf under the graph of the function cf can be obtained from the region
Rf under the graph of f by vertically stretching Rf by a factor c; we thus hope that
A.Rcf / D cA.Rf /. On the other hand if g W Œa; b ! R is another integrable and
nonnegative function, then, for each x0 2 Œa; b, the segment of the vertical line
x D x0 contained in Rf Cg (the region under the graph of f C g) has length equal
to the sum of the lengths of the segments of such a line which are contained in Rf
and Rg ; thus suggests that Rf Cg can be obtained by glueing Rg right above Rf and,
hence, that we should have A.Rf Cg / D A.Rf / C A.Rg /.
The next result shows that the two properties of the Riemann integral hinted by
the heuristic reasonings above are actually true. From now on, we shall refer to these
properties by saying that the Riemann integral is respectively linear and additive.
Proposition 10.13 Let f ; g W Œa; b ! R be integrable, and let c 2 R. Then:
Rb
Rb
(a) cf W Œa; b ! R is integrable, with a cf .x/dx D c a f .x/dx.
Rb
Rb
Rb
(b) f Cg W Œa; b ! R is integrable, with a .f .x/Cg.x//dx D a f .x/dxC a g.x/dx.
Proof (a) As in the proof of the previous proposition, let P D fa D x0 < x1 <
< xk D bg be a partition of Œa; b.
If c > 0, it is immediate to verify (see Problem 12, page 207) that
mj .cf / D inf .cf / D c inf f D c mj .f /
Œxj1 ;xj Œxj1 ;xj and, analogously, Mj .cf / D cMj .f /. Thus,
s.cf I P/ D
k
X
mj .cf /.xj xj1 / D c
jD1
k
X
mj .f /.xj xj1 / D c s.f I P/
jD1
and, analogously, S.cf I P/ D c S.f I P/. The integrability of f and the elementary
properties of the concepts of supremum and infimum (see Problem 12, page 207,
again) give us
Z
b
sup s.cf I P/ D sup.c s.f I P// D c sup s.f I P/ D c
P
P
P
f .x/dx
a
370
10 Riemann’s Integral
and
Z
b
inf S.cf I P/ D inf.c S.f I P// D c inf s.f I P/ D c
P
P
P
f .x/dx:
a
If c < 0, the deduction of the two relations above is entirely analogous; one just
needs to observe that mj .cf / D c Mj .f /, Mj .cf / D c mj .f /, s.cf I P/ D c S.f I P/ and
S.cf I P/ D c s.f I P/, so that
Z
sup s.cf I P/ D sup.c S.f I P// D c inf S.f I P/ D c
P
P
P
b
f .x/dx
a
Rb
and, in the same way, infP S.cf I P/ D c a f .x/dx.
Rb
In any case, we have infP S.cf I P/ D supP s.cf I P/ D c a f .x/dx. Therefore, cf
is integrable, with
Z
Z
b
b
cf .x/dx D inf S.cf I P/ D c
P
a
f .x/dx:
a
(b) Let P and Q be partitions of Œa; b such that P [ Q D fa D x0 < x1 < <
xk D bg. It readily follows from Problem 13, page 207, that
sup .f C g/ sup f C sup g;
Œxj1 ;xj Œxj1 ;xj Œxj1 ;xj or (with respect to the partition P [ Q) Mj .f C g/ Mj .f / C Mj .g/. Hence,
S.f C gI P [ Q/ D
k
X
Mj .f C g/.xj xj1 /
jD1
k
X
.Mj .f / C Mj .g//.xj xj1 /
jD1
D S.f I P [ Q/ C S.gI P [ Q/
S.f I P/ C S.gI Q/:
However, since infR S.f C gI R/ S.f C gI P [ Q/, the inequalities above give
inf S.f C gI R/ S.f I P/ C S.gI Q/;
R
for all partitions P, Q and R of Œa; b.
10.4 Operating with Integrable Functions
371
In the last inequality above, taking the infimum over all partitions P and Q of
Œa; b and invoking Problem 13, page 207, again, we get
inf S.f C gI R/ inffS.f I P/ C S.gI Q/I P; Q partitions of Œa; bg
R
D inf S.f I P/ C inf S.gI Q/
P
Z
Q
Z
b
b
f .x/dx C
D
g.x/dx:
a
(10.25)
a
Arguing as we have done up to this moment, we successively get mj .f C g/ mj .f / C mj .g/, s.f C gI P [ Q/ s.f I P/ C s.gI Q/ and supR s.f C gI R/ s.f I P/ C
s.gI Q/. Hence, by using Problem 13, page 207, yet another time, we obtain
sup s.f C gI R/ sup s.f I P/ C sup s.gI Q/
R
P
Z
Q
b
D
Z
b
f .x/dx C
g.x/dx:
a
(10.26)
a
Finally, since supR s.f C gI R/ infR S.f C gI R/, it follows from (10.25)
and (10.26) that
Z
inf S.f C gI R/ D sup s.f C gI R/ D
R
R
Thus, f C g is integrable, with
Z
b
f .x/dx C
a
Rb
a
b
g.x/dx:
a
.f .x/ C g.x//dx D
Rb
a
f .x/dx C
Rb
a
g.x/dx.
t
u
An obvious corollary of the previous proposition is that, if f ; g W Œa; b ! R are
integrable functions, then f g W Œa; b ! R is also integrable, with
Z
Z
b
a
Z
b
.f .x/ g.x//dx D
b
f .x/dx a
g.x/dx:
(10.27)
a
Indeed, item (a) guarantees the integrability of f , whereas item (b) guarantees that
of f g D f C .g/. Relation (10.27) now follows from the formulas of items (a)
and (b) of the previous proposition:
Z
Z
b
a
Z
b
.f .x/ g.x//dx D
b
f .x/dx C
Z
a
.g.x//dx
a
b
Z
f .x/dx C .1/
D
a
b
g.x/dx:
a
If f W Œa; b ! R is a continuous function, the chain rule assures that jf j D j j ı f W
Œa; b ! R is also continuous; in particular, jf j is also integrable. In what follows,
372
10 Riemann’s Integral
we establish the integrability of jf j by supposing that f is only integrable. We also
obtain a quite useful inequality relating the integrals of f and jf j, which is known as
the triangle inequality for integrals.
Proposition 10.14 If f W Œa; b ! R is an integrable function, then the function
jf j W Œa; b ! R is also integrable, and the following inequality holds:
ˇZ
ˇ
ˇ
ˇ
b
a
ˇ Z
ˇ
f .x/dxˇˇ b
jf .x/jdx:
(10.28)
a
Proof Let fC ; f W Œa; b ! Œ0; C1/ be given by
fC .x/ D maxff .x/; 0g and f .x/ D minff .x/; 0g:
It is immediate to verify that f D fC f and jf j D fC C f . Hence, if
fC is integrable, then the discussion immediately subsequent to Proposition 10.13
guarantees that f D fC f is also integrable; however, this being the case, item (b)
of that proposition assures that the same holds for jf j D fC C f .
Now, in order to prove (10.28), note first of all that the monotonicity of the
Rb
integral, together with the fact that f 0, gives a f .x/dx 0. In turn, by
successively applying (10.27), this inequality and the formula of item (b) of the
previous proposition we get
Z
Z
b
Z
b
f .x/dx D
a
b
fC .x/dx a
Z
b
Z
b
fC .x/dx C
a
Z
f .x/dx
a
f .x/dx
a
b
.fC .x/dx C f .x//dx
D
Z
a
b
D
jf .x/jdx:
(10.29)
a
(For another proof of (10.28) we refer the reader to Problem 7.) Analogously,
Z
Z
b
f .x/dx D
a
Z
b
.f /.x/dx a
Z
b
b
j.f /.x/jdx D
a
jf .x/jdx;
a
so that
Z
Z
b
a
which is equivalent to (10.28).
Z
b
jf .x/jdx b
f .x/dx a
jf .x/jdx;
a
10.4 Operating with Integrable Functions
373
We are left to showing the integrability of fC . To this end, let P D fa D x0 <
x1 < < xk D bg be a partition of Œa; b. If f .x/ 0 for some x 2 Œxj1 ; xj , then
Mj .fC / D Mj .f /; on the other hand, from fC f we get mj .fC / mj .f /, so that
Mj .fC / mj .fC / Mj .f / mj .f /:
If f .x/ < 0 for all x 2 Œxj1 ; xj , then fC D 0 in Œxj1 ; xj and, then,
Mj .fC / mj .fC / D 0 Mj .f / mj .f /:
In any case, we have Mj .fC / mj .fC / Mj .f / mj .f /, so that
S.fC I P/ s.fC I P/ D
k
X
.Mj .fC / mj .fC //.xj xj1 /
jD1
k
X
.Mj .f / mj .f //.xj xj1 /
jD1
D S.f I P/ s.f I P/:
Now, since f is integrable, Cauchy’s criterion for integrability assures that, given
> 0, we can choose the partition P in such a way that S.f I P/ s.f I P/ < .
However, this being the case, it follows from the last inequality above that S.fC I P/
s.fC I P/ < . Therefore, again by Cauchy’s criterion, we conclude that fC is
integrable.
t
u
Example 10.15 Given a function f W Œa; b ! R, it may well happen that jf j is
integrable but f is not. The classical example is furnished by f W Œ0; 1 ! R such that
f .x/ D
1; if x … Q
:
1; if x 2 Q
An argument analogous to that of Example 10.3 guarantees that f is not integrable.
On the other hand, jf j is constantly equal to 1, hence integrable.
The first and last inequalities in (10.29) make it clear the validity of the
Rb
triangle inequality for integrals: the first inequality guarantees that a f .x/dx can
be computed as the difference between the areas of the regions RC and R of the
cartesian plane, where RC is under the graph of f and above the horizontal axis,
while R is above the graph of f and below the horizontal axis; on the other hand,
Rb
the last equality gives a jf .x/jdx as the sum of the areas of RC and R .
Our next result guarantees that the integral is also additive with respect to the
domains Œa; c and Œc; b of two integrable functions. By heuristically reasoning with
a nonnegative function f W Œa; b ! R whose restrictions to the intervals Œa; c and
Œc; b are both integrable, we conclude that this is quite a plausible result. Indeed,
374
10 Riemann’s Integral
letting R, RjŒa;c and RjŒb;c denote the regions of the cartesian plane respectively
under the graphs of f and of its restrictions to Œa; c and Œc; b, we have R D RjŒa;c [
RjŒb;c , such that RjŒa;c and RjŒb;c have no interior points in common; therefore, one
expects that A.R/ D A.RjŒa;c / C A.RjŒb;c /.
Proposition 10.16 Let a function f W Œa; b ! R and a real number c 2 .a; b/ be
given. If the restrictions of f to the intervals Œa; c and Œc; b are both integrable, then
f is also integrable, with
Z c
Z b
Z b
f .x/dx D
f .x/dx C
f .x/dx:
a
a
c
Proof Let fjŒa;c and fjŒc;b denote the restrictions of f to the intervals Œa; c and Œc; b,
respectively. Given > 0, the integrabilities of fjŒa;c and fjŒc;b assure, by means of
the Cauchy criterion, the existence of a partition P of Œa; c and Q of Œc; b such that
S.fjŒa;c I P/ s.fjŒa;c I P/ <
and S.fjŒc;b I Q/ s.fjŒc;b I Q/ < :
2
2
If R D P [ Q, then R is a partition of Œa; b and it is immediate that
S.f I R/ D S.fjŒa;c I P/ C S.fjŒc;b I Q/ and s.f I R/ D s.fjŒa;c I P/ C s.fjŒc;b I Q/:
Hence,
S.f I R/ s.f I R/ D S.fjŒa;c I P/ s.fjŒa;c I P/ C S.fjŒc;b I Q/ s.fjŒc;b I Q/ < ;
so that, by invoking Cauchy’s criterion once more, we deduce the integrability of f .
Rc
Rb
Rb
For what is left to do, let D a f .x/dx a f .x/dx c f .x/dx. In the notations
of the above discussion, we have
Z c
Z b
S.f I R/ f .x/dx f .x/dx
a
c
Z
c
D S.fjŒa;c I P/ C S.fjŒc;b I Q/ a
Z
b
f .x/dx f .x/dx
c
S.fjŒa;c I P/ C S.fjŒc;b I Q/ s.fjŒa;c I P/ s.fjŒc;b I Q/
D S.fjŒa;c I P/ s.fjŒa;c I P/ C S.fjŒc;b I Q/ s.fjŒc;b I Q/ < I
analogously,
Z
Z
c
s.f I R/ b
f .x/dx a
f .x/dx
c
s.fjŒa;c I P/ C s.fjŒc;b I Q/ S.fjŒa;c I P/ S.fjŒc;b I Q/
D .S.fjŒa;c I P/ s.fjŒa;c I P// .S.fjŒc;b I Q/ s.fjŒc;b I Q// > :
Therefore, jj < . Finally, since > 0 was chosen arbitrarily, we conclude that
D 0.
t
u
10.4 Operating with Integrable Functions
375
The previous proposition shall let us present yet another class of examples of
integrable functions. However, before we can do so, we need the following auxiliary
result.
Lemma 10.17 If f ; g W Œa; b ! R are such that f is integrable and f D g in .a; b/,
Rb
Rb
then g is integrable and a f .x/dx D a g.x/dx.
Proof Let’s consider the case in which f D g in Œa; b/, leaving the general case to
the reader (cf. Problem 8).
If P D fa D x0 < x1 < < xk D bg is a partition of Œa; b, the coincidence of f
and g in Œx0 ; xk1 furnishes
k ˇ
ˇX
ˇ
ˇ
inf f inf g .xj xj1 /ˇ
js.f I P/ s.gI P/j D ˇ
jD1
Œxj1 ;xj Œxj1 ;xj ˇ
ˇ
ˇ
ˇ
D ˇ inf f inf g .xk xk1 /ˇ
Œxk1 ;xk Œxk1 ;xk jf .b/ g.b/j .b xk1 /I
analogously,
jS.f I P/ S.gI P/j jf .b/ g.b/j.b xk1 /:
It follows from the computations above and the triangle inequality that
S.gI P/ s.gI P/ D .S.gI P/ S.f I P// C .S.f I P/ s.f I P//
C .s.f I P/ s.gI P//
jS.gI P/ S.f I P/j C .S.f I P/ s.f I P//
C js.f I P/ s.gI P/j
2jf .b/ g.b/j.b xk1 / C .S.f I P/ s.f I P//:
Now, given > 0, the integrability of f guarantees, by means of the Cauchy
criterion, the existence of a partition P such that S.f I P/ s.f I P/ < 2 and (refining
P, if necessary) b xk1 < 4.jf .b/g.b/jC1/
. With such a partition P, the computations
above give
S.gI P/ s.gI P/ 2jf .b/ g.b/j.b xk1 / C .S.f I P/ s.f I P//
2jf .b/ g.b/j C < :
4.jf .b/ g.b/j C 1/
2
Therefore, by resorting to the Cauchy criterion once more, we conclude that g is
integrable.
376
10 Riemann’s Integral
For the equality of the integrals of f and g, note that (again by the computations
above)
Z b
f .x/dx S.f I P/ D .S.f I P/ S.gI P// C S.gI P/
a
jS.f I P/ S.gI P/j C S.gI P/
jf .b/ g.b/j.b xk1 / C S.gI P/:
Hence,
Z b
Z
a
Z
b
f .x/dx b
g.x/dx jf .b/ g.b/j.b xk1 / C S.gI P/ a
g.x/dx:
a
Choosing the partition P in such a way that b xk1 < 2.jf .b/g.b/jC1/
and
Rb
S.gI P/ a g.x/dx < 2 (this last choice being possible thanks to the Cauchy
criterion), we get
Z
Z
b
a
Z
b
f .x/dx b
g.x/dx jf .b/ g.b/j.b xk1 / C S.gI P/ a
g.x/dx
a
jf .b/ g.b/j C < :
2.jf .b/ g.b/j C 1/
2
However, since > 0 was chosen arbitrarily, it follows from the above that
Z
b
Z
b
f .x/dx a
g.x/dx 0:
a
Finally, changing the roles of f and g in the reasoning above (which is perfectly
valid, for we have already established the integrability of g), we get the opposite
inequality between the integrals, thus showing that they are equal.
t
u
We now need the coming
Definition 10.18 A function f W Œa; b ! R is piecewise continuous if there exist
real numbers a D x0 < x1 < < xk D b such that f is continuous in the
interval .xj1 ; xj /, for 1 j k, and the lateral limits limx!aC f .x/, limx!b f .x/
and limx!xj ˙ f .x/ exist, for 1 j < k.
Figure 10.4 sketches the graph of a piecewise continuous function f W Œa; b ! R,
which is discontinuous in exactly three points.
Proposition 10.19 If f W Œa; b ! R is piecewise continuous, then f is integrable, with
Z
b
a
f .x/dx D
k Z
X
jD1
xj
xj1
f .x/dx:
(10.30)
10.4 Operating with Integrable Functions
377
Fig. 10.4 A piecewise
continuous function
f W Œa; b ! R
y
a
x1
x2
b
x
Proof If fj W Œxj1 ; xj ! R is such that fj D f in .xj1 ; xj / and fj .xj1 / D
limx!xj1 C f .x/, fj .xj / D limx!xj f .x/, then fj is continuous, thus integrable.
Now, Lemma 10.17 Rassures that theR restriction of f to the interval Œxj1 ; xj is
xj
xj
also integrable, with xj1
f .x/dx D xj1
fj .x/dx. Finally, by repeatedly applying
Proposition 10.16, we conclude that f is integrable in Œa; b and that (10.30) is valid.
t
u
For the coming example, the reader may find it useful to review the concept of
characteristic function of a set, in Definition 10.11.
Example 10.20 Let a D x0 < x1 < < xk D b be a partition of the interval Œa; b,
and f W Œa; b ! R be defined by
f .x/ D
k
X
cj XŒxj1 ;xj ;
jD1
with cj 2 R for 1 j k. Since f is clearly piecewise continuous, the previous
proposition guarantees its integrability, with
Z
b
f .x/dx D
a
k Z
X
jD1
xj
f .x/dx:
xj1
On the other hand, Problem 2, page 357, gives
Z
xj
Z
f .x/dx D
xj1
xj
cj dx D cj .xj xj1 /:
xj1
Therefore, by combining both inequalities above, we get
Z
b
a
f .x/dx D
k
X
jD1
cj .xj xj1 /:
378
10 Riemann’s Integral
Our next result establishes the converse of Proposition 10.16 and will be of
crucial importance for the actual computation of integrals, in the next section.
Proposition 10.21 Let f W Œa; b ! R be integrable and a < c < b. Then, the
restrictions of f to the intervals Œa; c and Œc; b (which, whenever no danger of
confusion appears, will be denoted simply by f ) are also integrable, with
Z
Z
b
a
Z
c
f .x/dx D
f .x/dx C
a
b
f .x/dx:
(10.31)
c
Proof It suffices to prove the integrability of the restrictions of f to Œa; c and Œc; b.
Indeed, once we have done this, (10.31) will follow from Proposition 10.16.
For what is left to do, given > 0, Cauchy’s criterion assures the existence of a
partition R of Œa; b such that S.f I R/ s.f I R/ < . Letting R0 D R [ fcg, it follows
from Lemma 10.1 that S.f I R0 / s.f I R0 / S.f I R/ s.f I R/ < ; therefore, we can
suppose from the very beginning that c 2 R.
Let fjŒa;c denote the restriction of f to the interval Œa; c. If R D fa D x0 < x1 <
< xl D c < < xk D bg and P D fa D x0 < x1 < < xl D cg, then P is a
partition of Œa; c, such that
S.fjŒa;c I P/ s.fjŒa;c I P/ D
l
X
.Mj .f / mj .f //.xj xj1 /
jD1
k
X
.Mj .f / mj .f //.xj xj1 /
jD1
D S.f I R/ s.f I R/ < :
Hence, again by Cauchy’s criterion, fjŒa;c is integrable.
Analogously, the restriction of f to the interval Œc; b is integrable.
t
u
The last result of this section shows that the product of two integrable functions
is also integrable. Nevertheless, as Problem 10 shows, the value of the integral of
the product of the functions cannot be computed from the values of the integrals of
the factors. (However, see Problem 12.)
Proposition 10.22 If f ; g W Œa; b ! R are integrable functions, then fg W Œa; b ! R
is also integrable.
Proof Suppose we have established the integrability of the square of an integrable
function. Then, .f C g/2 , f 2 and g2 will be integrable, so that, since fg D 12 ..f C
g/2 f 2 g2 /, several applications of Proposition 10.13, together with the remark
right after it, guarantee the integrability of fg.
Let’s then show that f 2 is integrable. To this end, note first that, since f is
bounded, there exists a real constant c such that f C c 0 in Œa; b. Looking at
c as a constant function in Œa; b and writing f 2 D .f C c/2 2cf c2 , we conclude
10.4 Operating with Integrable Functions
379
(quite similarly as in the above reasoning) that f 2 is integrable if .f C c/2 has this
property. It then suffices to show that, if f is integrable and nonnegative, then f 2 is
also integrable.
For what is left to do, assume f to be integrable and nonnegative, so that f 2 is
surely bounded. Now, given a partition P D fa D x0 < x1 < < xk D bg of Œa; b,
it follows from Problem 14, page 207, that
sup f 2 D . sup f /2 D Mj2
Œxj1 ;xj Œxj1 ;xj and, analogously, infŒxj1 ;xj f 2 D m2j . Hence,
S.f 2 I P/ s.f 2 I P/ D
k
X
.Mj2 m2j /.xj xj1 /
jD1
D
k
X
.Mj C mj /.Mj mj /.xj xj1 /
jD1
2 sup f Œa;b
k
X
.Mj mj /.xj xj1 /
jD1
D 2 sup f .S.f I P/ s.f I P//:
Œa;b
Let > 0 be given. By invoking Cauchy’s criterion again, and thanks to the
integrability of f , we can suppose that S.f I P/ s.f I P/ < 2.sup f /C1 . Therefore,
Œa;b
the above computations give
S.f 2 I P/ s.f 2 I P/ < 2.sup f / Œa;b
2.supŒa;b f / C 1
< :
Hence, once more from Cauchy’s criterion, f 2 is integrable.
t
u
Problems: Section 10.4
1. Give an example of a differentiable function f W Œa; b ! R such that f 0 is not
bounded (hence, not integrable).
2. Let f W Œa; b ! R be continuous, with a finite number of zeros in the interval
Œa; b, say x1 < x2 < < xk . If
2
RC
f D f.x; y/ 2 R I a x b and 0 y f .x/g
380
10 Riemann’s Integral
and
2
R
f D f.x; y/ 2 R I a x b
and f .x/ y 0g;
prove that
Z
b
a
f .x/dx D A.RC
f / A.Rf /:
For the next problem, given continuous functions f ; g W Œa; b ! R, such that
f .x/ g.x/ for every x 2 Œa; b, let
Rfg D f.x; y/ 2 R2 I a x b and f .x/ y g.x/g
be the portion of the cartesian plane situated between the graphs of f and g. Let
the area of Rfg be defined by
Z
b
A.Rfg / D
.g.x/ f .x//dx:
a
3. Prove Cavalieri’s principle: for i D 1; 2, let fi ; gi W Œa; b ! R be continuous
functions such that fi .x/ gi .x/ for every x 2 Œa; b. If, for every x 2 Œa; b, the
length of the line segment joining .x; f1 .x// to .x; g1 .x// equals that of the line
segment joining .x; f2 .x// to .x; g2 .x//, then A.Rf1 g1 / D A.Rf2 g2 /.
4. Let bc W R ! RRbe the integer part function (cf. Problem 9, page 152). Given
n
n 2 N, compute 0 bxcdx.
5. Let fg W R ! R be the fractional part function (cf. Problem 10, page 152).
Given n 2 N, do the following items:
R nC1
R1
(a) Show thatR n fxgdx D 0 fxgdx.
n
(b) Compute 0 fxgdx.
6. * Let f ; g W Œa; b ! R be continuous functions such that f .x/ g.x/ for every
Rb
Rb
x 2 Œa; b. If a f .x/dx D a g.x/dx, prove that f D g.
7. Let f W Œa; b ! R be an integrable function. Assuming that jf j is also
integrable, use the fact that jf .x/j f .x/ jf .x/j for every x 2 Œa; b to
deduce (10.28).
8. * Complete the proof of Lemma 10.17, by examining the case in which f D g
in .a; b/.
Rb
Rb
9. Given 0P
a < b and n 2 N, compute a xn dx. Then compute a f .x/dx, where
n
f .x/ D jD0 aj xj , with a0 ; a1 ; : : : ; an 2 R and an ¤ 0.
10. Give examples of continuous and nonnegative functions f ; gI Œa; b ! R, such
Rb
Rb
Rb
that a f .x/dx > 0 and a g.x/dx > 0 but a f .x/g.x/dx D 0.
, show that
11. * Let f W Œa; b ! R be an integrable function. For 0 < < ba
2
Z
lim
Z
b
!0 aC
b
f .x/dx D
f .x/dx:
a
10.5 The Fundamental Theorem of Calculus
381
12. * Prove Cauchy’s inequality for integrals: if f ; g W Œa; b ! R are integrable
functions, then
Z b
ˇZ b
ˇ Z b
1=2 1=2
ˇ
ˇ
f .x/g.x/dxˇ f .x/2 dx
g.x/2 dx
:
ˇ
a
a
a
Moreover, if f and g are continuous, show that equality happens if and only if
there exists 2 R such that f .x/ D g.x/ for every x 2 Œa; b.
13. * If f W Œa; b ! R is a continuous and convex function, prove that
Z b
f .a/ C f .b/
1
f .x/dx:
2
ba a
10.5 The Fundamental Theorem of Calculus
The main purpose of this section is to turn (10.11) into a theorem. Along all that
follows, unless explicitly stated otherwise, we shall let I be an interval of the real
line and f W I ! R be integrable in each interval Œa; b I.
For c 2 I and Œa; b I, we set
Z c
Z a
Z b
f .x/dx D 0 and
f .x/dx D f .x/dx:
c
b
a
With these conventions at hand, and applying the second part of Proposition 10.21,
it is immediate to verify that
Z b
Z c
Z b
f .x/dx D
f .x/dx C
f .x/dx;
(10.32)
a
a
c
for every a; b; c 2 I (i.e., not only when a < b and c 2 .a; b/).
This being said, the following definition will be of paramount importance for
what is to come.
Definition 10.23 Let I R be an interval and f W I ! R be integrable in each
interval Œa; b I. For a fixed c 2 I, the indefinite integral of f based at c is the
function F W I ! R, given by
Z x
F.x/ D
f .t/dt:
c
If f ; F W I ! R are as in the previous definition, (10.32) promptly gives
Z a
Z b
Z b
f .t/dt f .t/dt D
f .t/dt;
(10.33)
F.b/ F.a/ D
c
for all a; b 2 I.
c
a
382
10 Riemann’s Integral
We can now state and prove the major result of this section, which is known in
mathematical literature as the fundamental theorem of Calculus (FTC).
Theorem 10.24 (FTC) Let I R be an interval, let f W I ! R be integrable in
each interval Œa; b I and F W I ! R be the indefinite integral of f based at c 2 I.
If f is continuous at x0 2 I, then F is differentiable at x0 , with F 0 .x0 / D f .x0 /.
Proof For x 2 I n fx0 g and successively applying (10.33), Problem 2, page 357, the
additivity of the integral and the triangle inequality for integrals, we get
ˇ F.x/ F.x /
ˇ
ˇ 1 Z x
ˇ
ˇ
0
ˇ
ˇ
ˇ
f .x0 /ˇ D ˇ
f .t/dt f .x0 /ˇ
ˇ
x x0
x x0 x 0
Z
ˇ
1 ˇˇ x
ˇ
D
.f .t/ f .x0 //dtˇ
ˇ
jx x0 j x0
Z
ˇ
1 ˇˇ x
ˇ
jf .t/ f .x0 /jdtˇ:
ˇ
jx x0 j x0
(In the last inequality
above, the modulus
Rx
R x outside the integral is due to the fact that,
if x < x0 , then x0 jf .t/ f .x0 /jdt D x 0 jf .t/ f .x0 /jdt.)
The continuity of f at x0 guarantees that, given > 0, there exists ı > 0 such that
t 2 I; jt x0 j < ı ) jf .t/ f .x0 /j < :
Hence, for 0 < jx x0 j < ı, we have jt x0 j < ı for every t belonging to the
interval with endpoints x0 and x, so that jf .t/ f .x0 /j < . It thus follows from
the monotonicity of the integral, and once more from the result of Problem 2, page
357, that
ˇZ x
ˇ ˇZ x
ˇ
ˇ
ˇ ˇ
ˇ
jf .t/ f .x0 /jdtˇ ˇ
dtˇ D jx x0 j:
ˇ
x0
x0
Finally, the above computations assure that, for x 2 I such that 0 < jx x0 j < ı,
we have
ˇ F.x/ F.x /
ˇ
1
0
ˇ
ˇ
jx x0 j D :
f .x0 /ˇ ˇ
x x0
jx x0 j
Hence, F is differentiable at x0 , with F 0 .x0 / D f .x0 /.
t
u
The FTC is the ingredient that was missing for us to get an easy way of computing
integrals of various elementary functions. However, before we can actually do that,
we need one more piece of terminology.
Definition 10.25 Let I R be an interval and f W I ! R be integrable in each
interval Œa; b I. A primitive for f in I is a differentiable function F W I ! R such
that F 0 D f in I.
10.5 The Fundamental Theorem of Calculus
383
In terms of the previous definition, the FTC guarantees that each indefinite
integral of a continuous function f W I ! R is a primitive of f in I. Indeed, letting
F W I ! R be the indefinite integral of F based at c 2 I, it follows from the FTC
and the continuity of f in I that F 0 .x/ D f .x/ for every x 2 I. Yet in another way, we
have that:
Z x
d
f continuous )
f .t/dt D f .x/:
(10.34)
dx c
The coming result assures that, even for functions which are merely integrable,
there are no other possible primitives. In other words, it guarantees that, even if f is
just an integrable function, there are no other possible solutions for (10.37).
Theorem 10.26 Let I R be an interval and f W I ! R be integrable in each
interval Œa; b I. If F W I ! R is a primitive for f , then, for a given x0 2 I, we
have
Z x
F.x/ D F.x0 / C
f .t/dt
(10.35)
x0
for every x 2 I.
Proof Suppose x > x0 (the case x D x0 is trivial and
R x the case x R<x x0 can be dealt
with in an analogous way, taking into account that x0 f .t/dt D x 0 f .t/dt).
Let P D fx0 < x1 < < xk D xg be a partition of Œx0 ; x. Since F is
differentiable in I, it is continuous in each interval Œa; b I. Hence, Lagrange’s
MVT gives j 2 .xj1 ; xj / such that
F.xj / F.xj1 / D F 0 .j /.xj xj1 / D f .j /.xj xj1 /;
for 1 j k. Then, we get
F.x/ F.x0 / D
k
X
.F.xj / F.xj1 //
jD1
D
k
X
f .j /.xj xj1 /
jD1
D †.f I PI /:
Now recall that, according to Riemann’s Theorem 10.7,
Z
x
f .t/dt D lim †.f I PI /;
x0
jPj!0
384
10 Riemann’s Integral
for every choice D .j / of intermediate points relative to P. Therefore, in view
of the computations in the previous paragraph, there is no other choice than to
have (10.35).
t
u
If f W I ! R has a primitive, then, thanks
to the previous result, one uses to
R
denote a generic primitive of f by writing f .t/dt. The reader must be careful not to
R
Rb
making confusion with the notations f .t/dt and a f .t/dt: while the first one refers
to a differentiable function from I to R whose derivative coincides with f in
R I, the
second denotes a real number. Also, note that the theorem guarantees that if f .t/dt
is a primitive of f in I, then the primitives of f in I are the functions of the form
Z
f .t/dt C C;
where C is a real constant.
With Theorem 10.26 at our disposal, the coming corollary assures that, if f W
Rb
Œa; b ! R is an integrable function, then, in order to effectively compute a f .t/dt,
it suffices to find a primitive for f . From now on, given a continuous function F W
Œa; b ! R, we denote
ˇxDb
ˇ
F.x/ˇ
D F.b/ F.a/:
xDa
Corollary 10.27 If f W Œa; b ! R is an integrable function and F W Œa; b ! R is a
primitive for f , then
Z b
ˇxDb
ˇ
f .t/dt D F.x/ˇ :
(10.36)
a
xDa
Proof Let x0 D a and x D b in (10.35).
t
u
As we anticipated before, the former corollary provides us with a general strategy
Rb
for computing the actual value of a f .t/dt: it suffices to get to visualize the
integrand function f as the derivative of some function F, then applying (10.36)
Rb
to obtain a f .t/dt D F.b/ F.a/. That is why we usually look at (10.36) as
Z b
ˇxDb
ˇ
F 0 .t/dt D F.x/ˇ :
a
xDa
Let us see a couple of examples.
P
Example 10.28 Let f W R ! R be such that f .x/ D njD0 aj xj for every x 2 R,
where a0 , a1 , . . . , an are given real numbers, with an ¤ 0. Since F.x/ D
P
aj jC1
n
is clearly a primitive of f , given reals a < b we have
jD0 jC1 x
Z
a
b
n
ˇxDb X
aj
ˇ
.bjC1 ajC1 /:
f .t/dt D F.x/ˇ
D
xDa
j
C
1
jD0
10.5 The Fundamental Theorem of Calculus
Example 10.29 Compute
R
sin x dx,
0
385
R
0
sin2 x dx and
R
0
cos2 x dx.
Solution Since cos is a primitive for sin, the previous corollary gives
Z
0
ˇxD
ˇ
sin x dx D cos xˇ
D cos C cos 0 D 2:
xD0
With respect to the second integral, since sin2 x D
2 cos 2x, we have (again by that result)
Z
sin2 x dx D
0
1
2
Z
0
.1 cos 2x/dx D
1
2 .1
cos 2x/ and
d
dx
sin 2x D
ˇ
1
1
ˇxD
x sin 2x ˇ
D :
xD0
2
2
2
Finally, since sin2 x C cos2 x D 1, we have
Z
0
cos2 xdx D
Z
0
.1 sin2 x/dx D
ˇxD Z
ˇ
D xˇ
xD0
0
Z
0
Z
1 dx sin2 x dx D sin2 x dx
0
D :
2
2
t
u
Another way of rephrasing (10.34), which is quite important for the theory of
ordinary differential equations3 is collected in the following result.
Proposition 10.30 Let I R be an interval and f W I ! R be a continuous
function. Given x0 2 I and y0 2 R, the initial value problem
y0 D f .x/
y.x0 / D y0
(10.37)
has as its only solution the function F W I ! R given by
Z
x
F.x/ D y0 C
f .t/dt:
(10.38)
x0
Proof A solution of (10.37) is a differentiable function F W I ! R such that F 0 .x/ D
f .x/ for every x 2 I and F.x0 / D y0 . Since f is continuous, (10.34) guarantees that
F, defined as in (10.38), is a solution of (10.37).
3
In spite of being quite an interesting and important subject, we shall not have occasion to give
a systematic development of the theory of ODE’s in these notes. For the interested reader, we
recommend the outstanding book of professor George Simmons, [23].
386
10 Riemann’s Integral
On the other hand, if F1 ; F2 W I ! R solve (10.37), then F10 D f D F20 . Therefore,
Corollary 9.47 guarantees that F1 F2 is constant. However, since .F1 F2 /.x0 / D
y0 y0 D 0, we conclude that F1 F2 D 0, i.e., F1 D F2 .
t
u
Most often, the task of explicitly finding a primitive of a given integrable
function (if possible) is not immediate. That’s why Calculus courses generally spend
considerable time in developing specific techniques for the computation of integrals,
which are generically referred to as integration techniques. We discuss the most
relevant parts of this toolkit in the rest of this section and along the next one, starting
with the formula for integration by parts.
Proposition 10.31 If f ; g W Œa; b ! R are differentiable functions with integrable
derivatives, then
Z
b
ˇxDb Z
ˇ
f 0 .x/g.x/dx D f .x/g.x/ˇ
xDa
a
b
f .x/g0 .x/dx:
(10.39)
a
Proof Firstly, note that the integrals in both sides of (10.39) are well defined. Indeed,
f and g, being differentiable, are continuous; on the other hand, since f 0 and g0 are
integrable, Proposition 10.22 guarantees that f 0 g and fg0 are also integrable.
Now, since .fg/0 D f 0 g C fg0 , Corollary 10.27 gives
Z
b
.f 0 .x/g.x/ C f .x/g0 .x//dx D
a
Z
b
ˇxDb
ˇ
.fg/0 .x/dx D f .x/g.x/ˇ :
xDa
a
Finally, in order to obtain (10.39), it suffices to apply the additivity of the integral to
the left hand side of the equality above.
t
u
R
R 2
Example 10.32 Compute 0 x sin x dx and 0 x cos x dx.
Solution For the first integral, letting f .x/ D cos x and g.x/ D x in (10.39), we
get
Z
Z
x sin x dx D
0
0
ˇxD Z
ˇ
x. cos x/dx D x cos xˇ
0
xD0
Z
DC
0
. cos x/dx
ˇxD
ˇ
cos x dx D C sin xˇ
D :
xD0
0
For the second, letting f .x/ D sin x and g.x/ D x2 in (10.39), we obtain
Z
0
x2 cos x dx D
Z
0
ˇxD Z
ˇ
x2 sin0 x dx D x2 sin xˇ
Z
D 2
xD0
0
2x sin x dx
x sin x dx D 2;
0
where we used the result of the first part in the last equality above.
t
u
10.5 The Fundamental Theorem of Calculus
387
Note that, in terms of primitives, the integration by parts formula gives
Z
Z
f 0 .x/g.x/dx D f .x/g.x/ f .x/g0 .x/dx:
In words, a primitive for f 0 g can be obtained by subtracting a primitive for fg0
from fg.
The following particular case of integration by parts is frequently useful.
Corollary 10.33 If f W Œa; b ! R is differentiable and f 0 is integrable, then
Z
b
ˇb Z
ˇ
f .x/dx D xf .x/ˇ a
a
b
xf 0 .x/dx:
(10.40)
a
Proof Changing the roles of f and g in (10.39), we get
Z
b
ˇb Z
ˇ
f .x/g0 .x/dx D f .x/g.x/ˇ a
a
b
f 0 .x/g.x/dx:
a
Now, it suffices to let g.x/ D x for every x 2 Œa; b.
t
u
Remark 10.34 The derivative of a differentiable function may not be integrable. For
an example, see Problem 1, page 379.
Our coming example will find a prominent role in the proof of Theorem 10.56,
in Sect. 10.7.
R =2
Example 10.35 Let n 2 ZC and In D 0 .cos x/n dx. Show that:
(a) nIn D .n 1/In2 for every integer n 2.
(b) I2k D .2.2k/Š
k kŠ/2 2 for every integer k 0, and I2k1 D
k 1.
.2k kŠ/2
2k.2k/Š
for every integer
Proof (a) For n 2, integration by parts gives
Z
In D
=2
0
Z
.cos x/n1 cos x dx D
ˇ=2 Z
ˇ
D .cos x/n1 sin xˇ Z
D .n 1/
=2
0
Z
D .n 1/
0
0
=2
=2
0
=2
0
.cos x/n1 sin0 x dx
.n 1/.cos x/n2 . sin x/ sin x dx
.cos x/n2 .1 cos2 x/dx
Z
.cos x/n2 dx .n 1/
D .n 1/In2 .n 1/In :
Hence, nIn D .n 1/In2 .
0
=2
cosn x dx
388
10 Riemann’s Integral
(b) It’s immediate that I0 D
R =2
0
.2m/Š
2
.2m mŠ/2
ˇxD ˇ 2
dx D xˇ
D
xD0
.
2
By induction hypothesis,
for some integer m 0. Letting n D 2m C 2 in
suppose that I2m D
the recurrence relation of item (a), we get
I2mC2 D
2m C 1
2m C 1 .2m/Š I2m D
2m C 2
2m C 2 .2m mŠ/2 2
D
.2m C 2/.2m C 1/ .2m/Š m 2
.2.m C 1//2
.2 mŠ/ 2
D
.2m C 2/Š
:
.2mC1 .m C 1/Š/2 2
Thus, the first part of (c) is valid for every k 0.
Finally, the proof of the formula for I2k1 is completely analogous, and will
be left as an exercise to the reader.
t
u
Let’s finish this section by using the FTC to present a proof of the irrationality
of . Our discussion somewhat follows part B:17 of another marvelous book of
professor G. Simmons [22], and (as quoted there) is an elaboration of ideas of
C. Hermite by I. Niven. We remark, however, that the irrationality of was first
established by the Swiss mathematician of the XVIII century J. H. Lambert.
is irrational.
Theorem 10.36 (Lambert)
2
Proof If were rational, say D pq , with p; q 2 N, then we would have 2 D pq2 ,
with p2 ; q2 2 N. Thus, suppose, for the sake of contradiction, that 2 D ab , with
a; b 2 N.
Let’s start by taking a twice continuously differentiable function g W Œ0; 1 ! R,
so that
d 0
.g .x/ sin.x/ g.x/ cos.x// D .g00 .x/ C 2 g.x// sin.x/:
dx
The FTC gives
ˇxD1
ˇ
.g.1/ C g.0// D .g0 .x/ sin.x/ g.x/ cos.x//ˇ
xD0
Z
1
D
Z
0
1
D
0
.g00 .x/ C 2 g.x// sin.x/dx
.g00 .x/ C 2 g.x// sin.x/dx:
(10.41)
10.5 The Fundamental Theorem of Calculus
389
Loosely speaking, Lambert’s ingenious idea was to use the fact that 2 is rational
to try to choose g in such a way that, on the one hand, g.0/; g.1/ 2 Z, while, on the
other,
ˇZ 1
ˇ
ˇ
ˇ
ˇ .g00 .x/ C 2 g.x// sin.x/dxˇ < 1:
ˇ
ˇ
0
In light of (10.41), this would clearly give us an absurd.
Actually, departing from a continuous function f and using the assumption that
2 D ab , set
g.x/ D bn . 2n f .x/ 2n2 f .2/ .x/ C 2n4 f .4/ .x/ /
D an f .x/ ban1 f .2/ .x/ C b2 an2 f .4/ .x/ ;
(10.42)
where n 2 N.
The second equality above assures that
f .j/ .0/; f .j/ .1/ 2 Z; 8 j 2 ZC ) g.0/; g.1/ 2 Z:
On the other hand, a straightforward computation using the first equality
in (10.42) gives
g00 .x/ C 2 g.x/ D bn 2nC2 f .x/ D
an
f .x/I
b
Hence, to reach the desired contradiction, f must be also such that
ˇZ
ˇ
ˇ
an ˇˇ 1
f .x/ sin.x/dxˇˇ < 1:
b ˇ 0
(10.43)
Since (from the triangle inequality for integrals)
ˇZ
ˇ
ˇ
ˇ
0
1
ˇ Z
ˇ
f .x/ sin.x/dxˇˇ 1
0
Z
jf .x/ sin.x/jdx 0
1
jf .x/jdx
but we have no hints on the size of a, in order to get (10.43) it is natural to choose f
in such a way that
maxfjf .x/jI 0 x 1g C
;
nŠ
for some positive constant C. Once we have done that, we will have
ˇZ
ˇ
n
ˇ
an ˇˇ 1
ˇ a ;
f
.x/
sin.x/dx
ˇ
ˇ
b 0
bnŠ
which, for a given a > 0, is less that 1 for a sufficiently large n.
390
10 Riemann’s Integral
We now have enough clues to search for f , and a moment’s thought shows
f .x/ D
1 n
x .1 x/n
nŠ
to be a strong candidate. Indeed, for such an f we obviously have jf .x/j 1 for
0 x 1, so that it suffices to check that f .j/ .0/; f .j/ .1/ 2 Z for every j 2 ZC .
What is left to do is this is relatively easy. First of all, it follows from the binomial
formula that
2n
f .x/ D
1 X k
ak x ;
nŠ kDn
with ak 2 Z for n k 2n. Therefore, if 0 j n or j > 2n, we obviously have
f .j/ .0/ D 0 and f .j/ .1/ D 0. On the other hand, if n < j 2n, then
f .j/ .x/ D
D
2n
1 X
k.k 1/ : : : .k j C 1/ak xkj
nŠ kDj
2n
X
kŠ
jŠ
ak xkj
nŠ
.k
j/ŠjŠ
kDj
!
k
D
j.j 1/ : : : .n C 1/
ak xkj ;
j
kDj
2n
X
so that f .j/ .0/ and f .j/ .1/ also belong to Z.
t
u
Problems: Section 10.5
1. In each of the items below, compute the given integral:
R 2
(a) 0 .1 cos t/2 dt.
R 2 p
(b) 0
1 cos t dt.
R1
(c) 0 x sin.x/dx.
2. In the two items below, compute the primitives:
R
(a) R .sin x/3 cos x dx.
(b) x2 sin x dx.
3. Given m; n 2 N, show that
Z
2
0
Z
sin.mx/ sin.nx/dx D
0
2
cos.mx/ cos.nx/dx D
0; if m ¤ n
:
; if m D n
10.5 The Fundamental Theorem of Calculus
391
4. In each of the following items, solve the indicated initial value problem in the
largest possible interval I R:
(a)
p
y0 D x
.
y.1/ D 2
(b)
y0 D sec2 x
.
y. 6 / D 1
5. * Let g; h W Œc; d ! R be differentiable functions and f W Œa; b ! R be
continuous. Show that
d
dx
Z
h.x/
f .t/dt D f .h.x//h0 .x/ f .g.x//g0 .x/:
g.x/
6. * Let f W R ! R be a continuous and periodic function, with period p > 0.
R aCp
Rp
Prove that a f .t/dt D 0 f .t/dt for every a 2 R.
7. Let f W Œa; b ! R be the restriction of an affine function to the interval Œa; b.
If f is nonnegative, the region Rf is either a right triangle or a right trapezoid.
Show that, in each of these cases, the value for the area of Rf obtained with
the aid of the FTC coincides with the one obtained by means of the ordinary
formulas of Euclidean Geometry for the areas of these polygons (cf. [4], for
instance).
8. Let f W .a; b/ ! R be continuous, nonnegative and increasing (resp.
decreasing). Prove that every primitive of f is strictly convex (resp. strictly
concave).
n
P
1
9. Given n 2 N, compute nkD0 .kC1/.kC2/
k in terms of n.
10. Prove the mean value theorem for integrals: given a continuous function f W
Œa; b ! R, there exists c 2 Œa; b such that
1
ba
Z
b
f .x/dx D f .c/:
a
11. Let f W Œa; b ! R be a continuously differentiable function, with f .a/ D
f .b/ D 0. Prove that
Z
b
2
f .x/ dx
a
2
4
Z
b
2
2
x f .x/ dx
a
Z
b
f 0 .x/2 dx :
a
12. (Romania) Let F be the set of continuous functions f W Œ0; ! R such that
Z
0
Z
f .x/ sin xdx D
0
f .x/ cos xdx D 1:
R
Compute inff 2F 0 f .x/2 dx.
R =2
13. * Let n 2 ZC and In D 0 .cos x/n dx. The purpose of this problem is to show
I
that limn!C1 nC1
In D 1. To this end, do the following items:
392
10 Riemann’s Integral
(a) Show that InC1 In for every integer n 0.
(b) Use
item (a) of Example 10.35 to conclude that each of the sequences
I2k1
I2k
and I2k1
is nondecreasing.
I2k2
k1
k1
(c) Conclude that both sequences of the previous item converge to limits `0
and `1 , say, such that `0 ; `1 > 0.
1
I
In
n
(d) Show that nC1
D
. Then, let n D 2k ! C1 to get `0 `1 D 1.
In
nC1 In1
(e) Conclude that `0 D `1 D 1 and, then, that limn!C1
InC1
In
D 1.
10.6 The Change of Variables Formula
Continuing with the development of the theory, we shall now discuss the change
of variables formula. As the subsequent discussion will make it evident, such a
formula will provide another useful integration technique.
Theorem 10.37 Let f W Œa; b ! R be continuous. If g W Œc; d ! Œa; b is
differentiable and g0 is integrable, then
Z g.d/
Z d
f .t/dt D
f .g.s//g0 .s/ds:
(10.44)
g.c/
c
Proof In order to compute the integral at the left hand side, we start by taking a
primitive F W Œa; b ! R for f , whose existence follows from the FTC. Then, thanks
to Corollary 10.27, we get
Z
Z
g.d/
g.d/
f .t/dt D
g.c/
ˇxDg.d/
ˇ
F 0 .t/dt D F.x/ˇ
xDg.c/
g.c/
ˇsDd
ˇ
D F.g.d// F.g.c// D .F ı g/.s/ˇ :
sDc
Now, since F and g are differentiable, the chain rule guarantees that the function
F ı g W Œc; d ! R is also differentiable, with
.F ı g/0 .s/ D F 0 .g.s//g0 .s/ D f .g.s//g0 .s/ D .f ı g/.s/g0 .s/:
On the other hand, since f and g are continuous, we have f ı g continuous, thus,
integrable; however, since g0 is integrable, it follows from Proposition 10.22 that
.f ı g/g0 integrable. Hence, applying Corollary 10.27 once more, we have
Z
ˇsDd Z d
ˇ
0
D
.F ı g/ .s/ds D
.F ı g/.s/ˇ
sDc
c
d
f .g.s//g0 .s/ds:
c
t
u
10.6 The Change of Variables Formula
393
Remark 10.38 For a variant of the previous result, applicable to the case of a
piecewise continuous f , see Problem 6.
The corollary below isolates a simple, yet quite useful, particular case of the
change of variables formula.
Corollary 10.39 If f W Œa; b ! R is continuous and 2 R , then
Z
Z b
1 b
f .s/ ds D
f .t/ dt:
a
a
(10.45)
Proof For > 0 it suffices to let g W Œa; b ! Œa; b be given by g.s/ D s in
Theorem 10.37. Indeed, with such a choice we get
Z b
Z g.b/
Z b
Z b
f .t/ dt D
f .t/ dt D
f .g.s//g0 .s/ ds D
f .s/ ds:
a
g.a/
a
a
For < 0, let g W Œa; b ! Œb; a be given by g.s/ D s. Then, arguing as in
Ra
Rb
R a
R b
the previous case and recalling that b D a and b D a , we obtain
Z b
Z a
Z g.a/
f .t/ dt D f .t/ dt D f .t/ dt
a
b
Z
D
a
g.b/
f .g.s//g0 .s/ ds D
Z
b
b
f .s/ ds:
a
t
u
We shall sometimes refer to (10.44) as the formula for integration by substitution. In what follows, we give some heuristics on such an expression.
Rb
Given a continuous function f W Œa; b ! R, in order to compute a f .t/dt with
the aid of the FTC we need to obtain a primitive for F. In this sense, and after a
careful examination of the expression of f .t/, we may eventually notice that it will
be quite simplified if we perform an adequate substitution of variables, changing t
by an expression g.s/, depending on a new variable s.
However, since f is defined in the interval Œa; b, for such a substitution to make
sense it is necessary that g.s/ belongs to the interval Œa; b when s varies in some
interval Œc; d. In general, the best way of identifying such an interval Œc; d is solving
inequalities a g.s/ b, thus finding the appropriate values of c and d. Then, by
formally differentiating the equality t D g.s/, we get dt D g0 .s/ ds, so that the
original integral must be corrected by the factor g0 .s/ before it can be transformed
into a new integral.
In short, the execution of both passages described below give us the formula for
integration by substitution in any particular case we encounter:
Z
a
b
substitting
tDg.s/
f .t/ dt HHHHH
Z
‹
‹
solving
0
ag.s/b
Z
d
f .g.s//g .s/ ds HHHHHHH
c
f .g.s//g0 .s/ ds:
394
10 Riemann’s Integral
Finally, we observe that for the last integral above to make sense, the function
s 7! f .g.s//g0 .s/ must be integrable. Hence (as we saw in the proof of the theorem)
it is natural to suppose that g is differentiable and g0 is integrable in Œc; d.
Let’s give a concrete example, showing how the informal discussion of the two
previous paragraphs simplifies the use of the change of variables formula in the
actual computation of integrals.
R 2p
Example 10.40 Compute p t sin.t2 / dt.
p
Solution Performing the substitution t D s, we get
Z 2p
Z
Z ‹
p
1 ‹
1
2
ds
D
t
sin.t
/
dt
D
s.sin
s/
sin s ds;
p
p
2 ‹
2 s
‹
where the correction factor
p
1
ds.
t D s, to get dt D 2p
s
1
p
2 s
was obtained by formally differentiating equality
p
p
the integration interval Π; 2 , we solve the inequalities
p
p In order
p to correct
s 2 , thus getting s 4. Hence,
Z 2p
Z
ˇsD4
1 4
1
ˇ
2
.
cos
s/
t
sin.t
/
dt
D
sin
s
ds
D
D 1;
ˇ
p
sD
2 2
t
u
where we used the FTC in the last equality.
Also with respect to the previous example (and in accordance to Theorem 10.37),
note that all of the performed
are valid. Indeed,
p substitutions
p
p what we really did was
to use g W Œ; 4 ! Œ ; 2 such that g.s/ D s, and such a function is
1
differentiable, with g0 .s/ D 2p
integrable in the interval Œ; 4.
s
R =2
.sin x/n dx.
R =2
Solution In Example 10.35, we computed In D 0 .cos x/n dx. Here, we shall use
R =2
integration by substitution to show that In D 0 .sin x/n dx too. To this end, start by
recalling that cos x D sin. 2 x/. Therefore, letting x D 2 y (so that dx D dy),
we get
Z =2
Z ‹
Z ‹
n
n
.sin x/ dx D
.sin.=2 y// .1/dy D .cos y/n dy:
Example 10.41 Given n 2 ZC , compute
0
0
‹
‹
The correct endpoints of the last integral can be easily found once we note that
0 2 y 2 if and only if 0 y 2 . Nevertheless, one has to be careful, for,
x D 0 ) y D 2 and x D 2 ) y D 0. Hence, we get
Z 0
Z =2
Z =2
.sin x/n dx D .cos y/n dy D
.cos y/n dy D In :
0
=2
0
t
u
10.6 The Change of Variables Formula
395
From a more formal viewpoint, the last substitution above reduces to using g W
Œ0; 2 ! Œ0; 2 such that g.x/ D 2 x. Since g.0/ D 2 , g. 2 / D 0 and g0 .x/ D x,
we have
Z
=2
0
Z
.sin x/n dx D
Z
g.0/
g. 2 /
Z
D
0
2
.sin x/n dx D
0
2
.sin g.x//n g0 .x/dx
Z
.sin.=2 x// dx D
2
n
0
.cos x/n dx:
One of the most important usages of integration by substitution is related to
trigonometric substitution. To get to the point, suppose we have a continuous
R1
function f W Œ1; 1 ! R and wish to compute 1 f .x/dx. Then, recalling that the
restriction of the sine function to the interval Π2 ; 2 is differentiable, has integrable
derivative and satisfies sin. 2 / D 1 and sin 2 D 1, it follows from (10.44) (with
g.t/ D sin t) that
Z
1
1
Z
f .x/dx D
sin
2
sin. 2 /
Z
f .x/ dx D
2
2
f .sin t/ cos t dt:
Depending on the algebraic expression that defines f .x/ in terms of x, the last
integral above may be much simpler to compute that the original one.
Although the discussion on the previous paragraph has a number of useful
variations (depending on the more adequate trigonometric function to be used),
we shall not try to list them all. Instead, we collect two relevant examples of
trigonometric substitutions.
R1 1
Example 10.42 Compute 0 1Cx
2 dx.
Proof Recalling, from Example 9.35, that
to get
Z
1
0
d
dx
arctan x D
1
,
1Cx2
we may use the FTC
ˇxD1
1
ˇ
dx
D
arctan
x
D arctan 1 D :
ˇ
2
xD0
1Cx
4
Alternatively, by performing the trigonometric substitution x D tan t and noticing
that 0 x 1 , 0 t 4 , we get, from the formula for integration by
substitution and Example 9.34,
Z
1
0
1
dx D
1 C x2
Z
Z
4
0
4
D
0
1
tan0 t dt D
1 C tan2 t
ˇtD ˇ 4
1 dt D tˇ
D :
tD0
4
Z
4
0
1
sec2 t dt
sec2 t
t
u
396
10 Riemann’s Integral
Example 10.43 Let
be the circle with center O and radius R, and choose a
cartesian
coordinate
system
such that O.0; 0/. If f W ŒR; R ! R is given by
p
f .x/ D R2 x2 , then the region Rf under the graph of f (and above the horizontal
axis is the half-disk bounded by and lying in the upper half-plane. Then,
Z
A.Rf / D
R
R
Z
f .x/dx D
R
p
R2 x2 dx:
R
The trigonometric substitution x D R cos t, with 0 t , gives
Z
A.Rf / D
Z
0
p
R2 R2 cos2 t.R sin t/dt
D
R2 sin2 tdt D
0
R2
;
2
where we used the result of Example 10.29 in the last equality.
A similar reasoning proves that the area of the half-disk bounded by and lying
2
in the lower half-plane also equals R2 . Hence, the area of the whole disk equals
2
2 R2 D R2 , as the reader surely predicted from earlier studies of Euclidean
Geometry.
Remark 10.44 As was pointed out in Remark 9.12, in Chap. 11 the sine and cosine
functions will be constructed without relying in any concepts or results of Euclidean
Geometry. In turn, this will allow us to compute the derivatives of the sine and cosine
functions also without reference to Euclidean Geometry. Once this has been done,
the previous example will provide us with a genuine way of computing the area of
a circle with radius R.
Back to the general discussion of integration by substitution, let intervals I and
J and functions g W I ! J and f W J ! R be given, such that f is continuous and
g is continuously differentiable. It’s immediate to verify that (10.44) furnishes the
following equality of primitives:
Z
Z
f .t/dt .g.x// D
f .g.x//g0 .x/dx C C;
(10.46)
where C is a real constant. We now examine a specific relevant case.
Example 10.45 Let I be an interval, f W I ! .0; C1/ be a continuously
differentiable function and r be a nonzero rational number. Since
d
f .x/r D rf .x/r1 f 0 .x/;
dx
we see that 1r f .x/r is a primitive for the function x 7! f .x/r1 f 0 .x/.
10.6 The Change of Variables Formula
In other words, we have
Z
397
f .x/r1 f 0 .x/dx D
1
f .x/r C C;
r
(10.47)
an equality that can be easily remembered from (10.46), with f in place of g and
x 7! xr1 in place of f : on the one hand, the FTC gives us
Z
f .x/
tr1 dt D
tr ˇˇf .x/
1
ˇ C C D f .x/r C CI
r
r
on the other, the variable substitution t D f .x/ gives
Z
Z
f .x/
t
r1
dt D
f .x/r1 f 0 .x/dx:
We conclude this section by observing that, apart from the problems collected
below, we shall see some other relevant examples of integration by trigonometric
substitution in the coming sections.
Problems: Section 10.6
1. In each of the items below, compute the given integral:
R1 1
(a) 0 pxC1
dx.
R 1=2 x
(b) 0 p 2 dx.
1x
R p2 x
(c) 1 1Cx
4 dx.
2. In each of the items below, compute the given primitives, together with their
maximal domains of definition:
R p
(a) x2 p1 C x3 dx.
R
(b) x5 p1 C x3 dx.
R 3
(c) R x p
1 x2 dx.
(d) cos x dx.
3. Let f W Œa; a ! R be a continuous function. Show that:
Z
a
a
4. Compute
R1
sin.x/
1 1Cx2
f .x/dx D
dx and
R
1
2
12
Ra
2 0 f .x/dx; if f is even
:
0; if f is odd:
cos.x/
1Cx3
dx.
398
10 Riemann’s Integral
For the coming example the reader might find it useful to review the basics on
ellipses, in Chap. 6 of [4], for instance.
5. Prove that the area of an ellipse of principal axes of lengths 2a and 2b equals
ab.
6. Prove the following version of Theorem 10.37: if f W Œa; b ! R is piecewise
continuous and g W Œc; d ! Œa; b is increasing (resp. decreasing) and
differentiable, with integrable derivative, then
Z
g.d/
Z
d
f .t/dt D
g.c/
f .g.s//g0 .s/ds:
c
7. If f W Œa; b ! Œa; b is an increasing and differentiable bijection, compute all
Rb
possible values of a .f .x/ C f 1 .x//dx.
For the next problem, the reader may find it convenient to review the concept
of reciprocal polynomial, in Problem 28, page 44.
8. Let p be a polynomial of degree n. Show that p is reciprocal if and only if
Z
x
1
Z
p.t/
t
dt D
n=2C1
1
p.t/
1=x
tn=2C1
dt;
for every real x > 1.
9. Let f W Œ1; C1/ ! R be a differentiable function such that f .1/ D 1 and f 0 .x/ D
1
, for every x 1. Prove that limx!C1 f .x/ does exist and is at most 1C 4 .
x2 Cf .x/2
10. Let I R be an interval and g W I ! I be a continuously differentiable function
of finite order, i.e., such that g.m/ D IdI for some m 2 N, where g.m/ D g ı : : : ı g,
the composition of g with itself m times. If x 2 I and f W I ! R is a continuous
function semi-invariant over g, in the sense that .f ı g/g0 D f , compute the
R g.x/
possible values of x f .t/dt.
10.7 Logarithms and Exponentials
This section brings a fairly standard presentation of the main properties of two of
the most important functions of Mathematics, namely, the natural logarithm and the
exponential functions, postponing to the next section some interesting and important
applications of them.
Definition 10.46 The natural logarithm function log W .0; C1/ ! R is
defined by
Z
x
log x D
1
1
dt:
t
10.7 Logarithms and Exponentials
399
y
y=
1
t
y=t
O
x
1
t
Fig. 10.5 Definition of log x for x > 1
It readily follows from the above definition that log 1 D 0, log x > 0 if x > 1
and log x < 0 if 0 < x < 1. Moreover, letting f W .0; C1/ ! R denote the inverse
proportionality function, so that f .x/ D 1x for every x > 0, then, in terms of the
area of the region of the first quadrant situated under the graph of f (cf. Fig. 10.5),
we have
8
< A.fjŒ1;x /; if x > 1
log x D
0; if x D 1 :
:
A.fjŒx;1 /; if x < 1
The FTC assures that log is a differentiable function, with
log0 x D
1
:
x
Therefore, log is infinitely differentiable and, since log0 > 0, the study of the first
variation of functions shows that log is increasing. Also, since
log00 x D 1
< 0;
x2
it follows from Corollary 9.70 that log is strictly concave.
The coming result and their corollaries bring more important properties of the
natural logarithm.
Proposition 10.47 For x; y > 0, we have
log.xy/ D log x C log y:
(10.48)
400
10 Riemann’s Integral
Proof For the first part, the additivity of the integral gives
Z xy
Z x
Z xy
Z xy
1
1
1
1
log xy D
dt D
dt C
dt D log x C
dt:
t
t
t
1
1 t
x
x
Now, if g W Œ1; y ! Œx; xy is given by g.t/ D xt, the change of variables
Theorem 10.37 furnishes
Z xy
Z g.y/
Z y
Z y
1
1
1
1
0
dt D
dt D
g .s/ds D
xds
t
t
g.s/
xs
x
g.1/
1
1
Z y
1
ds D log y:
D
1 s
t
u
Corollary 10.48 For x > 0 and r 2 Q, we have log xr D r log x.
Proof Applying (10.48) with y D x we get log x2 D 2 log x. If we assume that
log xk D k log x for every x > 0 and some k 2 N, then letting y D xk in (10.48)
we get
log xkC1 D log.x xk / D log x C log xk
D log x C k log x D .k C 1/ log x:
Therefore, it follows by induction that log xn D n log x for every x > 0 and every
n 2 N.
We now apply this particular case with x1=n in place of x to obtain
log x D log.x1=n /n D n log x1=n
or, which is the same, log x1=n D 1n log x.
In view of the particular cases of the two previous paragraphs and letting r D
with m; n 2 N, we have
log xr D log xm=n D log.x1=n /m D m log x1=n D m m
,
n
1
log x D r log x:
n
In order to extend this last equality for negative rational exponents, start by letting
y D x1 in (10.48) to get
0 D log 1 D log.x x1 / D log x C log x1 ;
so that log x1 D log x. Therefore, if r is a negative rational and s D r > 0,
we get
log xr D log.xs /1 D log xs D .s log x/ D r log x:
t
u
10.7 Logarithms and Exponentials
401
Corollary 10.49 limx!0C log x D 1 and limx!C1 log x D C1. In particular,
Im .log/ D R.
Proof Fix a > 1. Since log a > 0, it follows from the previous corollary that
n
n
log an D n log a ! C1 and log an D n log a ! 1:
On the other hand, since 0 <
1
a
< 1, Example 7.12 furnishes
lim an D lim .1=a/n D 0 and
n!C1
n!C1
lim an D lim
n!C1
n!C1
1
D C1:
an
The limits above, together with the fact that log is an increasing function,
guarantee that limx!0C log x D 1 and limx!C1 log x D C1. Hence, it follows
from Problem 7, page 290, that Im .f / D R.
t
u
We now relate the number e, defined in Example 7.41, to the natural logarithm
function.
Theorem 10.50 log e D 1.
Proof First of all, note (cf. Fig. 10.6, where we sketched the portion of the graph of
t 7! 1t from t D 1 to t D 1 C 1n ) that
Z 1C 1
Z 1C 1
ˇtD1C 1
n 1
n
1
1
1
n
ˇ
D
dt <
1 D
dt D tˇ
D 1C
log 1 C
tD1
n
t
n
n
1
1
and, analogously
Z 1C 1
Z 1C 1
n 1
n
1
1
log 1 C
D
dt >
n
t
1C
1
1
1
n
dt D
1
1
n
D
:
nC1 n
nC1
Hence,
1
1
1
< log 1 C
< ;
nC1
n
n
(10.49)
Fig. 10.6 Estimating
log 1 C 1n
1
n
n+1
1
1+
1
n
402
10 Riemann’s Integral
n
so that nC1
< n log 1 C 1n < 1. The squeezing theorem (Problem 6, page 218)
then gives limn!C1 n log 1 C 1n D 1 or, which is the same,
1 n
lim log 1 C
D 1:
n!C1
n
n
Now, since (according to Theorem 7.42) e D limn!C1 1 C 1n and log is
continuous, we have
1 n
1 n
log e D log lim 1 C
D lim log 1 C
D 1:
n!C1
n!C1
n
n
t
u
Gathering together the information obtained so far on the natural logarithm, we
sketch its graph on Fig. 10.7.
The sketch above of the graph of the logarithm function suggests that it grows
relatively slowly. The next result quantifies this suspicion.
Theorem 10.51 For an integer n > 1, we have limx!C1
1
Proof Firstly, observe that t > t1 2n for t > 1. Hence,
1
t
log
x
p
n x
<
for x > 1,
Z
x
log x D
1
D
1
2n
1
dt <
t
t
1=2n
Z
x
1
1
1 2n
Z
dt D
x
D 0.
1
1
t1 2n
for t > 1, so that,
1
t 2n 1 dt
1 t
1
ˇtDx
p
p
ˇ
D 2n. 2n x 1/ < 2n 2n x:
ˇ
tD1
y
y=x
log
1
O
1
Fig. 10.7 Graph of log W .0; C1/ ! R
e
x
10.7 Logarithms and Exponentials
403
Therefore, for x > 1 we have
p
2n
2n
log x
x
<
2n
D p
:
0< p
p
n
n
2n
x
x
x
Now, since limx!C1
guarantees that
log x
p
n x
2n
p
2n
x
D 0, the squeezing theorem (cf. Problem 6, page 218)
! 0 when x ! C1.
t
u
Since the natural logarithm function log W .0; C1/ ! R is a continuous and
increasing bijection, we can consider its inverse
exp W R ! .0; C1/;
(10.50)
called the exponential function. Theorem 8.35 and Problem 8, page 176, guarantee
that exp is also a continuous and increasing bijection. Moreover, for given x; y 2 R,
with x > 0, we have
log x D y , x D exp.y/I
in particular, it follows from log 1 D 0 and log e D 1 that exp.0/ D 1 and
exp.1/ D e.
On the other hand, since log0 x D 1x ¤ 0, Theorem 9.28 assures that exp is a
differentiable function such that, for x > 0 and y D log x,
exp0 .y/ D
1
1
D
D x D exp.y/:
log0 x
1=x
In short,
exp0 D exp;
which in particular shows that exp is infinitely differentiable. Also, since exp00 D
exp > 0, it follows from Corollary 9.70 that exp is a strictly convex function.
The coming result translates, for the exponential function, the properties of the
natural logarithm expressed in Proposition 10.47 and Corollary 10.48.
Proposition 10.52 For x; y 2 R, we have:
(a) exp.x C y/ D exp.x/ exp.y/.
(b) exp.r/ D er , if r 2 Q.
Proof
(a) Letting a D exp.x/ and b D exp.y/, we have a; b > 0 and log a D x, log b D y.
Hence,
x C y D log a C log b D log.ab/;
404
10 Riemann’s Integral
so that exp.x C y/ D ab. The above computations also give
exp.x/ exp.y/ D ab D exp.x C y/:
(b) Since log and exp are inverses of each other and log e D 1, it follows from
Proposition 10.47 that
log exp.r/ D r D log er :
Thus, the injectivity of log gives exp.r/ D er .
t
u
Thanks to the item (b) of the former proposition, from now on we shall write
exp.x/ D ex ;
stressing that the right hand side coincides with the usual meaning of ex when x 2 Q,
and defines ex when x … Q. In this new notation, item (a) of the previous result is
written as
exCy D ex ey ;
for all x; y 2 R.
More generally, the exponential function allows us to rigorously define what
is meant by the power ax , for positive a (and real x). Indeed, in the notation just
introduced, we set
ax D ex log a ;
(10.51)
so that
axCy D e.xCy/ log a D ex log a ey log a D ax ay
and
log ax D log.ex log a / D x log a
for all x; y 2 R.
Notice that if r 2 Q, say r D
m
n,
with m 2 Z and n 2 N, Corollary 10.48 gives
ar D e n log a D elog
m
(10.52)
p
n
am
D
p
n
am I
in other words, for x D r 2 Q definition (10.51) matches with the meaning we have
attributed to ar up to now. In particular, we have a0 D 1 and a1 D a.
10.7 Logarithms and Exponentials
405
If a D 1, it follows from (10.51) that ax D 1 for every x 2 R. On the other
hand, given a positive real a ¤ 1, the exponential function of basis a is fa W R !
.0; C1/ such that fa .x/ D ax .
Relation (10.51), together with the properties of the exponential function (of
basis e) and the chain rule, guarantee that fa is a differentiable bijection, with
derivative
fa0 .x/ D expx log a log a D ax log a:
(10.53)
In particular, since log a > 0 , a > 1, we get
fa is increasing , fa0 > 0 , a > 1I
accordingly, fa is decreasing if and only if 0 < a < 1.
The inverse of fa is defined to be the function loga W .0; C1/ ! R, which
is called the base a logarithm function, whose properties will be the object of
Problem 1. Thus, the natural logarithm function is just the base e logarithm function.
An important variant of the above discussion is the following: given an interval I
and functions f W I ! .0; C1/ and g W I ! R, we define f .x/g.x/ by letting
f .x/g.x/ D eg.x/ log f .x/ :
Observe that the requisite of f .x/ being positive gives sense to this definition. Also,
the chain rule applied to the right hand side of the equality above assures that the
function x 2 I 7! f .x/g.x/ is differentiable whenever f and g are so. In this respect,
see Problem 9.
We can now state and prove a useful extension of Theorem 10.50.
x
Theorem 10.53 e D limx!C1 1 C 1x .
Proof A reasoning analogous to that in the proof of Theorem 10.50 gives, for x > 0,
1
1
1
< log 1 C
< :
xC1
x
x
x
It thus follows from (10.52) that log 1 C 1x D x log.1 C 1x /.
Multiplying both inequalities above by x > 0, we get
1 x
x
< log 1 C
< 1;
xC1
x
so that the squeezing theorem yields
1 x
lim log 1 C
D 1:
x!C1
x
(10.54)
406
10 Riemann’s Integral
Finally, since the exponential function is continuous, we can write
1 x
e D exp lim log 1 C
x!C1
x
1 x
D lim exp log 1 C
x!C1
x
x
1
:
D lim 1 C
x!C1
x
t
u
We now use Theorem 10.51 to show that the exponential function increases more
rapidly than any polynomial when x ! C1.
Theorem 10.54 If p is a polynomial, then limx!C1
Proof Let p.x/ D am x C am1 x
m
m1
p.x/
ex
D 0.
C C a1 x C a0 , with am ¤ 0. Since
X
xk
p.x/
D
ak x ;
x
e
e
kD0
m
it suffices to show that limx!C1
we have x D log y and, then,
xn
ex
D 0 for every n 2 N. To this end, letting y D ex
.log y/n
xn
D
D
ex
y
log y
p
n y
n
:
Finally, since y ! C1 when x ! C1, Theorem 10.51 gives
xn
log y n
log y n
D
lim
D 0:
lim x D lim
p
p
n y
x!C1 e
y!C1
y!C1 n y
t
u
We close this section by deriving asymptotic estimates on the size of log n and
nŠ. In other words, we estimate the sizes of log n and nŠ for large values of n. Our
first result is due to Euler.4
Theorem 10.55 (Euler)
Given n 2 N, we have
lim
n!C1
n
X
1
kD1
k
!
log n D ;
(10.55)
where is a positive real constant.
4
The Swiss mathematician Leonhard Euler is one of the most prolific mathematicians of all times,
even if we stick only to relevant Mathematics. His contributions impressively vary from Geometry
to Combinatorics, passing through Number Theory, Calculus and Physics. In each of these areas
of knowledge there is at least one, if not several, celebrated Euler’s theorem.
10.7 Logarithms and Exponentials
407
Proof First of all, observe that
0<
n
X
1
kD1
k
log n D
n
X
1
kD1
D
n1 X
kD1
Now, since
1
n
k
n1
X
.log.k C 1/ log k/
kD1
1
1
1
log 1 C
C :
k
k
n
! 0 when n ! C1, it suffices to prove that the series
1
log 1 C
k
k
X 1
k1
(10.56)
converges to a positive sum .
Applying inequality (10.49) with k in place of n, we conclude that (10.56) is a
series of positive terms, such that
X
n1 1
1
1
1
log 1 C
<
D1 :
k
k
k kC1
n
kD1
n1 X
1
kD1
t
u
Therefore, Proposition 7.44 guarantees that it does converge.
The positive number defined by (10.55) is known in mathematical literature as
the Euler-Mascheroni constant, and its value with five correct decimal places is
Š 0; 57721. Up to this day it is unknown whether is rational or irrational, albeit
there is a strong suspicion that it should be irrational.
Notice that (10.55) can be written as
log n Š
n
X
1
kD1
k
;
(10.57)
with error less than 1n (cf. Problem 19).
In order to motivate what comes next, let n 2 N. The arithmetic-geometric mean
inequality (cf. Theorem 5.7) gives
nŠ D 1 2 : : : n <
1 C 2 C C n
n
n
thus, for sufficiently large n, we have
1
n < 1 C
n
n=2
nŠ
n
Š e:
D
n C 1
2
n
I
408
10 Riemann’s Integral
On the other hand, since n C 1 2.n 1/ < 3.n 2/ < for n 3, we get
.nŠ/2 D .1 n/.2 .n 1// : : : ..n 1/ 2/.n 1/
n
2
n2 .n C 1/n2 D
.n C 1/n I
nC1
therefore, again for sufficiently large n,
1
n nŠ
1C
n=2
n
nC1
n
n=2
Š
p
e:
The coming theorem refines the above naive estimates, giving a sharp asymptotic
estimate for nŠ. In order to properly state and prove it, it is now convenient to
introduce some notations. Given sequences .an /n1 and .bn /n1 of positive reals, we
write an
bn and an D o.bn / as shorthands for the following possible asymptotic
behaviors of abnn :
an
an
an
D 1 and an D o.bn / , lim
D 0:
n!C1 bn
n!C1 bn
bn , lim
In particular, an D o.1/ if limn!C1 an D 0. Also, if ˛ > 0 and ˛an
lim
n!C1
so that
bn ˛an
an
bn , then
bn ˛an
D 0;
an
D o.1/; consequently,
bn ˛an
an D .˛ C o.1//an :
bn D an C .bn an / D ˛ C
an
(10.58)
We are finally in position to prove the following result, due to the Scottish
mathematician of the XVIII century James Stirling and known in mathematical
literature as Stirling’s formula.
p
n
Theorem 10.56 (Stirling) nŠ
2n ne .
n n p
n e
n
Proof We have to show that limn!C1
D p1 or, which is the same, that
nŠ
2
p
nn en 2n
lim log
n!C1
nŠ
!
1
D log p :
2
Firstly, we shall show that the limit at the left hand side above actually exists. To
this end, note that
log
p nŠ
.nŠ/2
1
nn en n
D n log n n log p D .n log n n/ log
;
nŠ
2
n
n
10.7 Logarithms and Exponentials
409
with
n log n n D
n1
X
Œ..k C 1/ log.k C 1/ .k C 1// .k log k k/
kD1
D
n1 Z
X
kD1
kC1
log x dx 1
k
(we used the result of Problem 4 in the last equality) and
.nŠ/2
1
1
log
D .log.1 2/ C log.2 3/ C C log..n 1/n//
2
n
2
n1
X
log k C log.k C 1/
D
kD1
2
:
Hence,
p X
n1 Z kC1
log k C log.k C 1/
nn en n
log
D
1:
log x dx nŠ
2
k
kD1
(10.59)
R kC1
. Problem 13 (with a D k,
For k 1, let ak D k log x dx log kClog.kC1/
2
b D k C 1 and f D log, which is concave) furnishes
Z
ak D
k
kC1
log x dx log k C log.k C 1/
> 0:
2
On the other hand, by successively applying the results of Problems 4 and 20, we get
Z kC1
1
1
.k C1/ log.k C1/Ck log k C1 > 0:
log x dx D log k C
log k C
2
2
k
Therefore,
log k C log.k C 1/
1
ak < log k C
2
2
2
k C 12
1
1
1
D log
D log 1 C
2
k.k C 1/
2
4k.k C 1/
1
1 1
1
1
D
;
< 2 4k.k C 1/
8 k kC1
where we used the result of Problem 5 in the last inequality above.
410
10 Riemann’s Integral
We conclude that
P
k1
n1
X
ak is a series of positive terms, such that
ak <
kD1
n1
X
1 1
kD1
8
Thus, the comparison test assures that
from (10.59) that there exists
lim log
n!C1
1
k kC1
P
k1 ak
D
1
:
8
does converge, and it follows
p n1
X
X
nn en n
D lim
ak 1 D
ak 1:
n!C1
nŠ
kD1
k1
For the actual computation of the desired limit, let c > 0 be such that
1 D log c. Since
P
k1 ak
p nn en n
lim log
D log c;
n!C1
nŠ
we have
p nn en n
c De
D exp lim log
n!C1
nŠ
n n p n
n e
D lim exp log
n!C1
nŠ
n n p n
ne
:
D lim
n!C1
nŠ
log c
p
It follows from (10.58) (with an D nn en n, bn D nŠ and ˛ D 1c ) that
nŠ D
p
1
C o.1/ nn en n:
c
Now recall that, according to Example 10.35, if In D
I2k D
R =2
0
.cos x/n dx for n 2 N, then
.2k/Š .2k kŠ/2
and
I
D
2k1
.2k kŠ/2 2
2k.2k/Š
for every integer k 1. Hence,
10.7 Logarithms and Exponentials
I2k
I2k1
411
..2k/Š/2 2k 24k .kŠ/4
2
h
p i2
1
2k 2k
2k 2k c C o.1/ .2k/ e
D
h
p i4 2
24k 1c C o.1/ kk ek k
D
D
1
c
2
k
C o.1/
2
2 ! 2c :
k
I2k
! 1. Therefore, the above
However, Problem 13, page 391 showed that I2k1
2
computations give 2c D 1 or, which is the same, c D p1 .
t
u
2
Example 10.57 For n 2 N, the binomial number 2n
is the largest of the 2n C 1
n
coefficients of the binomial expansion of .x C y/2n . This gives
!
!
2n
X
2n
2n
D 22n ;
.2n C 1/
>
k
n
kD0
22n
> 2nC1
. Stirling’s formula considerably improves this estimate, for, with
so that 2n
n
the aid of it, we get
!
.2n/Š
2n
D
n
.nŠ/2
p
2 2n 2ne
p
2 2n ne
2n
2n
22n
D p :
n
Problems: Section 10.7
1. * Given 0 < a ¤ 1, prove that loga is an infinitely differentiable bijection, with
loga x D
log x
log a
and
log0a x D
1
;
x log a
(10.60)
for all 0 < a; x ¤ 1. Moreover, given positive reals a, b, c, x and y, with
a; b; c ¤ 1, prove also that:
(a)
(b)
(c)
(d)
(e)
log D loge .
loga is increasing if a > 1 and decreasing if 0 < a < 1.
loga .xy/ D loga x C loga y.
loga c D logb c loga b.
loga b logb c logc a D 1.
412
10 Riemann’s Integral
2.
3.
4.
5.
6.
Which of e or e is bigger? Justify your answer!
Find all a; bR2 N such that a ¤ b and ab D ba . Justify your answer!
* Compute log x dx D x log x x C C.
* Show that log.x C 1/ < x for every x > 0.
Use Corollary 10.48, together with the result of Problem 8, page 338, to give
another proof of the fact that log is strictly concave.
Find all negative values of a such that f W R ! R, given by f .x/ D ex C ax3 ,
has a single inflection point.
* Let ˛ ¤ 0 be a real number and f W .0; C1/ ! R be the function given by
f .x/ D x˛ . Show that f 0 .x/ D ˛x˛1 for every x > 0.
Given an interval I and differentiable functions f W I ! .0; C1/ and g W I ! R,
compute the derivative of the function h W I ! R defined by h.x/ D f .x/g.x/ ,
for every x 2 I.
Given a; b > 0, let f W Œ0; 1 ! R be defined by f .x/ D xa .1 x/b . Show that f
attains a maximum value and compute it.
The set of rationals is not closed with respect to powers; for instance, if a D
b D 12 , then a; b 2 Q, albeit ab D p1 , an irrational number. If a and b are
7.
8.
9.
10.
11.
2
12.
13.
14.
15.
Justify your answer.
positive irrationals, is it always true that abis irrational?
x
If a is a positive real, prove that limx!C1 1 C ax D ea .
x
Find all positive values of a for which limx!C1 xCa
D e.
xa
Let f W R ! R be a differentiable function such that f .0/ D 0 and f 0 .x/ > f .x/
for every x 2 R. Show that f .x/ > 0 for every x > 0.
* The purpose of this problem is to compute primitives for the function t 7!
sec3 t, where sec denotes the restriction
of the secant function to an interval
of the form 2 C k; 2 C k , with k 2 Z. To this end, do the following
items:
(a) Show that sec0 Rt D tan t sec t and dtd log j sec t tan tj D sec t.
(b) Conclude that sec t dt D log j sec t R tan tj C C.
(c) Write sec 2 t D 1 C tan2 t and integrate sec3 t dt by parts to get
Z
sec3 t dt D log j sec t tan tj C sec t tan t and, hence,
Z
sec3 t dt D
Z
sec3 t dt
1
.sec t tan t log j sec t tan tj/ C C:
2
16. * The hyperbolic sine and hyperbolic cosine, respectively denoted sinh; cosh W
R ! R, are defined by
sinh x D
ex ex
2
and
cosh x D
ex C ex
:
2
10.7 Logarithms and Exponentials
413
y
y
(cos t, sin t)
(cosh t, sinh t)
x
x
Fig. 10.8 Circular and hyperbolic sines and cosines
With respect to them, do the following items:
(a) Show that cosh2 x sinh2 x D 1, sinh0 x D cosh x and cosh0 x D sinh x.
(b) Show that cosh is an even and strictly convex function with image Œ1; C1/,
such that cosh x D 1 , x D 0.
(c) Show that sinh is an odd strictly increasing function with image R. Show
also that sinh is strictly convex in .0; C1/ and strictly concave in .1; 0/,
and that x D 0 is its only inflection point.
(d) Sketch the graphs of the hyperbolic sine and hyperbolic cosine in a single
cartesian system of coordinates; in this same cartesian system, sketch also
the graphs of x 7! 12 ex (x > 0) and x 7! 12 ex , x 7! 12 ex (x < 0).
(e) The graph of the hyperbolic cosine is known as the catenary.5 Given x0 2
R, show that the curvature of the catenary at x0 (cf. Problem 3, page 344)
equals k.x0 / D cosh12 x .
0
The next problem explains the names hyperbolic sine and hyperbolic cosine.
To put it in context, recall (cf. Fig. 10.8, left) that the area of the circular sector
of radius 1, contained in the first quadrant of the cartesian plane, centered at
the origin and bounded by the radii that join the origin to the points .1; 0/ and
.cos t; sin t/ equals 2t .
17. For all t 2 R, observe that the point .cosh t; sinh t/ belongs to the right branch
of the hyperbola x2 y2 D 1, as well as to the straight line y D .tanh t/x, where
sinh t
tanh t D cosh
. For a fixed t > 0, let A be the area of the bounded portion of the
t
cartesian plane bounded by the horizontal axis, the right branch of x2 y2 D 1
and the straight line y D .tanh t/x (cf. Fig. 10.8, right). Show that A D 2t .
18. Given > 0, show that limx!0C x log x D 0.
5
For the reader’s curiosity, the surface of revolution generated by the rotation of the catenary along
the horizontal axis is an example of a minimal surface, in this case a catenoid. Minimal surfaces
have very interesting properties, some of which can be found in the excellent book of professor
M. P. do Carmo [8].
414
10 Riemann’s Integral
19. * With respect to (10.57), show that
0 < log n n
X
1
kD1
k
<
1
:
n
20. * If f W Œ1; C1/ ! R is given by
f .x/ D log.x C 1=2/ .x C 1/ log.x C 1/ C x log x C 1;
prove that f is positive in Œ1; C1/.
n
21. Show that limn!C1 p
D e.
n
nŠ
22. Let 0 < ˛ < 1 be given and k; n 2 N be such that k > ˛n. Prove that
..kC1/e/n
p
.
kn
n
2n
10.8 Miscellaneous
This section gathers some interesting applications of the natural logarithm and
exponential function. We start by examining three examples which show that the
fact that we now have logarithms and exponentials at our disposal gives greater
flexibility to the material discussed so far.
Example 10.58 (Thailand) Find all functions f W N ! R satisfying the following
conditions:
(a) f is increasing.
(b) f .1/ D 1 and f .2/ D 4.
(c) f .xy/ D f .x/f .y/ for all x; y 2 N.
Solution An easy induction gives f .2k / D 22k , for every k 2 N.
Take an integer a > 2 and suppose, for the sake of contradiction, that f .a/ a2 1. Since Q is dense in R, we can choose m; n 2 N such that
log 22
m
log 22
<
;
<
log a2
n
log.a2 1/
so that 2n < am and .a2 1/m < 22n . Hence, successive applications of (a) and (c)
give us
f .2n / D 22n > .a2 1/m f .a/m D f .am / > f .2n /;
which is a contradiction.
A similar reasoning shows that f .a/ a2 C 1 also leads to a contradiction, so
that the only left possibility is f .a/ D a2 .
u
t
10.8 Miscellaneous
415
The next example shows how to use logarithms to guarantee the existence of
powers of a fixed (and almost arbitrary) given basis beginning (from the left) by
a prescribed set of algarisms. At this point, the reader might find it helpful to
review the material of Sect. 7.3, specially Corollary 7.33, as well as the statement of
Problem 1, page 411.
Example 10.59 Let a > 1 be an integer which is not a power of 10, and s be a
finite sequence of algarisms, the first of which is nonzero. Prove that there exists a
power of a whose decimal representation starts, from the left, with the sequence s
of algarisms.
Proof We want to show that there exist m; n 2 N for which
s 10m an .s C 1/ 10m :
Taking logarithms in base 10, this is the same as showing that there exist naturals m
and n for which log10 s C m n log10 a log10 .s C 1/ C m or, which is the same,
log10 s m C n log10 a log10 .s C 1/:
If we show that log10 a is irrational, the proof will be finished thanks to Corollary 7.33 of Kronecker’s lemma.
Suppose, by contradiction, that log10 a 2 Q, say log10 a D pq , with p; q 2 N
relatively prime. Then aq D 10p , and the Fundamental Theorem of Arithmetic (cf.
introduction to Chap. 1), allow us to write a D 2k 5l b, for some nonnegative integers
k; l and some natural b relatively prime with 10. But these two facts give the equality
2kq 5lq bq D 2p 5p , from which we deduce that b D 1, kq D p and lq D p. Thus, we
conclude that k D l, which would give us a D 2k 5k D 10k , a contradiction.
t
u
We now refine the kind of argument of the previous example to establish the best
possible converse of Problem 16, page 11.
Example 10.60 Prove that there exist infinitely many naturals n such that the
decimal representations of 2n and 5n start (at left) with a 3.
Proof If 2n and 5n are to begin with a 3, it is necessary and sufficient that there exist
naturals k and l such that 3 10k < 2n < 4 10k and 3 10l < 5n < 4 10l or, taking
logarithms in base 10, that
log10 3 < k C n log10 2 < log10 4 and log10 3 < l C n log10 5 < log10 4:
In turn, this is equivalent to asking that the fractional parts6 of both n log10 2 and
n log10 5 belong to the interval .log 3; log 4/.
6
Recall that the fractional part of x 2 R, denoted fxg, is defined as fxg D x bxc, and belongs to
the interval Œ0; 1/.
416
10 Riemann’s Integral
Since n log10 2 C n log10 5 D n and fng D 0, such fractional parts are at the same
distance from 12 . Finally, since 12 2 .log10 3; log10 4/, we conclude that it suffices to
show that there exist infinitely many naturals k and n such that
log10 3 < k C n log10 2 < log10 4:
However, as we have shown in the solution to the previous example, log10 2 is
irrational, so the existence of such k and n follows again from Corollary 7.33 of
Kronecker’s lemma.
t
u
We now turn to inequalities. Up to this moment, we have given or hinted at three
different proofs of the inequality between the arithmetic and geometric means (one
in Theorem 5.7, another in Problem 13, page 140, and a third one in Example 9.59).
Jensen’s inequality applied to the natural logarithm function provides a fourth one,
which is particularly straightforward. However, before we present it, it is worth to
point out the following generalization of (10.48), which can be easily established by
induction: for x1 ; x2 ; : : : ; xn > 0, we have
log.x1 x2 : : : xn / D log x1 C log x2 C C log xn :
(10.61)
Example 10.61 Let n 2 be an integer and x1 , x2 , . . . , xn be positive reals. Since
log W .0; C1/ ! R is strictly concave, the version of Corollary 9.75 for strictly
concave functions gives us
log x1 C log x2 C C log xn
x 1 C x2 C C xn
log
:
n
n
If we successively apply (10.61) and Corollary 10.48 (with x D x1 x2 : : : xn and
r D 1n ) to the left hand side, we get
p
x 1 C x2 C C xn
n
;
log x1 x2 : : : xn log
n
(10.62)
with equality if and only if x1 D x2 D D xn . But since log is an increasing
function, it follows from here that
p
n
x1 x2 : : : xn x 1 C x2 C C xn
;
n
with equality if and only if x1 D x2 D D xn .
Essentially the same idea as that of the proof of the previous example allows one
to obtain the important Young’ inequality.7
7
After William H. Young, English mathematician of the XIX and XX centuries.
10.8 Miscellaneous
417
Proposition 10.62 (Young ) Let p and q be positive reals such that
Given a; b > 0, we have
ab bq
ap
C ;
p
q
1
p
C
1
q
D 1.
(10.63)
with equality if and only if ap D bq .
Proof Given positive reals x1 , x2 , t1 and t2 > 0 such that t1 Ct2 D 1, strictly concave
character of the natural logarithm function in .0; C1/, together with Jensen’s
inequality, gives
log.t1 x1 C t2 x2 / t1 log x1 C t2 log x2 ;
with equality if and only if x1 D x2 .
Letting t1 D 1p , t2 D 1q , x1 D ap and x2 D bq , we get
log
bq
ap
C
p
q
1
1
log ap C log bq D log ab;
p
q
with equality if and only if ap D bq .
Finally, since log is strictly increasing, it follows from the above inequality that
bq
ap
C
ab;
p
q
with equality if and only if ap D bq .
t
u
The inequality of the coming result is a consequence of Young’s inequality and
is known as Hölder’s inequality.8 Note that, when p D q D 2, such an inequality
reduces to Cauchy’s inequality (cf. Theorem 5.13). Nevertheless, it is only after
defining ax for reals a; x > 0 that we can make sense of it.
Theorem 10.63 (Hölder) Let a1 ; a2 ; : : : ; an and b1 ; b2 ; : : : ; bn be given positive
reals, and p; q > 0 be such that 1p C 1q D 1. Then,
n
X
ai bi iD1
n
X
iD1
!1=p
p
ai
n
X
!1=q
q
bi
;
iD1
with equality if and only if
p
p
a2
apn
a1
q D q D ::: D q:
b1
b2
bn
8
After Otto Hölder, German mathematician of the XIX and XX centuries.
418
10 Riemann’s Integral
Proof Letting A D
p 1=p
Pn
iD1
ai
n
X
and B D
ai bi AB ,
iD1
Now, letting xi D
ai
A
n
X
p
and yi D
xi D
iD1
and want to prove that
inequality:
n
X
iD1
xi yi bi
,
B
iD1
bi
, we have
n
X
ai bi
1:
A B
iD1
we have
n
n
n
X
1 X p
1 X q
q
a
D
1;
y
D
b D1
i
Ap iD1 i
Bq iD1 i
iD1
Pn
iD1 xi yi
1. But this is straightforward from Young’s
q
n p
X
x
y
C i
p
q
i
iD1
1X p 1X q
D
x C
y D 1:
p iD1 i
q iD1 i
p
p
ai
Ap
q 1=q
Pn
n
n
q
To have equality, we must have xi D yi for 1 i n or, which is the same,
D
q
bi
Bq
for 1 i n. Yet in another way, we must have
p
p
a2
apn
Ap
a1
q D q D ::: D q D q:
B
b1
b2
bn
Conversely, it’s immediate to verify that if the above condition is satisfied then we
do have equality.
t
u
A version of Hölder’s inequality for integrals will be the object of Problem 8 of
the next section.
We finish this section by discussing two more elaborate applications of Jensen’s
inequality, the first of which appeared as a problem at the 2001 IMO. We divide it
into two examples.
Example 10.64 Sketch the graph of the function f W R ! R given by
1
f .x/ D p
:
1 C 8ex
Solution Firstly, note that 0 < f .x/ < 1 for every x 2 R. Moreover, since
limx!C1 ex D C1 and limx!1 ex D 0, we have
lim f .x/ D 0
x!C1
and
lim f .x/ D 1:
x!1
It follows that the graph of f is entirely contained in the strip of the cartesian plane
bounded by the horizontal lines y D 0 and y D 1, and that these lines are asymptotes
10.8 Miscellaneous
419
to the graph. On the other hand, since f .0/ D 13 , the graph intersects the vertical axis
at the point 0; 13 .
Now, an immediate computation with the aid of the chain rule gives
f 0 .x/ D 4ex
< 0;
.1 C 8ex /3=2
for every x 2 R, so that f decreases along the whole real line. Also, computing f 00
(again with the aid of the chain rule) we get
f 00 .x/ D 4ex
.1 4ex /:
.1 C 8ex /5=2
Therefore, the sign of f 00 coincides with that of 4ex 1, so that f is strictly convex
in .2 log 2; C1/, strictly concave in .1; 2 log 2/ and has x0 D 2 log 2 as its
only inflection point.
Finally, noticing that 2 log 2 Š 1:38, f .2 log 2/ D p1 Š 0:58 and gathering
3
together the above information, we get Fig. 10.9 as a reasonably accurate sketch for
the graph of f .
t
u
We now come to the promised IMO problem.
Example 10.65 (IMO) Prove that, for every positive reals a, b and c, one has
a
b
c
p
Cp
Cp
1:
a2 C bc
b2 C ac
c2 C ab
Proof Note, first of all, that the expression at the left hand side of the inequality
equals
q
1
1 C 8 abc2
Cq
1
1 C 8 bac2
Cq
1
1 C 8 ab
c2
:
y
y=1
0, 58
−1, 38
Fig. 10.9 Graph of the function x 7!
p 1
1C8ex
x
420
10 Riemann’s Integral
Letting
prove that
bc
a2
D ex ,
ac
b2
D ey and
ab
c2
D ez we have x C y C z D 0 and wish to
f .x/ C f .y/ C f .z/ 1;
where f W R ! R is the function of the previous example. To what is left to do,
there are three possibilities:
(i) x; y; z > 2 log 2: Jensen’s inequality gives
f .x/ C f .y/ C f .z/ 3f
xCyCz
3
D 3f .0/ D 1:
(ii) x; y 2 log 2 < z: then
2
f .x/ C f .y/ C f .z/ f .x/ C f .y/ 2f .2 log 2/ D p > 1:
3
(iii) x 2 log 2 < y; z: again by Jensen’s inequality, we get
f .x/ C f .y/ C f .z/ f .x/ C 2f
yCz
2
x
D f .x/ C 2f :
2
If g W .1; 2 log 2 ! R is given by
x
g.x/ D f .x/ C 2f ;
2
observe that
lim g.x/ D lim f .x/ C 2 lim f .x/ D 1
x!1
x!1
x!C1
and
1
2
g.2 log 2/ D p C p > 1:
3
17
On the other hand,
x
4ex
4ex=2
D
g0 .x/ D f 0 .x/ f 0 C
;
x
3=2
2
.1 C 8e /
.1 C 8ex=2 /3=2
so that g0 .x/ < 0 if and only if 2 log 7 < x < 2 log 2. Hence,
x 2 .1; 2 log 7 ) g.x/ > 1
10.8 Miscellaneous
421
and
x 2 .2 log 7; 2 log 2/ ) g.x/ > g.2 log 2/ > 1:
t
u
9
The subsequent discussion derives an inequality which widely generalizes that
between the arithmetic and geometric means and is known as the inequality of
power means. We start by defining what one means for such power means.
Definition 10.66 Given positive reals a1 ; a2 ; : : : ; an and ˛ 2 R, we define the ˛-th
power mean of a1 , a2 , . . . , an as the positive real number M˛ D M˛ .a1 ; : : : ; an /,
such that
8
< a˛1 Ca˛2 CCa˛n 1=˛
; if ˛ ¤ 0
n
M˛ D p
:
: n a a : : : a ; if ˛ D 0
1 2
n
Theorem 10.67 Let a1 ; a2 ; : : : ; an be given positive reals. If ˛ < ˇ are any
reals, then
min fai g M˛ Mˇ max fai g;
1in
(10.64)
1in
with equality in any of the inequalities above if and only if all of the ai ’s are equal.
Proof We can assume, without loss of generality, that not all of the ai ’s are equal.
Let’s momentarily assume that (10.64) is true (with strict inequalities) for all
0 < ˛ < ˇ. Given reals x < y < 0, we have x > y > 0; since a11 ; : : : ; a1n are also
positive and not all equal, it follows from our assumptions that
min
1in
1
ai
< My
1
1
;:::;
a1
an
< Mx
1
1
;:::;
a1
an
< max
1in
1
:
ai
Now, an easy computation gives
Mt
1
1
;:::;
a1
an
D Mt .a1 ; : : : ; an /1
for every t 2 R, so that the above inequalities are equivalent to
1
max fai g
1in
< My .a1 ; : : : ; an /
1
< Mx .a1 ; : : : ; an /
1
<
1
min fai g
1in
9
Another such remarkable generalization, due to Newton and McLaurin, is the object of Sect. 17:2
of [5].
422
10 Riemann’s Integral
or, which is the same,
min fai g < Mx .a1 ; : : : ; an / < My .a1 ; : : : ; an / < max fai g:
1in
1in
Hence, we conclude that it suffices to analyse the cases 0 < ˛ < ˇ and 0 D ˛ < ˇ.
Also without loss of generality, assume that a1 D max1in fai g. Since not all of
the ai ’s are equal, we get
ˇ
Mˇ .a1 ; : : : ; an / D
ˇ
ˇ
a1 C a2 C C an
n
!1=ˇ
ˇ
<
na1
n
!1=ˇ
D a1 D max fai g
1in
and, analogously, min1in fai g < M˛ . Thus, we are left to showing that
0 < ˛ < ˇ ) M0 < M˛ < Mˇ :
The first inequality above is equivalent to
p
a˛1 C a˛2 C C a˛n
> n a˛1 a˛2 : : : a˛n ;
n
which, in turn, follows immediately from the inequality between the arithmetic and
geometric means.
In order to prove the second inequality, letting Mˇ D K it suffices to show that
1
K
a˛1 C a˛2 C C a˛n
n
1=˛
<1
or, which is the same,
.a1 =K/˛ C .a2 =K/˛ C C .an =K/˛
n
1=˛
< 1:
Letting bi D .ai =K/ˇ , we conclude that not all of the bi ’s are equal, but satisfy
1
b1 C b2 C C bn
D ˇ
n
K
˛=ˇ
moreover, since .ai =K/˛ D bi
˛=ˇ
b1
ˇ
ˇ
ˇ
a1 C a2 C C an
n
!
D 1I
, we want to show that
˛=ˇ
C b2
˛=ˇ
C C bn
n
!1=˛
< 1:
(10.65)
10.8 Miscellaneous
423
To what is left to do, let f W .0; C1/ ! R be given by f .x/ D x˛=ˇ . Since
0 < ˛=ˇ < 1, two applications of the result of Problem 8, page 412, give
˛
f .x/ D
ˇ
00
˛
1 x˛=ˇ2 < 0;
ˇ
so that f is strictly concave. It thus follows from Jensen’s inequality that
f .b1 / C C f .bn /
<f
n
b1 C C bn
n
D f .1/ D 1;
t
u
which is exactly (10.65).
Given a1 ; a2 ; : : : ; an > 0, it is worth noticing that the function f W R ! R,
defined for ˛ 2 R by
f .˛/ D M˛ .a1 ; : : : ; an /;
is continuous. Actually, f is obviously continuous at any ˛ ¤ 0, so that it suffices to
show that
lim f .˛/ D f .0/:
˛!0
To this end, the continuity of x 7! ex and x 7! log x assures that we need only
show that
lim log f .˛/ D log f .0/
˛!0
or, which is the same,
˛
a1 C a˛2 C C a˛n
log.a1 a2 an /
1
D
:
lim log
˛!0 ˛
n
n
(10.66)
However, since
lim log
˛!0
a˛1 C a˛2 C C a˛n
n
D log 1 D 0;
an application of l’Hôpital’s rule to the left hand side of (10.66), together
with (10.53), gives
424
10 Riemann’s Integral
˛
˛
a1 C a˛2 C C a˛n
a1 C a˛2 C C a˛n
1
d
D lim
log
lim log
˛!0 ˛
˛!0 d˛
n
n
.log a1 /a˛1 C C .log an /a˛n
˛!0
a˛1 C C a˛n
D lim
log a1 C C log an
n
log.a1 a2 : : : an /
:
D
n
D
Problems: Section 10.8
1. Prove that equation a2 D 2a has exactly three real roots, one of which is
irrational.
2. (Israel) Find all real solutions of the system of equations
p
8
< x C log.x C px2 C 1/ D y
y C log.y C p y2 C 1/ D z :
:
z C log.z C z2 C 1/ D x
3. (Croatia) Prove that there exists no polynomial function p W R ! R such that
p.x/ D log x, for every x 2 R.
4. Given n 2 N, show that the decimal representation of n has exactly blog10 ncC1
algarisms.
5. (Brazil) Let m 2 N and a1 ; a2 ; : : : ; am 2 f0; 1; 2; : : : ; 9g, with a1 ¤ 0. Given
k 2 N, prove that there exists n 2 N such that the decimal representation of nk
starts (from the left) with the sequence of algarismos a1 a2 : : : am .
6. (Russia) Show that, for any reals 1 < a < b < c, one has
loga .loga b/ C logb .logb c/ C logc .logc a/ > 0:
7. (Brazil) Give an example of a function f W Œ0; C1/ ! R such that f .0/ D 0
and f .2x C 1/ D 3f .x/ C 5 for every x 2 R.
8. (South Korea) Find all functions f W ZC ! ZC satisfying the following
conditions:
(a) 2f .m2 C n2 / D f .m/2 C f .n/2 , for all m; n 2 ZC .
(b) If m; n 2 ZC , with m n, then f .m2 / f .n2 /.
9. (Romania) Given a > 0, show that
Z
lim n
n!C1
0
1
xn
aC1
dx D log
:
xn C a
a
10.8 Miscellaneous
425
10. Let x, y and z be positive reals whose sum of squares
equals 8. Prove that the
q
2
sum of their cubes is greater than or equal to 16 3 .
11. Prove the following generalization of the inequality between the arithmetic
and geometric means, known as the weighted arithmetic-geometric means
inequality: given positive reals a1 , a2 , . . . , an and naturals k1 , k2 , . . . , kn such
that k11 C k12 C C k1n D 1, we have
a k2
ak11
a kn
C 2 C C n a1 a2 : : : an ;
k1
k2
kn
with equality if and only if a1 D a2 D : : : D an .
12. (USA) Given positive real numbers a, b and c, prove that
a b c a b c
aCbCc
3
aCbCc
;
with equality if and only if a D b D c.
13. (Leningrad) Let x1 , x2 , . . . , xn be nonnegative reals with sum equal to 12 . Show
that
n
Y
1 xj
2n 1 n
:
1 C xj
2n C 1
jD1
14. Let 0 < x1 ; x2 ; x3 ; x4 < be real numbers such that x1 C x2 C x3 C x4 D 2.
Prove that
4
Y
sin xi
iD1
xi
16
:
4
15. (IMO shortlist) Let a1 , a2 , . . . , an be real numbers greater than or equal to 1.
Prove that
1
1
1
n
C
CC
p
:
n a a :::a C 1
a1 C 1
a2 C 1
an C 1
1 2
n
16. (IMO shortlist) Let a1 , a2 , . . . , an be real numbers in the interval 0; 12 , with
a1 C a2 C C an < 1. Prove that
1
a1 a2 : : : an Œ1 .a1 C a2 C C an /
nC1 :
.a1 C a2 C C an /.1 a1 /.1 a2 / : : : .1 an /
n
426
10 Riemann’s Integral
17. Prove Minkowski’s inequality10 : given an integer k > 1 and positive reals a1 ,
a2 , . . . , an , b1 , b2 , . . . , bn , we have
v
v
v
u n
u n
u n
uX
uX
uX
k
k
k
k
k
t
t
.ai C bi / ai C t
bki ;
iD1
with equality if and only if
iD1
a1
b1
D
a2
b2
D D
iD1
an
.
bn
10.9 Improper Integration
The purpose of this section is to extend Riemann’s integral to functions which are
not necessarily bounded and are defined in arbitrary intervals.
If I R is an interval, then I is of one of the forms ŒA; B, .A; B/, ŒA; B/, .A; B,
.A; C1/, ŒA; C1/, .1; A/, .1; A or .1; C1/ D R, with A; B 2 R. In
order to avoid the need of examining each one of these cases separately, and for
general notational convenience, we:
(i) shall refer to the real numbers A, B, or to the symbols 1, C1 as the endpoints
of I (albeit 1 and C1 are just formal symbols, and not real numbers), which
will be generically denoted by ˛ and ˇ;
(ii) impose that 1 < A; B < C1, for all A; B 2 R.
Thus, if I has endpoints ˛ and ˇ, with ˛ < ˇ, then
.˛; ˇ/ I and I n .˛; ˇ/ f˛; ˇg:
For instance, if I has endpoints ˛ D A and ˇ D C1, we have .A; C1/ I and
I n .A; C1/ fA; C1g, so that I D .A; C1/ or ŒA; C1/.
Let I be a fixed interval and f W I ! R be a given function, such that f is
integrable in each interval Œa; b I. (In particular, if I D Œa; b, then we assume
that f is integrable in I.) This implies that f is bounded in each interval Œa; b I,
albeit it may well happen that f is not bounded in I. (Nevertheless, recall that if f is
continuous in I, then, even if f is unbounded in I, Weierstrass’ Theorem 8.26 assures
that it is bounded in each interval Œa; b I.)
Definition 10.68 Let I R be an interval with endpoints ˛ and ˇ, with ˛ < ˇ,
and f W I ! R be integrable in each interval Œa; b I. We say that f is integrable
10
After Hermann Minkowski, German mathematician of the XIX and XX centuries. Minkowski
made several seminal contributions to different branches of Mathematics. His geometric method
on Algebraic Number Theory proved to be quite fruitful for the development of that theory. Also,
Minkowski’s four dimensional space is the correct setting for the geometric description of special
relativity.
10.9 Improper Integration
427
in I if, for some x0 2 .˛; ˇ/, the limits
Z
x0
lim
x!˛
Z
y
f .t/ dt and lim
f .t/ dt
y!ˇ
x
(10.67)
x0
do exist. This being the case, we define the improper integral of f in I, denoted
Rˇ
˛ f .t/ dt, by setting
Z
Z
ˇ
x!˛
˛
and say that the improper integral
Z
x0
f .t/ dt D lim
f .t/ dt;
y!ˇ
x
Rˇ
y
f .t/ dt C lim
x0
f .t/ dt converges.
˛
It is important to realize that the former definition does not depend on the chosen
x0 2 I. More precisely, suppose that x0 2 I is chosen so that the limits (10.67) do
exist, and let x00 be another point in I. Then,
Z
x00
Z
f .t/ dt D
x
so that the limit limx!˛
R x00
x
x00
lim
Analogously, limy!ˇ
Z
lim
Z
x00
lim
x!˛
R x0
0
x0
f .t/ dt D lim
Z
y
f .t/ dt D
f .t/ dt C
R x0
x00
Z
f .t/ dt C lim
y!ˇ
x
x0
Z
x00
f .t/ dt C
x
f .t/ dt:
x0
f .t/ dt does exist, with
x00
y!ˇ
Therefore, since
Z
x!˛
x00
f .t/ dt;
x0
f .t/ dt also exists, with
x
Ry
x00
f .t/ dt C
x
Z
x!˛
Z
x0
x0
x00
Z
y!ˇ
f .t/ dt:
x0
f .t/ dt D 0, we get
Z
y
x00
y
f .t/ dt C lim
x0
f .t/ dt D lim
x!˛
Z
y
f .t/ dt C lim
y!ˇ
x
f .t/ dt:
x0
On the other hand, in case I D Œa; b, we claim that the definition of integral given
in Chap. 10 coincides with the present one. In order to check that, let f W Œa; b ! R
be (bounded and) integrable in Œa; b, and fix x0 2 .a; b/. Then f is integrable in
Œa; x0 and, for x 2 .a; x0 /, we have
Z
x0
x
Z
x0
f .t/ dt D
a
Z
x
f .t/ dt f .t/ dt:
b
428
10 Riemann’s Integral
Letting M be an upper bound for jf j in Œa; b, it follows from the triangle inequality
for integrals and Proposition 10.12 that
ˇZ x
ˇ Z x
ˇ
ˇ
ˇ
ˇ
f
.t/
dt
jf .t/jdt M.x a/;
ˇ
ˇ
a
a
Rx
so that the squeezing principle gives limx!a a f .t/ dt D 0. Therefore,
Z x0
Z x
Z x0
Z x0
f .t/ dt D
f .t/ dt lim
f .t/ dt D
f .t/ dt:
lim
x!a x
x!a a
a
a
Since f is also integrable in Œx0 ; b, a similar reasoning allows us to conclude that
Z y
Z b
Z b
Z b
lim
f .t/ dt D
f .t/ dt lim
f .t/ dt D
f .t/ dt:
y!b x
0
y!b y
x0
x0
Finally, by invoking Proposition 10.21, we get
Z
lim
x!a x
x0
Z
Z
y
f .t/ dt C lim
y!b x
0
x0
f .t/ dt D
Z
Z
b
f .t/ dt C
x0
a
b
f .t/ dt D
f .t/ dt:
a
Thanks to the above discussion, we can restrict the study of improper integrals
to the cases in which I D .A; B, ŒA; B/, .1; B or ŒA; C1/, with A; B 2 R. Along
the rest of this section, we stick to the cases I D ŒA; B/ and ŒA; C1/, leaving to the
reader the task of formulating the analogous results when I D .A; B or .1; B.
Let’s start by looking at two examples.
R C1
Example 10.69 The improper integral 0 et dt does converge. More precisely,
R C1 t
we have 0 e dt D 1.
Proof For x > 0, it follows from the FTC that
Z
0
x
ˇtDx
ˇ
et dt D et ˇ
D 1 ex :
tD0
R C1
t
u
Now, Theorem 10.54 gives limx!C1 ex D 0, so that 0 et dt D 1.
R C1 ˛
Example 10.70 The improper integral 1 t dt converges if and only if ˛ < 1.
Proof For x > 1 and ˛ ¤ 1, note that
Z
x
1
t˛ dt D
1 ˛C1 ˇˇtDx
1
t ˇ
.x˛C1 1/:
D
tD1
˛C1
˛C1
Now, since x˛C1 D e.˛C1/ log x , we have (again by Theorem 10.54) that
lim x˛C1 D
x!C1
0; if ˛ C 1 < 0
:
C1; if ˛ C 1 > 0
(10.68)
10.9 Improper Integration
429
Hence, the right hand side of (10.68) has a (finite) limit when x ! C1 if and only
if ˛ C 1 < 0, i.e., if and only if ˛ < 1.
Finally, if ˛ D 1, then
Z
x
1
ˇtDx
ˇ
t1 dt D log tˇ
D log x ! C1
tD1
when x ! C1.
t
u
Also with respect to the previous example, since
x˛
D lim x˛C1 D 0
x!C1 x1
x!C1
lim
R C1
if and only if ˛ < 1, we use to say that 1 t˛ dt converges if and only if x˛
decays to zero more rapidly than x1 when x ! C1. See also Problem 3, which
R1
deals with the case 0 t˛ dt.
The coming theorem is the fundamental result on the convergence of improper
integrals. Note that it is the analogue, for improper integrals, of Theorem 7.27. For
its statement, as well as for remark just following its proof, the reader may find it
convenient to read again Problem 8, page 218.
Theorem 10.71 Let f W ŒA; C1/ ! R be integrable in each interval Œa; b ŒA; C1/. The following assertions are equivalent:
R C1
(a) A f .t/ dt converges.
ˇR
ˇ
ˇ x
ˇ
(b) For every > 0, there exists M > A such that x1 ; x2 > M ) ˇ x12 f .t/ dtˇ < .
(c) There exists L 2 R such that, for every sequence
R x .xn /n1 in ŒA; C1/ and
satisfying limn!C1 xn D C1, one has limn!C1 A n f .t/ dt D L.
Rx
Proof Let F W ŒA; C1/ ! R be defined by F.x/ D A f .t/ dt.
R C1
R C1
(a) ) (b): assume that A f .t/ dt converges, with A f .t/ dt D L. By
definition, we have limx!C1 F.x/ D L, so that, given > 0, there exists M > 0 for
which
x > M ) jF.x/ Lj <
:
2
Hence, for x1 ; x2 > M, triangle inequality gives
ˇZ
ˇ
ˇ
ˇ
x2
x1
ˇ
ˇ
f .t/ dtˇˇ D jF.x2 / F.x1 /j
jF.x2 / Lj C jL F.x1 /j
<
C D :
2
2
430
10 Riemann’s Integral
(b) ) (c): assume that condition (b) holds, and let .xn /n1 be a sequence such
that xn A for every n 1 and limn!C1 xn D C1. Given > 0, take M as in the
statement and n0 2 N such that n > n0 ) xn > M. Then, for m; n > n0 , we have
xm ; xn > M, so that
ˇZ xm
ˇ
ˇ
ˇ
ˇ
jF.xm / F.xn /j D ˇ
f .t/ dtˇˇ < :
xn
Therefore, .F.xn //n1 is a Cauchy sequence, thus convergent, by Theorem 7.27.
If L D limn!C1 F.xn /, we claim that limn!C1 F.x0n / D L for every sequence
0
.xn /n1 satisfying x0n A for n 1 and limn!C1 x0n D C1. Indeed, the discussion
in the previous paragraph (with .x0n /n1 in place of .xn /n1 ) assures the existence of
L0 D limn!C1 F.x0n /. However, if it was L0 ¤ L, then the sequence .x00n /n1 , given
for k 1 by x002k1 D x2k1 and x002k D x02k , would also converge to C1 and be
such that
lim F.x002k1 / D L and
k!C1
lim F.x002k / D L0 :
k!C1
Since L ¤ L0 , the limit limn!C1 F.x00n / would not exist, which contradicts the
argument of the previous paragraph (this time with .x00n /n1 in place of .xn /n1 ).
(c) ) (a): suppose that there exists L 2 R such that, for every sequence
.xn /n1 in ŒA; C1/ with limn!C1 xn D C1, we have limn!C1 F.xn / D L. If
limx!C1 F.x/ does not exist or is different from L, then there exist > 0 and, for
every B > A, a real number xB > B such that jF.xB / Lj . Taking B successively
equal to A C 1, A C 2, A C 3, . . . , we would get a sequence x1 , x2 , x3 , . . . satisfying
xn > A C n and jF.xn / Lj for every n 1. Therefore, limn!C1 xn D C1
but .F.xn //n1 does not converge to L, which is an absurd.
t
u
Remark 10.72 It is immediate to adapt the previous result to the case of a function
f W ŒA; B/ ! R which is integrable in each interval ŒA; b with A < b < B.
More precisely, and as the reader can easily verify, the following assertions are
equivalent:
RB
(a) A f .t/ dt converges.
(b) For
ˇR every ˇ > 0, there exists ı > 0 such that B ı < x1 ; x2 < B )
ˇ x2
ˇ
ˇ x1 f .t/ dtˇ < .
(c) There exists L 2 R such that, for everyR sequence .xn /n1 in ŒA; B/ and satisfying
x
limn!C1 xn D B, one has limn!C1 A n f .t/ dt D L.
We now establish an important consequence of the previous theorem, which is
known in mathematical literature as the comparison test for improper integrals.
Proposition 10.73 Let f ; g W ŒA; C1/ ! R be such that jf .x/j g.x/ for every
x 2 ŒA; C1/. If f and g are integrable in every interval Œa; b ŒA; C1/ and
R C1
R C1
g.t/ dt converges, then A f .t/ dt also converges and
A
ˇZ C1
ˇ Z C1
ˇ
ˇ
ˇ
f .t/ dtˇˇ g.t/ dt:
ˇ
A
A
10.9 Improper Integration
431
R C1
Proof Firstly, notice that g.x/ 0 for every x A, so that A g.t/ dt 0. Then,
R C1
given > 0, the convergence of A g.t/ dt, together with Theorem 10.71, assures
the existence of M > A such that
Z x2
x 1 ; x2 > M )
g.t/ dt < :
x1
Now, the triangle inequality for integrals, together with Proposition 10.12,
guarantees that, also for x1 ; x2 > M, we have
ˇZ
ˇ
ˇ
ˇ
x2
x1
ˇ Z
ˇ
f .t/ dtˇˇ x2
Z
jf .t/jdt x1
x2
g.t/ dt < :
x1
Then, once more from Theorem 10.71, we get the convergence of the integrals
R C1
R C1
f .t/ dt and A jf .t/j dt.
A
Finally, it follows from what we did above, together with the properties of limits
of functions (cf. Sect. 9.2) that
ˇZ
ˇ
ˇ
ˇ
A
C1
ˇZ x
ˇ
ˇ
ˇ
ˇ
Z x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
f .t/ dtˇ D ˇ lim
f .t/ dtˇ D lim ˇ
f .t/ dtˇˇ
x!C1 A
x!C1 A
Z x
Z x
lim
jf .t/jdt lim
g.t/ dt
x!C1 A
Z
x!C1 A
C1
D
g.t/ dt:
A
t
u
Remark 10.74 As in the previous remark, the former proposition can be easily
adapted to deal with the case I D ŒA; B/. More precisely, if f ; g W ŒA; B/ ! R
are integrable in each interval ŒA; b with A < b < B, and such that jf .x/j g.x/
RB
RB
for every x 2 ŒA; B/, then the convergence of A g.t/ dt implies that of A f .t/ dt;
moreover, one also has
ˇZ
ˇ
ˇ
ˇ
B
A
ˇ Z
ˇ
f .t/ dtˇˇ B
g.t/ dt:
A
Example 10.75 We shall use Proposition 10.73 to establish the convergence of
Dirichlet’s integral
Z
C1
0
sin t
dt:
t
432
10 Riemann’s Integral
To this end, start by observing that the fundamental trigonometric limit (cf.
Lemma 9.11) assures that the function t 7! sint t extends continuously to 0.
R C1 sen t
Therefore, it suffices to establish the convergence of 1
t dt.
To what is left, let x > 1. Integrating by parts, we get
Z x
Z x
cos t ˇˇtDx
sin t
cos t
dt D dt
ˇ tD1
t
t
t2
1
1
Z x
cos x
cos t
D cos 1 dt:
x
t2
1
R C1 sin t
Since j cos xj 1 for every x, the computations above guarantee that 1
dt
t
R C1 cos t
converges if and only if 1
dt converges. Hence, we have reduced our work
t2
to the convergence of this last improper integral.
ˇ xˇ
ˇ 12 and
Finally, it is enough to apply the comparison test, noticing that ˇ cos
x2
x
R C1 1
dt
does
converge.
(according to Example 10.70) 1
2
t
Problem 10, page 436, brings another proof of the convergence of Dirichlet’s
integral. On the other hand, it is possible to show (cf. [10, 20] or [27], for
instance) that
Z C1
sin t
dt D :
t
2
0
In turn, this computation is of fundamental importance to the theory of Fourier
series.11 Figure 10.10 sketches the graph of x 7! sinx x .
Back to the development of the theory, Proposition 10.73 assures that if f W
R C1
ŒA; C1/ ! R is integrable in each interval Œa; b ŒA; C1/ and A jf .t/j dt
R C1
converges, then A f .t/ dt also converges and
ˇZ C1
ˇ Z C1
ˇ
ˇ
ˇ
f .t/ dtˇˇ jf .t/j dt
(10.69)
ˇ
A
A
(recall that jf j is also integrable in each interval Œa; b ŒA; C1/, by ProposiR C1
tion 10.14). In such a case, we say that A f .t/ dt is absolutely convergent. Not
every convergent improper integral is absolutely convergent; actually, Problem 11,
R C1 sin t
page 436, shows that 0
dt is not absolutely convergent.
t
We finish this section by showing that the link between the theory of convergent
series and convergent improper integrals is actually deeper. More precisely, the
coming result establishes an equivalence between the convergence of the integral
of a positive function and that of a certain series. For this reason, it is known as the
integral test for absolute convergence of series.
11
After Jean-Baptiste-Joseph Fourier, French engineer and mathematician of the XIX century. For
more on Fourier see the footnote on page 469. A glimpse on the theory of Fourier Series is the
content of Problems 11 to 17, page 468.
10.9 Improper Integration
433
1.0
0.5
−80
−60
−40
−20
20
40
60
80
−0.5
Fig. 10.10 Graph of x 7!
sin x
x
Theorem 10.76 Let n0 2 N and f W Œn0 ; C1/ ! R be a decreasing function,
integrable in each interval Œa; b Œn0 ; C1/ and such that limx!C1 f .x/ D 0.
Then
Z
X
C1
f .t/ dt converges ,
n0
f .k/ converges:
kn0
Proof Let g; h W Œn0 ; C1/ ! R be the functions given, in the interval Œk; k C 1/ (for
each k n0 ), by g.x/ D f .k C 1/ and h.x/ D f .k/.
The conditions on f guarantee that 0 g.x/ f .x/ h.x/ for every x n0 .
Hence, the comparison test for improper integrals assures that
Z
Z
C1
C1
f .t/ dt converges )
g.t/ dt converges
n0
n0
and
Z
Z
C1
C1
h.t/ dt converges )
n0
f .t/ dt converges:
n0
On the other hand, for an integer n > n0 , we have
Z
n
g.t/ dt D
n0
n
X
kDn0 C1
Z
n
f .k/ and
h.t/ dt D
n0
n1
X
kDn0
f .k/:
434
10 Riemann’s Integral
Finally, let’s apply equivalence (a) , (c) of Theorem 10.71, with xk D k for
every k n0 :
R C1
R C1
(i) If n0 f .t/ dt converges, then n0 g.t/ dt converges and, hence
Z
Z
C1
n!C1 n0
n0
n
X
n
g.t/ dt D lim
g.t/ dt D lim
n!C1
X
f .k/ D
kDn0 C1
f .k/:
kn0 C1
P
P
P
Therefore,
kn0 f .k/ converges, with
kn0 f .k/ D f .n0 / C
kn0 C1 f .k/.
P
(ii) If kn0 f .k/ converges, then
Z
Z
C1
h.t/ dt D lim
n
n!C1 n
0
n0
n1
X
h.t/ dt D lim
n!C1
f .k/ D
kDn0
X
f .k/;
kn0
R C1
h.t/ dt converges. Hence, what we did before assures the converR C1
gence of n0 f .t/ dt.
t
u
so that
n0
Example 10.77 The integral test allows us to easily establish
P the divergence of the
harmonic series, as well as the convergence of the series k1 k1r , for every real
number r > 1. For the harmonic series, letting f W Œ1; C1/ ! R be given by
f .x/ D 1x , we have
Z
C1
1
1
dt D lim
x!C1
t
Z
x
1
1
dt D lim log x D C1:
x!C1
t
Analogously, taking f W Œ1; C1/ ! R given by f .x/ D
Z
C1
1
1
dt D lim
x!C1
tr
Z
1
x
1
1
dt D lim
r
x!C1
t
1r
Let’s now see one more interesting example.
Example 10.78 Examine the convergence of the series
1
,
xr
with r > 1, we have
1
xr1
P
1 D
1
:
r1
1
k2 k log k .
1
Solution Let f W Œ2; C1/ ! R be given by f .x/ D x log
x , so that f is continuous,
P
1
decreasing and such that limx!C1 f .x/ D 0. The integral test says that k2 k log
k
R C1
converges if and only if 2 f .t/ dt also does.
10.9 Improper Integration
435
To what is left, note that the substitution of variable s D log t (cf. Problem 2,
page 435) gives
Z
Z
C1
2
f .t/ dt D
C1
2
1
dt D
t log t
Z
C1
log 2
1
ds
s
D lim .log s log 2/ D C1:
s!C1
Hence,
R C1
2
f .t/ dt diverges, so that
P
1
k2 k log k
also diverges.
t
u
Problems: Section 10.9
1. Examine the convergence of the following improper integrals:
(a)
(b)
R
2
2
R C1
0
(c)
sec t dt.
et sin t dt.
(d)
R1
1
log t
R0C1
dt.
1
1 1Ct2
dt.
2. * Prove the following version of the change of variables theorem for
improper integrals: let f W ŒA; C1/ ! R be continuous and g W ŒB; C1/ !
ŒA; C1/ be differentiable, with limx!C1 g.x/ D C1 and g0 integrable in
R C1
each interval Œa; b ŒB; C1/. Then, g.B/ f .t/ dt converges if and only if
R C1
f .g.s//g0 .s/ ds converges. Moreover, if this is so, we also have
B
Z
Z
C1
C1
f .t/ dt D
g.B/
f .g.s//g0 .s/ ds:
B
R1
3. * Show that 0 t˛ dt converges if and only if ˛ > 1. Then, for ˛ < 0, conclude
R1 ˛
that 0 t dt converges if and only if x˛ grows to C1 more slowly than x1 as
x ! 0C.
R1
R1
4. (Berkeley) Prove that both integrals 0 cos.x2 /dx and 0 sin.x2 /dx converge.
5. (Berkeley—adapted) Show that:
R
R
(a) R0 log.sin x/dx andR 0 log j cos xjdx converge.
(b) 0 log.sin x/dx D 0 log j cos xjdx D log 2.
6. Examine the convergence of the following series, where ˛ > 0 is a given real
number:
P
1
(a)
˛.
Pk2 .log1k/
(b)
˛.
Pk2 k.log k/ 1
(c)
k2 k.log k/.log log k/˛ .
436
10 Riemann’s Integral
R C1 log t
7. Show that 0
dt converges and compute its value.
1Ct2
The next problem extends the results of Problem 12, page 381, and of
Theorem 10.63.
8. * Let I R be an interval with endpoints ˛ and ˇ, f ; g W I ! .0; C1/ be
integrable functions and p; q > 0 such that 1p C 1q D 1.
Rˇ
Rˇ
Rˇ
(a) If ˛ f .t/p dt D ˛ g.t/q dt D 1, show that ˛ f .t/g.t/ dt 1.
(b) Deduce Hölder’s inequality for integrals:
Z
ˇ
˛
Z
f .t/g.t/ dt ˇ
˛
!1=p Z
f .t/ dt
!1=q
ˇ
p
:
q
g.t/ dt
˛
9. In the proof of
Theorem 10.55,
P
we saw that in order to guarantee the existence
n
1
of limn!C1
log
n
it suffices to assure the convergence of the series
kD1 k
X 1
k1
k
log.k C 1/ C log k :
To this end, let f W Œ1; C1/ ! R be given by f .x/ D
x 1, and do the following items:
1
x
log.x C 1/ C log x for
(a) Show that f is decreasing, with limx!C1 f .x/ D 0.
(b) Apply the integral test to establish the convergence of the series.
10. The purpose of this problem is to give another proof of the convergence of the
R C1 sin t
improper integral 0
dt. To this end, do the following items:
t
P
(a) Prove that it suffices to show the convergence of the series k1 sink k .
(b) In turn, use Abel’s criterion (cf. Problem 18, page 243) to show that this
amounts
P to establishing the boundedness of the sequence .sn /n1 , such that
sn D nkD1 sin k for n 2 N.
(c) Apply the result of Problem 10, page 359, to find sn D
hence, jsn j 1
sin 12
sin
sin nC1
2
sin 12
n
2
and,
.
11. Do the following items:
(a) Show that at least one of the numbers j sin nj, j sin.n C 1/j or j sin.n C 2/j
is always greater than or equal to 12 .
P
(b) Conclude that the series k1 j sink kj diverges.
R C1 sin t
dt is not absolutely convergent.
(c) Show that 0
t
10.10 Two Important Applications
437
12. If p1 D 2 < p2 D 3 < p3 D 5 < P
is the sequence of prime numbers,
a theorem of Euler states that the series k1 p1k diverges.12 In order to show
this, do the following items:
(a) Show that .1 x/1 e2x for 0 x 12 .
(b) For n 2 N, let p1 < p2 < < pl be the prime numbers less that or equal
to n. Show that
n
l X
1
1
1 Y
1C
C 2 C :
k kD1
pk
pk
kD1
(c) Use the results of (a) and (b) to prove that
P
(d) Conclude that k1 p1k diverges.
Pn
1
kD1 k
e
Pl
2
kD1 pk
.
10.10 Two Important Applications
We recall that our initial motivation for the study of the integral was geometric,
namely, to define and compute the area of the region of the cartesian plane under the
graph of a given nonnegative (integrable) function. We could also use the concept
of integral to deal with the following geometric problems:
(I) How to define and compute the length of the graph of f W Œa; b ! R?
(II) In case f is positive, how to define and compute the volume of the solid of
revolution generated by the rotation of the region under the graph of f around
the horizontal axis?
(III) Yet in the case of a positive f , how could one define and compute the area of
the surface of revolution generated by the rotation of the graph of f around the
horizontal axis?
In this section we analyse Problem (I), leaving the analysis of Problems (II) and
(III) to [4]. As a byproduct of our discussion, we will present a thorough definition
and computation of the length of a circle.
Then, let f W Œa; b ! R be continuous. Fix (cf. Fig. 10.11) an interval Œc; d .a; b/, a partition P D fc D x0 < x1 < < xk D dg of Œc; d and let xj D xj xj1
for 1 j k.
For jPj sufficiently small, it is a reasonable guess to think of the line segment
joining points .xj1 ; f .xj1 // and .xj ; f .xj // as a good approximation for the portion
of the graph of f situated between the vertical lines x D xj1 and x D xj . If this is
so, it is also reasonable to assume that the length `j of such a segment should serve
as a good approximation for the length of this portion of the graph of f , whichever
way we define it.
12
For a sligthly different proof, see Chap. 9 of [5].
438
10 Riemann’s Integral
f
(xj−1 , f (xj−1 ))
(xj , f (xj ))
xj−1
c = x0
x
xj
d = xk
Fig. 10.11 Approximating the length of a graph
This being said, the formula for the distance between two points of the cartesian
plane gives
q
`j D .xj xj1 /2 C .f .xj / f .xj1 //2
s
f .xj / f .xj1 / 2
D 1C
xj :
xj xj1
Now, if f is continuously differentiable in .a; b/, then Lagrange’s MVT guarantees
f .x /f .x /
the existence of j 2 .xj1 ; xj / such that xj j xj1j1 D f 0 .j /; this way,
`j Š
q
1 C f 0 .j /2 xj D
q
1 C f 0 .j /2 .xj xj1 /:
Adding up the above expression for `j for 1 j k, we conclude that
k q
X
1 C f 0 .j /2 .xj xj1 /
(10.70)
jD1
would be a reasonable approximation for what we would think of as the length of
the graph of the restriction of f to the interval Œc; d. Moreover, we could equally
hope that such an approximation would get better and better as jPj ! 0.
On the other hand, since (10.70) coincides with the Riemann sum
p
†. 1 C .f 0 /2 I PI /;
it follows from Riemann’s Theorem 10.7 that
k q
X
jD1
as jPj ! 0.
Z
d
1 C f 0 .j /2 xj !
c
p
1 C f 0 .x/2 dx
10.10 Two Important Applications
439
Rdp
The discussion above allows us to conclude that c 1 C f 0 .x/2 dx is the only
reasonable way of defining the length of the graph of the restriction of f to the
interval Œc; d. If we now notice that the union of the graphs of such restrictions (as
Œc; d varies over all bounded and closed subintervals of .a; b/) coincides with the
graph of f but the points .a; f .a// and .b; f .b//, we are naturally led to state the
following
Definition 10.79 Let f W Œa; b ! R be continuous in Œa; b and continuously
differentiable in .a; b/. If the improper integral
Z
b
`D
p
1 C f 0 .x/2 dx;
(10.71)
a
does exist, then we say that the graph of f is rectifiable and has length `.
Yet with respect to the former definition, recall that
Z
b
Z
p
1 C f 0 .x/2 dx D lim
b
!0 aC
a
p
1 C f 0 .x/2 dx:
Therefore, if f is continuously differentiable in Œa; b, then Problem 11, page 380,
assures that (10.71) holds in the ordinary sense (i.e., not as an improper integral).
Example 10.80 Let be a circle of center O and radius R, and choose a cartesian
system
of coordinates such that O.0; 0/. If f W ŒR; R ! R is given by f .x/ D
p
R2 x2 , then the graph of f is the semicircle of situated on the upper halfplane of the cartesian plane. Note that f is continuous in ŒR; R and continuously
differentiable in .R; R/. Hence, according to (10.71), the length of equals twice
the value of the improper integral
Z
R
R
p
1 C f 0 .x/2 dx;
provided it exists.
To see that this is indeed the case, first note that
p
1 C f 0 .x/2 D
s
1C
x
p
R 2 x2
2
R
D p
:
2
R x2
Now, performing the trigonometric substitution x D R cos t and observing that RC
x R , arccos.1 C R / t arccos.1 R /, we get
440
10 Riemann’s Integral
Z
R
RC
Z
p
1 C f 0 .x/2 dx D
Z
D
R
RC
R
p
dx
2
R x2
arccos.1 R /
arccos.1C R /
Z
D
R
p
.R sin t/dt
R2 R2 cos2 t
arccos.1 R /
arccos.1C R /
R dt
arccos 1 D R arccos 1 C
R
R
:
Finally, since the arccos function arccos W Œ1; 1 ! Œ0; is continuous, we obtain
Z
lim
!0
p
arccos 1 1 C f 0 .x/2 dx D lim R arccos 1 C
!0
R
R
RC
R
D R .arccos.1/ arccos.1// D R:
Finally, since essentially the same argument holds for f , we conclude (as
expected) that the length of equals 2R.
In spite of the former example, the integral in (10.71) is generally difficult
(or even impossible—cf. Problem 2) to compute exactly. Let’s see an illustrative
example.
Example 10.81 Given b > 0, let’s compute the length ` of the portion of the
parabola y D x2 situated between the points .0; 0/ and .b; b2 /.
With f .x/ D x2 in (10.71) we get
Z
`D
0
b
p
1 C 4x2 dx:
In order to compute such an integral, we apply the trigonometric substitution x D
1
d
2
2
2
2 tan t. Since 1 C tan t D sec t, dt tan t D sec t and 0 x b , 0 t arctan.2b/, it follows from the formula of integration by substitution that
Z
`D
arctan.2b/ p
0
1 C tan2 t sec2 t dt D
Z
arctan.2b/
sec3 t dt:
0
Now, item (c) of Problem 15, page 412, gives
Z
sec3 t dt D
1
.sec t tan t log j sec t tan tj/ C C:
2
10.10 Two Important Applications
441
p If we set ˛ D arctan.2b/, then tan ˛ D 2b and, hence, sec ˛ D
1 C 4b2 . Therefore,
` D
1
p
1 C tan2 ˛ D
ˇ˛
ˇ
.sec t tan t log j sec t tan tj/ C C ˇ
2
p
p
1
D b 1 C 4b2 log. 1 C 4b2 2b/:
2
0
We now turn to a completely different topic, using Proposition 10.73 to introduce
and discuss some of the amazing properties of the Gamma function.
The Gamma function is the function W .0; C1/ ! .0; C1/ defined by
Z
.x/ D
C1
et tx1 dt:
0
Since
Z
C1
0
et tx1 dt D
Z
1
et tx1 dt C
0
Z
C1
et tx1 dt;
1
in order to see that .x/ is well defined it suffices to establish the convergence of
both improper integrals at the right hand side above.
To what is left to do, initially observe that, since x > 0, Problem 3, page 435,
R1
guarantees the convergence of 0 tx1 dt. Since 0 < et tx1 < tx1 for 0 < t 1,
Remark 10.74 (modified in the obvious way to be applied to the interval .0; 1)
R1
assures the convergence of 0 et tx1 dt.
On the other hand, for a fixed x > 0, we claim that et tx1 Cet=2 for t 1,
where C is a positive constant that only depends on x. Indeed, letting n > x C 1 be
a natural number, it follows from Theorem 10.54 that
tn
D 0:
t!C1 et=2
lim
Therefore, there exists C > 0 such that
so, then
tn
et=2
C for every t 1. However, if this is
et tx1 et tn Cet=2
(10.72)
for every t 1, as we wished to show.
Finally, note that an obvious variation of Example 10.69 guarantees the converR C1
gence of the improper integral 1 et=2 dt. Hence, inequality (10.72) allows us to
apply Proposition 10.73 (with A D 1, f .t/ D et tx1 and g.t/ D Cet=2 ) to conclude
R C1
that 1 et tx1 dt does converge.
As a quick look through the classics [3] and [17] makes it clear, the Gamma
function has many important applications in Mathematics, notably in Mathematical
442
10 Riemann’s Integral
Physics and Analytic Number Theory. Roughly, one can say that this is due to the
fact that it extends the notion of factorial of natural numbers to the positive reals.
In order to check the above claim, let’s start by noting that Example 10.69 gives
Z
.1/ D
C1
0
et dt D 1:
On the other hand, for x > 0 the integration by parts formula furnishes
Z
.x/ D lim
1
s!0 s
et tx1 dt C lim
Z
s
et tx1 dt
s!C1 1
Z
1 t x ˇˇtD1 1 1 t x
e tˇ C
e t dt
s!0 x
tDs
x s
Z
ˇ
1 s t x
1 t x ˇtDs
e tˇ C
e t dt
C lim
s!C1 x
tD1
x 1
Z
Z
1 C1 t x
1 1 t x
1
e t dt C
e t dt C lim es sx :
D
s!C1
x 0
x 1
x
D lim
Now, according to Theorem 10.54, we have lims!C1 1x es sx D 0. Therefore, it
follows from the above computations that
.x/ D
1
x
Z
1
0
et tx dt C
1
x
Z
C1
et tx dt D
1
1
x
Z
0
C1
et tx dt D
1
.x C 1/:
x
Then,
.x C 1/ D x .x/
for every x > 0, and an easy induction starting from .1/ D 1 gives .n/ D .n1/Š
for every n 2 N.
We now show that is continuous and log W .0; C1/ ! R is convex in
.0; C1/. Actually, since
D elog
in view of Proposition 9.68 it suffices to establish the second claim above.
The convexity of log amounts to showing that
log ..1 s/x C sy/ .1 s/ log .x/ C s log .y/;
for all x; y > 0 and 0 < s < 1. Letting s D 1q and 1 s D 1p , we have p; q > 0 such
that 1p C 1q D 1. Then, the fact that log is an increasing function that takes products
10.10 Two Important Applications
443
into sums shows that the validity of the above inequality is equivalent to that of
y
x
C
p
q
.x/1=p .y/1=q ;
(10.73)
for all x; y > 0 and all p; q > 0 such that 1p C 1q D 1.
In order to establish (10.73), we first observe that
y
x
C
p
q
Z
C1
D
t
e t
y
x
p C q 1
Z
C1
dt D
0
y1
x1
et=p t p et=q t p dt:
0
x1
Therefore, letting f .t/ D et=p t p , g.t/ D et=q t
inequality (cf. Problem 8, page 436), we get
y
x
C
p
q
Z
Z
C1
D
f .t/g.t/dt 0
Z
C1
D
et tx1 dt
and applying Hölder’s
1=p Z
C1
f .t/p dt
0
1=p Z
0
y1
q
C1
et ty1 dt
1=q
C1
g.t/q dt
0
1=q
0
D .x/1=p .y/1=q :
Remark 10.82 One can actually show that is infinitely differentiable, and that
the expression for its kth derivative can be computed by differentiating under the
integral that defines . In particular, we must have
0
Z
.x/ D
C1
et tx1 log t dt:
0
The convergence of the improper integral at the right hand side above is the object of
Problem 5, and the proof of the continuous differentiability of the Gamma function,
together with the above formula for 0 .x/, will be the object of Problems 18 and 19,
page 471.
Problems: Section 10.10
1. For x0 ¤ 0, compute the length of the portion of the catenary situated between
the points of abscissas 0 and x0 .
2. Let E be an ellipse of major axis AA0 and minor axis BB0 , with AA0 D 2a and
BB0 D 2b. If we choose a cartesian system in which A0 .a; 0/ and B0 .0; b/, it is
2
2
well known that E has equation ax2 C by2 D 1. In this respect, do the following
items:
444
10 Riemann’s Integral
p
(a) If f W Œa; a ! R is given by f .x/ D ba a2 x2 , show that the graph of f
is the portion of E situated in the upper halfplane. Then, conclude that the
length of E is given by twice the value of the improper integral
Z
s
a
1C
a
b 2 x2
dx:
x2 /
a2 .a2
p
(b) Let c D a2 b2 and x D a cos t, with 0 t . Use the formula of
integration by substitution to show that the length of E is given by
Z
2b
r
1C
0
c
b
2
sin2 t dt:
(10.74)
The definite integral of item (b) cannot be computed exactly. More generally, it
is possible to show (albeit this is well beyond the scope of these notes) that the
indefinite integral
Z p
1 C sin2 t dt;
(10.75)
where is a positive real, cannot be explicitly computed in terms of elementary
functions.13 Thanks to (10.74), one says that (10.75) is an elliptic function or a
highly transcendental function. Such functions were brought into prominence
by the German mathematician of the XIX century Carl Gustav Jacob Jacobi, and
for this reason (10.75) is also known as Jacobi’s function.
3. We are given in the plane a straight line r and a circle of radius 1, such that r
and are tangent to each other. The cycloid generated by (cf. Fig. 10.12) is
the curve described by a point P on as rolls along r without sliding. In this
respect, do the following items:
(a) Fix a position of and let O denote its point of tangency with r at this
position. Then, choose a cartesian system xOy having r as horizontal axis
P
Γ
A
t
O
t
T
Fig. 10.12 The cycloid generated by
13
A proof of this fact can be found in [12].
π
2π
x
10.10 Two Important Applications
445
and such that is in the upper half-plane. Show that, after rolls in the
positive direction (with respect to the horizontal axis) by t radians starting
from O, this point will be at position P.x.t/; y.t//, with x.t/ D t sin t and
y.t/ D 1 cos t.
(b) Note that function t 7! x.t/, from Œ0; 2 into itself, is a differentiable
bijection with inverse x 7! t.x/ differentiable in .0; 2/. Therefore, if f W
Œ0; 2 ! R is the composite f .x/ D y.t.x//, the graph of f is the portion
of the cycloid situated in the vertical strip of the cartesian plane defined by
the inequalities 0 x 2. Such a portion is called a step of the cycloid.
Compute the length of it.
p
R C1 2
4. It is possible to show that 0 es ds D 2 (see, for instance, Chap. 8 of
p
[20]). Use this fact to compute . 12 / D .
R C1
5. Show that, for x > 0, the integral 0 et tx1 log t dt is absolutely convergent.
6. Let k > 1 and n1 ; n2 : : : ; nk be naturals such that k j .n1 C n2 C C nk /. Prove
that
p
n1 C n2 C C nk
Š k n1 Šn2 Š : : : nk Š:
k
Then, if k is odd, conclude that
p
kC1
k
Š 1Š2Š : : : kŠ:
2
Chapter 11
Series of Functions
This last chapter presents the adequate notion of convergence for sequences and
series of functions, with an emphasis on the study of power series. We start by
examining a power series naturally attached to an infinitely differentiable function,
called its Taylor series. Then, in Sects. 11.2 and 11.3 we discuss the basic results on
uniformly convergent series of functions and on power series. Finally, in Sect. 11.4,
we present some applications of the theory to sequences defined by linear recurrence
relations.
11.1 Taylor Series
This section begins by introducing another important generalization of Lagrange’s
MVT, namely, Taylor formula1 with Lagrange remainder. In order to properly
state it, recall (cf. Sect. 9.3) that a function f W Œa; b ! R is n times continuously
differentiable in Œa; b if f is n times differentiable in Œa; b and f .n/ W Œa; b ! R is
continuous.
Theorem 11.1 (Taylor) Let I be an interval and f W I ! R be a function n times
differentiable in I. Given distinct x0 ; x 2 I, there exists c between x0 and x such that
f .x/ D
n1 .k/
X
f .x0 /
kD0
1
kŠ
.x x0 /k C
f .n/ .c/
.x x0 /n :
nŠ
(11.1)
After Brook Taylor, English mathematician of the XVIII century.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6_11
447
448
11 Series of Functions
Proof Firstly, assume that x0 < x and let g W Œx0 ; x ! R be given by
g.t/ D f .x/ n1 .k/
X
f .t/
kŠ
kD0
.x t/k ˛
.x t/n ;
nŠ
with ˛ 2 R chosen in such a way that g.x0 / D 0 (evidently, such a choice is always
possible).
Since f is n times differentiable in I and Œx0 ; x I, several applications of the
results of Problem 3, page 300, guarantee that g is differentiable in Œx0 ; x. Since
g.x0 / D g.x/ D 0, Rôlle’s theorem assures the existence of c 2 .x0 ; x/ such that
g0 .c/ D 0. On the other hand, a simple computation furnishes
˛ f .n/ .t/
.x t/n1 ;
.n 1/Š
g0 .t/ D
so that for c 2 .x0 ; x/ we have g0 .c/ D 0 if and only if ˛ D f .n/ .c/. Hence,
0 D g.x0 / D f .x/ n1 .k/
X
f .x0 /
kŠ
kD0
.x x0 /k f .n/ .c/
.x x0 /n ;
nŠ
as we wished to show.
Now, let x < x0 and J D x0 C x I D fx0 C x tI t 2 Ig. Then J is an interval and
I D x0 CxJ, so that we can define h W J ! R by h.s/ D f .x0 Cxs/. Corollary 9.42
assures that h is n times differentiable in J, with h.k/.s/ D .1/k f .k/ .x0 C x s/ for
0 k n. Hence, by applying (11.1) to h and to s0 < s in J, we obtain
h.s/ D
n1 .k/
X
h .s0 /
kD0
kŠ
.s s0 /k C
h.n/ .c/
.s s0 /n
nŠ
for some c 2 .s0 ; s/. In turn, this gives
f .x0 C x s/ D
n1
X
.1/k f .k/ .x0 C x s0 /
kD0
kŠ
.s s0 /k C
.1/n f .n/ .c/
.s s0 /n
nŠ
for some c 2 .s0 ; s/. Finally, letting s0 D x and s D x0 (so that s0 < s), we get (11.1)
for x < x0 .
t
u
If f W I ! R is a twice differentiable function, then, for distinct x0 ; x 2 I, Taylor’s
formula with Lagrange remainder guarantees that
00
f .x/ D f .x0 / C f 0 .x0 /.x x0 / C
for some c between x0 and x.
f .c/
.x x0 /2
2
(11.2)
11.1 Taylor Series
449
Fig. 11.1 The trapezium rule
f
a = x0
xk−1
xk
b = xn
In what follows, given a positive and twice differentiable function f W Œa; b ! R,
we use the above relation to estimate the difference between the area of the region
R under the graph of f and that of the polygonal approximation of R by the union
of the gray trapezoids of Fig. 11.1. For this reason, the coming result is known in
mathematical literature as the trapezium rule.
Example 11.2 Let f W Œa; b ! R be a twice differentiable (but not necessarily
positive) function. If fa D x0 < x1 < < xn D bg is a uniform partition of
interval Œa; b, then
ˇZ
ˇ
n
ˇ b
X
1
b a ˇˇ .b a/3
ˇ
.f .xk / C f .xk1 //
f .x/ dx sup jf 00 j:
ˇ
ˇ
2
ˇ a
ˇ
2
n
4n
Œa;b
kD1
Proof For 1 k n, let gk W Œxk1 ; xk ! R be given by
Z
gk .x/ D
1
f .t/dt .f .x/ C f .xk1 //.x xk1 /:
2
xk1
x
Letting denote the expression at the right hand side of the stated inequality, the
additivity of the integral, together with the fact that xk xk1 D ba
n for 1 k n,
gives
ˇ n Z
ˇˇ
ˇX
xk
1
ˇ
ˇ
Dˇ
f .x/ dx .f .xk / C f .xk1 //.xk xk1 / ˇ
ˇ
ˇ
2
xk1
kD1
ˇ n
ˇ
n
ˇX
ˇ X
ˇ
ˇ
Dˇ
gk .xk /ˇ jgk .xk /j;
ˇ
ˇ
kD1
(11.3)
kD1
where we used triangle inequality in the last passage above.
Now, since f is twice differentiable, the same happens with gk . Moreover,
gk .xk1 / D 0 and easy computations furnish
1
1
g0k .x/ D f .x/ f 0 .x/.x xk1 / .f .x/ C f .xk1 //
2
2
450
11 Series of Functions
and
1
g00k .x/ D f 00 .x/.x xk1 /;
2
so that g0k .xk1 / D g00k .xk1 / D 0. Hence, Taylor’s formula with Lagrange remainder
guarantees the existence of ck 2 .xk1 ; xk / such that
gk .xk / D gk .xk1 / C g0k .xk1 /.xk xk1 / C
g00k .ck /
.xk xk1 /2
2
1
D f 00 .ck /.ck xk1 /.xk xk1 /2 :
4
However, since xk xk1 D
jgk .xk /j D
ba
n ,
it follows from the above computations that
1 00
jf .ck /jjck xk1 j.xk xk1 /2
4
1 00
.b a/3
jf .ck /jjxk xk1 j3 D jf 00 .ck /j
:
4
4n3
Finally, by substituting this last expression in estimate (11.3) for , we get
n
X
kD1
jgk .xk /j n
X
jf 00 .ck /j
kD1
n
.b a/3
.b a/3 X 00
D
jf .ck /j
4n3
4n3 kD1
.b a/3
.b a/3
n sup jf 00 j D
sup jf 00 j:
3
2
4n
4n
Œa;b
Œa;b
t
u
In words, the result of the previous example gives a numerical approximation,
with controlled error, for the computation of an integral. As the discussion of the
computation of the length of an ellipse shows (cf. Problem 2, page 443), sometimes
this is the best one can get. In this respect, see also Problem 3 of this section.
For more on the use of Calculus methods to get useful numerical approximations,
see Chap. 15 of [2] or Part B of [7].
Back to the development of the theory, let f W I ! R be infinitely differentiable.
For x0 2 I, we say that
X f .k/ .x0 /
k0
kŠ
.x x0 /k
is the Taylor series of f at x0 .
The coming result gives a necessary and sufficient condition for the Taylor series
of f at x0 to converge to f in I.
11.1 Taylor Series
451
Proposition 11.3 Let I be an interval and f W I ! R be an infinitely differentiable
function. If there exists a constant C 0 such that jf .n/ .x/j Cn for every n 1
and every x 2 I, then, given x0 2 I, we have
f .x/ D
X f .k/ .x0 /
k0
kŠ
.x x0 /k
for every x 2 I.
Proof Fixed x 2 I and n 2 N, it follows from Taylor formula with Lagrange
remainder that
f .x/ D
n1 .k/
X
f .x0 /
kŠ
kD0
.x x0 /k C
f .n/ .c/
.x x0 /n
nŠ
(11.4)
for some c between x0 and x. Now,
ˇ
ˇ .n/
ˇ .Cjx x0 j/n n
ˇ f .c/
nˇ
ˇ
.x
x
/
! 0;
0
ˇ
ˇ nŠ
nŠ
where we used the result of Problema 3, page 218, in the last passage above.
Therefore, it suffices to let n ! C1 in (11.4) to get the desired result.
t
u
Example 11.4 Since j sin.2j/ xj D j sin xj 1 and j sin.2j1/ xj D j cos xj 1, the
previous proposition gives
sin x D
X sin.k/ 0
k0
kŠ
xk D
X .1/j1
x2j1
.2j
1/Š
j1
for every x 2 R. Analogously,
cos x D
X .1/j
j0
.2j/Š
x2j ;
for every x 2 R. In Sect. 11.5, these series expansions will be the departure points
for the rigorous construction of the sine and cosine functions.
We now apply Taylor formula with Lagrange remainder to show that the Taylor
series of the exponential function converges to such a function along the whole real
line.
Theorem 11.5 For x 2 R, one has
ex D
X 1
xk :
kŠ
k0
(11.5)
452
11 Series of Functions
Proof For a fixed x 2 R, it follows from Theorem 11.1 (with x0 D 0) that
ex D
n1 k
X
x
kD0
kŠ
xn
;
nŠ
C ec (11.6)
for some c between 0 and x. Now, using the fact that the exponential function is
increasing, together with the result of Problem 3, page 218, we obtain
ˇ c ˇ
n
n
ˇ e nˇ
n
ˇ x ˇ D ec jxj maxfe0 ; ex g jxj !
0:
ˇ nŠ ˇ
nŠ
nŠ
n
Hence, the squeezing principle gives limn!C1 ec xnŠ D 0. Finally, letting n ! C1
in (11.6), we get the desired result.
t
u
Note that the previous result largely generalizes Example 7.41. For future
reference, we observe that changing x by ax in (11.5), we obtain
eax D
X ak
k0
kŠ
xk
(11.7)
for every x 2 R.
.n/
Sometimes, estimates of the error f nŠ.c/ .x x0 /n in (11.1) as simple as those
we did above do not suffice to guarantee that the Taylor series of an infinitely
differentiable function converges to such a function at all points it could do. We
now look at such an example.
Example 11.6 Function x 7! log.1 C x/ is defined in the whole interval .1; 1/
and is infinitely differentiable there. Letting f .x/ D log.1 C x/ for jxj < 1, it’s
k1
immediate to verify that f .k/ .x/ D .1/.1Cx/.k1/Š
for every k 2 N. Hence, f .k/ .0/ D
k
.1/k1 .k 1/Š, and it immediately follows that the Taylor series of log.1 C x/
centered at 0 is given by
X .1/k1
k1
k
xk :
(11.8)
ˇ
ˇ
P
k1
ˇ
ˇ
Since ˇ .1/k xk ˇ jxjk and k1 jxjk converges when jxj < 1, the comparison
test for series assures that (11.8) converges whenever x 2 .1; 1/. Nevertheless,
if we try to estimate the error in (11.1) as we did before, we will only be able to
conclude that
log.1 C x/ D
X .1/k1
k1
k
xk
11.1 Taylor Series
453
for x 2 Π12 ; 1/. Indeed, it follows from (11.1) that, for 0 < jxj < 1, we have
log.1 C x/ D
n1
X
.1/k1
k
kD0
xk C
.1/n1
xn
n.1 C c/n
for some c between 0 and x.
If 0 < x < 1, then 0 < c < x < 1 and we get
ˇ
ˇ
ˇ .1/n1
ˇ
nˇ
n n
ˇ
ˇ n.1 C c/n x ˇ x ! 0:
If 1 < x < 0, then 1 < x < c < 0, so that 1 C c > 1 C x > 0. Therefore,
ˇ
ˇ
ˇ
ˇ .1/n1
1
jxjn
jxjn
1
1
jxj n
nˇ
ˇ
D
x
D
:
ˇ
ˇ n.1 C c/n
n .1 C c/n
n .1 C x/n
n 1Cx
For 1 < x < 0, one easily concludes that
interval, the last estimate above gives
jxj
1Cx
< 1 , 12 x < 0; yet in this
ˇ
ˇ
ˇ .1/n1
ˇ 1 n
nˇ
ˇ
x
ˇ n.1 C c/n
ˇ n ! 0:
On the other hand, for 1 < x < 12 we have
1
n
jxj
1Cx
n
jxj
1Cx
> 1, so that
n
! C1I
ˇ
ˇ
n1
ˇ
ˇ
xn ˇ ! 0
hence, in the interval 1; 12 we are not able to conclude that ˇ .1/
n.1Cc/n
when n ! C1.
We shall remedy situations like that of the previous example in Sect. 11.3. For the
time being, let us show that it may well happen that the Taylor series of an infinitely
differentiable function f W R ! R, centered at some x0 2 R, does not converge to f
in any open interval centered at x0 .
Example 11.7 Let f W R ! R be given by
f .x/ D
e1=x ; if x > 0
:
0; if x 0
Then, f is infinitely differentiable in R, with f .k/ .0/ D 0 for every k 2 ZC . In
particular, the Taylor series of f centered at 0 vanishes identically.
454
11 Series of Functions
Proof Evidently, limx!0C e1=x D 0, so that f is continuous at 0, thus, in the whole
real line. Also evidently, f is infinitely differentiable in .0; C1/ and in .1; 0/,
with f .n/ .x/ D 0 if x < 0 and n 2 ZC .
We claim that, for n 2 ZC , there exists a polynomial pn such that
f .n/ .x/ D pn .1=x/e1=x
(11.9)
for every x > 0. Let’s check this by induction on n 0, the case n D 0 being
obvious. By induction hypothesis, assume that for some integer k 0 there exists
a polynomial pk such that f .k/ .x/ D pk .1=x/e1=x for every x > 0. Then, the chain
rule gives, for x > 0,
f
.kC1/
1
.x/ D 2
x
1
1
0
pk
e1=x D pkC1 .1=x/e1=x ;
pk
x
x
with pkC1 .x/ D x2 .pk .x/ p0k .x//, so that pkC1 is again a polynomial.
Now that we have (11.9), let’s make one more induction to prove that f is
infinitely differentiable at 0, with f .n/ .0/ D 0 for every n 2 ZC . By induction
hypothesis, assume that f is k times differentiable at 0, with f .k/ .0/ D 0. Since f is
infinitely differentiable in Rnf0g, in order to show that f is kC1 times differentiable
at 0, with f .kC1/ .0/ D 0, it suffices to show that f .kC1/ .0/ D 0. To this end, note that
for h > 0 we have
pk .1=h/e1=h
f .k/ .h/ f .k/ .0/
D
D qk .1=h/e1=h ;
h
h
where qk .x/ D xpk .x/, a polynomial; then, again by Theorem 10.54, we have
f .k/ .h/ f .k/ .0/
qk .1=h/
qk .y/
D lim
D lim
D 0:
1=h
h!0C
h!0C
y!C1 ey
h
e
lim
.k/
On the other hand, for h < 0 we have limh!0 f .h/f
h
f .k/ .0/ D 0. Therefore, f .kC1/ .0/ exists and equals 0.
.k/ .0/
D 0, since f .k/ .h/ D
t
u
Problems: Section 11.1
1. Let I be an open interval and f W I ! R be convex and twice differentiable in I.
If x0 2 I and r denotes the tangent to the graph of f at the point .x0 ; f .x0 //, show
that no point on the graph of f lies under r.
2. Let I be an open interval, f W I ! R be n times continuously differentiable and
x0 2 I be such that f 0 .x0 / D f 00 .x0 / D D f .n1/ .x0 / D 0. If n is even and
11.1 Taylor Series
455
f .n/ .x0 / > 0 (resp. f .n/ .x0 / < 0), show that x0 is a point of strict local minimum
(resp. maximum) for f , i.e.,
x 2 I n fx0 g sufficiently close to x0 ) f .x/ > f .x0 / .resp.; f .x/ < f .x0 //:
3. Let `.E/ be the length of the ellipse E of major axis 2a, minor axis 2b and focal
distance 2c. Show that, for every n 2 N, one has
ˇ
0
1ˇ
s
ˇ
ˇ
n1
X
ˇ
j Aˇˇ 2. C 1/b 3
ˇ`.E/ 2b @1 C
1 C 2 sin2
;
ˇ
ˇ
n
n
n2
ˇ
ˇ
jD1
where D bc .
4. Obtain the Taylor series of the functions sinh and cosh. In each case, show that it
converges to the corresponding function in the whole real line.
5. Let I be an interval and f W I ! R be n times differentiable and such that f .n/ is
constant. Prove that f either vanishes identically or is a polynomial of degree at
most n. More precisely, if x; x0 2 I, show that
f .x/ D
n
X
f .j/ .x0 /
jD0
jŠ
.x x0 /j :
6. Use (11.6) to give another proof of Theorem 10.54.
The next problem revisits the analysis of Newton’s method for numerical
approximations of roots of differentiable functions (cf. Problem 9, page 300).
7. Let f W Œa; b ! R be continuous in Œa; b and twice continuously differentiable
in .a; b/, with f 0 ; f 00 > 0 in .a; b/. Assume that f .a/ < 0 < f .b/, and let ˛ be the
only root of f in the interval Œa; b. Do the following items:
(a) If ˇ 2 .˛; b and D ˇ f .ˇ/
f 0 .ˇ/ ,
show that ˛ < ˇ.
(b) If .an /n1 is such that a1 2 .˛; b and anC1 D an ff0.a.ann// for every n 1,
show that .an /n1 converges and an ! ˛ as n ! C1.
(c) Refine the analysis of item (b) in the following way:
(i) Show that there exists n 2 .˛; an / such that f .an / D f 0 .an /.an ˛/ 1 00
f .n /.an ˛/2 .
2
00
f 00 .n /
n/
2
2
(ii) Conclude that 0 anC1 ˛ D 2ff 0.
.an / .an ˛/ 2f 0 .n / .an ˛/ .
(iii) Suppose that a < c < d < b satisfy d c < 1 and f .c/ < 0 < f .d/.
00
If D maxŒc;d 2ff 0 and we start with a1 D d, use (ii) to successively
n
conclude that 0 anC1 ˛ .an ˛/2 and 0 an ˛ n .d c/2
for every n 1.
Yet with respect to the previous problem, note that Problem 10, page 219,
n
guarantees that n .d c/2 ! 0 as n ! C1. Hence, the result of item (iii)
above estimates the error with which an approaches ˛. Note also that we can
get results similar to those of the previous problem by assuming that f 0 > 0 and
456
11 Series of Functions
f 00 < 0 in .a; b/; it suffices to start with a1 2 Œa; ˛/. The same is true if we assume
that f 0 < 0 and f 00 < 0 (resp. f 00 > 0) in .a; b/; it suffices to start with a1 2 .˛; b
(resp. a1 2 Œa; ˛/).
8. The polynomial f .x/ D x3 2x 5 is such that f .2/ D 1 and f . 52 / D 92 .
Therefore, the IVT guarantees the existence of a root ˛ 2 .2; 52 / for f . Apply the
results of the previous problem to estimate ˛ with five correct decimal places.
11.2 Series of Functions
The material on this section is a prelude to the study of power series and extends, to
sequences and series of functions, some concepts and results of Sects. 7.2 and 7.4.
We start by defining the concept of limit of a sequence of functions.
In all that follows, unless stated otherwise, I denotes an interval.
Definition 11.8 For each n 2 N, let a function fn W I ! R be given. If
limn!C1 fn .x/ exists for each x 2 I, we define the pointwise limit of the sequence
of functions .fn /n1 as the function f W I ! R such that
f .x/ D lim fn .x/
n!C1
(11.10)
for every x 2 I. In this case, we also say that .fn /n1 is a pointwise convergent
sequence of functions, or that .fn /n1 converges pointwise to f .
The pointwise limit of a sequence fn W I ! R of functions, if exists, is unique.
Indeed, if .fn /n1 converges pointwise to f W I ! R, it follows from (11.10) that
for each x 2 I the sequence of real numbers .fn .x//n1 converges to f .x/; hence,
the uniqueness of the limit of a convergent sequence of reals (cf. Proposition 7.14)
assures that there exists only one possible value for f .x/.
Example 11.9 We collect here two examples illustrating the fact that the pointwise
limit of a pointwise convergent sequence of functions is not necessarily a well
behaved function.
(a) For each n 2 N, let fn W Œ0; 1 ! R be such that fn .x/ D xn for every x 2 Œ0; 1. It
follows from Example 7.12 that the pointwise limit of .fn /n1 exists and is the
function f W Œ0; 1 ! R such that
f .x/ D
0; if x 2 Œ0; 1/
:
1; if x D 1
In particular, this example shows that the (pointwise) limit of a pointwise
convergent sequence of continuous (actually, even infinitely differentiable)
functions can be discontinuous.
11.2 Series of Functions
457
(b) For each n 2 N, let fn W Œ0; 1 ! R be given by fn .x/ D nx.1 x2 /n for every
x 2 Œ0; 1. Since 0 < x < 1 ) 0 < 1 x2 < 1, Problem 4, page 218, assures
that limn!C1 nx.1 x2 /n D 0 if 0 < x < 1. Also, since fn .0/ D fn .1/ D 0
for every n 2 N, we conclude that the pointwise limit of the sequence .fn /n1
exists and equals the function f W Œ0; 1 ! R that vanishes identically in Œ0; 1;
R1
in particular, f is integrable, with 0 f .x/dx D 0. Now, note that the FTC gives
Z
1
0
Hence, limn!C1
fn .x/dx D R1
0 fn .x/dx
D limn!C1
Z
lim
n.1 x2 /nC1 ˇˇxD1
n
:
D
ˇ
2.n C 1/ xD0
2.n C 1/
n!C1 0
1
n
2.nC1/
Z
fn .x/dx ¤
1
D 12 , so that
lim fn .x/dx:
0 n!C1
We now introduce a notion of convergence for sequences of functions which is
stronger than that of pointwise convergence. As we shall see right after the coming
definition, under such more restrictive notion of convergence pathologies like those
of the previous will not take place.
Definition 11.10 A sequence .fn /n1 of functions fn W I ! R converges uniformly
for a function f W I ! R if the following condition is satisfied: given > 0, there
exists n0 2 N such that
n n0 ) jfn .x/ f .x/j < ; 8 x 2 I:
(11.11)
In words, a sequence .fn /n1 of real functions defined in I converges uniformly
to f W I ! R if the choice of a sufficiently large index n (n n0 , in the notations of
the former definition) makes jfn .x/ f .x/j < for every x 2 I.
x
n
Example 11.11 For n 2 N, let fn W R ! R be given by fn .x/ D . 1Cx
2 / . Then,
.fn /n1 converges uniformly to the function that vanishes identically in R.
jxj
1
1
Proof Since 1Cx
2 2 for every x 2 R, we have jfn .x/ 0j D jfn .x/j 2n for every
x 2 R. In view of this inequality and given > 0, choose n0 2 N such that 21n < for n n0 . Then, for n n0 we have jfn .x/ 0j < for every x 2 R, as wished. u
t
The coming corollary is an immediate consequence of Definition 11.10.
Corollary 11.12 If a sequence .fn /n1 of functions fn W I ! R converges uniformly
to f W I ! R, then .fn /n1 converges pointwise to f .
The concept of uniform convergence has the following geometric interpretation:
since
jfn .x/ f .x/j < , f .x/ < fn .x/ < f .x/ C ;
458
11 Series of Functions
Fig. 11.2 Geometric
interpretation of uniform
converge
y
f+
f
f−
fn
a
b
x
we conclude that .fn /n1 converges uniformly to f W I ! R if, given arbitrarily
> 0, there exists n0 2 N such that, for n n0 , the graph of fn is contained in
the strip of the plane of width 2 centered in the graph of f (the gray region of
Fig. 11.2).
The actual importance of the concept of uniform convergence lies in the coming
Convergence Theorems 11.13 and 11.14. In particular, comparing such results with
the sequences of functions collected in Example 11.9, we conclude that the converse
of Corollary 11.12 is not necessarily true. In other words, we conclude that there
exist sequences of functions fn W I ! R such that .fn /n1 converges pointwise, but
not uniformly, to some function f W I ! R.
Theorem 11.13 If a sequence .fn /n1 of continuous functions fn W I ! R converges
uniformly to f W I ! R, then f is continuous.
Proof Given x0 2 I and n 2 N, the triangle inequality gives
jf .x/ f .x0 /j jf .x/ fn .x/j C jfn .x/ fn .x0 /j C jfn .x0 / f .x0 /j:
Now, for a given > 0, the uniform convergence of .fn /n1 to f guarantees the
existence of n0 2 N such that n n0 ) jfn .x/ f .x/j < 4 for every x 2 I. Hence,
by writing the previous inequality for n D n0 we get
jf .x/ f .x0 /j 2 C jfn0 .x/ fn0 .x0 /j D C jfn0 .x/ fn0 .x0 /j:
4
2
On the other hand, the continuity of fn0 assures the existence of ı > 0 such that
x 2 I; jx x0 j < ı ) jfn0 .x/ fn0 .x0 /j <
:
2
Therefore, for x 2 I such that jx x0 j < ı, we have
jf .x/ f .x0 /j C jfn0 .x/ fn0 .x0 /j < C D ;
2
2
2
so that f is continuous at x0 .
Finally, since x0 was chosen arbitrarily in I, we conclude that f is continuous in I.
t
u
11.2 Series of Functions
459
Theorem 11.14 Let .fn /n1 be a sequence of continuous functions fn W Œa; b ! R,
converging uniformly to f W Œa; b ! R. If gn ; g W Œa; b ! R are defined by
Z x
Z x
gn .x/ D
fn .t/dt and g.x/ D
f .t/dt;
a
a
then .gn /n1 converges uniformly to g. In particular,
Z
Z
b
b
f .x/dx D lim
fn .x/dx:
n!C1 a
a
(11.12)
Proof For x 2 Œa; b, it follows from the triangle inequality for integrals (cf.
Proposition 10.14) that
ˇZ x
ˇ Z x
ˇ
ˇ
ˇ
jgn .x/ g.x/j D ˇ .fn .t/ f .t//dtˇˇ jfn .t/ f .t/jdt:
a
a
Now, given > 0, the uniform convergence of .fn /n1 to f gives n0 2 N such that
n n0 ) jfn .t/ f .t/j ba
for every t 2 Œa; b. Hence, for n n0 and x 2 Œa; b, the above inequalities allow us
to estimate
Z x
jgn .x/ g.x/j dt D
.x a/ .b a/ D :
b
a
b
a
b
a
a
For the second part, recall that uniform convergence implies pointwise convergence. Therefore, gn .b/ ! g.b/ as n ! C1, which is the same as (11.12).
t
u
Problem 5 extends the above result to the realm of integrable functions. Actually,
the following important remark holds true.
Remark 11.15 In the notations of the previous result, a much more general result
is available. In order to state it properly, let I R be an interval with endpoints
˛ and ˇ (in the sense of Sect. 10.9) and fn W I ! R be given integrable functions
(in the improper sense, if I or f is unbounded), with .fn /n1 converging pointwise
to an integrable function f W I ! R. If there exists an integrable function F W I !
Œ0; C1/ such that
jfn .x/j F.x/; 8 x 2 I; n 2 N;
then it is possible to show that
Z
ˇ
˛
Z
f .x/dx D lim
n!C1 ˛
ˇ
fn .x/dx:
(11.13)
460
11 Series of Functions
This is the content of Lebesgue’s dominated convergence theorem (we abbreviate DCT), whose proof is beyond the scope of these notes. For the interested
reader, we refer to [20] or [27].
Also with respect to Lebesgue’s DCT, it is worth observing that, if I D Œa; b, it
suffices to assume that there exists L > 0 such that
jfn .x/j L; 8 x 2 I; n 2 N:
On the other hand, item (b) of Example 11.9 shows that such a condition is necessary
for Lebesgue’s DCT to hold true.
Finally, Problem 4 shows that proving (11.12) in a particular case and without
the aid of Lebesgue’s DCT can be a somewhat difficult task. Problem 18 asks you
to prove a very important consequence of Lebesgue’s DCT.
We continue our study of uniform convergence by presenting the famous
Weierstrass approximation theorem, that states that every continuous function
defined in a closed and bounded interval can be uniformly approximated by a
sequence of polynomials.
Theorem 11.16 (Weierstrass) Every continuous function f W Œa; b ! R is the
uniform limit of a sequence of polynomial functions pn W Œa; b ! R.
Before we jump into the proof, it is worth to do some heuristics to motivate it.
For n 2 N, the binomial theorem gives
!
n
X
n
n k
f .x/ D f .x/ x C .1 x/ D f .x/
x .1 x/nk
k
kD0
Letting a0 D a < a1 < a2 < : : : < an D b, with ak D a C nk .b a/, the uniform
continuity of f guarantees that, for large n, the values f .x/ for x 2 Œak1 ; ak do not
differ too much from f .ak /. Since every x 2 Œa; b belongs to one of the intervals
Œak1 ; ak , we expect that
!
n k
f .x/ Š
f .ak /
x .1 x/nk
k
kD0
n
X
for large n, and the right hand side is a polynomial function. This is precisely what
we shall prove.
Proof To make the forthcoming computations a little simpler, assume first that a D
0 and b D 1, and let
!
n
k n
X
pn .x/ D
xk .1 x/nk :
f
n
k
kD0
11.2 Series of Functions
461
Since f is uniformly continuous in Œ0; 1, given > 0 we can choose ı > 0
such that jf .x/ f .y/j < whenever x; y 2 Œ0; 1 satisfy jx yj < ı. Letting
M D maxfjf .t/jI t 2 Œ0; 1g, we have for a fixed x 2 Œ0; 1 that
!
ˇ
n k
ˇ
x .1 x/nk ˇ
k
!
k ˇ n
X ˇˇ
ˇ
xk .1 x/nk
ˇ
ˇf .x/ f
k
n
0kn
n ˇX
k
ˇ
jf .x/ pn .x/j D ˇ
f .x/ f
n
kD0
jx nk j<ı
!
k ˇ n
X ˇˇ
ˇ
xk .1 x/nk
C
ˇ
ˇf .x/ f
n
k
0kn
jx nk jı
X
0kn
jx nk j<ı
!
!
X
n k
n k
nk
x .1 x/
C
2M
x .1 x/nk :
k
k
0kn
jx kn jı
In the last line above, the first summand does not exceed
!
n
X
n
n k
x .1 x/nk D x C .1 x/ D :
k
kD0
On the other hand, the difficulty in estimating the second summand lies in
estimating how many integers 0 k n satisfy the inequality jx nk j ı. We
2
overcome this by inserting the factor ı12 x nk 1 under the † sign to get
2M
X
0kn
jx nk jı
!
n
k
2M X n k
nk
x .1 x/
x
2
k
ı kD0
n
2
!
n k
x .1 x/nk :
k
Letting
n X
k
SD
x
n
kD0
2
!
n k
x .1 x/nk ;
k
we proved that
jf .x/ pn .x/j C
for every x 2 Œ0; 1.
2M
S
ı2
(11.14)
462
11 Series of Functions
2
Substituting x nk D x2 2k
x
n
C
k 2
n
we compute
!
!
n
n
X
X
k n k
n k
nk
x .1 x/
x .1 x/nk
2x
S Dx
k
n
k
kD0
kD0
!
n
X k2 n
xk .1 x/nk
C
2 k
n
kD0
!
!
n
n
X
X
k n1 k
n 1 k1
2
2
nk
x .1 x/
x .1 x/nk
D x 2x
C
k
1
n
k
1
kD1
kD1
!
!
n
n
X k1 n1
X
1 n1 k
2
2
k
nk
x .1 x/
x .1 x/nk
D x 2x C
C
n
k
1
n
k
1
kD1
kD1
!
n
n 1 X
x
k1 n1 k
x .1 x/nk C
D x2 C
n
n1 k1
n
kD1
!
n
n 1
X
x
n 2 k2
D x2 C
x .1 x/nk C
x2
k2
n
n
kD2
2
D x2 C
n 1
n
x2 C
1
x
D x x2 ;
n
n
1
on the interval Œ0; 1.
so that S 4n
Therefore, back to (11.14) we get
jf .x/ pn .x/j C
M
;
2nı 2
for every x 2 Œ0; 1. Now, it suffices to see that this last expression is less than 2,
M
provided n > 2ı
2.
For the general case,let f W Œa; b ! R be a continuous function. Let g W Œ0; 1 !
R be given by g.x/ D f .1 x/a C xb , so that g is also continuous. What we did so
far guarantees the existence of a sequence .qn /n1 of polynomial functions such that
ya n
qn ! g uniformly on Œ0; 1. If pn .y/ D qn ba
, then pn is obviously a polynomial
function and, for y 2 Œa; b,
ˇ y a
y a ˇ
ˇ
ˇ
jf .y/ pn .y/j D ˇg
qn
ˇ:
ba
ba
n
Thus, pn ! f uniformly on Œa; b.
t
u
11.2 Series of Functions
463
Example 11.17 Let f W Œa; b ! R be a continuously differentiable function. Prove
n
that there exists a sequence .pn /n1 of real polynomials such that pn ! f and
n
p0n ! f 0 uniformly on Œa; b.
Proof By the Weierstrass approximation theorem, we can choose a sequence
n
.qn /n1 of polynomial functions such that qn ! f 0 uniformly on Œa; b. Letting
Z
x
pn .x/ D
qn .t/dt C f .a/
a
for x 2 Œa; b, we get a polynomial function satisfying pn .a/ D f .a/ and p0n D qn for
every n 2 N.
Hence, we can use the FTC again to write
Z
f .x/ pn .x/ D
x
f 0 .t/dt Z
a
Z
x
x
qn .t/dt D
a
f 0 .t/ qn .t/ dt
a
for every x 2 Œa; b, so that
Z
x
jf .x/ pn .x/j jf 0 .t/ qn .t/jdt a
Z
b
jf 0 .t/ qn .t/jdt:
a
Since qn ! f 0 uniformly as n ! C1, given > 0 we can find n0 2 N such
that jf 0 .t/ qn .t/j < for every t 2 Œa; b and every n > n0 . Therefore, for n > n0 ,
we have
Z
b
jf .x/ pn .x/j dt D .b a/
a
for every x 2 Œa; b.
t
u
We now turn our attention to the study of series of functions.
Definition 11.18 Given
P a sequence .fn /n1 of functions fn W I ! R, we define the
series of functions k1 fP
k as a shorthand for the sequence .sn /n1 of functions
sn W I ! R, such that sn D nkD1 fk for every n 1.
P
In the notations of the former definition, we say that k1 fk converges pointwise
(resp. uniformly) in I to f W I ! R if the sequence .sn /n1 converges pointwise (resp.
uniformly) to f . In this case, we write
f D
X
fk
k1
and note that f .x/ D
P
k1 fk .x/
for every x 2 I.
464
11 Series of Functions
We shall generally apply Theorems 11.13 and 11.14 to uniformly convergent
series of functions. In this sense, the coming corollary is of paramount importance
for us. Note that, in words, its item (b) says that uniformly convergent series of
continuous functions can be integrated termwise.
CorollaryP
11.19 For each n 1, let fn W Œa; b ! R be a continuous function. If
the series k1 fk converges uniformly to f W Œa; b ! R, then:
(a) Rf isP
continuous.
P Rb
b
(b) a k1 fk .x/dx D k1 a fk .x/dx.
P
Proof (a) If sn D nkD1 fk , then sn is a finite sum of continuous functions, so that it
is itself continuous. Since sn ! f uniformly, Theorem 11.13 guarantees that f
is a continuous function.
(b) Since sn ! f uniformly, Theorem 11.14, together with the additivity of the
integral, gives
Z
Z
b
a
n!C1 a
D lim
n!C1
Z
b
f .x/dx D lim
n!C1 a
n Z
X
kD1
b
sn .x/dx D lim
a
b
fk .x/dx D
XZ
k1
n
X
fk .x/dx
kD1
b
fk .x/dx:
a
t
u
The former corollary will only be useful if we have an efficient way of finding
out, in cases of interest, whether or not a given series of functions is uniformly
convergent. The coming result, known in mathematical literature as Weierstrass Mtest, provides a simple sufficient condition for the uniform convergence of a series
of functions. Note that the “M” stands for majorization.
P
Theorem 11.20 (Weierstrass M-Test) Let k1 fk be a series of functions defined
in an interval I and satisfying the following conditions:
(a) For each k P
1, there exists Mk > 0 such that jfk .x/j Mk for every x 2 I.
(b) The series k1 Mk converges.
P
uniformly in I. In particular, if all of
Then, the series of functions k1 fk converges
P
the fk ’s are continuous in I, then so is k1 fk .
P
Proof Given x 2 I, since jfk .x/j Mk for every n 1 and k1 Mk converges,
the comparison test for series of real
P numbers (cf. Proposition 7.44) guarantees the
absolute convergence of the series k1 fk .x/. Hence, we get a well defined function
f W I ! R such that
f .x/ D
X
k1
fk .x/; 8 x 2 I:
11.2 Series of Functions
465
P
P
In order to establish the uniform convergence of k1 fk , let sn D nkD1 fk . For
x 2 I and n 2 N, we have
ˇ
ˇ
ˇX
ˇ X
X
ˇ
ˇ
fk .x/ˇ jfk .x/j Mk :
(11.15)
jf .x/ sn .x/j D ˇ
ˇ
ˇ
k>n
k>n
k>n
Pn
P
Now, given > 0, the convergence of the sequence
kD1 Mk n1 to
k1 Mk
assures the existence of n0 2 N such that
ˇ
ˇ
ˇ
ˇ
n
X
ˇX
ˇ
ˇ
ˇ
M
M
k
k ˇ < ; 8 n n0 :
ˇ
ˇ k1
ˇ
kD1
ˇ
ˇP
Hence, ˇ k>n Mk ˇ < for every n n0 .
Back to (11.15) we conclude that
jf .x/ sn .x/j X
Mk < k>n
for n n0 and every x 2PI. Therefore, .sn /n1 converges uniformly to f , and this is
the same as saying that k1 fk converges uniformly to f .
The last part follows at once from Theorem 11.13.
t
u
Example 11.21
The Weierstrass M-test can be used to show that the series of
P
functions k1 k12 sin.kx/ converges uniformly in R. Indeed, on the one hand, since
j sin.kx/j 1 for every k 1 and x 2 R, we get
ˇ
ˇ
ˇ1
ˇ
ˇ sin.kx/ˇ 1
ˇ k2
ˇ k2
for every k 1 and x 2 R. On the other, just note that the series
convergent.
P
1
k1 k2
is
Accidentally,
the previous example also shows that, given a uniformly
P
P convergent
series k1 fk of differentiable functions fn W I ! R, the series k1 fk0 of their
derivatives is not necessarily convergent, even pointwise. Indeed,
X
X d 1
1
cos.kx/;
sin.kx/ D
2
dx
k
k
k1
k1
which doesn’t converge pointwise in any real x of the form x D 2`, with ` 2 Z.
We finish this section by using the material developed here to give an example of
a continuous function which is not differentiable at any point. For what comes next,
the reader might find it helpful to read the statement of Example 8.23 again. Up to
details, we follow the discussion in Sect. 9.7 of [9].
466
11 Series of Functions
Example 11.22 For y 2 R, let d.y/ denote the distance from x to the nearest integer.
If f W R ! R is given by
f .x/ D
X 9
10
k1
k
d.10k x/;
then f is well defined and continuous, albeit not differentiable at any x 2 R.
Proof Example 8.23 guarantees the continuity of x 7! d.10k x/, for every k 1.
9 k
9 k
P 9 k
On the other hand, since 0 10
d.10k x/ 12 10
and k1 10
converges, the
Weierstrass M-test assures the well definiteness and continuity of f .
Fix x > 0 (the case of x 0 is analogous and will be left to the reader). We shall
prove that f is not differentiable at x by constructing sequences .an /n2 and .bn /n2
n
such that an x < bn and an ; bn ! x, but
ˇ f .b / f .a / ˇ n
n ˇ
ˇ n
ˇ
ˇ ! C1:
bn an
This will contradict an obvious slight modification of Problem 8, page 300 (by
allowing an x < bn , instead of an < x < bn , there).
Given x > 0 and an integer n 2, let mn 2 N be such that mn 10n x <
mn C 1. If an D 10n mn and bn D 10n .mn C 1/, we obviously have an x < bn
n
k
k
ˇand bkn an D k10 ˇ. If k n is also integer, then 10 an ; 10 bn 2 N, so that
ˇd.10 bn / d.10 an /ˇ D 0; on the other hand, if k < n, then Example 8.23 gives
ˇ
ˇ
ˇd.10k bn / d.10k an /ˇ 10kn :
(11.16)
Therefore,
f .bn / f .an / D
n1 X
9 k
d.10k bn / d.10k an / :
10
kD0
(11.17)
n C1
Since 10n1 an D m10n and 10n1 bn D m
by separately considering
the cases
ˇ
ˇ 10 , 1
1
. Therefore,
0 < x < 1 and x 1 one easily sees that ˇd.10 bn d.101 an /ˇ D 10
the triangle inequality in (11.17), together with (11.16), gives
11.2 Series of Functions
467
jf .bn / f .an /j 19
10 10
n1
n2 ˇX
ˇˇ
9 k
ˇ
d.10k bn / d.10k an / ˇ
ˇ
10
kD0
n2
ˇ
9n1 X 9 k ˇˇ
d.10k bn / d.10k an /ˇ
n
10
10
kD0
9n1 X 9
10n
10
kD0
D
7 9n1 C 1
7 9n1 C 1
jbn an j:
D
8 10n
8
n2
k
10kn D
n2
9n1
1 X k
9
10n
10n kD0
Thus,
ˇ f .b / f .a / ˇ 7 9n1 C 1 n
n ˇ
ˇ n
! C1:
ˇ
ˇ
bn an
8
t
u
Problems: Section 11.2
P
x
k
1. Prove that the series of functions k1 . 1Cx
2 / converges uniformly in R and
compute its sum.
2. Prove that the Taylor series of the functions ex , sin x and cos x converge
uniformly to such functions in each interval Œa; a, with a > 0.
3. For n 2 N, let fn W Œ0; 1 ! R be given by fn .x/ D nxenx . Prove that .fn /n1
converges pointwise, albeit not uniformly, 0. Also, show (without invoking
R1
Lebesgue’s DCT) that 0 fn .x/ dx ! 0 as n ! C1.
4. For n 0 integer, let fn W Œ0; 1 ! R be given by fn .x/ D xn ex . Do the
following items:
(a) Show that .fn /n0 converges pointwise to f W Œ0; 1 ! R, with f .x/ D
0; if x < 1
:
e1 ; if x D 1
(b) Show that, for each integer n 0, there exist natural numbers an and bn
R1
such that 0 fn .x/ dx D an bn e1 , with a0 D b0 D 1, an D nan1 and
bn D nbn1 C 1 for every n 2 N. P
(c) Conclude that an D nŠ and bn D nŠ nkD0 kŠ1 for every n 2 N.
R1
(d) Use the result of the previous items to show that 0 fn .x/ dx ! 0 as
n ! C1.
468
11 Series of Functions
5. Let fn W Œa; b ! R be a sequence of integrable functions, converging uniformly
to a function f W Œa; b ! R. Show that f is integrable and
Z
Z
b
f .x/dx D lim
n!C1 a
a
b
fn .x/dx:
6. Let .fn /n1 be a sequence of differentiable functions
P fn W Œa; b ! R, such that
jfn0 .x/j Mn for every n 2 N and x 2 P
Œa; b. If k1 Mk is a convergent series
and there exists x0 2 Œa; bPsuch that k1 fk .x0 / converges absolutely, show
that the series of functions k1 fk converges uniformly in the interval Œa; b to
a differentiable function, such that
X
d X
fk .x/ D
fk0 .x/
dx k1
k1
for every x 2 Œa; b.
Rb
7. (Putnam) Let f W Œa; b ! R be a continuous function such that a f .x/xk dx D 0
for every k 2 ZC . Prove that f vanishes identically.
8. (Berkeley) Does there exist a continuous function f W Œ0; 1 ! R such that
R1
R1
k
0 f .x/xdx D 1 and 0 f .x/x dx D 0 for every nonnegative integer k ¤ 0?
Justify your answer.
9. (Berkeley) Let 'n W Œ0; 1 ! R be a sequence of nonnegative continuous
functions such that
Z
1
lim
n!C1 0
xk 'n .x/dx
exists, for each k 2 ZC . Prove that, for any given continuous function f W
Œ0; 1 ! R, the limit
Z
lim
n!C1 0
1
f .x/'n .x/dx
also exists.
P
10. With respect to the series of functions k1 1k sin.kx/, do the following items:
(a) Use Abel’s criterion (cf. Problem 18, page 243), together with the discussion of items (b) and (c) of Problem 20, page 244, to show that the given
series converges pointwise in the interval .0; 2/.
(b) Revisit the proof of Abel’s criterion, as sketched in the hints given to
Problem 18, page 243, to show that Abel’s identity guarantees that the
convergence of item (a) is uniform in every interval of the form Œı; 2 ı,
for 0 < ı < .
11.2 Series of Functions
469
11. Let f W R ! R be a piecewise continuous function, periodic with period 2.
The Fourier series2 of f is the series of functions
a0 .f / X
C
.ak .f / cos.kx/ C bk .f / sin.kx// ;
2
k1
where
ak .f / D
1
Z
f .x/ cos.kx/dx and bk .f / D
1
Z
f .x/ sin.kx/dx
for all k. In this respect, do the following items:
(a) If f is continuously differentiable and k 1, prove that
1
1
ak .f / D bk .f 0 / and bk .f / D ak .f 0 /:
k
k
Then, conclude that ak .f /; bk .f / ! 0 as k ! C1 (this is a special
case of Riemann-Lebesgue’s lemma, to be proved in greater generality in
Problem 15).
(b) If f is twice continuously differentiable and k 1, prove that
ak .f / D 1
ak .f 00 /
k2
and bk .f / D 1
bk .f 00 /:
k2
Then, conclude that the Fourier series of f converges uniformly in R and,
hence, defines a continuous function in R.
(c) Yet assuming f to be twice continuously differentiable, let g W R ! R be
the continuous function given by
g.x/ D
a0 .f / X
C
.ak .f / cos.kx/ C bk .f / sin.kx// :
2
k1
Prove that al .g/ D al .f / for l 0 and bl .g/ D bl .f / for l 1.
For the next two problems, we say that a function f W I ! R is piecewise
continuously differentiable if there exists reals a1 < a2 < : : : < an in I such
that f is continuously differentiable in all of the intervals .1; a1 \I, Œai ; aiC1 for 1 i < n and Œan ; C1/ \ I. We shall assume without proof the validity of
Fourier’s convergence theorem,3 which states that if f W R ! R is piecewise
2
Fourier’s studies on heat conduction, collected in his famous book Théorie Analytique de la
Chaleur, laid the groundwork for modern Mathematical Physics, which in turn has greatly
influenced the development of the theory of Partial Differential Equations.
3
In spite of being known by the name of Fourier, such a result is actually due to Dirichlet.
470
11 Series of Functions
continuous, piecewise continuously differentiable and periodic of period 2,
then its Fourier series is pointwise convergent in the whole real line, converging
at x0 2 R to
1
lim f .x/ C lim f .x/ :
x!x
x!x
C
0
2
0
For a proof of Fourier’s convergence theorem, together with a discussion of
several other interesting properties of Fourier series, see [10, 20] or [24].
12. Let f W R ! R be periodic of period 2 and given by f .x/ D x2 in the interval
Œ; .
(a) Compute its Fourier series.
P
(b) Use Fourier’s theorem to show that k1
1
k2
D
2
6 .
13. Let f W R ! R be periodic of period 2 and such that
8
< 0; if x < 0
f .x/ D 1; if 0 x < :
:
(a) Compute its Fourier series.
(b) Use Fourier’s theorem to deduce Leibniz formula4 for :
1
1 1
D 1 C C :
4
3
5 7
14. (Miklós-Schweitzer—adapted) Let p > 0 be a real number and f ; g W R ! R
be continuous functions, with g being periodic of period p.
(a) Prove that
n Z
X
kD1
kp
n
.k1/p
n
f
kp
n
Z
1 X kp
f
n kD1
n
0
Z
Z
p
p
n 1
!
f .x/dx
g.y/dy :
p 0
0
g.nx/dx D
n
p
g.y/dy (b) Prove the following theorem of Fejér5 :
Z
lim
n!C1 0
4
5
p
1
f .x/g.nx/dx D
p
Z
p
0
f .x/dx
Z
p
0
g.x/dx :
We shall give a self-contained proof of such a formula in Problem 9, page 484.
After Lipót Fejér, Hungarian mathematician of the XX century.
11.2 Series of Functions
471
15. Prove the Riemann-Lebesgue lemma: let f W R ! R be a continuous function,
periodic with period 2. If ak .f / and bk .f / denote the Fourier coefficients of f ,
as defined in Problem 11, prove that ak .f /; bk .f / ! 0 as k ! C1.
For the next problem, let f W R ! R be continuous and periodic with period
2. For n 2 ZC , let
a0 X C
ak cos.kx/ C bk sin.kx/
2
kD1
n
Sn f .x/ D
denote the n-th partial sum of the Fourier series of f and
n f .x/
D
S0 f .x/ C S1 f .x/ C C Sn f .x/
:
nC1
n
Another theorem of Fejér states that n f ! f uniformly on R.
16. Let f ; g W R ! R be continuous and periodic with period 2. If ak .f / D ak .g/
and bk .f / D bk .g/ for every k, prove that f D g.
17. (Berkeley)
p Find all continuous functions f W R ! R such that f .x/ D f .xC1/ D
f .x C 2/ for every x 2 R.
18. * This problem establishes a fairly general version of Leibniz’ rule of
differentiation under the integral sign, assuming the validity of Lebesgue’s
DCT. To this end, let I; J R be intervals and f W I J ! R be continuous
in each variable separately.6 Let ˛ and ˇ be the endpoints of J (in the sense of
Sect. 10.9).
(a) Assume that for every x0 2 I and every sequence .xn /n1 in I converging to
x0 , there exists an integrable function g W J ! R such that jf .xn ; t/j g.t/
for every n 1 and every t 2 J. Then, t 7! f .x; t/ is integrable for each
x 2 I. Moreover, if F W I ! R is given by
Z ˇ
F.x/ D
f .x; t/dt;
˛
then F is continuous.
(b) Let the function x 7! f .x; t/ be continuously differentiable for each t 2 J,
@f
and let @x
denote its derivative. Assume that for every x0 2 I and every
sequence .xn /n1 in I converging to x0 , there exist integrable
functions
gW
ˇ
ˇ
ˇ @f
ˇ
.xn ; t/ˇ G.t/
J ! R and G W J ! R such that jf .xn ; t/j g.t/ and ˇ @x
@f
for every n 1 and every t 2 J. Then, t 7! f .x; t/ and t 7! @x
.x; t/
are integrable for each x 2 I and F W I ! R, defined as in item (a), is
continuously differentiable in I, with
Z ˇ
@f
F 0 .x/ D
.x; t/dt:
(11.18)
@x
˛
6
I.e., such that both t 7! f .x; t/ and x 7! f .x; t/ are continuous functions.
472
11 Series of Functions
19. * Show that the Gamma function is continuously differentiable, with
0
Z
.x/ D
C1
et tx1 log t dt:
0
11.3 Power Series
In Sect. 11.1 we saw several examples of infinitely differentiable functions defined
in open intervals and which coincide with their Taylor series in such intervals. The
Taylor series of an infinitely differentiable function is a particular case of a power
series, i.e., of a series of functions of the form
X
ak .x x0 /k ;
(11.19)
k0
where a0 ; a1 ; a2 ; : : : are given real numbers. In this case, as it happens with number
series, we say that an .x x0 /n is the n-th term of the power series and that an is the
n-th coefficient of it.
In this section we develop the basic aspects of the theory of power series, starting
with the following central result.
P
Theorem 11.23 Given a power series k0 ak .x x0 /k , there exists 0 R C1
such that the series:
(a) Converges absolutely in the interval .x0 R; x0 C R/ and diverges in
R n Œx0 R; x0 C R.
(b) Converges uniformly in the interval Œx0 r; x0 C r, 8 0 < r < R.
P
Proof Firstly, note that the series k0 ak .x x0 /k converges absolutely in the
P
interval .x0 R; x0 C R/ if and only if the series k0 ak xk converges absolutely
P
in the interval .R; R/. Accordingly, k0 ak .x x0 /k converges uniformly in
P
Œx0 r; x0 C r if and only if k0 ak xk converges uniformly in Œr; r. Therefore,
we can assume that x0 D 0.
Let’s first deal with item (a).
P
Claim 4 if k0 ak xk converges at x D ˛ ¤ 0, then it converges absolutely at any
x 2 .˛; ˛/.
Indeed, for such an x, we have
X
k0
P
jak xk j D
ˇ x ˇk
ˇ ˇ
jak ˛ k j ˇ ˇ :
˛
k0
X
(11.20)
ak ˛ k converges, Proposition 7.36 guarantees the existence of n0 2 N
ˇ ˇk
such that n n0 ) jan ˛ n j < 1; therefore, it follows from (11.20) that jak xk j ˇ ˛x ˇ Since
k0
11.3 Power Series
473
ˇ x ˇk
ˇ x ˇk
ˇ ˇ
P
ˇ ˇ for k n0 . Since the geometric series
ˇ ˇ is convergent (for, ˇ x ˇ < 1),
kn0 ˛
˛
˛
P
the comparison test assures that the same holds true for kn0 jak xk j and, thus, for
P
k
k0 jak x j.
P
Claim 5 if k0 ak xk diverges at x D ˇ ¤ 0, then it also diverges at any x 2 R
such that jxj > jˇj. P
For such an x, if k0 ak xk converged, the previous claim would assure the
P
(absolute) convergence of the series k0 ak ˇ k , which is an absurd.
P
Claims 4 and 5 guarantee that, with respect to k0 ak xk , one of the three
following possibilities does happen: (i) it converges only at x D 0; (ii) it is absolutely
convergent at any x 2 R; (iii) there exist ˛; ˇ ¤ 0 such that j˛j < jˇj and the series
converges absolutely when jxj < ˛ and diverges when jxj > jˇj.
If either (i) or (ii) happens, there is nothing left to do. If (iii) happens, let
R D supfr > 0I
X
ak uk converges absolutely when juj < rg:
k0
P
We claim that k0 ak xk converges absolutely when jxj < R and diverges when
jxj > R. To this end, let’s look at two separate cases:
P
(i) If jxj < R, take r such that jxj < r < R and k0 ak uk converges absolutely
P
when juj < r. Then, in particular k0 ak xk is absolutely convergent.
P
(ii) If jxj > R and k0 ak xk converged, then, taking RQ satisfying R < RQ < jxj,
P
k
when
it would follow from Claim 4 that
k0 ak u converges absolutely
P
Q But this obviously contradicts the definition of R. Therefore, k0 ak xk
juj < R.
diverges.
P
For item (b), given 0 < r < R, it follows from what we did above that k0 jak rk j
for jxj r, that jak xk j jak rk j D
converges. Hence, letting Mk D jak rk j, we have,P
Mk . Therefore, Weierstrass M-test assures that k0 ak xk converges uniformly in
the interval Œr; r.
t
u
In the notations of the previous result, we say that 0 R C1 is the radius
of convergence of the power series (11.19), and .x0 R; x0 C R/ is its interval of
convergence.
Problems 10 and 11 give a general formula for the radius of convergence in
terms of the coefficients of a given power series. For the time being, the following
consequence of the previous result allows us to easily compute it in a number of
interesting examples. At this point, we suggest that the reader runs through the proof
of the ratio test (cf. Proposition 7.51) once more, just to note that it remains valid if
(in the corresponding notations) l D C1.
Corollary 11.24 Let .an /n0 be a sequence
ˇ
ˇ of nonzero real numbers. IfPthere exists
ˇ an ˇ
0 R C1 such that limn!C1 ˇ anC1 ˇ D R, then the power series k0 ak .x x0 /k has radius of convergence equal to R.
474
11 Series of Functions
Proof Since
ˇ
ˇ
ˇ anC1 .x x0 /nC1 ˇ
ˇ
ˇ
ˇ a .x x /n ˇ D
n
0
jx x0 j n jx x0 j
ˇ
ˇ !
;
ˇ an ˇ
R
ˇ anC1 ˇ
P
0j
< 1 and
the ratio test assures that k0 ak .x x0 /k converges absolutely if jxx
R
jxx0 j
jxx0 j
diverges if R > 1. Since R < 1 if and only if jx x0 j < R, the previous
result guarantees that R is precisely the radius of convergence of the given power
series.
t
u
Example 11.25
ˇ
ˇ
P
ˇ 1=n ˇ
(a) The power series k1 1k xk has radius of convergence 1, for ˇ 1=.nC1/
!
ˇ D nC1
n
1 as n ! C1.
ˇ
ˇ
P
ˇ nŠ ˇ
1
(b) The series k1 kŠxk has radius of convergence 0, for ˇ .nC1/Š
! 0.
ˇ D nC1
ˇ
ˇ
P
ˇ 1=nŠ ˇ
1
k
(c) The series k1 kŠ .x 2/ has radius of convergence C1, for ˇ 1=.nC1/Š ˇ D
n C 1 ! C1P
as n ! C1. Actually, we already know from the material of
Sect. 11.1 that k1 kŠ1 .x 2/k D ex2 for every real x.
P
k
(d) The former corollary does not apply to the power series k1 kŠ1 x2 , for it has
infinitely many vanishing coefficients. See, however, Problem 2. P
k
(e) Given ˛ ¤ 0, the previous corollary assures
ˇ n ˇ that the power series k0 .˛x/
ˇ
ˇ
1
˛
1
has radius of convergence j˛j
, for ˇ ˛nC1
for every n 0. By
ˇ D j˛j
Proposition 7.38, for jxj <
1
j˛j
we have
X
k0
.˛x/k D
1
:
1 ˛x
(11.21)
We now collect another important consequence of Theorem 11.23, which will be
of crucial importance for the proof of the subsequent result.
P
Proposition 11.26 If the power series k0 ak .x x0 /k has radius of convergence
R > 0, then:
P
(a) The function f W .x0 R; x0 C R/ ! R, given by f .x/ D k0 ak .x x0 /k , is
continuous.
Rx
P
ak
(b) For every x 2 .x0 R; x0 C R/, we have x0 f .t/dt D k0 kC1
.x x0 /kC1 . In
particular, the radius of convergence of the series at the right hand side above
is greater than or equal to R.
Proof LetP
fk W .x0 R; x0 C R/ ! R be given by fk .x/ D ak .x x0 /k . Theorem 11.23
gives f D k1 fk in .x0 R; x0 CR/, the convergence being uniform in Œx0 r; x0 Cr,
for every 0 < r < R. Since fk is continuous for every k, items (a) and (b) follow
immediately from items (a) and (b) of Corollary 11.19.
t
u
The coming theorem is the central result in the theory of power series.
11.3 Power Series
475
Theorem 11.27 If the power series
R > 0, then:
P
k0
ak .x x0 /k has radius of convergence
P
(a) The function f W .x0 R; x0 C R/ ! R, given by f .x/ D k0 ak .x x0 /k , is
infinitely differentiable.
P
kŠ
(b) For every n 2 N, we have f .n/ .x/ D kn .kn/Š
ak .x x0 /kn for every x 2
.n/
.x0 R; x0 C R/, and the series defining f also has radius of convergence
equal to R.
Proof As in the proof of Theorem 11.23 we can assume, without any loss of generality, that x0 D 0. P
We shall first prove that, letting RQ be thePradius of convergence
of the power series k1 kak xk1 , then RQ D R and f 0 .x/ D k1 kak xk1 for every
x 2 .R; R/.
Note first that jkak xk j P
jak xk j for every integer k 1 and every x 2 R.
Therefore, if the power series k1 kak xk1 converges absolutely at some x, then the
P
P
same holds true for the power series k1 kak xk and, thus, for k0 ak xk . Hence,
RQ R.
Now, for 0 < x < R and 0 < h < R x, we have
1
0
X
X .x C h/k xk f .x C h/ f .x/
1 @X
D
:
ak .x C h/k a k xk A D
ak
h
h k1
h
k1
k1
On the other hand, Lagrange’s MVT gives, for each k 1, some ck 2 .x; x C h/
k
k
such that .xCh/h x D kck1
k . Therefore,
X
f .x C h/ f .x/
D
kak ck1
k ;
h
k1
P
so that the power series k1 kak ck1
converges absolutely. Since jkak xk1 j k
P
k1
jkak ck1
also converges absok j, the comparison test guarantees that
k0 kak x
lutely. Analogously, such a series converges absolutely if R < x < 0, so that
RQ R.
up to this point assures that g W .R; R/ ! R, given by g.x/ D
P Our discussion
k1
ka
x
,
is
a
well defined function. By item (b) of the previous proposition,
k
k0
we have
Z x
X kak
X
xk D
g.t/dt D
ak xk D f .x/ a0 :
k
0
k1
k1
Hence, the FTC gives, for jxj < R,
f 0 .x/ D g.x/ D
X
k0
kak xk1 :
476
11 Series of Functions
For what is left to do assume, by induction hypothesis, that
already
P we have
kŠ
shown f to be m times differentiable in .R; R/, with f .m/ .x/ D km .km/Š
ak .x x0 /km ; assume also that this last power series has radius of convergence equal to
R.
Then,
the first part, f .m/ is a differentiable function, the power series defining
0 by.mC1/
.m/
Df
has radius of convergence R and
f
f .mC1/ .x/ D
D
X
kŠ
.k m/ak .x x0 /km1
.k
m/Š
kmC1
X
kŠ
ak .x x0 /k.mC1/ :
.k
.m
C
1//Š
kmC1
This completes the inductive argument and, thus, the proof of (b).
t
u
In the notations of item (b) of the previous result, we observe that the power
series expansion of f .n/ is obtained by termwise differentiating, n times, the power
dn
series expansion of f . Indeed, letting dx
n denote the n-th derivative of a function (if
it exists), an immediate computation gives
dn
kŠ
.x x0 /kn :
.x x0 /k D
n
dx
.k n/Š
We now collect two useful consequences of the previous theorem, the first of
which refines the analysis of item (b) of Proposition 11.26.
P
Corollary 11.28 If the power series k0 ak .x x0 /k has radius ofPconvergence
ak
R > 0, then the radius of convergence of the integrated power series k0 kC1
.x x0 /kC1 is also equal to R.
P
ak
d P
kC1
, item (b) of the previous
Proof Since k0 ak .x x0 /k D dx
k0 kC1 .x x0 /
theorem guarantees that both of the given series have the same radius of convergence.
t
u
P
Corollary 11.29 Assume that the power series k0 ak .x x0 /k has radius of
convergence R > 0, and let f W .x0 R; x0 C R/ ! R be given by f .x/ D
P
k
k0 ak .x x0 / . Then:
.n/
(a) P
an D f nŠ.x0 / , for every n 0.
k
(b)
k0 ak .x x0 / is the Taylor series of f .
Proof Item (b) follows immediately from (a). For (a), item (b) of the previous result
.n/
nŠ
gives f .n/ .x0 / D .nn/Š
an D nŠan ; therefore, an D f nŠ.x0 / .
t
u
We shall now see that the results presented so far allow us to reobtain, by means
of a unified approach, some of the Taylor expansions discussed in Sect. 11.1. In this
respect, see also Problem 3.
11.3 Power Series
477
P
1
Example 11.30 Item (e) of Example 11.25 assures that 1Cx
D k0 .1/k xk , with
radius of convergence equal to 1. Hence, item (b) of Proposition 11.26 gives, for
jxj < 1,
Z
x
log.1 C x/ D
0
X
1
dt D
1Ct
k0
Z
x
0
.1/k tk dt D
X .1/k
k0
kC1
xkC1 :
(11.22)
Example 11.31 Note that
1
D 1 x2 C x4 x 6 C 1 C x2
for jxj < 1. Therefore, again by item (b) of Proposition 11.26, we have, for jxj < 1,
Z
x
arctan x D
0
D x
X
1
dt
D
1 C t2
k0
Z
x
0
.1/k t2k dt
x5 x7
x3
C
C :
3
5
7
(11.23)
Problems 8 and 9 will show that (11.22) and (11.23) remain true for x D 1.
Example
11.32 It follows easily from Corollary 11.24 P
that the power series
P
1 k
1 k
x
converges
in
the
whole
real
line.
If
f
.x/
D
k0 kŠ
k0 kŠ x for x 2 R,
then item (b) of Theorem 11.27 furnishes
f 0 .x/ D
X
k1
X 1
1
xk1 D
xk D f .x/:
.k 1/Š
kŠ
k0
Therefore,
d x
.e f .x// D ex .f .x/ C f 0 .x// D 0;
dx
so that x 7! ex f .x/ is a constant function. Finally, since f .0/ D 1, we get ex f .x/ D
1 or, which is the same, f .x/ D ex .
We now apply Theorem 11.27 to show that, for any ˛ 2 R n f0g and jxj < 1, one
can write .1 C x/˛ as a convergent power series. To this end, given
˛ 2 R and
an
integer n 0, we define the generalized binomial number ˛n by letting ˛0 D 1
and, for n 1,
!
˛.˛ 1/.˛ 2/ : : : .˛ n C 1/
˛
D
:
(11.24)
n
nŠ
478
11 Series of Functions
The following lemma establishes some useful properties of generalized binomial
numbers. As in Sect. 4.2, the property of item (a) is also known as Stifel’s relation.
Lemma 11.33 Given ˛ 2 R and n 2 N, we have:
˛1
C
.
(a) ˛n D ˛1
n n1
(b) ˇ˛n ˛n D ˛1
for
every
˛ ¤ 0.
n1
˛ˇ
ˇ
ˇ
(c) n 1 when j˛j 1.
Proof Item (a) is an easy computation:
!
!
1
˛
˛1
D ˛.˛ 1/.˛ 2/ : : : .˛ n C 1/
nŠ
n
n
1
.˛ 1/.˛ 2/ : : : .˛ n/
nŠ
1
.˛ 1/.˛ 2/ : : : .˛ n C 1/.˛ .˛ n//
nŠ
1
D
.˛ 1/.˛ 2/ : : : .˛ n C 1/
.n 1/Š
!
˛1
D
:
n1
D
Item (b) follows immediately from (11.24).
Finally, for item (c), if j˛j 1 then (11.24) and the triangle inequality give
ˇ !ˇ
ˇ ˛ ˇ j˛j.j˛j C 1/.j˛j C 2/ : : : .j˛j C n 1/
1 2 ::: n
ˇ
ˇ
D 1:
ˇ
ˇ
ˇ n ˇ
nŠ
nŠ
t
u
The coming result is known in mathematical literature as the binomial series
theorem, or simply as the binomial
theorem, and is due to Newton. Notice
that (11.25) generalizes (4.11), for ˛k D 0 if ˛ 2 N and k > ˛.
Theorem 11.34 (Newton)
For ˛ ¤ 0 and jxj < 1, we have
!
X ˛
xk :
.1 C x/ D
k
k0
˛
Proof Firstly, assume 0 < j˛j 1. Since
ˇ ˇ
ˇ ˛ ˇ
nC1 n
ˇ n ˇ
! 1;
ˇ ˛ ˇ D
ˇ nC1 ˇ
j˛ nj
(11.25)
11.3 Power Series
479
P Corollary 11.24 assures that k0 ˛k xk has radius of convergence equal to 1.
Hence,
k Theorem 11.27, the function f W .1; 1/ ! R given by f .x/ D
P ˛by
k0 n x is differentiable. Also from that result,
!
!
˛ k1 X ˛ 1 k1
k
x
D
˛
x ;
f .x/ D
k
k1
k1
k1
X
0
(11.26)
where we used item (c) of the previous lemma in the last equality above. Therefore,
!
X ˛1
xk1
.1 C x/f .x/ D ˛.1 C x/
k
1
k1
0
!
! 1
X ˛1
X ˛1
xk1 C
xk A
D ˛@
k
1
k
1
k1
k1
0
0
1
!
!
X ˛1
X ˛1
xk1 C
xk1 A
D ˛ @1 C
k
1
k
2
k2
k2
0
D ˛ @1 C
X
k2
1
!
!
X ˛
˛
k1 A
x
xk
D˛
k1
k
k0
D ˛f .x/:
If g.x/ D .1 C x/˛ f .x/, then we have for jxj < 1 that
g0 .x/ D ˛.1 C x/˛1 f .x/ C .1 C x/˛ f 0 .x/
D .1 C x/˛1 ˛f .x/ C .1 C x/f 0 .x/ D 0:
Therefore, g is constant in .1; 1/. Since g.0/ D 1, we get .1 C x/˛ f .x/ D 1 for
jxj < 1, as wished.
P For the general case, suppose that .1 C x/˛ D k0 ˛k xk for some ˛ ¤ 0 and
every x 2 .1; 1/. Let us show that similar formulas hold true for ˛ 1 and ˛ C 1
(and every x 2 .1; 1/):
(a) Theorem 11.27, together with item (b) of Lemma 11.33, gives, for jxj < 1,
.1 C x/
˛1
!
1 d
1 X ˛ k1
˛
D .1 C x/ D
k
x
˛ dx
˛ k1 k
!
!
X ˛1
X ˛1
k1
x
xk :
D
D
k
1
k
k1
k0
480
11 Series of Functions
(b) Item (a) of lemma 11.33 gives us
.1 C x/
˛C1
!
!
!
X ˛
X ˛
X ˛
k
k
x D
x C
xkC1
D .1 C x/
k
k
k
k0
k0
k0
!
!!
!
X
X ˛C1
˛
˛
C
xk D 1 C
xk
D1C
k
k
1
k
k1
k1
!
X ˛C1
xk :
D
k
k0
By induction, we conclude that (11.25) is true for all ˛ ¤ 0 and jxj < 1.
t
u
Corollary 11.35 Given ˛; ˇ ¤ 0, we have
!
X ˛
.ˇx/k
.1 C ˇx/ D
k
k0
˛
for every x 2 R such that jxj <
1
jˇj .
Proof It suffices to apply (11.25) with ˇx instead of x, noticing that jˇxj < 1 ,
1
jxj < jˇj
.
u
t
Example 11.36 As an application of the corollary above, note that for jxj <
we have
!
X 1=2
1=2
.2x/k ;
D
.1 2x/
k
k0
with
!
1
1
1
1
1=2
1 ::: n C 1
D
nŠ
2
2
2
n
1 3 5
2n 1
2 2 2
2
.2n/Š
n
2 2 4 .2n/
.1/n
nŠ
.1/n
D
nŠ
D
!
.1/n 2n
.1/n .2n/Š
:
D
D
n
nŠ 2n 2n nŠ
4n
Hence, for jxj <
1
2
we have
1=2
.1 2x/
!
X 1 2k
xk :
D
k
k
2
k0
1
2
11.3 Power Series
481
We finish this section by presenting a result on the extension of a power series to
the endpoints of its interval of convergence. To this end, we first need to discuss a
lemma which is interesting in itself.
P
Lemma 11.37 Let f .x/ D k0 ak xk be defined in the interval .R; R/, with ak P
0 for every k 0. If k0 ak Rk converges, then
X
a k Rk :
lim f .x/ D
x!R
k0
Proof Since ak 0 for each k 0, we have, for 0 < x < R,
X
X
ak Rk f .x/ D
ak .Rk xk / 0:
k0
k0
On the other hand,
X
ak Rk f .x/ D
k0
n
X
ak .Rk xk / C
kD0
n
X
X
ak .Rk xk /
k>n
X
a k Rk xk C
a k Rk ;
kD0
k>n
where we used again the fact that ak 0 to obtain
P the above inequality.
Now, given > 0, the convergence of k0 ak Rk assures the existence of
P
k
such that
n0 2 N P
k>nak R < for n > n0 . Fix such an n > n0 . Since
n
k
k
limx!R kD0 ak R x D 0, there exists ı > 0 such that
x 2 .R ı; R/ )
n
X
ak Rk xk < :
kD0
With such choices and in view of the above computations, we conclude that
x 2 .R ı; R/ ) 0 n
X
ak Rk f .x/ < 2:
kD0
t
u
Remark 11.38 In the notations of the statement of the previous lemma, Problem 7
shows that the assumption on the sign of the ak ’s can be dropped.
It is somewhat surprising that the converse of the previous lemma also holds true.
Such a result is known in mathematical literature as a tauberian theorem after the
work of Alfred Tauber, Austrian mathematician of the XIX and XX centuries. For
the sake of simplifying the notation, we assume that R D 1, leaving to the reader
the (easy!) task of dealing with the general case.
482
11 Series of Functions
P
k
Theorem 11.39 Let f .x/ D
the interval .1; 1/, with ak 0 for
k0 ak x in P
every
k
0.
If
lim
f
.x/
does
exist,
then
x!1
k0 ak converges and limx!1 f .x/ D
P
a
.
k0 k
Proof The last part follows from the previous lemma. For what is left to do, note
that, for 0 x < 1,
xf 0 .x/ D
X
X 2 X1
jC1
kak xk D
k1
kD2j
j0
X 2 X1
kak xk
jC1
2 j a k x2
jC1 1
kD2j
j0
jC1
X 2 X1
jC1
j
D
2
ak x2 1 :
kD2j
j0
In particular, for a fixed natural number l and 0 < x < 1, we have
Z
x
0
Z x
l
2jC1
X
X1
jC1
j
tf .t/dt 2
ak
t2 1 dt
0
jD0
0
kD2j
jC1
l
2jC1
X
X1
x2
j
2
ak jC1
D
2
j
jD0
kD2
l
2
1
1 X X
jC1
a k x2 :
2 jD0
j
jC1
D
kD2
Now, let a D limx!1 f .x/. Since f is increasing and nonnegative, we get
Z
x
0
tf 0 .t/dt D xf .x/ Z
x
0
f .t/dt a:
Taking these two estimates together, we arrive at
l 2X
1
X
jC1
jD0
a k x2
jC1
2a
kD2j
for every 0 x < 1, so that making x ! 1 we find
2lC1
X1
kD1
l 2X
1
X
jC1
ak D
jD0
kD2j
ak 2a:
11.3 Power Series
483
Finally, since the sequence of partial sums of the series
ing, this last inequality gives
X
ak a0 C 2a:
P
k0
ak is nondecreas-
k0
t
u
Example 11.40 By applying the formula of Problem 11 for the radius of convergence,
we conclude that .1; 1/ is the interval of convergence of the power series
P
2j
x
. Alternatively, we can observe that, for jxj < 1,
j0
0
X
x2 j
j0
X
jxjk < C1:
k1
On the other hand, since the sequence of the coefficients of this series clearly
diverges, the tauberian theorem assures that
lim
x!1
X
x2 D C1:
j
j1
For a more refined estimate on the growth of
Problem 12.
P
x2 as x ! 1, see
j
j1
Problems: Section 11.3
1. Compute the radii of convergence of the power series given below:
P
1
xk .
(a)
Pk0 2kC1
k2 k
e x.
(b)
Pk0
k 2k
(c)
k0 .1/ x .
P
k
2. Compute the radius of convergence of the power series k0 kŠ1 x2 .
3. Use the approach of Example 11.32, together with the result of Example 9.48,
to show that
sin x D
X .1/k1
X .1/k
x2k1 and cos x D
x2k :
.2k
1/Š
.2k/Š
k0
k0
4. * For k 2 N and jxj < 1, prove that
!
X kCn1
1
xn :
D
n
.1 x/k
n0
484
11 Series of Functions
5. Show that for jxj < 1 we have
arcsin x D
X
k0
!
1
2k 2kC1
x
:
4k .2k C 1/ k
6. The formula of Example 11.30 can be modified in order to compute log a for
every a > 0. To this end, do the following items:
(a) Show that, for x 2 .1; 1/, one has
x3
x5
x7
1Cx
1
log
DxC
C
C
C :
2
1x
3
5
7
(b) Show that x 7! 1Cx
defines a bijection from .1; 1/ onto .0; C1/.
1x
(c) Use the formula of item (a) to compute log 3 with four correct decimal
places.
P
7. * Let f W .R; R/ ! R be given by f .x/ D k0 ak xk . The purpose of this
P
problem is to show that, if k0 ak Rk converges, then
lim f .x/ D
x!R
X
a k Rk :
To this end, do the following items:
P
(a) Let rn D kn ak Rk and, for jxj < R, let y D
numbers n < m,
m
X
(11.27)
k0
x
R.
Show that, for natural
m
X
n
m
ak x D rn y rmC1 y C
rk yk yk1 :
k
kDn
kDnC1
(b) Show that, for natural numbers n < m,
m
ˇ
ˇX
ˇ
ˇ
ak xk ˇ jrn j C jrm j C sup jrk j yn ym :
ˇ
k>n
kDn
P
(c) Show that k0 ak xk converges uniformly on the interval Œ0; R. Then,
conclude that (11.27) is true.
8. Use the result of the previous problem to show that
log 2 D 1 1 1
1
C C :
2
3 4
11.3 Power Series
485
9. * Use the material of this section to give a self-contained proof of the Leibniz
formula for :
1
1 1
D 1 C C :
4
3
5 7
For the next two problems, we extend the notion of supremum by saying that
a sequence .ak /k0 has supremum C1 if it is unbounded from above; in this
case, we write
supfak I k 0g D C1:
Accordingly, we extend the notion of infimum by saying that a sequence .ak /k0
has infimum 1 if it is unbounded from below; in this case, we write
inffak I k 0g D 1:
10. Let .ak /k0 be a sequence of real numbers. Define its limit superior, denoted
lim sup ak , by letting
lim sup ak D lim supfaj ; ajC1 ; : : :g:
j!C1
(a) Prove that lim sup ak is a well defined concept and that, if sj D
supfaj ; ajC1 ; : : :g 2 R [ fC1g, then lim sup ak D inffsj I j 0g 2
R [ f˙1g.
(b) If lim sup ak 2 R, prove that it is the only real number M satisfying the two
following conditions:
i. Given > 0, there exists infinitely many n 2 N such that an > M .
ii. Given > 0, there exists at most finitely many n 2 N such that an M C .
P
11. Let k0 ak .x x0 /k be a power series with radius of convergence R. The
purpose of this problem is to prove that
RD
1
lim sup
p
;
k
jak j
where p
the right hand side is to be interpreted as being equal to 0 if
lim sup k jak j D C1. To this end, assume without loss of generality that
x0 D 0, let R be given as above and do the following items:
(a) Fix 0 < r < R and take r < s < R. Show that there exists k0 2 N such that
jak j < s1k for every k k0 . Then, show that
jxj r )
X
kk0
jak xk j X r
kk0
s
k
< C1:
486
11 Series of Functions
(b) Fix jxj > R and show that there exist infinitely
many indices k 2 N such
P
that jak xk j > 1. Conclude that the series k0 ak xk diverges.
(c) Conclude that R, defined as above, is indeed the convergence radius of the
given series.
P
j
12. As we saw in Example 11.40, the power series j1 x2 defines a differentiable
function f W .1; 1/ ! R. In this respect, prove that:
P
x2
(a) xf 0 .x/ > k2 xk D 1x
, for 0 x < 1.
(b) f .x/ > log.1 x/ x, for 0 x < 1.
11.4 Some Applications
In this section, we briefly discuss the use of power series in Algebra and differential
equations. For further applications of power series to Algebra and Combinatorics,
we refer the reader to [5].
We begin with the following
Definition 11.41 The (ordinary) generating function7 of a sequence of real
numbers .an /n0 is the power series
X
a k xk :
(11.28)
k0
The previous definition suggests that the main difference between the theory of
power series and the method of generating function lies in a change of point of
view. In the first case, we are primarily interested in examining the properties of the
function f defined by the power series at the right hand side of (11.28); in the second
(as we shall see next), we want to use the properties of f to infer conclusions about
the terms of the sequence .an /n0 .
Let’s illustrate this by revisiting Problem 20, page 220 with the aid of generating
functions.
Example 11.42 (TT) The set of naturals is partitioned into m disjoint, infinite and
nonconstant arithmetic progressions, of common ratios d1 , d2 , . . . , dm . Prove that
1
1
1
C
CC
D 1:
d1
d2
dm
P
Proof If f .x/ D k1 xk and ai is the initial term of the i-th AP (that of common
ratio di ), then the given condition, together with item (e) of Example 11.25, gives
7
In opposition to exponential generating functions, cf. Chap. 3 of [5], for instance.
11.4 Some Applications
487
for jxj < 1 that
f .x/ D
m
m
X
X
.xai C xai Cdi C xai C2di C / D
iD1
iD1
xai
:
1 xdi
Since 1 xdi D .1 x/.1 C x C x2 C C xdi 1 /, multiplying both sides of the
equality above by 1 x we get
xD
m
X
iD1
1CxC
x2
xai
C C xdi 1
(11.29)
for jxj < 1. Now, both sides of this last equality define functions continuous in Œ0; 1
and which coincide in Œ0; 1/; hence,Psuch functions coincide for x D 1 too, so that,
1
letting x D 1 in (11.29), we obtain m
t
u
iD1 di D 1.
The use of generating functions is particularly useful in the study of sequences
.an /n0 satisfying a given linear recurrence relation. Here, as in Sect. 3.2, we shall
treat the cases of (certain) linear recurrence relations of orders 2 and 3, postponing
the analysis of the general case to Chap. 20 of [5]. P
The idea is to consider the generating function k0 ak xk corresponding to the
given sequence and, then, follow the various stages below, which comprise a sort
receipt for several similar problems:
I. Use the initial terms of the sequence, as well as the recurrence relation it
satisfies, to conclude that the given generating function converges at some
interval of the form .r; r/.
II. Again with the aid of the initial terms and the given recurrence relation,
perform
appropriate (generally algebraic) operations with the equality f .x/ D
P
k
k0 ak x to get a formula for f .x/.
III. Develop the formula obtained in item II in Taylor series.
IV. Use the uniqueness of the power series representation of f , given by Corollary 11.29, to conclude that an equals the coefficient of xn in the power series
.n/
expansion of stage III (i.e., an D f nŠ.0/ ).
In order to go through stage I, the following result is frequently useful.
Lemma 11.43 Let .an /n0 be a sequence of real numbers. If there exist
P positive
reals c and M such that jan j cM n for every n 0, then the power series k0 ak xk
converges in the interval M1 ; M1 .
Proof
Since jan xn j D jan jjxjn cM n jxjn D cjMxjn and the geometric series
P
1
k
(cf. item (e) of Example 11.25), the
k0 jMxj converges when jxj < M P
comparison test for series guarantees that k0 ak xk converges when jxj < M1 . u
t
The following example implements stages I through IV above to get a positional
formula for the n-th Fibonacci number.
488
11 Series of Functions
Example 11.44 Let .Fn /n1 be the Fibonacci sequence, so that F1 D 1, F2 D 1 and
FkC2 D FkC1 C Fk for every integer k 1. Compute Fn as a function of n.
P
k
Solution Let f .x/ D
k1 Fk x be the generating function corresponding to the
Fibonacci sequence, and let’s run through the previously listed stages I to IV.
Step I: note first that an easy induction gives Fn P2n for every n 1.
Hence, Lemma 11.43 guarantees that the power series k1 Fk xk converges in
the interval 12 ; 12 .
Step II: for x 2 . 12 ; 12 /, we can write
f .x/ D F1 x C F2 x2 C
X
F k xk
k3
X
D x C x2 C
.Fk1 C Fk2 /xk
k3
D x C x2 C x
X
Fk1 xk1 C x2
k3
X
Fk2 xk2
k3
D x C x2 C x.f .x/ F1 x/ C x2 f .x/
D x C .x C x2 /f .x/:
Then, for x 2 . 12 ; 12 / we have
f .x/ D
x
:
1 x x2
Step III: writing 1xx2 D .1˛x/.1ˇx/, with ˛; ˇ 2 R, we have ˛ Cˇ D 1,
˛ˇ D 1 and
f .x/ D
x
x
:
D
1 x x2
.1 ˛x/.1 ˇx/
Imposing, with no loss of generality, that ˛ > ˇ, we get ˛ D
p
p
1 5
5 and, thus,
2 , so that ˛ ˇ D
1
f .x/ D p
5
Now, developing
1
1˛x
and
1
1ˇx
p
1C 5
2
1
1
:
1 ˛x 1 ˇx
as geometric series, we obtain
0
1
X
X ˛k ˇk 1 @X
k
kA
f .x/ D p
.˛x/ .ˇx/ D
p
xk ;
5 k0
5
k0
k1
and ˇ D
11.4 Some Applications
as long as jxj < min
489
n
1 1
1
2 ; j˛j ; jˇj
o
D 12 . Then,
X ˛k ˇk Fk x D
p
xk ;
5
k1
k1
X
k
for every x 2 . 12 ; 12 /.
Step IV: finally, Corollary 11.29 assures that the power series expansion of f is
unique, so that
the
last equality above, together with the initial definition of f ,
n
n
gives Fn D ˛ pˇ for every n 1.
5
t
u
Proceeding similarly to the above example, we now use generating functions to
give another proof of Theorem 3.16 (see also Problem 6).
Theorem 11.45 Given u; v 2 R, with v ¤ 0, let .an /n1 be such that akC2 D
uakC1 C vak for every k 1. Moreover, assume that the characteristic equation
x2 ux v D 0 has real roots ˛ and ˇ.
(a) If ˛ ¤ ˇ, then an D A˛ n1 C Bˇ n1 for every n 1, where A and B are the
A C B D a1
.
solutions of the linear system
˛A C ˇB D a2
(b) If ˛ D ˇ, then an D.A C Bn/˛ n1 for n 1, where A and B are the solutions
A C B D a1
of the linear system
.
.A C 2B/˛ D a2
Proof We start by showing that there exists q > 0 such that jan j qn for every
n 1. Indeed, assuming that jak j qk and jakC1 j qkC1 , it follows from the
triangle inequality that
jakC2 j D juakC1 C vak j jujjakC1j C jvjjak j jujqkC1 C jvjqk ;
so that jakC2 j qkC2 provided jujqkC1 C jvjqk qkC2 or, which is the same,
jujqCjvj q2 . Since the greatest root of the second degree equation x2 jujxjvj D
p
0 is x0 D 12 juj C u2 C 4jvj , we get jujq C jvj q2 whenever q > x0 . Hence,
2
jan j qn for every n 1, as long
p as ja1 j q, ja2 j q and q > x0 , for which it
suffices to choose q > maxfja1 j; ja2 j; x0 g.
Fix such a q, so that jan j qn for every n 1. Lemma11.43 guarantees that
P
the generating function k1 ak xk converges in the interval 1q ; 1q , thus defining
P
f W 1q ; 1q ! R by f .x/ D k1 ak xk .
In order to get a simpler expression for f .x/, we argue as in the previous example,
using the recurrence relation satisfied by the sequence .ak /k0 of the coefficients
of f :
490
11 Series of Functions
f .x/ D
X
a k xk D a 1 x C a 2 x2 C
k1
X
a k xk
k3
X
D a 1 x C a 2 x2 C
.uak1 C vak2 /xk
k3
2
D a1 x C a2 x C ux
X
ak1 xk1 C vx2
k3
X
ak2 xk2
k3
D a1 x C a2 x2 C ux.f .x/ a1 x/ C vx2 f .x/:
Therefore, .1 ux vx2 /f .x/ D a1 x C .a2 ua1 /x2 for every x 2 1q ; 1q .
Now, since ˛ and ˇ are the roots of x2 ux v D 0, we get ˛ C ˇ D u and
˛ˇ D v. Hence, 1 ux vx2 D .1 ˛x/.1 ˇx/, so that
f .x/ D
a1 x C .a2 ua1 /x2
.1 ˛x/.1 ˇx/
(11.30)
1
1
; jˇj
g.
for jxj < minf q1 ; j˛j
Let’s look separately at cases (a) and (b):
(a) If ˛ ¤ ˇ, we decompose the right hand side of (11.30) in partial fractions,8
Ax
Bx
i.e., we take A; B 2 R such that f .x/ D 1˛x
C 1ˇx
. Obviously, such A and B
A C B D a1
must satisfy
, and one promptly notes that such a linear
ˇA C ˛B D ua1 a2
system has a single solution, exactly because ˛ ¤ ˇ.
1
1
Expanding 1˛x
and 1ˇx
in geometric series, we get
f .x/ D Ax
X
X
.˛x/k C Bx
.ˇx/k
k0
k0
X
D
.A˛ k C Bˇ k /xkC1
k0
X
D
.A˛ k1 C Bˇ k1 /xk :
k1
P
Finally, comparing this last expression with the fact that f .x/ D k1 ak xk , we
obtain ak D A˛ k1 C Bˇ k1 for every k 1.
2
2 ua1 /x
(b) If ˛ D ˇ, then 2˛ D u, ˛ 2 D v and f .x/ D a1 xC.a
. This time we
.1˛x/2
try to decompose the right hand side of (11.30) in partial fractions by writing
8
A general theorem on the existence of partial fraction decomposition for quotients of polynomials
will be seen in [5]. For the time being, we shall content ourselves to describe what such a theorem
says in the simple cases we consider here.
11.4 Some Applications
491
Ax
Bx
f .x/ D 1˛x
C .1˛x/
2 , for some A; B 2 R. Obviously, such A and B must
A C B D a1
satisfy
, and again this linear system has a unique solution,
˛A D ua1 a2
for v ¤ 0 ) ˛ ¤ 0.
1
and
Expanding 1˛x
of Problem 4), we get
1
.1˛x/2
i power series (for the second fraction with the aid
f .x/ D Ax
X
.˛x/k C Bx
k0
D
X
X
.k C 1/.˛x/k
k0
.A C B.k C 1//˛ k xkC1
k0
D
X
.A C Bk/˛ k1 xk :
k1
Finally, arguing as in the previous case we get ak D .A C Bk/˛ k1 for every integer
k 1.
t
u
Generating functions can also be used to deal with linear recurrence relations
with nonconstant coefficients. To present such an example, we first need the
following result.
Proposition
0 C R/ ! R have power series expansions
P 11.46 If f ; g W .x0 R; xP
f .x/ D k0 ak .x x0 /k and g.x/ D l0 bl .x x0 /l , thenPthe function fg W .x0 n
R; x0 C
series expansion .fg/.x/ D
n0 cn .x x0 / , with
P R/ ! R hasPpower
n
cn D kClDn ak bl D kD0 ak bnk .
Proof Theorem 11.23 assures that the power series defining f and g converge
absolutely in the interval .x0 R; x0 C R/. Hence, by Theorem 7.53 and for every
x 2 .x0 R; x0 C R/, we have
0
f .x/g.x/ D @
X
10
ak .x x0 /k A @
k0
D
D
D
X
n0
kClDn
X
X
n0
kClDn
n0
1
bl .x x0 /l A
l0
X
X
X
!
ak .x x0 / bl .x x0 /
k
!
ak bl .x x0 /n
cn .x x0 /n :
l
492
11 Series of Functions
Finally, note that Theorem 7.53 guarantees that the convergence of this last series is
also absolute in the interval .x0 R; x0 C R/.
t
u
Example 11.47 Let .an /n0 be the sequence such that a0 D 1, a1 D 1 and
ak D ak1
C 2ak2
k
for every integer k 2. Compute an as a function of n.
P
k
Solution Denoting by f .x/ D
k0 ak x the generating function of the given
sequence, let’s once again run through the previously described stages I to IV.
Step I: once more we shall try to apply the comparison test for series, in the
form of Lemma 11.43. Assuming that jak2 j ˛ k2 , jak1 j ˛ k1 , the triangle
inequality gives
jak j jak1 j
˛ k1
C 2jak2 j C 2˛ k2 I
k
k
hence, we shall have jak j ˛ k provided
same,
˛ k1
k
C 2˛ k2 ˛ k or, which is the
˛
C 2 ˛2 :
k
Since such an inequality is true for ˛ D 2 and every k 2, and ja0 j 20 ,
ja1 j 21 , it follows by induction that jan j 2n for every integer n 1. Hence,
the above mentioned lemma, together with the theory of power
series, guarantees
that f is infinitely differentiable in the whole interval 12 ; 12 .
Step II: writing the given recurrence relation as kak D ak1 C 2kak2 for k 2,
we get
f 0 .x/ D
X
kak xk1 D a1 C
k1
D a1 X
k2
X
k2
ak1 xk1 C 2x
kak xk1 D a1 C
X
X
.ak1 C 2kak2 /xk1
k2
kak2 xk2
k2
0
1
X
X
D a1 .f .x/ a0 / C 2x @ .k 2/ak2 xk2 C 2
ak2 xk2 A
k2
k2
D a1 C a0 f .x/ C 2x.f 0 .x/ C 2f .x//:
Taking into account that a1 C a0 D 0, we get f 0 .x/ D .4x 1/f .x/ C 2xf 0 .x/ or,
which is the same,
.2x 1/f 0 .x/ D .4x 1/f .x/:
11.4 Some Applications
493
In order to solve such a differential equation, first note that f is positive in the
interval .r; r/, for some 0 < r 12 (for, f .0/ D a0 D 1 > 0 and f continuous
imply f positive in some neighborhood of 0). Therefore, for jxj < r we can write
4x 1
1
f 0 .x/
D
D 2 f .x/
2x 1
2x 1
so that, for jxj < r 12 ,
ˇtDx Z
ˇ
D
log f .x/ D log f .t/ˇ
tD0
Z x
2C
D
0
D 2x Thus, for jxj < r 1
2
x
0
1
2t 1
f 0 .t/
dt
f .t/
dt
1
log.1 2x/:
2
we have
f .x/ D e2x .1 2x/1=2 :
(11.31)
Step III: note that the power series expansion of e2x is given by (11.7), with
a D 2, and holds in the whole real line:
e2x D
X .2/k
k0
kŠ
xk :
Hence, it follows from Example 11.36 and Proposition 11.46 that, for f given as
in (11.31) and jxj < r 12 , we have
0
f .x/ D @
! 1
X 1 2l
xl A
xk A @
l
l
kŠ
2
l0
X .2/k
k0
10
!!
X .2/k 1 2l
xn
l
D
l
kŠ
2
n0 kClDn
!!
n
X X
.2/nl 1 2l
xn
l
D
l
.n
l/Š
2
n0 lD0
!!
n
nl
X
X
.1/
2l
2n
xn :
D
l .n l/Š
4
l
n0
lD0
X
494
11 Series of Functions
P
Step IV: comparing the last expression above with f .x/ D n0 an xn , it follows
once more from the uniqueness of power series expansions that
!
n
X
.1/nl 2l
n
:
an D 2
4l .n l/Š l
lD0
t
u
All of the above results can be seen as sorts of discrete versions of the following
result on ordinary differential equations:
Theorem 11.48 Let x0 2 R, R > 0 and p; q W .x0 R; x0 CR/ ! R be given by their
Taylor series centered at x0 . For ˛; ˇ 2 R, there exist a unique twice differentiable
function f W .x0 R; x0 C R/ ! R such that
00
f C pf 0 C qf D 0
:
(11.32)
f .x0 / D ˛; f 0 .x0 / D ˇ
Moreover, f is actually infinitely differentiable and given by its Taylor series
centered at x0 .
Interesting (and important) applications of the above result can be found, for
instance, in Chap. 6 of [2], or in Chap. VI of [19]. We now discuss a particular case
of it, referring to Problem 11.48 for a proof of the general case. See, also, Problem 4,
page 510.
Given real numbers a and b and a twice differentiable function f W R ! R
such that
f 00 C af 0 C bf D 0;
(11.33)
we say that the second degree equation
2 C a C b D 0
(11.34)
is the characteristic equation of (11.33). The discriminant D a2 4b of (11.34)
is said to be the discriminant of (11.33).
Theorem 11.49 Let a and b be given real numbers and f W R ! R be a twice
differentiable function satisfying (11.33). In the notations of the discussion above,
there exist real constants A and B such that
f .x/ D Af1 .x/ C Bf2 .x/
for every real x, where:
(a) f1 .x/ D e 2 and f2 .x/ D xe 2 if D 0.
(b) f1 .x/ D e˛x and f2 .x/
D eˇx if > 0 and ˛ andpˇ are the roots of (11.34).
px ax
ax
(c) f1 .x/ D e 2 cos
and f2 .x/ D e 2 sin x
if < 0.
2
2
ax
ax
In particular, for given values of f .0/ and f 0 .0/, (11.33) has exactly one solution.
11.4 Some Applications
495
Proof First of all, writing f 00 D af 0 bf it is immediate that any solution of (11.33)
is infinitely differentiable.
Claim 1 f is given by its Taylor series in the whole real line.
Fix M > 0 and choose C > 0 such that jf j C and jf 0 j C2 in ŒM; M. Since
f .kC2/ C af .kC1/ C bf .k/ D 0 for every k 2 ZC , we get
jf .kC2/ j jajjf .kC1/j C jbjjf .k/ j:
Thus, if jf .j/ j CjC1 in ŒM; M for 0 j k C 1, then
jf .kC2/ j jajCkC1 C jbjCk CkC2 ;
provided C2 jajC C jbj. Therefore, if we choose such a C from the beginning, we
conclude that
jf .n/ j CnC1 in ŒM; M; 8 n 0:
Now, for x 2 ŒM; M, Taylor’s formula gives
f .x/ D
n1 .k/
X
f .0/
kD0
kŠ
xk C
f .n/ .c/ n
x ;
nŠ
(11.35)
for some c 2 ŒM; M. However, since
ˇ
ˇ .n/
ˇ f .c/ n ˇ CnC1 n
.CM/n n
ˇ
ˇ
x
M
D
C
! 0;
ˇ
ˇ nŠ
nŠ
nŠ
P
f .k/ .0/ k
letting n ! C1 in (11.35) we conclude that f .x/ D
k0 kŠ x in ŒM; M.
Actually, since M > 0 was arbitrarily chosen, this holds for every x 2 R.
Claim 2 for given values of f .0/ and f 0 .0/, there is at most one solution for (11.33).
(This claim will also establish the
P uniqueness part of the theorem.)
For simplicity, write f .x/ D k0 akŠk xk . It follows from Theorem 11.27 that
f 0 .x/ D
X
k1
and, analogously, f 00 .x/ D
af 0 C bf D 0, we find
P
k0
X akC1
ak
xk1 D
xk
.k 1/Š
kŠ
k0
akC2 k
x.
kŠ
Substituting these expressions into f 00 C
X
xk
.akC2 C aakC1 C bak / D 0;
kŠ
k0
496
11 Series of Functions
and the uniqueness of power series expansions gives akC2 C aakC1 C bak D 0 for
every k 0. However, since a0 D f .0/ and a1 D f 0 .0/, the sequence .ak /k0 , and
hence the function f , is completely determined.
Finally, to finish the proof it suffices to check, in each of the cases (a), (b) and
(c), that: (i) the given functions f1 and f2 satisfy (11.33); (ii) for any given values for
f .0/ and f 0 .0/, one is able to find real values for A and B such that
Af1 .0/ C Bf2 .0/ D f .0/
:
Af10 .0/ C Bf20 .0/ D f 0 .0/
Once this has been done, the uniqueness of solution, as established in Claim 2,
allows us to conclude that f D Af1 C Bf2 .
Checking items (i) and (ii) in each of (a), (b) and (c) is actually quite simple. For
item (c), for instance, item (ii) amounts to showing that the system of equations
a
A D f .0/; A C
2
p
B D f 0 .0/
2
always has a solution, which is immediate. As for (i), one just needs to compute fi0
and fi00 for i D 1; 2, then showing that fi00 Cafi0 Cbfi D 0. We leave this as an exercise
to the reader.
t
u
Remark 11.50 Note that the prescription of the values of f .0/ and f 0 .0/, instead of
f .x0 / and f 0 .x0 / for some x0 2 R, is irrelevant. Indeed, letting g.x/ D f .x C x0 /, we
have g.0/ D f .x0 /, g0 .0/ D f 0 .x0 / and g00 C ag0 C bg D 0, so that we can apply the
previous result to find g and then find f .x/ D g.x x0 /.
Example 11.51 Find all differentiable functions f ; g W R ! R such that f .0/ D 0,
g.0/ D 1, f 0 D f C g and g0 D 2f .
Solution Since f 0 D f C g, which is a differentiable function, we conclude that f is
twice differentiable. Then f 00 D f 0 C g0 D f 0 C 2f or, which is the same,
f 00 f 0 2f D 0:
Now, the second degree equation 2 2 D 0 has real roots 2 and 1, so that
the previous theorem gives f .x/ D Ae2x C Bex for every real x. Since f .0/ D 0 and
f 0 .0/ D f .0/ C g.0/ D 1, we get A C B D f .0/ D 0 and 2A B D f 0 .0/ D 1.
Solving such a system of equations, it follows that A D 13 and B D 13 , so that
t
u
f .x/ D 13 e2x ex and g.x/ D f 0 .x/ f .x/ D 23 e2x C 2ex .
11.4 Some Applications
497
Problems: Section 11.4
1. Compute the number of nonnegative integer solutions of the equation
a1 C a2 C C ak D m;
where k and m are given natural numbers.
2. Compute the number of nonnegative integer solutions of the equation a1 C a2 C
a3 C a4 D 20, such that a1 2 and a3 7.
3. Generalize the result of Proposition
11.46, showing that if f W .x0 R; x0 CR/ !
P
R
is
given
by
f
.x/
D
a
.x
x0 /n , then for k 2 N we have f .x/k D
n
n0
P
n
n0 cn .x x0 / , where
cn D
X
ai1 ai2 : : : aik
and the above sum extends over all k-tuples .i1 ; : : : ; ik / of nonnegative integers
satisfying i1 C C ik D n.
4. In the notations of the discussion of Example 11.42, prove that
a1
a2
am
mC1
:
C
CC
D
d1
d2
dm
2
5. Use generating functions to find an as a function of n, where .an /n1 is given
by a1 D 2 and akC1 D ak C .k C 1/ for k 1.
6. Use the methods of this section to give another proof to Theorem 3.19.
7. The sequence .an /n0 is given by a0 D 1 and anC1 D 2an C n for n 0. In
order to compute an as a function of n, do the following items:
(a) If an ˛ n , show that anC1 ˛ nC1 as long as ˛ n .˛ 2/ n; then, conclude
that an 3n for every n 0.
(b) Show that the generating function of .an /n0 converges in the interval
1 1
12xC2x2
3 ; 3 and is given by f .x/ D .1x/
2 .12x/ .
(c) Find real constants A, B and C such that
A
C
B
1 2x C 2x2
D
C
:
C
.1 x/2 .1 2x/
.1 x/2
1x
1 2x
(d) Expand each of the functions of the right hand side above in power series
to conclude that an D 2nC1 .n C 1/ for n 0.
8. Let .an /n0 be given by a0 D 1, a1 D 3, a2 D 5 and akC3 D akC2 C 12 akC1 1
a for every integer k 0. The purpose of this problem is to show that .an /n0
2 k
converges and to compute the corresponding limit. To this end, do the following
items:
498
11 Series of Functions
n
(a) Show that jan j 5P
for every integer n 0. Then, conclude that the radius
of convergence of k0 ak xk is at least 15 .
P
(b) If f W . 15 ; 15 / ! R is given by f .x/ D k0 ak xk , use the given recurrence
15x2 8xC2
.
.x1/.x2 2/
A
Bp
Cp
x1 C x 2 C xC 2 ,
relation to get f .x/ D
(c) Writing f .x/ D
1
1 px
in
2
1
1C px
show that A D 9. Then, expand
1
1x ,
power series to get
2
X
B C .1/k1 C k
9 x:
p
f .x/ D
. 2/kC1
k0
(d) Conclude that an D 9 BC.1/n1 C
p
. 2/nC1
n
for n 0, and hence that an ! 9.
9. (Putnam) Let u; v; w W R ! R be given by the power series expansions
u.x/ D 1 C
x6
x9
x3
C
C
C
3Š
6Š
9Š
v.x/ D x C
x4
x7
x10
C
C
C
4Š
7Š
10Š
w.x/ D
x5
x8
x11
x2
C
C
C
C
2Š
5Š
8Š
11Š
Prove that, for every real x, one has
u.x/3 C v.x/3 C w.x/3 D 3u.x/v.x/w.x/ C 1:
10. The purpose of this problem is to give a proof of Theorem 11.48. To this end,
let
X
X
p.x/ D
bk .x x0 /k and q.x/ D
ck .x x0 /k
k0
k0
in the interval .x0 R; x0 C R/ and do the following items:
P
(a) If f .x/ D k0 ak .x x0 /k in the interval .x0 R; x0 C R/ and f satisfies
f 00 C pf 0 C qf D 0, show that
.k C 2/.k C 1/akC2 C
k
X
.j C 1/bkj ajC1 C ckj aj D 0
jD0
for every k 0. Then, conclude that a2 , a3 , a4 , . . . are uniquely determined
by a0 D f .x0 /, a1 D f 0 .x0 /, p and q.
11.5 A Glimpse on Analytic Functions
499
(b) Let a0 D ˛, a1 D ˇ and 0 < r < R be given. If jx x0 j < r, D 1r jx x0 j
and Ak D jak .x x0 /k j for k 0, show that
.k C 2/.k C 1/AkC2
k
X
.j C 1/jbkj jrkjC1 kj AjC1
jD0
C
k
X
jckj jrkj kjC2 Aj :
jD0
(c) Let B; C > 0 such that jbk jrk B and jck jrk C for every k 0. If
M D maxfBr; Cg, show that
.k C 2/.k C 1/AkC2 M
kC1
X
.j C 1/kC1j Aj :
jD0
(d) Let .AQ k /k0 be defined by AQ 0 D A0 , AQ 1 D A1 and
.k C 2/.k C 1/AQ kC2 D M
kC1
X
.j C 1/kC1j AQ j :
jD0
Show that Ak AQ k for every k 0.
(e) If AQ 0 ¤ 0 or AQ 1 ¤ 0, show that AQ k > 0 for every k 2. Then, show that
k.k C 1/ C M.k C 2/ k
AQ kC2
D
! < 1:
.k C 2/.k C 1/
AQ kC1
P
(f) Conclude that k0 AQ k converges, and that this implies the convergence of
P
P
k
k0 Ak . Then, note that this is the same as saying that
k0 ak .x x0 /
is absolutely convergent in .x0 r; x0 C r/. Finally, show that f , given as in
(a), is well defined and does satisfy (11.32).
11.5 A Glimpse on Analytic Functions
In this section, we show that a function defined by a power series is analytic, i.e., is
given by a convergent power series around each point of its interval of convergence.
We also study the set of zeros of an analytic function and use the results we get to
give purely analytic derivations of the properties of the sine and cosine functions.
Our purpose in doing so is to show the reader that all of the arguments we
have done so far concerning these functions are actually independent of any
500
11 Series of Functions
considerations on the geometry of circles. As we anticipated in Remark 9.12 and
now reinforce in a slightly different way, this is not a simple pedantism, for:
(i) the proof we have given for the formula sin0 x D cos x relied upon the
fundamental trigonometric limit (cf. Lemma 9.11);
(ii) in turn, the proof of that result made use of the formula for the area of a circular
sector;
(iii) then, in Example 10.43 we computed the area of a circle of radius R by using
the change of variables formula to reduce it to a simple application of the FTC,
which relies upon the fact that sin0 x D cos x.
Hence, if we are not able to free the properties of the sine and cosine functions
from the geometry of the circle, we will be forced to conclude that all of the
arguments related to items (i), (ii) and (iii) above—and, therefore, to a large part
of this book—are totally fallacious.
We begin our presentation with the following
Definition 11.52 Let I R be an open interval. A function f P
W I ! R is analytic
if, for every x0 2 I, there exists R > 0 and a power series k0 ak xk such that
P
.x0 R; x0 C R/ I and f .x/ D k0 ak xk for every x 2 .x0 R; x0 C R/.
As we pointed out in the beginning of this section, a function defined by a power
series in an open interval .x0 R; x0 C R/ (possibly with R D C1) is analytic. In
order to prove this, we need the following result on double series which is interesting
in itself.
P
Proposition 11.53 For each P
j; k 2 P
ZC , let ajk 2 R be given, such that k0 jajk j
converges for each j 0 and j0 k0 jajk j also converges. Then:
P
P
ajk converges P
for each
(a) Pk0 P
P j 0 and j0 ajk converges for each k 0.
(b)
j0
k0 ajk and
k0
j0 ajk converge and
XX
j0 k0
ajk D
XX
ajk :
(11.36)
k0 j0
Proof First of all, given m; n 2 N, our hypothesis gives
n X
m
X
kD0 jD0
jajk j D
m X
n
X
jajk j jD0 kD0
m X
X
jD0 k0
jajk j XX
j0 k0
Hence, letting m ! C1 we get
n X
X
kD0 j0
jajk j XX
j0 k0
jajk j < C1;
jajk j < C1:
11.5 A Glimpse on Analytic Functions
501
P
for every n 2 N. Therefore, j0 jajk j < C1 for every k 0 and, letting n ! C1
in the above inequality, we get
XX
jajk j k0 j0
XX
jajk j < C1:
j0 k0
P P
Thus, k0 j0 jajk j also converges.
P
Now we can prove (a) and (b), except for (11.36): since j0 jajk j converges for
ˇ
ˇP
P
ˇ
ˇ
every k 0, it is also true that j0 ajk converges for every k 0, with ˇ j0 ajk ˇ P
j0 jajk j. Summing this inequality over k 0, we get
X ˇˇ X ˇˇ X X
ajk ˇ jajk j < C1:
ˇ
k0
j0
k0 j0
P P
Therefore, k0 j0 ajk is absolutely convergent, hence convergent.
P Arguing in
an
entirely
analogous
way,
we
conclude
that
both
of
the
series
k0 ajk and
P P
a
are
(absolutely)
convergent.
jk
k0
j0
ˇ
ˇP P
P P
ˇ
ˇ
In order to prove (11.36), let D ˇ j0 k0 ajk k0 j0 ajk ˇ. Then, for
m; n 2 N, we have
m X
n X
ˇX
XX
X X ˇˇ
X
ˇ
Dˇ
ajk C
ajk ajk ajk ˇ
jD0 k0
j>m k0
kD0 j0
k>n j0
m X
n X
ˇX
ˇ ˇXX ˇ ˇXX ˇ
X
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇ
ajk ajk ˇ C ˇ
ajk ˇ C ˇ
ajk ˇ:
jD0 k0
kD0 j0
j>m k0
k>n j0
P P
P P
Let > 0 be given. Since j0 k0 jajk j and k0 j0 jajk j converge, we
can choose m0 ; n0 2 N such that
XX
XX
m m0 )
jajk j < and n n0 )
jajk j < :
j>m k0
k>n j0
ˇ
ˇP P
ˇ
ˇP P
ˇ
ˇ
ˇ
ˇ
In turn, this gives ˇ j>m k0 ajk ˇ < and ˇ k>n j0 ajk ˇ < , so that, with
m m0 and n n0 , we get in the above estimate for 502
11 Series of Functions
m
n
m X
n X
ˇ
ˇXX
ˇ
ˇX
XX
X
ˇ
ˇ
ˇ
ˇ
ˇ
ajk ajk ˇ C 2 D ˇ
ajk ajk ˇ C 2
jD0 k0
kD0 j0
k0 jD0
j0 kD0
m
n
n X
m X
m
n
ˇX
ˇ ˇXX
ˇ ˇXX
ˇ
X
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇ
ajk ajk ˇ C ˇ
ajk ˇ C ˇ
ajk ˇ C 2
kD0 jD0
m
XX
k>n jD0
jD0 kD0
jajk j C
k>n jD0
n
XX
jajk j C 2 j>m kD0
XX
j>m kD0
jajk j C
k>n j0
XX
jajk j C 2 < 4:
j>m k0
t
u
We are now in position to prove the following
Theorem 11.54 If f W .x0 R; x0 C R/ ! R is given by the power series f .x/ D
P
k
for a given y0 2 .x0 R; x0 CR/,
k0 ak .xx0 / , then f is analytic. More precisely,
P
there exist b0 ; b1 ; b2 ; : : : 2 R such that f .x/ D k0 bk .x y0 /k in .y0 r; y0 C r/,
where r D R jy0 x0 j > 0.
Proof First of all, note that .y0 r; y0 C r/ .x0 R; x0 C R/. Now, for x 2
.y0 r; y0 C r/, we can write
f .x/ D
X
ak .x x0 /k D
k0
D
X
!
ak
k0
X
k
ak .x y0 / C .y0 x0 /
k0
k
X
k
.y0 x0 /kj .x y0 /j :
j
jD0
If we can change the order of the sums in the last expression above, we will get
f .x/ D
XX
j0
Letting bj D
P
k
kj ak j .y0
ak
kj
!
k
.y0 x0 /kj .x y0 /j :
j
x0 /kj , we will have
f .x/ D
X
bj .x y0 /j ;
j0
thus finishing the proof.
Therefore, we are left to justifying the equality
!
!
k
XX
X
k
k
kj
j
.y0 x0 / .x y0 / D
.y0 x0 /kj .x y0 /j :
ak
ak
j
j
kj
k0
jD0
j0
X
11.5 A Glimpse on Analytic Functions
503
According to the last proposition, this will be true provided
!
k
jak j
jy0 x0 jkj jx y0 jj < C1:
j
jD0
k
XX
k0
However, this expression equals
X
k
jak j jy0 x0 j C jx y0 j ;
k0
which is finite, for jy0 x0 j C jx y0 j < jy0 x0 j C r D R and
absolutely for u 2 .R; R/.
P
k0
ak uk converges
t
u
We show next that the set of zeros of an analytic function f W I ! R do not
accumulate on I. To this end, we first need a formal definition of this concept.
Definition 11.55 Given X R and a 2 R, we say that a is an accumulation point
of X if there exists a sequence .an /n1 of pairwise distinct elements of X such that
n
an ! a.
Theorem 11.56 Let I R be an open interval, f W I ! R be an analytic function
and Z D fx 2 II f .x/ D 0g. The following conditions are equivalent:
(a) Z has no accumulation point in I.
(b) There exists x0 2 I such that f .k/ .x0 / D 0 for every integer k 0.
(c) f vanishes identically.
Proof (a) ) (b): let x0 2 I be an accumulation point of Z, say x0 D limn!C1 xn ,
where .xn /n1 is a sequence of pairwise distinct elements of Z. Since f .xn / D 0 for
every n 1, the continuity of f gives f .x0 / D 0.
By contradiction, assume that there exists an integer k0 1 such that
f .k0 / .x0 / ¤ 0. Without loss of generality, we can assume that k0 is minimum
with such a property. By the analyticity of f , we can choose R > 0 such that
.x0 R; x0 C R/ I and, in such an interval,
f .x/ D
X f .k/ .x0 /
X f .k/ .x0 /
.x x0 /k D .x x0 /k0
.x x0 /kk0 :
kŠ
kŠ
kk
kk
0
0
P
f .k/ .x0 /
kk0
If g.x/ D
, then the series which defines g converges in
kk0
kŠ .x x0 /
f .x/
.x0 R; x0 C R/ n fx0 g (for g.x/ D .xx
k in such an interval). Since it obviously
0/ 0
converges at x0 , we conclude that g is continuous in .x0 R; x0 C R/, with g.x0 / D
f .k0 / .x0 /
¤ 0. By the sign preserving lemma for continuous functions, there exists
k0 Š
0 < ı R such that g ¤ 0 on .x0 ı; x0 C ı/. Since f .x/ D g.x/.x x0 /k0 in this
last interval, we conclude that f ¤ 0 in .x0 ı; x0 C ı/ n fx0 g. But this contradicts
the fact that xn 2 .x0 ı; x0 C ı/ n fx0 g for every sufficiently large n.
504
11 Series of Functions
(b) ) (c): let x0 2 I be such that f .k/ .x0 / D 0 for every integer k 0, and let
R be the radius of convergence of the power series defining f around x0 . Then, for
x 2 .x0 R; x0 C R/ we have
f .x/ D
X f .k/ .x0 /
.x x0 /k D 0;
kŠ
kk
0
i.e., f vanishes identically in .x0 R; x0 C R/.
Let’s use this fact to show f vanishes identically in I \ Œx0 ; C1/. By contradiction, assume that there would exist a 2 I \ Œx0 ; C1/ such that f .a/ ¤ 0. If
A D fx 2 I \ Œx0 ; C1/I f .x/ ¤ 0g;
then A ¤ ; (since a 2 A) and x0 C R is a lower bound for A; therefore, A has an
infimum x00 , which obviously satisfies x0 C R x00 a. Hence, x00 2 I and, since
x00 is the greatest lower bound of A, we conclude that Œx0 ; x00 / \ A D ;. This is the
same as saying that f vanishes identically on Œx0 ; x00 /, hence on Œx0 ; x00 . In turn, this
gives that f .k/ .x00 / D 0 for every integer k 0. Now, an argument analogous to the
one in the previous paragraph guarantees the existence of R0 > 0 such that f .x/ D 0
for every x 2 .x00 R0 ; x00 C R0 /. Therefore, x0 could not be the infimum of A, which
is an absurd.
Finally, by the same token one can prove that f vanishes identically in I \
.1; x0 , so that f vanishes identically in I.
(c) ) (a): obvious!
t
u
We now collect two straightforward consequences of the previous result which
will be useful for what comes next.
Corollary 11.57 Let I R be an open interval and f ; g W I ! R be analytic
functions. If f .j/ .x0 / D g.j/ .x0 / for some x0 2 I and every integer j 0, then f D g.
Proof Letting h D f g, we get h.j/ .x0 / D 0 for every integer j 0. Since h is also
analytic on I, the previous result guarantees that h vanishes identically on I, and this
is the same as having f D g.
t
u
Corollary 11.58 Let f W .R; R/ ! R be an analytic function, possibly with R D
C1. If f vanishes somewhere in .0; R/ but doesn’t vanish identically, then f has a
smallest positive zero. More precisely, there exists a 2 .0; R/ such that f .a/ D 0 but
f doesn’t change sign on .0; a/.
Proof Otherwise, there would exist a sequence .xn /n1 of zeros of f in .0; R/, with
x1 > x2 > . If x0 D limn!C1 xn , then x0 would be an accumulation point of
the set of zeros of f in Œ0; R/, hence in its interval of convergence. By the previous
result, f would vanish identically, which is a contradiction.
t
u
We are now in position to undertake a purely analytical development of
Trigonometry. To this end, first notice that if sin and cos are to satisfy sin0 D cos
11.5 A Glimpse on Analytic Functions
505
and cos0 D sin, then they are infinitely differentiable and ultimately must be given
by their Taylor series centered at 0, as in Example 11.4.
The key to free Trigonometry from Euclidean Geometry is to turn this around,
defining sin and cos by means of those power series. Therefore, we start by letting
f .x/ D
X .1/j1
x2j1
.2j
1/Š
j1
and g.x/ D
X .1/j
j0
.2j/Š
x2j
(11.37)
for every x 2 R.
The analytic foundations of Trigonometry are the content of the four coming
results.
Proposition 11.59 With notations as above, we have that:
(a) f and g are well defined and analytical, with f odd and g even.
(b) f .4k/ D f , f .4kC1/ D g, f .4kC2/ D f , f .4kC3/ D g and g.4k/ D g, g.4kC1/ D f ,
g.4kC2/ D g, g.4kC3/ D f , for every integer k 0.
(c) f and g are nonconstant and such that f .x/2 C g.x/2 D 1 for every x 2 R.
(d) f .x C y/ D f .x/g.y/ C f .y/g.x/ and f .x y/ D f .x/g.y/ f .y/g.x/ for every
x; y 2 R. In particular, f .2x/ D 2f .x/g.x/ for every x 2 R.
P
k
Proof (a) The convergence of k0 jxjkŠ , together with the comparison test for
series, assures the absolute convergence, in the whole real line, of both power
series in (11.37). Then, Theorem 11.54 shows that they are analytical functions.
Now, the oddness of f and the evenness of g follow from their very definitions.
(b) Theorem 11.27 gives
f 0 .x/ D
X .1/j1
X .1/i
x2j2 D
x2i D g.x/I
.2j
2/Š
.2i/Š
j1
i0
analogously, g0 D f . Now, f 00 D g0 D f , g00 D .f /0 D g and the given
relations easily follow by induction.
(c) It follows from (b) that .f 2 Cg2 /0 D 2ff 0 C2gg0 D 2fgC2g.f / D 0. Therefore,
f 2 C g2 is constant and, since f .0/ D 0 and g.0/ D 1, we get f .x/2 C g.x/2 D 1
for every x 2 R. Also, f 0 .0/ D g.0/ D 1 and g00 .0/ D g.0/ D 1 assure that
f and g are nonconstant.
(d) For a fixed y 2 R, let h.x/ D f .x C y/ and l.x/ D f .x/g.y/ C f .y/g.x/. By the
chain rule, h; l W R ! R are infinitely differentiable. Moreover, by item (b) and
the chain rule again, these functions satisfy
h00 C h D 0 on R
and
h.0/ D f .y/; h0 .0/ D f 0 .y/
l00 C l D 0 on R
l.0/ D f .y/; l0 .0/ D g.y/
Since f 0 .y/ D g.y/, the uniqueness part of the proof of Theorem 11.49 (which
did not depend on any properties of sin and cos) guarantees that h D l.
506
11 Series of Functions
Now, the first formula, together with the fact that f is odd and g is even, gives
f .x y/ D f .x/g.y/ C f .y/g.x/ D f .x/g.y/ f .y/g.x/:
Finally, the expression for f .2x/ also follows from the first formula, once we let
y D x.
u
t
Proposition 11.60 The function f has a smallest positive zero.
Proof Start by observing that, with item (b) of the above proposition at our disposal,
the results of Problem 4, page 325 remain true for f and g in place of sin and cos,
respectively, so that
x
x3
f .x/ x
3Š
and g.x/ 1 x2
2Š
for every x 0. In turn, this gives
f .x/ x x3
x5
C
3Š
5Š
and g.x/ 1 x2
x4
C
2Š
4Š
for every x 0. For instance, for x 0 one has
x2 x4
x3
d
g.x/ 1 C
D f .x/ C x 0;
dx
2Š 4Š
3Š
2
4
so that g.x/ 1 C x2Š x4Š g.0/ 1 D 0.
p
3
Now, the maximum of x x3Š for x 0 is attained at x D 2 (check it!).
Therefore, it follows from the above that
p
p
f . 2/ 2 On the other hand, the minimum of 1 2
(check it too!). Since f .x/ x 1 x2Š C
p
p 3
2 2
2
D
:
3Š
3
x2
2Š
x4
4Š
C
x4
4Š
for x 0 is attained at x D
p
10
for x 0, this gives
p 4
p
p 2
p
p 10
10
10
C
D
:
f . 10/ 10 1 2Š
4Š
6
p
p p
the IVT (applied to the intervals Œ0; 2 and Œ 2; 10
p
p
p
assures the existence of 0 < ˛ < 2 < ˇ < 10 such that f .˛/ D f .ˇ/ D 610 .
Then, the MVT of Lagrange gives 2 .˛; ˇ/ such that f 0 . / D 0.
It follows from the last part of item (d) of the previous proposition that
Since
p
2 2
3
>
p
10
6 ,
f .2 / D 2f . /g. / D 2f . /f 0 . / D 0;
11.5 A Glimpse on Analytic Functions
507
so that f has a positive root. Since f doesn’t vanish identically, Corollary 11.58
shows that it has a smallest positive zero, as wished.
t
u
From now on, we let denote the smallest positive zero of f . The last part of
item (d) of Proposition 11.59 gives
;
2
and since f is positive on .0; / this shows that g 2 D 0. Item (c) of Proposition 11.59 allows us to write
r
2
0<f
D ˙ 1g
D ˙1;
2
2
so that f 2 D 1. Then, item (d) of Proposition 11.59 gives
0 D f ./ D 2f
f
2
x Df
2
2
g.x/ f .x/g
g
2
D g.x/:
(11.38)
We are finally in position to go one step further.
Proposition 11.61 With notations as above, we have that:
(a)
(b)
(c)
(d)
(e)
g W Œ0; ! Œ1; 1 is a decreasing bijection with continuous inverse.
g1 W .1; 1/ ! .0; / is differentiable.
Im.f / D Im.g/ D Œ1; 1.
f .x C / D f .x/ and g.x C / D g.x/.
f and g are periodic, with period 2.
Proof (a) Since g0 D f < 0 in .0; /, we conclude that g is decreasing in Œ0; .
Since
p
1 D g.0/ > g./ D ˙ 1 f ./2 D ˙1;
we conclude that g./ D 1. The rest follows from Theorem 8.35.
(b) This follows directly from Theorem 9.28. Indeed, since f ¤ 0 in .0; /, we get
for x 2 .0; / that g0 .x/ D f .x/ ¤ 0.
(c) Item (c) of Proposition 11.59 gives Im.g/ Œ1; 1. Then, item (a) above gives
Im.g/ D Œ1; 1, from where (11.38) gives Im.f / D Œ1; 1.
(d) An application of item (d) of Proposition 11.59, together with g./ D 1, gives
f .x C / D f .x/g./ C f ./g.x/ D f .x/:
Therefore,
f
3
2
Df
2
C D f
2
D 1
508
11 Series of Functions
and, hence,
g
3
2
r
D ˙ 1f
3
2
2
D 0:
In turn, (11.38) gives
D f x C
g.x C / D f x 2
2
f
g.x/ D g.x/:
D f .x/g
2
2
(e) Firstly, two applications of (d) give
f .x C 2/ D f .x C / C D f .x C / D f .x/
and, analogously, g.x C 2/ D g.x/.
Now, let p > 0 be such that f .x C p/ D f .x/ for every x 2 R. Then
f .p/ D f .0/ D 0, so that p (for, is the smallest positive zero of f ). Since
f .x C / D f .x/ and f is not identically zero, we also have p ¤ . If < p < 2,
then 0 < p < and
f .p / D f .p/g./ f ./g.p/ D 0;
so that p . Therefore, p 2, and 2 is the smallest p > 0 such that
f .x C p/ D f .x/ for every x 2 R.
Finally, we leave to the reader the (analogous) task of checking that 2 is the
smallest p > 0 such that g.x C p/ D g.x/ for every x 2 R.
t
u
Our final result shows that f and g parametrize the unit circle of the cartesian
plane.
Proposition 11.62 For every point .x; y/ in the unit circle x2 C y2 D 1 of the
cartesian plane, there exists a unique 2 Œ0; 2/ such that x D g. / and y D f . /.
Moreover, if y0 0, then can be chosen in the interval Œ0; . Also, as varies
from 0 to 2, the point .g. /; f . // turns around the circle exactly once.
Proof For the existence part, let .x0 ; y0 / belong to the unit circle, so that x20 Cy20 D 1.
Since g is periodic of period 2 and Im.g/ D Œ1; 1, there exists ˛ 2 Œ0; 2/ such
that x0 D g.˛/. Therefore,
y20 D 1 x20 D 1 g.˛/2 D f .˛/2 ;
so that y0 D ˙f .˛/. If y0 D f .˛/, let D ˛. If y0 D f .˛/, then the evenness g,
the oddness of f and their periodicity give
x0 D g.˛/ D g.˛/ D g.2 ˛/
;
y0 D f .˛/ D f .˛/ D f .2 ˛/
and it suffices to let
D 2 ˛.
11.5 A Glimpse on Analytic Functions
509
For the uniqueness part, assume, for the sake of contradiction, that there exist
0 1 < 2 < 2 such that .x0 ; y0 / D .g. 1 /; f . 1 // D .g. 2 /; f . 2 //. Then,
f.
2
1/
D f . 2 /g. 1 / f . 1 /g. 2 / D 0;
with 0 < 2 1 < 2. Since is the smallest positive zero of f and the next one
is 2, this gives 2 1 D . However, if this is so, then
x0 D g. 2 / D g. 1 C / D g. 1 / D x0
;
y0 D f . 2 / D f . 1 C / D f . 1 / D y0
so that x0 D y0 D 0. This is a contradiction to the fact that .x0 ; y0 / belongs to the
unit circle.
We leave the rest as an (easy) exercise to the reader.
t
u
At this point, the reader might perhaps want to look again at Example 10.80 to
realize that, in view of the four propositions above, that result gives a genuine proof
of the fact that the unit circle has length 2.
On the one hand, this validates the whole construction of the trigonometric
functions, as usually done in elementary school Trigonometry. On the other, Leibniz
formula for (cf. Problem 9, page 484) provides a purely analytical way of getting
numerical approximations for , which can be used to get the usual estimate
Š 3:14159, with five correct decimal places.
Problems: Section 11.5
1. Prove that, for every integer k 1,
2k
2kC1
X
X .1/j1
.1/j1 2j1
x
x2j1
sin x .2j
1/Š
.2j
1/Š
jD0
jD0
and
2k1
X
jD0
2k
X .1/j
.1/j 2j
x cos x x2j :
.2j/Š
.2j/Š
jD0
2. One can prove that if I; J R are open intervals and g W I ! J and f W J ! R are
analytic, then so is f ı g W I ! R. Use this fact to show that log W .0; C1/ ! R
is analytic and to find its Taylor series expansion around x0 > 0.
510
11 Series of Functions
3. Prove the following theorem of S. Bernstein9 : let I R be an open interval,
f W I ! R be infinitely differentiable and such that f .n/ .x/ 0 for every x 2 I
and every sufficiently large natural n. Then, f is analytic.
4. Let I R be an open interval and p; q W I ! R be analytic functions. The
purpose of this problem is to show that, for ˛; ˇ 2 R and x0 2 I, there exists a
unique analytic function f W I ! R such that
f 00 C pf 0 C qf D 0
:
f .x0 / D ˛; f 0 .x0 / D ˇ
(11.39)
To this end, do the following items:
(a) Show that (11.39) has at most one solution.
(b) Show that there exist some open interval J I, containing x0 , and an analytic
function fJ W J ! R satisfying (11.39).
(c) If J is as in (b) and J ¤ I, show that there exist an interval J 0 I, properly
containing J, and an analytic function fJ 0 W J 0 ! R satisfying (11.39) and
such that fJ 0 D fJ along J.
(d) Finish the proof of the problem.
9
After S. N. Bernstein, Russian mathematician of the XX century.
Bibliography
1.
2.
3.
4.
S. Abbott, Understanding Analysis (Springer, New York, 2001)
T. Apostol, Calculus, vol. 2 (Wiley, New York, 1967)
T. Apostol, Introduction to Analytic Number Theory (Springer, New York, 1976)
A. Caminha, An Excursion Through Elementary Mathematics II - Euclidean Geometry
(Springer, New York, 2018)
5. A. Caminha, An Excursion Through Elementary Mathematics III - Discrete Mathematics and
Polynomial Algebra (Springer, New York, 2018)
6. L.W. Cohen, G. Ehrlich, The Structure of the Real Number System (Literary Licensing,
Whitefish, 2012)
7. K. Davidson, A. Donsig, Real Analysis with Real Applications (Prentice Hall, Inc., Upper
Saddle River, 2002)
8. M.P. do Carmo, Differential Geometry of Curves and Surfaces (Prentice-Hall, Inc., Englewood
Cliffs, 1976)
9. D.G. de Figueiredo, Análise I (in Portuguese) (LTC, Rio de Janeiro, 1996)
10. D.G. de Figueiredo, Análise de Fourier e Equações Diferenciais Parciais (in Portuguese)
(IMPA, Rio de Janeiro, 2012)
11. D.G. de Figueiredo, Números Irracionais e Transcendentes (in Portuguese) (Brazilian Mathematical Society, Rio de Janeiro, 2002)
12. O. Foster, Lectures on Riemann Surfaces (Springer, New York, 1981)
13. C. Goffman, Real Functions (Rinehart & Company, Inc., New York, 1953)
14. C.R. Hadlock, Field Theory and Its Classical Problems (MAA, Washington, 1978)
15. P.R. Halmos, Naive Set Theory (Springer, New York, 1974)
16. D. Hilbert, Foundations of Geometry (Open Court Publ. Co., Peru, 1999)
17. H. Hochstadt, The Functions of Mathematical Physics (Dover, Mineola, 1986)
18. K. Hoffman, R. Kunze, Linear Algebra (Prentice-Hall, Boston, 1971)
19. D. Kreider, R. Küller, D. Ostberg, F. Perkins, An Introduction to Linear Analysis (AddisonWesley Publ. Co., Boston, 1966)
20. W. Rudin, Principles of Mathematical Analysis (McGraw-Hill, Inc, New York, 1976)
21. E. Scheinerman, Mathematics, a Discrete Introduction (Cengage Learning, Boston, 2012)
22. G.F. Simmons, Calculus Gems. Brief Lives and Memorable Mathematics, 3rd edn. (McGrawHill, Inc., New York, 1992)
23. G.F. Simmons, Differential Equations with Applications and Historical Notes, 3rd edn.
(Chapman and Hall/CRC, Boca Raton, 2016)
24. E. Stein, R. Shakarchi, Fourier Analysis, an Introduction (Princeton University Press, Princeton, 2003)
25. I. Stewart, Why Beauty Is Truth: A History of Symmetry (Basic Books, New York, 2008)
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6
511
512
Bibliography
26. M.B.W. Tent, The Prince of Mathematics: Carl Friedrich Gauss (A.K. Peters Ltd, Wellesley,
2006)
27. R. Wheeden, A. Zygmund, Measure and Integral: An Introduction to Real Analysis
(Chapman & Hall, New York, 1977)
Appendix A
Glossary
Problems tagged with a country’s name refer to any round of the corresponding
national mathematical olympiad. For example a problem tagged “Brazil” means that
it appeared in some round of some edition of the Brazilian Mathematical Olympiad.
Problems proposed in other mathematical competitions, or which appeared in
mathematical journals, are tagged with a specific set of initials, as listed below:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
APMO: Asian-Pacific Mathematical Olympiad.
Austrian-Polish: Austrian-Polish Mathematical Olympiad.
BMO: Balkan Mathematical Olympiad.
Berkeley: Berkeley Preliminary Examination.
Baltic Way: Baltic Way Mathematical Contest.
Crux: Crux Mathematicorum, a mathematical journal of the Canadian Mathematical Society.
EKMC: Eötvös-Kürschák Mathematics Competition (Hungary).
IMO: International Mathematical Olympiad.
IMO shortlist: problem proposed to the IMO, though not used.
Israel-Hungary: Binational Mathematical Competition Israel-Hungary.
Miklós-Schweitzer: The Miklós-Schweitzer Mathematics Competition
(Hungary).
NMC: Nordic Mathematical Contest.
OCM: State of Ceará Mathematical Olympiad.
OCS: South Cone Mathematical Olympiad.
OBMU: Brazilian Mathematical Olympiad for University Students.
OIM: Iberoamerican Mathematical Olympiad.
OIM shortlist: problem proposed to the OIM, though not used.
OIMU: Iberoamerican Mathematical Olympiad for University Students.
Putnam: The William Lowell Mathematics Competition.
TT: The Tournament of the Towns.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6
513
Appendix B
Hints and Solutions
Section 1.1
1. Write ab D dc D r, thus obtaining a D br, c D dr and, then, a ˙ c D r.b ˙ d/.
2. Let x D 0:a1 a2 a3 : : : and suppose that the sequence .a1 ; a2 ; a3 ; : : :/ is periodic
from some point on, as in (1.3), say. If y 2 N is the integer with decimal
representation b1 b2 : : : bp , conclude that 10lCp x D yC10l x, so that x D 10lCpy10p ,
a rational number. Conversely, let x D ab , with a; b 2 N and 0 < a b. If yk 2 N
is the integer with decimal representation a1 a2 : : : ak , use the division algorithm
to conclude that 10k a D byk C rk , with 0 rk < b. Then, use the fact that there
is only a finite number of possibilities for rk to assure the existence of natural
numbers l and p such that rlCp D rl . From this point on, conclude that the list
.a1 ; a2 ; a3 ; : : :/ is as in (1.3), with b1 D alC1 , b2 D alC2 , . . . , bp D alCp .
3. Adapt, to the present case, the proof of the uniqueness of additive inverses,
changing C by and 0 by 1.
4. Start by observing that 0 D 0 a D .1 C .1//a D 1 a C .1/a D a C .1/a;
then, use the uniqueness of additive inverses.
5. Among the given decimal representations, the only one that does not correspond
to a rational number is that of item (d); in order to prove this claim, observe
that, in the given decimal representation, there will be arbitrarily long sequences
of zeros. In order to write the numbers of items (a), (b) and (c) as irreducible
fractions, follow the steps suggested in the hint to Problem 2.
© Springer International Publishing AG 2017
A. Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume I,
Problem Books in Mathematics, DOI 10.1007/978-3-319-53871-6
515
516
B Hints and Solutions
Section 1.2
1. For items (b) and (c), use (7’) and (a); item (g) follows from items (b) and (c).
2. with equality if and only if a D b D c D 0. Adapt the proof of Corollary 1.5.
3. For item (a), observe that .rs/n D .rs/ .rs/ D .„ƒ‚…
r r / .„ƒ‚…
s s / D rn sn . For
„ ƒ‚ …
n
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
n
n
the other items, apply analogous arguments.
Consider separately the cases m < n, m D n and m > n. In each case, make
judicious use of the property of item (b) of the last problem.
nk nl
If b D 2k 5l , with k; l 2 ZC , and n D maxfk; lg, then ab D a2 10n5 .
If n D 2k, with k 2 N, write xn D .x2 /k and, then, apply item (g) of
Proposition 1.2. The case of an odd n can be treated similarly.
Start by showing that 12 13 C 14 15 > 15 .
aC2
Compute the difference aC1
bC1 bC2 .
Letting S denote the sum of the ten given numbers, show that 4S can be written
as the sum of ten real numbers, each of which equals the sum of four of the ten
given numbers.
Argue as in Example 1.4, using the fact that 31 < 32 D 25 and 17 > 16D 24 .
n
For items (a) and (b), begin by observing that an < bn if and only if ab < 1;
then, apply the result of Corollary
For item
(c), begin by noticing that
1.3.
a n
b n
an C bn < .a C b/n if and only if aCb
C aCb
< 1; then, apply the result
of Corollary 1.3.
2
Use the fact that a3 < ca2 and b3 < cb
then,
theorem.
1 and,
n apply
Pythagoras’
n
n
The given inequality is equivalent to 7 C 47 C 97 2. This last one
holds trivially for n D 1 and n D 2; for n 3, apply item (b) of Corollary 1.3
n 3
to conclude that 97 97 > 2.
Start by observing that, if a 4, then 0 < 1a C 1b C 1c 14 C 14 C 14 < 1, so that
1
C 1b C 1c cannot be an integer.
a
If a > 5 is an integer, show that 4.a 4/ > a; from this, conclude that it is not
worth to have summands greater that 5. Then, by performing similar changes
of summands, discard all of those which are equal to 4 or 5. Finally, show that
is more advantageous to have more summands equal to 3 than summands equal
to 2.
If we let a be the common leftmost digit, show that there exist nonnegative
integers k and l such that a10k < 2n < .aC1/10k and a10l < 5n < .aC1/10l ;
then, multiply these inequalities.
Suppose that all three inequalities are true; the second one can be written as
.a d/.c b/ < 0, so that the first one gives a d < 0 < c b. Now, use the
third inequality to get ad.c b/ < bc.a d/ and, then, arrive at a contradiction.
B Hints and Solutions
517
Section 1.3
p p
p
2. For item (a), for instance, show that . n xy/n D xy and . n x n y/n D xy; then, use
the definition of the n-th root.
3. By contraposition,1 show that, if b ¤ 0, then r is rational.
4. Reduce this problem to the previous one.
5. Argue by contraposition. p
p
p
6. If ab ¤ 0, compute .a C b 2/2 D .c 3/2 to conclude that 2 should be
rational, which is an absurd. Therefore, ab D 0 and, thus, a D 0 or b D 0. Then,
apply the result of Problem 3.
p
7. By the sake of contradiction, write 2 D ab , with a and b relatively prime natural
numbers, so that 2b2 D a2 . Then, successively show that a is even and b is even,
thus reaching a contradiction.
8. Adapt, to the present case, the hint given to the previous problem.
Section 2.1
1. For item (a), it suffices to develop the right hand side to get
.x y/.x C y/ D x.x C y/ y.x C y/
D .x2 C xy/ .xy C y2 /
D x2 y 2 :
In what concerns (b), we have
.x ˙ y/2 D .x ˙ y/.x ˙ y/ D x.x ˙ y/ ˙ y.x ˙ y/
D .x2 ˙ xy/ ˙ .xy ˙ y2 / D x2 ˙ 2xy C y2 :
Finally, item (c) is also obtained by expanding the right hand side, and we leave
this to the reader.
The standard ways to prove a proposition of the form A ) B (i.e., If A, then B) are either directly,
by contraposition or by contradiction. In the first case, we assume the validity of assertion A and
deduce the validity of assertion B directly; in the second case, we assume that assertion B is false
and deduce, directly, that assertion A is also false; finally, in the third case, we assume that assertion
A is true and assertion B is false and, from this, directly deduce a contradiction (i.e., deduce that an
assertion that is obviously false should be true, which is something impossible to happen). For a
detailed discussion on the fundamentals of Logic and methods of proof, we refer the reader to [21].
1
518
B Hints and Solutions
2. It suffices to notice that
n
p
m2 C n2 C p2
m
C
C
D
np
mp
mn
mnp
D
.m C n C p/2 2.mn C mp C np/
:
mnp
3. The statement of the problem, together with item (a) of Proposition 2.1, gives
us
2 b
1b 2
D
, .b.1 a//2 .a.1 b//2 D 0
a
1a
, .b.1 a/ a.1 b//.b.1 a/ C a.1 b// D 0
, .b a/.a C b 2ab/ D 0:
However, since a ¤ b, it follows that a C b D 2ab. Hence, we finally arrive at
1
aCb
2ab
1
C D
D
D 2:
a
b
ab
ab
4. Successively applying items (b) and (a) of Proposition 2.1, we get
2
2
1 xy
1 yx
p
p
p
p
p D
.x 2 xy C y/ C 2 xy
. x y/2 C 2 xy
D
.x y/.x C y/
xy
x2 y2
D
D 2 :
2
2
x .x C y/
x .x C y/
x
5. Applying item (a) of Proposition 2.1, we obtain
.x3 C y3 C z3 /2 .x3 y3 z3 /2
D
yCz
D
Œ.x3 C y3 C z3 / .x3 y3 z3 /Œ.x3 C y3 C z3 / C .x3 y3 z3 /
yCz
D
2.y3 C z3 / 2x3
yCz
D
2.y C z/.y2 yz C z2 / 2x3
yCz
D 4x3 .y2 yz C z2 /:
B Hints and Solutions
519
1
a
6. Since ab D 1 ,
Db,
1
b
D a, we have
a 1a b C 1b
.a b/ .b C a/
a2 b2
D
D
D 1;
a2 b2
a2 b2
a2 b2
where we have used item (a) of Proposition 2.1 in the next to last equality above.
7. We have .y x/.y C x/ D 192 , with y x and y C x integers such that
0 < y x < y C x. Therefore, the only possible choice is to have y x D 1 and
y C x D 361, so that y D 181 and x D 180.
8. A judicious application of item (b) of Proposition 2.1 gives
a4 C b4 D .a2 C b2 /2 2a2 b2 D Œ.a C b/2 2ab2 2.ab/2
D .m2 2n/2 2n2 D m4 4m2 n C 2n2 :
9. Item (c) of Proposition 2.1 gives
a6 C b6 D .a2 /3 C .b2 /3 D .a2 C b2 /.a4 a2 b2 C b4 /
D .a4 C b4 / a2 b2 D .a2 C b2 /2 3a2 b2
D 1 3.ab/2;
so that
13.ab/2
a6 Cb6
D
a6 Cb6
a6 Cb6
D 1.
10. Start by observing that
.ac C bd/2 C .ad bc/2 D
D .a2 c2 C 2acbd C b2 d2 / C .a2 d2 2adbc C b2 c2 /
D a2 .c2 C d2 / C b2 .d 2 C c2 / D .a2 C b2 /.c2 C d2 /:
11. Adding 1 to both sides of the given equation, we get
112 D 121 D x C y C xy C 1 D .x C 1/.y C 1/:
However, since x C 1; y C 1 > 1, the only possible choice is to have x C 1 D
y C 1 D 11.
p
p
12. For item (a), it follows from item (a) of Proposition 2.1, with x and y in
place of x and y, respectively, that
p
p
x y
1
p
p D p
p p
p
x˙ y
. x ˙ y/. x y/
p
p
p
p
x y
x y
:
D
D p 2
p
. x/ . y/2
xy
520
B Hints and Solutions
p
p
As for (b), we apply item (c) of Proposition 2.1, with 3 x and 3 y in place of
p
p
p
3
x and y, respectively, and observe that . 3 x/2 D x2 and . 3 x/3 D x (and
analogously for y). This way, we get
p
p
p
3 2
x 3 xy C 3 y2
1
p
p
p D p
p p
p
3
3
x˙ 3 y
. 3 x ˙ 3 y/. x2 3 xy C 3 y2 /
p
p
p
3 2
x 3 xy C 3 y2
:
D
xy
Finally, item (c) follows immediately from (b).
13. Item (a) of the previous problem gives
2
p
p
1
.n C 1/ n
2
2
n C 1 n D 2p
p Dp
p < p D p :
2 n
n
nC1C n
nC1C n
The other inequality can be proved in a similar way.
14. Applying twice item (a) of Problem 12, we obtain
p
p
p
p
1
2C 2 3
2C 2 3
D
p
p D
p
p
p
2C 2C 3
.2 C 2/2 . 3/2
3C4 2
p
p
p
.2 C 2 3/.3 4 2/
p
D
32 .4 2/2
p
p
p
1
D .2 C 2 3/.3 4 2/:
23
p
15. First
of all,p
note that,
for all real x, we have x3pC 3 D x3 C . 3 3/3 D .x C
p
p
3
3/.x2 x 3 3 C 3 9/. Therefore, making x D 2, we get
p
p
p
p
p p
3
3
3
2 2 C 3 D . 2 C 3/.2 2 3 C 9/
and, hence,
p p
p
2 3 3C 3 9
p
p D
p
2C 3 3
3C2 2
p p
p
p
2 2 3 3C 3 9 32 2
D
p
p
3C2 2
32 2
p
p p
p
3
3
D .3 2 2/.2 2 3 C 9/:
1
2
16. It suffices to write y C z D x, x C z D y and x C y D z to realize that each
of y C z, x C z and x C y is also nonzero. Now, making use of these equalities
in (a), we obtain
B Hints and Solutions
521
y2
z2
x2
y2
z2
x2
C
C
D
C
C
D 3:
.y C z/2
.x C z/2
.x C y/2
.x/2
.y/2
.z/2
Analogously,
x3
.yCz/3
C
y3
.xCz/3
C
z3
.xCy/3
D 3.
17. Applying item (a) of Proposition 2.1 several times, we get
a64 b64 D .a32 C b32 /.a32 b32 /
D .a32 C b32 /.a16 C b16 /.a16 b16 /
D .a32 C b32 /.a16 C b16 /.a8 C b8 /.a8 b8 /
D .a32 C b32 /.a16 C b16 /.a8 C b8 /.a4 C b4 /.a4 b4 /
D .a32 C b32 /.a16 C b16 / : : : .a2 C b2 /.a2 b2 /
D .a32 C b32 /.a16 C b16 / : : : .a2 C b2 /.a C b/.a b/;
so that
.a C
b/.a2
C
a64 b64
D .a32 C b32 /.a C b/:
C b4 /.a8 C b8 /.a16 C b16 /
b2 /.a4
18. For item (a), it suffices to see that
.a b/.an1 C an2 b C an3 b2 C C abn2 C bn1 / D
D a.an1 C an2 b C an3 b2 C C abn2 C bn1 /
b.an1 C an2 b C an3 b2 C C abn2 C bn1 /
D .an C an1 b C an2 b2 C C a2 bn2 C abn1 /
.an1 b C an2 b2 C an3 b3 C C abn1 C bn /
D an bn :
With respect to (b), observe that the changes of sign make sense precisely
because n is odd:
.a C b/.an1 an2 b C an3 b2 abn2 C bn1 / D
D a.an1 an2 b C an3 b2 abn2 C bn1 /
C b.an1 an2 b C an3 b2 abn2 C bn1 /
D .an an1 b C an2 b2 a2 bn2 C abn1 /
C .an1 b an2 b2 C an3 b3 abn1 C bn /
D an C bn :
522
B Hints and Solutions
19. More generally, let us factorise x4n C 4y4n , where n 2 N, inspiring ourselves in
the formula for .a C b/2 :
x4n C 4y4n D .x2n /2 C .2y2n /2
D Œ.x2n /2 C .2y2n /2 C 2x2n 2y2n 2x2n 2y2n
D .x2n C 2y2n /2 .2xn yn /2
D .x2n C 2y2n C 2xn yn /.x2n C 2y2n 2xn yn /:
20. It follows from Example 2.7 that
.a C b C c/3 D a3 C b3 C c3 C 3.a C b/.a C c/.b C c/:
Now, since .a C b/ C .a C c/ C .b C c/ D 2.a C b C c/, which is an even
integer, we conclude that at least one of a C b, a C c or b C c is also even, so
that 3.a C b/.a C c/.b C c/ is a multiple of 6. Therefore,
6 j .a C b C c/ , 6 j .a C b C c/3
, 6 j Œa3 C b3 C c3 C 3.a C b/.a C c/.b C c/
, 6 j .a3 C b3 C c3 /:
21. We present two different solutions. For the first one, just note that
a C b C c D 0 ) .a C b/3 D .c/3
) a3 C b3 C 3ab.a C b/ D c3
) a3 C b3 C 3ab.c/ D c3
) a3 C b3 C c3 3abc D 0:
Another possibility is to use the result of Example 2.7:
a C b C c D 0 ) .a C b C c/3 D 0
) a3 C b3 C c3 C 3.a C b/.a C c/.b C c/ D 0:
Now, from a C b C c D 0 we get a C b D c, a C c D b, b C c D a and,
then,
3.a C b/.a C c/.b C c/ D 3.c/.b/.a/ D 3abc:
p
p
22. Since a C a2 b a a2 b, both sides of the equality we wish to prove
are nonnegative real numbers. Hence, it suffices to prove that their squares are
equal, i.e., that
B Hints and Solutions
523
p
a˙ b D
aC
!
p
a2 b
C
2
a
!
p
a2 b
2
v
!
!
p
p
u
2b
u a C a2 b
a
a
˙ 2t
2
2
s
D a˙2
p
a2 . a2 b/2
:
4
In view of the above, this is pretty clear.
23. By the sake of contradiction, suppose that such x, y and z did exist. Then,
1
writing xCyCz
1x D 1y C 1z and performing the additions on both sides, we
would easily conclude that x.x C y C z/ D yz, or, which is the same, that
x2 D .xy C xz C yz/. Similar reasoning would give us x2 D y2 D z2 D
.xy C xz C yz/. Now, from x2 D y2 we would get x D y or x D y; however,
since x C y ¤ 0, we should have x D y. Analogously, we should also have
1
x D z, so that the given equality would reduce to 3x
D 3x . This is obviously
impossible.
1
1
1
24. Making x D bc
, y D ca
and z D ab
, we have
xy C xz C yz D
D
1
1
1
C
C
.b c/.c a/
.b c/.a b/
.c a/.a b/
1
Œ.a b/ C .c a/ C .b c/ D 0:
.b c/.c a/.a b/
Now, if x; y; z are real numbers such that xy C xz C yz D 0, then we have
x2 C y2 C z2 D .x C y C z/2 . Therefore, in our case we have
1
1
1
C
C
D
2
2
.b c/
.c a/
.a b/2
Finally, just note that
25. Initially, note that
1
bc
C
1
ca
C
1
ab
1
1
1
C
C
bc
ca
ab
is a rational number.
p
p
p
p
p
3
3
3
. 3 a C b/3 D a C b C 3 ab. 3 a C b/
and, hence,
p
p
p
. 3 a C 3 b/3 .a C b/
3
ab D
p
2 Q:
p
3. 3 a C 3 b/
2
:
524
B Hints and Solutions
From this, we conclude that
p
p
ab
ab
3
3
a bD p
D p
p
p
p
p 2 Q:
3
3
3
3
2
2
. a/ C ab C . b/
. a C 3 b/2 3 ab
Finally,
p
3
aD
p
p
p
1 p
3
3
Œ. 3 a C b/ C . 3 a b/ 2 Q
2
p
and, analogously, 3 b 2 Q.
26. Raise both members of the equality a C c C d D .b C e C f / to the third power
and apply the formula of Example 2.7 to get
a3 Cc3 Cd 3 C3.aCc/.aCd/.cCd/ D .b3 Ce3 Cf 3 /3.bCe/.bCf /.eCf /:
Now, since a3 C b3 C c3 C d3 C e3 C f 3 D 0, it follows from the above equality
that
.a C c/.a C d/.c C d/ D .b C e/.b C f /.e C f /:
Analogously, starting from a C e C f D .b C c C d/ and arguing as above, we
find that
.a C e/.a C f /.e C f / D .b C c/.b C d/.c C d/:
Finally, multiplying the two relations thus obtained and cancelling out the
common factor .c C d/.e C f /, we arrive at the desired relation.
27. Start by observing that
b3 < b3 C 6ab C 1 b3 C 6b2 C 1 < b3 C 6b2 C 12b C 8 D .b C 2/3 :
Therefore, if b3 C 6ab C 1 is a perfect cube, one has
b3 C 6ab C 1 D .b C 1/3 D b3 C 3b2 C 3b C 1;
so that 2a D b C 1. Substituting b D 2a 1 into a3 C 6ab C 1, we conclude
that a3 C 6a.2a 1/ C 1 D a3 C 12a2 6a C 1 is a perfect cube. However, it
is immediate to verify that
a3 < a3 C 12a2 6a C 1 < .a C 4/3 ;
so that the only possibilities are
a3 C 12a2 6a C 1 D .a C 1/3 ; .a C 2/3 or .a C 3/3 :
B Hints and Solutions
525
These possibilities give, respectively, the equations 9a2 a 9a D 0, 6a2 18a D 7 and 3a2 33a D 26. The first one gives a D 1; the second has
no integer solutions, for the left hand side is even, while the right hand one is
odd; the third also has no integer solutions, for the left hand side is a multiple
of 3, while the right hand one is not.
Section 2.2
1. For the first inequality, interpret jx aj as the distance from x to a in the real line
(alternatively, apply the definition of modulus, separately considering the cases
x a 0 and x a < 0). For the other three inequalities, adapt the hint given to
the analysis of the first one.
2. Separately analyse the cases (i) x; y 0, (ii) x 0 > y, (iii) y 0 > x and (iv)
x; y < 0, showing that equality occurs in all of them.
3. For item (a), consider the cases x 0 and x < 0. For item (b), argue as in
Example 2.9.
4. Separately analyse cases x < 0, 0 < x < 1 and x > 1. The solution set is
.1; 0/ [ .1; C1/.
5. Since jyj y and j yj D jyj for all real y, we have jx aj x a and
jx bj D jb xj b x. Therefore, in order for the equation to have a real root
x, we must necessarily have
c D jx aj C jx bj .x a/ C .b x/ D b a:
Thus, a necessary condition for the existence of solutions is that c b a.
In other words, if c < b a, then the given equation will have no solutions. Let
us see what happens if c b a (i.e., let us see whether this condition also
suffices to the existence of solutions). Since every real number x satisfies one of
x a, a < x b or x > b, we shall analyse these cases separately:
• x a: then, x a 0 and x b 0 (why?), so that the given equation
reduces to .x a/ .x b/ D c, whose root is x D 12 .a C b c/. Verify that
such a root indeed satisfies the condition x a.
• a < x b: the given equation reduces to .x a/ .b x/ D c, or b a D c.
If such an equality is true, then every real number x satisfying a < x b
will be a solution of the equation; if it is false, then the equation will have no
solutions x satisfying the inequalities a < x b.
• x > b: arguing as in the previous cases, we easily arrive at the solution
x D 12 .a C b C c/, which does satisfy condition x > b.
526
B Hints and Solutions
6. It suffices to see that
ˇ
ˇ
ˇ
ˇ
ˇ ˇ
p ˇ ˇ 1
ˇr C 2 p ˇ
ˇ 1
1 ˇˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ r C 1 2ˇ D ˇ r C 1 C 1 2ˇ D ˇ r C 1 p2 C 1 ˇ
p
p
j 2 rj
1
D
p
< jr 2j;
2
.r C 1/. 2 C 1/
p
for r 0 and 2 > 1.
y
7. For item (a), compute 1Cy
x1 C x. For item (b), start by using the result of (a),
jaCbj
jajCjbj
1CjajCjbj
.
together with the triangle inequality jaCbj jajCjbj, to get 1CjaCbj
8. For the first part, adapt the proof of Example 2.13. It may help you to notice that
..n C 1/ C .n C 2/ C C 2n/ .1 C 2 C C n/ D n2 C .1 C 2 C C n/ .1 C 2 C C n/ D n2 :
For the second part, show that every x 2 .n; n C 1/ is a root of the equation.
Section 2.3
1. If x D ˛ is a root of the given equation, then ˛ 2 C bj˛j C c D 0. But, then, we
also have .˛/2 C bj ˛j C c D ˛ 2 C bj˛j C c D 0, so that x D ˛ is also a
root of the equation. Therefore, since ˛ C .˛/ D 0, the sum of the roots has
to be equal to 0.
p
2. For item (a), squaring both sides of x C 2 D 10 x we get x C 2 D .10 x/2 ,
or x2 21x C 98 D 0. Since 7 C 14 D 21 and 7 14 D 98, the roots of this
last equation are x D 7 and x D 14. However, only
p x D 7 satisfies the original
equation, for, substituting x D 14 in it, we find 16 D 4, which is an absurd.
In what concerns (b), we initially observe that, for the square roots to have a
meaning in R, we must have 32 x 13 ; on the other hand,
p
p
p
x C 10 D 2x C 3 C 1 3x ,
p
p
, x C 10 D . 2x C 3 C 1 3x/2
p
, x C 10 D 4 x C 2 .2x C 3/.1 3x/
p
, x C 3 D .2x C 3/.1 3x/
, .x C 3/2 D .2x C 3/.1 3x/
, 7x2 C 13x C 6 D 0;
so that x D 1 or x D 67 . Since both of these values belong to the interval
3 1
2 ; 3 , we conclude that they are the solutions of the given equation. Finally,
B Hints and Solutions
527
2
for item (c), we note that the substitution
p of variables y D x C 18x transform
the given equação into y C 30 D 2 y C 45. Squaring both sides, we easily
arrive at y2 C 56y C 720 D 0, an equation whose roots are p
y D 36 and
y D 20. From these, only y D 20 serves us, for y C 30 D 2 y C 45 gives
y C 30 0 and ypC 45 0, i.e., y 30. Therefore, x2 C 18x C 20 D 0 and,
then, x D 9 ˙ 61.
p
3. Start by observing that, in order for 2x 1 to have a meaning in the set of
reals, we must have x 12 . By a similar reason, it must be the case that x p
2x 1. However, since
x
p
2x 1 , x2 2x 1 , x2 2x C 1 0;
which is always true, the conditions for the existence of the square roots in the
given equation reduce to x 12 . Now,
2
q
A D
p
x C 2x 1 C
q
2
p
x 2x 1
q
p
p
D 2x C 2 .x C 2x 1/.x 2x 1/
p
D 2x C 2 x2 .2x 1/ D 2x C 2jx 1j:
We shall consider here only item (b), leaving the other two items as exercises to
the reader. If A D 1, we have to solve the equation 2x C 2jx 1j D 1, under the
condition x 12 . If x 1, that equation reduces to 2x C 2.x 1/ D 1, so that
x D 34 (which does not satisfy the condition x 1). If 12 x < 1, the equation
reduces to 2x C 2.1 x/ D 1, an equality which is false for every x 2 12 ; 1 .
Therefore, there are no solutions in item (b).
4. Let ax2 C bx C c D 0 be the equation of the first quiz, a0 x2 C bx C c D 0 be
that of the second one and ax2 C bx C c0 D 0 be that of the third one. Since
the roots of the second quiz are 2 and 3, we have ab0 D 5 and ac0 D 6, so
that bc D 56 ; on the other hand, since the roots of the third quiz are 2 and 7,
we have ba D 5. Therefore, the second degree equation of the first quiz
has the form ax2 C 5ax 6a D 0, and, thus, has the same roots as those of
x2 C 5x 6 D 0, i.e., 6 and 1.
5. It follows from Proposition 2.16 that a C b D a and ab D b. The second
equality gives a D 1, and substituting this value into the first equality we get
b D 2.
6. Again by Proposition 2.16 (applied to both equations), we have a D ˛2 C
ˇ 2 D .˛ C ˇ/2 2˛ˇ D 13 2 9 D 5 and b D ˛ 2 ˇ 2 D .˛ˇ/2 D 92 D 81.
Therefore, a C b D 86.
528
B Hints and Solutions
7. The simplest possibility is x2 Sx C P D 0, where
S D u3 C v 3 D .u C v/.u2 uv C v 2 /
D Œ.u C v/2 3uv
D Œ.1/2 3.1/ D 4
and P D u3 v 3 D .uv/3 D .1/3 D 1.
8. A possibility is the equation x2 S0 x C P0 D 0, where S0 D .˛S C P/ C
.ˇS C P/ D .˛ C ˇ/S C 2P D S2 C 2P and P0 D .˛S C P/.ˇS C P/ D
˛ˇS2 C .˛ C ˇ/SP C P2 D 2S2 P C P2 . Here, we used the relations ˛ C ˇ D S
and ˛ˇ D P. p
p
9. If ˛ D 7 C 4 3 and ˇ D 7 4 3, then ˛ C ˇ D 14 and ˛ˇ D 1, so that
˛ and ˇ are the roots of the second degree equation x2 14x C 1 D 0. Hence,
˛ 2 D 14˛ 1 and ˇ 2 D 14ˇ 1 and, starting from such equalities, we get
˛ kC2 C ˇ kC2 D 14.˛ kC1 C ˇ kC1 / .˛ k C ˇ k /:
Now, make k successively equal to 0, 1, 2 and 3 to compute, also successively,
˛2 C ˇ 2 , ˛ 3 C ˇ 3 , ˛ 4 C ˇ 4 and ˛ 5 C ˇ 5 .
10. Substituting x by ˛, we get the equality ˛ 2 D ˛ C 1 and, from it, ˛ 4 D .˛ 2 /2 D
.˛ C 1/2 D ˛ 2 C 2˛ C 1 D 3˛ C 2 and ˛ 5 D ˛ ˛ 4 D ˛.3˛ C 2/ D 3˛ 2 C 2˛ D
3.˛ C 1/ C 2˛ D 5˛ C 3. Hence, ˛ 5 5˛ D 3.
11. Since the roots of the equation must be integers whose product equals 5, we
conclude that they must be (i) 1 and 5; or (ii) 1 and 5. In the first case
m D .1 C 5/ D 6, whereas, in the second, m D ..1/ C .5// D 6.
12. The given equation is equivalent to x2 .b C c C 2a2/x C Œbc C a2 .b C c/ D 0.
In order for the roots of this last equation to be real and distinct, it suffices to
show that > 0, where is its discriminant. In fact, since a ¤ 0, we have
D .b C c C 2a2 /2 4Œbc C a2 .b C c/
D .b C c/2 C 4a4 4bc D .b c/2 C 4a4 > 0:
13. Making a D x 1x , we get x2 ax 1 D 0 .1/ and, hence, 1x D x a.
p
p
Therefore, the original
equation
becomes x D a C 1 .x a/ or, which
p
p
is the same,
p x a D 1 C a x .2/.
p Squaring both sides of .2/, we get
x2 C a 2 ax D 1 C a px, or x2 .2 a 1/x 1 D 0 .3/. Subtracting .3/
from .1/, it follows that .2 a 1 a/x D 0. Now, since x ¤ 0, it must be that
p
p
2 a 1 a D 0, from which a D 1. Then, x 1x D 1, so that x D 1˙2 5 .
p
p
p
However, .2/ gives x 1 D x a D 1 C a x 0, so that x D 1C2 5 .
14. If ˛ is a common root of both equations, then a˛ 3 ˛ 2 ˛ .aC1/ D 0 . / and
a˛ 2 ˛.aC1/ D 0. Multiply the second equality by ˛ to get a˛ 3 ˛ 2 .aC1/
˛ D 0, and subtract this result from . / to arrive at .a C 1/˛ ˛ .a C 1/ D 0,
B Hints and Solutions
529
so that ˛ D 1 C 1a . Therefore, if the two given equations are to have a common
root, this must be equal to 1 C 1a . It now suffices to verify that 1 C 1a actually is
a root of both equations, and we leave this simple task to the reader.
15. For item (a), it suffices to see that
x3 3x2 C 5x D .x3 3x2 C 3x 1/ C .2x 2/ C 3
D .x 1/3 C 2.x 1/ C 3:
Now, it follows from (a) that the equalities in the statement of the problem can
be rewritten as
.x 1/3 C 2.x 1/ C 3 D 1 and .y 1/3 C 2.y 1/ C 3 D 5I
adding these, we arrive at
.x 1/3 C .y 1/3 C 2.x C y 2/ D 0:
In the above relation, substitute the factorisation
.x 1/3 C .y 1/3 D .x C y 2/Œ.x 1/2 .x 1/.y 1/ C .y 1/2 to get
.x C y 2/Œ.x 1/2 .x 1/.y 1/ C .y 1/2 C 2 D 0:
Now, there are two possibilities: x C y 2 D 0 or .x 1/2 .x 1/.y 1/ C
.y 1/2 C 2 D 0. In order to show that the second one doesn’t happen, make
x 1 D a and y 1 D b and observe that
3b2
b 2 3b2
b2
C
C2D a
C 2 > 0:
a ab C b C 2 D a ab C
C
4
4
2
4
2
2
2
17. Compare the coefficients at both sides of the equality
x3 C ax2 C bx C c D .x ˛/.x2 C b0 x C c0 /
to get the equalities b0 ˛ D a, c0 ˛b0 D b and ˛c0 D c. Now, use the first
and third relations to get b0 D a C ˛ and c0 D ˛c , showing, then, that these
values for b0 and c0 also satisfy the relation c0 ˛b0 D b (at this step you shall
need to use the fact that
˛ 3 Cpa˛ 2 C b˛ Cpc D 0.p
p
3
3
18. For item (a), let a D 2 C 5 and b D 2 5, so that ˛ D a C b. Then,
use the fact that
˛ 3 D a3 C b3 C 3ab.a C b/ D a3 C b3 C 3ab˛;
530
19.
20.
21.
23.
B Hints and Solutions
together with the relation ab D 1. For (b), use the result of the previous
problem to show that x3 C 3x 4 D .x 1/.x2 C x C 4/, so that x D 1 is the
only real root of x3 C 3x 4 D 0.
Use the result of Problem 21 to compare the coefficients of a3 x3 C a2 x2 C a1 x C
a0 D 0 and a3 .x x1 /.x x1 /.x x3 /.
In order to compute the value of the first sum of powers, use relations (2.21),
together with the identity ˛ 2 C ˇ 2 C 2 D .˛ C ˇ C /2 2.˛ˇ C ˛ C ˇ /;
for the other two sums of powers, start by observing that, since ˛ is a root of
the equation, we have ˛ 3 D 3˛ 1, and analogously for ˇ and . Then add the
left and right hand sides of these three identities.
Expand the products in .y C d/3 C a.y C d/2 C b.y C d/ C c and impose that
the coefficient of y2 equals 0.
It suffices to see that
1
1
1
3
2
x C 3 D xC
x 1C 2
x
x
x
!
1 2
1
xC
3 :
D xC
x
x
24. Just note that
!2
1
1 2
1 2
2
x C 4 D x C 2 2D
xC
2 2:
x
x
x
4
25. First, note that the equality x2 x 1 D 0 is equivalent to x 1x D 1. Then, use
the identity
x3 1
D
x
x3
D x
1
x2 C 1 C 2
x
!
2
1
1
x
C3 :
x
x
1
x
26. The equality x2 4xC1 D 0 is equivalent to xC 1x D 4. Now, use the expression
for x3 C x13 , deduced in the hint to Problem 23, and observe that x6 C x16 D
2
3
x C x13 2.
28. Initially, consider the cases n D 2, 4 and 6; then, adapt the reasoning of these
particular cases to the general one.)
29. For item (a), see the hint to Problem 23. For item (b), argue as was done in the
text for biquadratic equations.
30. For item (a), let’s consider two cases: (i) if a; b > 0 then c C d D a < 0
and cd D b > 0, so that c; d < 0; but then ef D d < 0, so that e > 0 > f .
(ii) If a; b < 0, then, in the notations of (i), c > 0 > d. Therefore, starting
B Hints and Solutions
531
with the second equation, if necessary, we can assume from the beginning that
a > 0 > b. Then, letting c d be the roots of x2 C ax C b D 0, we get
c C d D a < 0 and cd D b < 0, so that c > 0 > d. For (b), we have
D m2 4n and 0 D p2 4q. Compute
0
D
D
m C
2
p !2
m 4
2
p !
p
p 1 2
m 2m C C 8m C 8 4
and notice that, since m > 0,
p
p
0 < , m2 2m 4 C 8 3
p
p
p
, <m< 3 8 ,m< 3 8
, .m C 8/2 < 9.m2 4n/
, 2.m 1/2 > 9.n C 2/:
For (c), note that the process cannot continue indefinitely if the inequality 0 <
always holds. Then, show that at some point we reach a situation in which
n D 1 or m D 1, n D 2. If n D 1, conclude that the roots of x2 Cmx1 D 0
should be 1 and 1, so that m should be equal to 0, which is not the case. Then,
we are left to m D 1 and n D 2. In this case, argue backwards to show that
a D 1, b D 2.
Section 2.4
1. Execute the elimination algorithm in each of the items above.
2. Apply the elimination algorithm, in the way done in Example 2.26.
3. Apply the variable substitution a D 1x , b D 1y , c D 1z to transform the given
system into a linear system of three equations in three unknowns.
4. Show that xj D a1j b1 C a2j b2 C a3j b3 , for 1 j 3.
5. For item (a), it suffices to see that x1 D x2 D D xn D 0 always is a solution.
In what concerns (b), suppose that x1 D ˛1 , x2 D ˛2 , . . . , xn D ˛n and x1 D ˇ1 ,
x2 D ˇ2 , . . . , xn D ˇn are two distinct solutions of (2.25). If t 2 R is arbitrary,
show that xi D t˛i C .1 t/ˇi , for 1 i n, is also a solution. Finally, for item
(c), assume that the system has only one solution when b1 D b2 D D bm D 0,
so that this solution is x1 D x2 D D xn D 0. Let b1 , b2 , . . . , bm be arbitrary
real numbers, and x1 D ˛1 , x2 D ˛2 , . . . , xn D ˛n and x1 D ˇ1 , x2 D ˇ2 ,
. . . , xn D ˇn be two solutions of (2.25). Then, it is immediate to check that
x1 D ˛1 ˇ1 , x2 D ˛2 ˇ2 , . . . , xn D ˛n ˇn is a solution of (2.25) when
b1 D b2 D D bm D 0, so that our assumption gives ˛i ˇi D 0, for
1 i n.
532
B Hints and Solutions
Sections 2.5
1. Equations x C yz D 2 and y C xz D 2 give .x y/.1 z/ D 0, so that either
x D y or z D 1. Do the same for the other two pairs of equations, and compare
the different possibilities that arise.
2. Letting z D 1=.x C y/, we obtain the second degree system
zCx D aC1
:
zx D a 1
Hence, there are two possibilities: .z; x/ D .a; 1/ or .z; x/ D .1; a/. Those give
rise to the systems
x C y D a1
and
xD1
xCyD 1
;
x D a1
both of which can be immediately solved.
3. Set u D xCy
and v D xy
, and show that the given system is equivalent to the
xy
xy
second degree system
uCv D5
;
uv
D6
so that u D 2 and v D 3, or vice-versa.
4. Add the three given equations and write the result as
.ax21 C .b 1/x1 C c/ C .ax22 C .b 1/x2 C c/ C .ax23 C .b 1/x3 C c/ D 0:
Then, apply the result of Lemma 2.14 to the second degree trinomial ax2 C
.b 1/x C c and analyse separately itens (a) and (b).
5. Suppose, without loss of generality, that x y. Then, use the equations of the
system to conclude that x D y D z. Finally, note that x D 1 is a root of the third
degree polynomial equation x3 2xC1 D 0, and apply the result of Problem 21,
page 43.
6. Write the left hand side as a sum of squares and apply Lemma 2.31, with n D 3.
7. Substituting y D 14 .a z 3x/ into the first equation, we get
x2 C
1
.a z 3x/2 D 4z
16
or, which is the same,
25x2 6x.a z/ C .a z/2 64z D 0:
B Hints and Solutions
533
For this last equation to have a single solution, the discriminant of the second
degree trinomial (in x) of the left hand side must be equal to 0, i.e., we should
have
36.a z/2 D 100Œ.a z/2 64z:
or, which is the same, .a z/2 D 100z. If this is so, then (from the second
3
degree equation in x above) x D 25
.a z/ and (from the second equation of the
4
system) y D 25 .a z/. Therefore, the system has a single solution if and only if
the equation (in z) .a z/2 D 100z has a single solution. Since it is equivalent
to z2 2z.a C 50/ C a2 D 0, its discriminant must also be equal to 0, i.e., we
should have .a C 50/2 D a2 . Thus, a D 25.
8. Start by writing
xC
p
p
3
3
2 x C 2 D .x 1/ C 1 C
2 xC2
x1
x1
p
xC2
D .x 1/ C
2 xC2
x1
!2
r
p
xC2
x1
D
x1
Then, apply Lemma 2.31 to conclude that it suffices to solve the system of
equations
p
x1
r
p
xC2
D y1
x1
s
p
yC2
D z1
y1
r
zC2
D 0;
z1
p
so that x D y D z D 3C2 13 .
p
p
9. Letting a D 3 x C 5 and b D 3 4 x, we get a C b D 3 and a3 C b3 D 9.
Hence
9 D a3 C b3 D a3 C .3 a/3 D 27 27a C 9a2
or, which is the same, a2 3a C 2 D 0. Therefore, a D 1 or 2, from where
x D 4 or x D 3.
p
10. Argue as in the previous problem, letting y D 5 x to transform the given
equation into a system of equation in x and y.
534
B Hints and Solutions
11. First of all, observe that
2
x2
1
2
Dx C 1
x C
.x C 1/2
xC1
2
D x2 C 1 C
1
2
2
.x C 1/
xC1
D .x2 C 2x C 1/ C
D .x C 1/2 C
1
2
2x .x C 1/2
xC1
1
2
C 2:
2.x C 1/ 2
.x C 1/
xC1
Now, make the variable substitution y D x C 1 C
.x C 1/2 C
1
xC1
to get
1
D y2 2;
.x C 1/2
so that the given equation simplifies to the second degree equation
y2 2y 3 D 0, whose roots are 3 and 1. Then, we have
xC1C
1
D 3 or 1;
xC1
and each of these possibilities gives rise to a second degree equation in x.
12. Firstly, observe that, if x D 0, then y D z D 0 (and analogously if y D 0 or
z D 0). Therefore, we can assume that xyz ¤ 0. Setting u D 1x , v D 1y and
w D 1z , we get
vD
1 C 4x2
1
1
u2
D
C 1:
D 2 C1D
2
y
4x
4x
4
Transforming the other two equations in analogous ways, we get the system
below, which is equivalent to the original one:
u2 C 4 D 4v; v 2 C 4 D 4w; w2 C 4 D 4u:
Adding these three equations, we arrive at the equality
.u 2/2 C .v 2/2 C .w 2/2 D 0;
and Lemma 2.31 gives u D v D w D 2 and, hence, x D y D z D 12 . Thus, the
solutions of the system are x D y D z D 0 or x D y D z D 12 .
B Hints and Solutions
535
13. Add the equations of the system to conclude that it suffices to solve the system
of equations
8
a C b C c D 3d
ˆ
ˆ
<
b C c C d D 3a
:
ˆ c C d C a D 3b
:̂
d C a C b D 3c
Then, note that a D b D c D d D 14 .a C b C c C d/.
14. Write the given equation as .E1 F1 / C .E2 F2 / C C .En Fn / D 0,
observing that Ei Fi 0 for 1 i n.
15. For u 2 R, note that
4u2 u4 D 4 .4 4u2 C u4 / D 4 .2 u2 /2 :
Hence,
p
p
p
4x2 x4 3 C 4y2 y4 C 4z2 z4 C 5
p
p
p
D 1 .2 x2 /2 C 4 .2 y2 /2 C 9 .2 x2 /2
p
p
p
1 C 4 C 9 D 1 C 2 C 3;
6 D
and it follows
that 2 x2 D 2 y2 D 2 z2 D 0,
p from thepprevious problem
p
i.e., x D ˙ 2, y D ˙ 2, z D ˙ 2.
4
x
4
16. If x 4, write 5x
4 x D xC 4 x to conclude that z 4; then, argue similarly
with the second equation to conclude that y 4. Then, either x; y; z 4 or
x; y; z 4. In either case, add the three equations of the system to get
4
4
4
x
y
z
C C D C C ;
x
y
z
4
4
4
and use the result of Problem
p 14.
p
17. For item (a), write x C 2x 2 2 D 1x .x 2/2 . For item (b), start by observing
that xC 2x D 2y gives that x and y are both positive or both negative; analogously,
the same is true for y and z. Therefore, either x, y and z are all positive or
all negative. Now, note that .x; y; z/ is a solution of the system if and only if
.x; y; z/ is also a solution, so that we can restrict to the case x; y; z > 0.
Now, the result of item (a) gives
2y D x C
p
2
2 2;
x
536
B Hints and Solutions
p
p
so that y 2; analogously, x; z 2. Adding the three equations of the
system, we get
xCyCzD
Finally, note that u 2
u
for u p
2
2
2
C C :
x
y
z
2, and use the result of Problem 14.
Section 3.1
2. For item (a), let an D n, for odd n 1, and an D .1/n1 n, for even n 1.
n
For (b), let an D nC1
, for n 1. Finally, for item (c), let an D n, for odd n 1,
1
and an D n for even n 1.
1
3. It suffices to note that bkC1 D akC1
D 3aak C1
D 3 C a1k D 3 C bk .
k
4. For item (a), we can write a1 D a2 D a3 D 1 and ak D ak1 C ak2 C ak3 , for
k 4. For item (b), we can write a1 D 1 and akC1 D 2ak , for k 1.
5. Observe that the entries in the n-th line form and AP of initial term n, common
difference 2 and 2n 1 terms.
6. Write 11
: : :…
1 as a sum of powers of 10, then use the formula for the sum of the
„ ƒ‚
n
terms of a GP.
7. Use the result of the previous problem.
8. It suffices to show that bkC1 bk does not depend on k; the common value we
shall obtain will be the common difference of the AP .bk /k1 . By definition, we
have
bkC1 bk D .a2kC2 a2kC1 / .a2kC1 a2k /
D .akC2 akC1 /.akC2 C akC1 / .akC1 ak /.akC1 C ak /
D r.akC2 C akC1 / r.akC1 C ak /
D r.akC2 ak / D r 2r D 2r2 :
9. Apply the formula for the general term of an AP to ap C aq and au C av .
10. Let a1 D rm, with m 2 ZC . By the formula for the general term of AP’s, we
have
ak C al D 2a1 C .k C l 1/r D a1 C .m C k C l 1/r D amCkCl :
11. The formula for the general term of AP’s gives us the relations ˛ D ap D
a1 C .p 1/r, ˇ D aq D a1 C .q 1/r and apCq D a1 C .p C q 1/r, where r
is the common difference of the AP. By viewing the first two relations above as
a linear system of equations in a1 and r, express them in terms of ˛, ˇ, p and q;
then, substitute these expressions for a1 and r into the formula for apCq .
B Hints and Solutions
537
12. Letting 2n 1 denote the smallest of such integers and k their quantity, the
formula for the general term of an AP assures that the biggest of the numbers
equals .2n 1/ C 2.k 1/ D 2n C 2k 3. Therefore, their sum is equal to
1
3
3
2 Œ.2n 1/ C .2n C 2k 3/k D 7 , so that .2n C k 2/k D 7 . Now, since
2
2n C k 2 k > 1, the only possibility is 2n C k 2 D 7 and k D 7, so that
n D 22.
13. Let r be the common difference of the AP and let a2 D m, so that r 2 N and
m > 1. Then, amkC2 D a2 C ..mk C 2/ 2/r D m C mkr D m.1 C kr/, which
is clearly a composite number.
14. Item (b) follows immediately from (a). For item (a), separately consider the
cases n even and n odd.
p
p
p
15. By contradiction, suppose that am D 2, an D 3 and ap D 5, where
.ak /k1 is an AP and m, n and p are pairwise distinct natural numbers. Then,
use the formula for the general term of AP’s to compute nm
pm .
16. Apply, to the left hand side, the formula for the general term of GP’s.
17. Adapt, to the present case, the solution given to the Example 3.14.
18. Adapt, to the present case, the solution given to the Example 3.14.
19. Suppose that there exists such an AP .ak /k1 , of common difference r, i.e.,
such that ar1 … Q but a2n D am ap , for some distinct indices m, n and p. Use the
formula for the general term of an AP to reach a contradiction.
20. Show that the sequence .bk /k1 , such that bk D ak 1, is a GP.
22. Suppose that .ak /k1 is a GP of common ratio q, such that am D 2, an D 3 and
ap D 5. Then, a1 qm1 D 2, a1 qn1 D 3 and a1 qp1 D 5, so that qmn D 23 and
qmp D 25 . Therefore,
mp
mn
2
2
.mn/.mp/
Dq
D
3
5
or, which is the same, 2np 5mn D 3mp . Now, get a contradiction.
Section 3.2
1. Apply the result of Theorem 3.16.
2. Apply the result of Theorem 3.16.
3. Start by observing that akC2 akC1 D .2akC1 1/ .2ak 1/ D 2akC1 2ak
and, hence, that akC2 3akC1 C 2ak D 0, for every integer k 1. Now, apply
the result of Theorem 3.16.
4. Use the fact that x2 C rx C s D .x ˛/2 .
7. For item (a), apply Theorem 3.16. For item (b), use the formula of (a). Finally,
for (c), use the formulas of item (a) and (3.10).
538
B Hints and Solutions
8. For item (a), the given recurrence relation furnishes
4b2kC1 D 4.1 C 24akC1/ D 4 C 6.1 C 4ak C
1
D 4 C 6 1 C .b2k 1/ C bk
6
p
1 C 24an /
D b2k C 6bk C 9 D .bk C 3/2 :
For item (b), it follows from (a) that 2bnC2 bnC1
pD 3 D 2bnC1 bn and, thus,
2bnC2 3bnC1 C bn D 0. Since b1 D 5 and b2 D 1 C 24a2 D 4, Theorem 3.16
guarantees that bn D 3 C 24n . At last, for item (c), it follows from items (a) and
(b) that
!
4 2
1 2
1
an D
.b 1/ D
3C n 1
24 n
24
2
4
4
1
2C n
4C n
D
24
2
2
1
1
1
1 C n1
1C n :
D
3
2
2
Sections 3.3
1. If .ak /k1 is a second order AP, then .akC1 ak /k1 is a non constant AP, so
that there exists r ¤ 0 for which .akC2 akC1 / .akC1 ak / D r, for every
k 1. In particular, for every k 1, we have akC2 2akC1 C ak ¤ 0 and
.akC3 akC2 / .akC2 akC1 / D .akC2 akC1 / .akC1 ak /:
The converse can be prove in an analogous way.
Write akC1 ak D 3k 1 and use telescoping sums.
Write akC1 ak D 8k and use telescoping sums.
1
1
Observe that .k1/k
D k1
1k and use telescoping sums.
The previous problem is a particular case of the present one.
1
Start by noting that k12 < .k1/k
for every integer k > 1; then, apply the result
of Problem 4.
7. Adapt to the present case the hint to Problem 4, writing
2.
3.
4.
5.
6.
1
1
D
.4k 1/.4k C 3/
4
1
1
:
4k 1 4k C 3
B Hints and Solutions
539
8. For item (a), it suffices to observe that .2k C 1/3 .2k 1/3 D 24k2 C 2 D
16k2 C 8k2 C 2 D 16k2 C .2k 1/2 C .2k C 1/2 . For item (b), start by using
the formula for telescoping sums to write
.2n C 1/3 2 D
n
X
Œ.2k C 1/3 .2k 1/3 C .33 2/
kD2
D
n
X
Œ.4k/2 C .2k C 1/2 C .2k 1/2 C 25
kD2
n
X
Œ.4k/2 C .2k C 1/2 C .2k 1/2 C 42 C 32 :
D
kD2
Then, observe that the last expression above is a sum of .n 1/ 3 C 2 D 3n 1
perfect squares.
k
1
9. Note that .kC1/Š
D .kC1/1
D kŠ1 .kC1/Š
and use telescoping sums.
.kC1/Š
10. For item (a), observe that .kC1/2 Ck2 Ck2 .kC1/2 D k4 C2k3 C3k2 C2kC1 D
.k2 C k C 1/2 . In what concerns (b), it follows from (a) that
s
1
1
C
C1D
2
k
.k C 1/2
s
k2 C k C 1
.k2 C k C 1/2
:
D
k2 .k C 1/2
k.k C 1/
It now suffices to write
1
1
1
k2 C k C 1
D1C
D1C k.k C 1/
k.k C 1/
k kC1
and use the formula for telescoping sums or the result of Problem 4.
F
F
Fk
1
11. Write Fk FkC1
D FkC2
D F1k FkC2
and, then, use the formula for telescoping
k FkC2
kC2
sums.
12. Rationalise the fraction pak C1pakC1 —see Problem 12, page 25—and, then, use
the formula for telescoping sums.
13. For item (a), just note that x4 C x2 C 1 D .x4 C 2x2 C 1/ x2 D .x2 C 1/2 x2 D
.x2 C 1 x/.x2 C 1 C x/. For (b), use the result of (a) to write
k4
k
k
D 2
2
Ck C1
.k C k C 1/.k2 k C 1/
1
1
1
:
D
2 k2 k C 1 k2 C k C 1
Then, use the formula for telescoping
P sums.
14. Rationalise the fraction under the
notation—see Problem 12, page 25—and,
then, use the formula for telescoping sums.
540
B Hints and Solutions
15. Adapt, to the present case, the hint given to the previous problem.
16. Initially, observe that
n n n Y
Y
1
1 Y
1
1 2 D
1
1C
j
j jD2
j
jD2
jD2
D
n n Y
j1 Y jC1
jD2
j
jD2
j
:
Now, take ak D k for k 1. It follows from Proposition 3.29 that
n n n Y
Y
1
aj1 Y ajC1
1 2 D
j
aj
aj
jD2
jD2
jD2
D
a1 anC1
nC1
1 nC1
D
:
D an a2
n
2
2n
17. If S is the sum we wish to compute, then
SD
101
X
kD0
Since 1 xk D 1 k
101
x3k
:
.1 xk /3 C x3k
D x101k , we get
SD
101
X
x3k
3
x
C x3k
kD0 101k
and can write
2S D
101
X
kD0
D
101
X
kD0
D
101
X
x3k
x3k
C
3
3
3
x101k C xk
x
C x3k
kD0 101k
101
X x3
x3k
101k
C
3
x3101k C x3k
x
C x3k
kD0 101k
101 3
X
x C x3
k
101k
3
x
C x3k
101k
kD0
D
101
X
1 D 102:
kD0
Hence, S D 51.
18. StartQby factorising both the numerator and denominator of the fraction under
the notation. If ak D k2 k C 1, show that akC1 D k2 C k C 1 and, then, use
the formula for telescoping products.
B Hints and Solutions
541
19. We first show that n must be even. To this end, suppose that A D A1 [ [ An ,
with A1 , . . . , An satisfying the conditions of the problem. For 1 k n, let
xk 2 Ak be the
Pelement of Ak that equals the arithmetic mean of the other three.
Then 4xk D x2Ak x, and it follows that
4
n
X
xk D
n X
X
kD1 x2Ak
kD1
xD
X
x D 1 C 2 C C 4n D 2n.4n C 1/:
x2A
Hence, n.4n C 1/ must be even and, since 4n C 1 is odd, it follows that n must
be even. Now, let n D 2k, with k 2 N, and write
f1; 2; 3; : : : ; 8kg D A1 [ [ A2k ;
where
A2j1 D 8.j 1/ C f1; 3; 4; 8g and A2j D 8.j 1/ C f2; 5; 6; 7g:
(Here, for X R, we define the set X C t by X C t D fx C tI x 2 Xg). Since
3 4 D 1 C 3 C 8 and 3 5 D 2 C 6 C 7, it is immediate to verify that the sets
Ai , defined as above, satisfy the conditions of the problem.
20. By the result of Problem 13, page 26, we have
p
nC1
p
p
p p
n
n D n2 C n n2
s
1 2 p 2
1
nC
<
n D
2
2
p
p
and, thus, 2. n C 1 n/ < p1n . Analogously, we get
Taking these two inequalities together, we get
10000
X
kD2
p1
n
p p
< 2. n n 1/.
10000
10000
X 1
X p
p
p
p
kC1 k <
k k1 :
p <
k
kD2
kD2
However,
10000
X
p
p
p
p
k C 1 k D 2 10001 2 2
kD2
and
10000
X
p
p
p
k k1 D2
10000 1 D 198:
kD2
542
B Hints and Solutions
p
p
Hence, if we show that 2 10001 2 2 > 197, it will come that the greatest
integer which is less than or equal to S equals 198. To what is left to do, just see
that
p
p
p
2 10001 > 2 2 C 197 , 2 10001
p
2
> 2 2 C 197
p
, 40004 > 788 2 C 38817
p
, 788 2 < 1187 , 7882 2 < 11872
2
, 1241888 < 1408969;
which is indeed true.
21. For item (a), factorising .k C 1/pC1 kpC1 with the aid of the result of
Problem 18, page 26, we get
.k C 1/
pC1
k
pC1
pC1
X
D
.k C 1/pC1j kj1
jD1
<
pC1
X
.k C 1/pC1j .k C 1/j1
jD1
D .p C 1/.k C 1/p :
Analogously,
.k C 1/
pC1
k
pC1
D
pC1
X
.k C 1/pC1j kj1
jD1
X
pC1
>
kpC1j kj1 D .p C 1/kp :
jD1
For item (b), it follows from (a) and from the formula for telescoping sums that
n1
X
kD1
k <
p
n1
X
.k C 1/pC1 kpC1
pC1
kD1
D
npC1 1
npC1
<
pC1
pC1
and
n
X
kD1
kp >
n
X
kpC1 .k 1/pC1
kD1
pC1
D
npC1
:
pC1
B Hints and Solutions
543
Section 4.1
1. It follows from the induction hypothesis that
1 C 2 C C k C .k C 1/ D
.k C 1/.k C 2/
k.k C 1/
C .k C 1/ D
:
2
2
2. It follows from the induction hypothesis that
k
X
j3 D
jD1
2
k.k C 1/
2
C .k C 1/3 D
.k C 1/.k C 2/
2
2
:
3. It follows from the induction hypothesis that
1
1
1
1
1
1
C C
C
D
2
3
2k 1 2k
2k C 1
1
1
1
1
1
CC
C
D C
k
kC1
2k 1 2k
2k C 1
1
1
1
1
CC
C
C
:
D
kC1
2k 1
2k
2k C 1
4. By induction hypothesis, we have
.k C 1/ C h.1/ C h.2/ C h.3/ C C h.k 1/ C h.k/ D 1 C kh.k/ C h.k/;
Therefore, it suffices to prove that 1 C .k C 1/h.k/ D .k C 1/h.k C 1/ or, which
is the same, that .k C 1/.h.k C 1/ h.k// D 1. However,
.k C 1/.h.k C 1/ h.k// D .k C 1/ 1
D 1:
kC1
5. It follows from the induction hypothesis that
k
X
j.j C 1/ D
jD1
D
1
.k 1/k.k C 1/ C k.k C 1/
3
1
k.k C 1/.k C 2/:
3
6. It follows from the induction hypothesis that
kC1
X
1
.2j 1/2 D .2k 1/2k.2k C 1/ C .2k C 1/2
6
jD1
D
1
.2k C 1/2.k C 1/.2k C 3/:
6
544
B Hints and Solutions
8. In trying to appeal to the induction hypothesis, we observe that 4kC1 C
15.kC1/1 D .4k C15k1/C3.4k C5/. Then, since 4k C15k1 is a multiple
of 9 (by that induction hypothesis), it suffices to show that 4k C 5 is a multiple
of 3. To this end, make another proof by induction: 4lC1 C 5 D .4l C 5/ C 3 4l ;
since 4l C 5 (by the new induction hypothesis) and 3 4l are multiples of 3, it
follows that 4lC1 C 5 is also a multiple of 3.
9. Appealing to the induction hypothesis, we observe that .k C 1/3 .k C 1/ D
.k3 k/ C 3.k2 C k/. Now, since k3 k (by that induction hypothesis) and
3.k2 C k/ are multiples of 3, it comes that .k C 1/3 .k C 1/ is also a multiple
of 3.
k1
10. If 43 1 D 3k q, for some q 2 N, then
43 1 D .43
k
k1
/3 1 D .43
k1
D 3k qŒ..43
k1
D 3k qŒ.4
3k1
1/Œ.43
k1
/2 1/ C .43
k1
1/.4
D 3k qŒ3k q.43
k1
3k1
/2 C 4 3
k1
C 1
1/ C 3
C 1/ C 3k q C 3
C 1/ C 3k q C 3
D 3kC1 qŒ3k1 q.43
k1
C 1/ C 3k1 q C 1:
11. Suppose we have n1 ways of choosing an object of type 1, n2 ways of choosing
an object of type 2, . . . , nk ways of choosing an object of type k, nkC1 ways of
choosing an object of type k C 1. In order to choose an object of each of these
k C 1 types, start by choosing an object of each of the types from 1 to k. By
induction hypothesis, there are exactly n1 n2 : : : nk possible distinct choices. On
the other hand, for each of these choices, we have nkC1 ways of choosing an
object of type k C 1. Therefore, the number of ways of choosing an object of
each of the types from 1 to k C 1 is equal to
n n : : : nk C n1 n2 : : : nk C C n1 n2 : : : nk D n1 n2 : : : nk nkC1 :
„1 2
ƒ‚
…
nkC1 times
12. By induction hypothesis, suppose that any set with k elements has exactly 2k
subsets. If A D fa1 ; a2 ; : : : ; ak ; akC1 g is a set with k C 1 elements. It has two
distinct kinds of subsets: those which are subsets of B D fa1 ; a2 ; : : : ; ak g and
those which contain akC1 . By the induction hypothesis, A has exactly 2k subsets
of the first type. On the other hand, the subsets of the second type are of the form
B0 [ fakC1g, where B0 is a subset of B. Again by induction hypothesis, there are
also 2k subsets of this second type. Hence, A has 2k C 2k D 2kC1 subsets.
13. To the first case, by using the induction hypothesis (strong induction!), we get
ukC2 D rukC1 suk D rakC1 sak D akC2 . The second case is completely
analogous.
B Hints and Solutions
545
14. By induction hypothesis, suppose that .a1 ; a2 ; : : : ; ak / is an AP of common
difference r. Then,
X 1
X 1
1
k1
1
k
D
D
C
D
C
a1 akC1
aa
aa
ak akC1
a1 ak
ak akC1
jD1 j jC1
jD1 j jC1
k
k1
and, hence, kak D .k 1/akC1 C a1 . However, ak D a1 C .k 1/r, so that
.k 1/akC1 D k.a1 C .k 1/r/ a1 D .k 1/.a1 C kr/:
Therefore, akC1 D a1 C kr, and .a1 ; a2 ; : : : ; ak ; akC1 / is also an AP.
15. For item (a), the induction hypothesis gives
akC1 D a2k ak C 1 D ak .ak 1/ C 1
D ak .a1 : : : ak1 / C 1 D a1 : : : ak1 ak C 1:
For (b), it follows from the induction hypothesis and the result of (a) that
kC1
k
X
X
1
1
1
1
1
D
C
D2
C
a
a
a
a
a
:
:
:
a
a
j
j
kC1
1
2
k
kC1
jD1
jD1
D2
akC1 a1 a2 : : : ak
1
D2
:
a1 a2 : : : ak akC1
a1 a2 : : : ak akC1
16. By induction hypothesis, we have 22
D .22 /2 > k2k , so that it suffices to
2 k
k
show that k2k .k C 1/kC1 or, which is the same, kC1
k C 1. Such an
inequality is true for every integer k 3, for we have
kC1
k2
kC1
k
1
D k1C
kC1
k
k
> .k 1/k .k 1/3 k C 1:
17. For n 2 Z, let an D xn C xn . We first show that an 2 Z for every n 2 N. The
hypothesis of the problem gives a1 2 Z, so that
a 2 D x2 C
1
1 2
D
x
C
2 D a21 2 2 Z:
x2
x
Now suppose, by induction hypothesis, that a1 ; a2 ; : : : ; ak 2 Z, for a certain
integer k 2. Then,
546
B Hints and Solutions
1
akC1 D xkC1 C kC1
x
1
1
1
k
k1
x
xC
D x C k
C k1
x
x
x
D ak a1 ak1 2 Z;
and the induction step is complete. To what was left to show, it suffices to see
that a0 D 2 and, if n < 0 is an integer, then an D an 2 Z.
18. It follows from induction hypothesis that
x2kC1 D .x3k 3xk /2 D x6k 6x4k C 9x2k
D .yk C 2/3 6.yk C 2/2 C 9.yk C 2/
D y3k 3yk C 2 D ykC1 C 2:
19. It suffices to prove that the quotient
immediate by induction, for
xnC1 Cxn1
xn
does not depend on n. This is
xnC2 xn C x2n x2nC1 xn1 xnC1
xnC2 C xn xnC1 C xn1
D
xnC1
xn
xnC1 xn
D
.x2nC1 1/ C x2n x2nC1 .x2n 1/
D 0:
xnC1 xn
20. For item (a) we have, by induction hypothesis, F1 C F2 C C Fk C FkC1 D
.FkC2 1/ C FkC1 D FkC3 1. For (b) (and also by induction hypothesis),
2
2
D Fk FkC1 C FkC1
D FkC1 FkC2 :
F12 C F22 C C Fk2 C FkC1
Items (c), (d) and (e) can be dealt with in an analogous way.
21. We make induction on n 1. For n D 1, the relation in the statement of the
problem is simply the recursive definition of FmC1 . In general, it follows from
the induction hypothesis that
FmCkC1 D FmCk C FmCk1
D .Fm FkC1 C Fm1 Fk / C .Fm Fk C Fm1 Fk1 /
D Fm .FkC1 C Fk / C Fm1 .Fk C Fk1 /
D Fm FkC1 C Fm1 FkC1 :
22. In both items, make induction on n, for a fixed m 1. For the initial cases
2FmC1 D Fm C Lm and 2LmC1 D 5Fm C Lm , you will also have to make
inductive arguments.
B Hints and Solutions
547
23. For (a), observe that F1 D 1, F2 D 1, F3 D 2, F4 D 3, F5 D 5, F6 D 8,
F7 D 13, F8 D 21, F9 D 34, F10 D 55, F11 D 89, F12 D 144, F13 D 233,
F14 D 377. Therefore, F1 D 12 , F12 D 122 and F13 > 132 , F14 > 142 . On
the other hand, if Fk > k2 and FkC1 > .k C 1/2 , then FkC2 D Fk C FkC1 >
k2 C .k C 1/2 > .k C 2/2 for every k > 3. Hence, we get by induction that
Fn > n2 for every n 13. For the first part of item (b), note that ˛ is a root of
the second degree equation x2 x 1 D 0, so that ˛ 2 D ˛ C 1. Thus, we have
by induction that
˛ kC1 D ˛ k ˛ D .Fk ˛ C Fk1 /˛ D Fk ˛ 2 C Fk1 ˛
D Fk .˛ C 1/ C Fk1 ˛ D .Fk C Fk1 /˛ C Fk
D FkC1 ˛ C Fk ;
Now, note that ˛ n n2 ˛ D Fn ˛ C Fn1 n2 ˛ D .Fn n2 /˛ C Fn1 . Since ˛
is irrational, it follows from this equality that ˛ n n2 ˛ will be an integer if and
only if Fn n2 D 0. Hence, the solution of ii. follows from that of the item (a).
24. Use induction to construct a subsequence .ank /k1 of .an /n1 such that ank >
nk ; a1 ; a2 ; : : : ; ank1 .
25. If aj D j for 1 j k, then we get from
a31 C a32 C C a3k C a3kC1 D .a1 C a2 C C ak C akC1 /2
that
13 C 23 C C k3 C a3kC1 D .1 C 2 C C k C akC1 /2 :
According to the result of Problem 2, this last relation gives
1
Œk.k C 1/2 C a3kC1 D
4
1
k.k C 1/ C akC1
2
2
:
Performing the computations at the right hand side, we arrive at a3kC1 D a2kC1 C
k.k C1/akC1 or, which is the same, a2kC1 akC1 k.k C1/ D 0. Since akC1 > 0,
it immediately
follows that akC1 D k C 1.
p
26. If xk > a, then
p
p
p
1
1
a
xk 2 a C
D
xkC1 a D
.xk a/2 > 0:
2
xk
2xk
To the other inequality, we start by observing that, if x y >
y C ay (*); in fact,
p
a, then x C ax 548
B Hints and Solutions
a
.x y/
a
1 1
yC
D .x y/ C a
D
.xy a/ 0;
xC
x
y
x y
xy
for xy p 2
p
a D a. Now, if xk a C
xkC1
1
,
2k1
then, from (*), we get
!
1 p
1
a
1
a
xk C
D
a C k1 C p
1
2
xk
2
2
a C 2k1
p
1 p
1
a
1
D aC k:
<
a C k1 C p
2
2
2
a
27. If k 6 and a1 , a2 , . . . , ak are positive integers such that
1
1
1
C 2 C C 2 D 1;
2
a1
a2
ak
then
1
1
1
1
1
1
1
3
C C C
C
CC
D C D 1:
2
2
2
4
4
4
.2a1 /
.2a2 /
.2ak /
4
4
Hence, by a simple variant of the strong form of the principle of mathematical
induction, it suffices to consider the cases k D 6, k D 7 and k D 8. Before we
look at them, note that
1
1
1
1
C C
D
9
9
36
4
and
1
1
1
1
C C C D 1:
4
4
4
4
1
Therefore, for k D 6, we have 14 C 14 C 14 C 19 C 19 C 36
D 1; for k D
1
1
1
1
1
1
1
7, we have 16 C 16 C 16 C 16 C 9 C 9 C 36 D 1; for k D 8, we have
1
1
1
C 14 C 19 C 19 C 36
C 19 C 19 C 36
D 1.
4
28. Argue as in the proof of Example 4.12.
29. For the existence of a representation for n, if .k C 1/Š n < .k C 2/Š, let akC1
be the greatest natural number such that akC1 .k C 1/Š n < .akC1 C 1/.k C 1/Š.
Then, the inequalities akC1 .k C1/Š n < .k C2/Š D .k C2/ .k C1/Š guarantee
that akC1 k C 1; on the other hand, since
n akC1 .k C 1/Š < .akC1 C 1/.k C 1/Š akC1 .k C 1/Š D .k C 1/Š;
it follows from the induction hypothesis that
n akC1 .k C 1/Š D a1 1Š C a2 2Š C C ak kŠ;
for certain integers a1 , a2 , . . . , ak , such that 0 aj j, for 1 j k. For the
uniqueness of the representation of n, suppose we have established it for every
natural n such that n .k C 1/Š. Take n 2 N such that .k C 1/Š n < .k C 2/Š,
and assume that
B Hints and Solutions
549
n D a1 1Š C a2 2Š C C akC1 .k C 1/Š D b1 1Š C b2 2Š C C bkC1 .k C 1/Š;
with 0 aj ; bj j for every j 1. Then, by the result of Example 3.26, we get
akC1 .k C 1/Š n D b1 1Š C b2 2Š C C bkC1 .k C 1/Š
1 1Š C 2 2Š C C k kŠ C bkC1 .k C 1/Š
D .bkC1 C 1/ .k C 1/Š 1 < .bkC1 C 1/ .k C 1/Š;
(B.1)
so that akC1 bkC1 . Analogously, bkC1 akC1 and, thus, akC1 D bkC1 .
Therefore, we have
n akC1 .k C 1/Š D a1 1Š C a2 2Š C C ak kŠ D b1 1Š C b2 2Š C C bk kŠ;
with (again by the result of Example 3.26)
n akC1 .k C 1/Š 1 1Š C 2 2Š C C k kŠ D .k C 1/Š 1:
It thus follows from the induction hypothesis that aj D bj , for 1 j k.
30. For the existence of a representation as asked by the problem assume, as
induction hypothesis, that for some natural n 3 every m Fn can be uniquely
written as asked. Now, take a natural m such that Fn < m FnC1 . If m D FnC1 ,
there is nothing to do. Otherwise, 0 < m Fn < FnC1 Fn D Fn1 < Fn .
Therefore, by induction hypothesis, there exist an integer r 1 and
nonconsecutive indices 1 < t1 < < tr < n such that mFn D Ft1 C CFtr .
Then, m D Ft1 C CFtr CFn , and we claim that ntr > 1. In fact, if ntr D 1,
we would have m Fn C Ftr D Fn C Fn1 D FnC1 , thus contradicting the
fact that m < FnC1 . For the uniqueness of representation assume, as induction
hypothesis, that for some natural n 3 every natural m Fn can be uniquely
represented as asked. Now, take m 2 N such that Fn < m FnC1 , and consider
two cases separately:
(a) m D FnC1 : there are two subcases. First, suppose that FnC1 D Ft1 C Ft2 C
C Ftr , with r 1, 1 < t1 < t2 < < tr < n and tjC1 tj > 1 for
1 j < r. Then,
FnC1 D Ft1 C Ft2 C C Ftr < Ft1 1 C Ft1 C Ft2 C C Ftr
D Ft1 C1 C Ft2 C C Ftr < Ft2 1 C Ft2 C Ft3 C C Ftr
D Ft2 C1 C Ft3 C C Ftr < Ft3 1 C Ft3 C C Ftr
D < Ftr C1 Fn < FnC1 ;
550
B Hints and Solutions
which is an absurd. Now, if FnC1 D Ft1 C C Ftr C Fn , then
Ft1 C C Ftr D FnC1 Fn D Fn1
and, by the uniqueness of the representation of Fn1 , we would conclude
that any other representation of FnC1 would be FnC1 D Fn C Fn1 , which
does not satisfy the conditions of the statement of the problem.
(b) Fn < m < FnC1 : let us show that Fn is necessarily one of the summands
of the representation. In fact, if m D Ft1 C C Ftr , with tr < n, then a
reasoning analogous to that of the first subcase of item (a) would give us the
absurd conclusion m Fn . Therefore, for a representation of m as in the
statement of the problem, we should surely have m D Ft1 C C Ftr C Fn ,
with tr < n 1. Thus, since m Fn < FnC1 Fn D Fn1 , the uniqueness of
the representation of m follows from the uniqueness of the representation
of m Fn .
Section 4.2
2n1
2n 2n1
It suffices to note that 2n
n D n n1 D 2 n1 .
Adapt, to the
present
n case, the idea of the solution
of Example 4.18.
Compare nk to kC1
using the definition of nj .
Make a proof by induction on n.
Apply Stiefel’s relation to the numerator of the fraction and, then, use telescoping
sums.
6. Make a proof by induction, applying Stiefel’s relation in the induction step.
7. Let an be defined by the expression at the right hand side of the equality we wish
to establish. Show that a1 D 1, a2 D 1 and (using Stiefel’s relation) anC2 D
anC1 C an , for every integer n 1.
8. Make induction on n; more precisely, show that, for each n 2 N, the set
f1; 2; : : : ; ng has nk subsets of k elements each, for every 0 k n. For the
induction step, note that each of the subsets of k elements of f1; 2; : : : ; n C 1g is
of one of two possible types: those not containing n C 1 – thus, being subsets of k
elements of f1; 2; : : : ; ng – and those containing n C 1 – thus, being equal to sets
of the type A [ fn C 1g, where A is a subset of k 1 elements of f1; 2; : : : ; ng.
1.
2.
3.
4.
5.
Section 4.3
1. Use Newton’s binomial formula to expand 4n D .1 C 3/n .
B Hints and Solutions
551
2. It follows from the previous problem that A D 4n . Moreover, by an analogous
8
reasoning to that of the previous problem, we get B D 12n1 . Therefore, 34 D
n1
n1
B
12
3
A D 4n D 4 , so that n D 9.
1 n
3. Compute the first three terms in the expansion of .1; 1/n D 1 C 10
.
4. The sum we wish to compute equals
1
11Š
11Š
11Š
11Š
11Š
11Š
11Š
C
C
C
C
C
D
1Š10Š 3Š8Š
5Š6Š
7Š4Š
9Š2Š
11Š0Š
!
!
!
!
!
!!
1
11
11
11
11
11
11
C
C
C
C
C
D
11Š
1
3
5
7
9
11
D
1
2111 :
11Š
5. The general term of such expansion is
65 1
k
3k
. Now, observe that
!
!
65
1
1
1
1
65 1
<
,
<
, 65 k > 3.k C 1/ , k 15:
k C 1 3kC1
65 k
kC1 3
k 3k
6.
7.
8.
9.
1
65 1
Analogously, 65
, k 16, so that the maximal term is
k >
k
kC1 3kC1
3
65 1
16 316 .
The given condition is equivalent to a2 C b2 C ab D c2 C d2 C cd. On the other
hand, it follows from the binomial expansion that a4 Cb4 C.aCb/4 D c4 Cd 4 C
.cCd/4 if and only if a4 Cb4 C2a3 bC3a2 b2 C2ab3 D c4 Cd 4 C2c3 d C3c2 d2 C
2cd3 . Finally, observe that a4 C b4 C 2a3 b C 3a2 b2 C 2ab3 D .a2 C b2 C ab/2
and, analogously,c4 C d4 C 2c3 d C 3c2 d2 C 2cd3 D .c2 C d2 C cd/2 .
Use the result of Problem 3, page 103—with 2n in place of n—together with
the theorem on the lines of the Pascal triangle.
Adapt, to the present case, the idea of the solution of Example 4.27.
It suffices to see that
!
!
n
n
X
n j X
n 1 j1
ja D
an
a D an.1 C a/n1
j
j
1
jD0
jD1
and
!
!
!
!
n
n
n
n
X
X
X
X
n 2
n1
n1
n1
j D
n
jD
n
.j 1/ C
n
j
j1
j1
j1
jD0
jD1
jD1
jD1
!
!
n
n
X
X
n1
n2
n.n 1/
Cn
D
j1
j
2
jD2
jD1
D n.n 1/ 2n2 C n 2n1 :
552
B Hints and Solutions
10. Use the formula for the general term of an AP, together with the result of the
previous problem.
11. Write 1q D 1 C a, with a > 0, and use the fact that .1 C a/n 1 C na.
p
p
p
12. Making ˛ D a C n b, we get .˛ a/n D b. Now, arguing as in the proof
of the Example 4.23, we conclude that
.˛ p n
a/ D .˛ n C an2 ˛ n2 C an4 ˛ n4 C /C
p
C .an1 ˛ n1 C an3 ˛ n3 C / a;
for certain integers a0 , a1 , a2 , . . . , an2 , an1 . Hence,
Œ.˛ n C an2 ˛ n2 C an4 ˛ n4 C / b2 D
D a.an1 ˛ n1 C an3 ˛ n3 C /2 :
so that ˛ is the root of the polynomial equation of degree 2n
Œ.xn C an2 xn2 C an4 xn4 C / b2 a.an1 xn1 C an3 xn3 C /2 D 0:
13. First of all, let’s find out which is the (constant) value of the given expression.
To this end, substituting x D 2 and y D z D 1 (observe that xCyCz D 0), we
x5 Cy5 Cz5
get x5 C y5 C z5 D 30 and xyz.xy C yz C zx/ D 6, so that xyz.xyCyzCzx/
D 5.
5
5
5
x Cy Cz
Therefore, we ought to prove that xyz.xyCyzCzx/
D 5, for all nonzero reals x, y
and z such that x C y C z D 0. To this end, note that
x5 C y5 C z5 D x5 C y5 C .x y/5 D x5 C y5 .x C y/5
D x5 C y5 .x5 C 5x4 y C 10x3 y2 C 10x2 y3 C 5xy4 C y5 /
D .5x4 y C 5xy4 C 10x3 y2 C 10x2 y3 /
D 5xyŒ.x3 C y3 / C 2xy.x C y/
D 5xy.x C y/Œ.x2 xy C y2 / C 2xy
D 5xy.z/Œ.x2 C 2xy C y2 / xy
D 5xyzŒ.x C y/.x C y/ xy D 5xyzŒ.x C y/.z/ xy
D 5xyz.xy C yz C zx/:
B Hints and Solutions
553
14. Applying Newton’s expansion formula twice, we get
!
n
X
n
.x C y/nl zl
.x C y C z/ D
l
lD0
! nl
!
n
X
n X n l nlk k l
x
D
yz
l kD0
k
lD0
!
!
nl
n X
X
n n l nlk k l
x
D
y z:
l
k
lD0 kD0
n
nŠ
Now, observe that nl nl
D .nkl/ŠkŠlŠ
, so that it suffices to make j D n k l
k
to get the formula in the statement of the problem.
10
10 j k 1 l
16. The general term in the expansion of 1 C x C 6x
1x x
is j;k;l
D
10 kl
x , with j C k C l D 10. Therefore, we should have k C l D 10 j
j;k;l
and k l D 0, so that k D l D 5 2j and, hence, j must be even. Now, make
j successively
equal to 0, 2, 4, 6, 8 and 10 and add all of the corresponding
summands j;5 10
j
j .
2 ;5 2
n
P
D .1 C 1 C 1/n and
17. Use the trinomial expansion formula to get jCkClDn j;k;l
P
n
l
n
jCkClDn .1/ j;k;l D .1 C 1 1/ .
18. Make induction on n. To the induction step you will need to use Stiefel’s
relation, together
that FkC2 D FkC1 C Fk , for every k 1.
Pwith the fact
k 1991 1991k
19. Let’s show that 995
D 1. To this end, start by writing
kD0 .1/ 1991k
k
!
1991
1991 k
D
.1/
1991 k
k
kD0
! 995
!
995
X
X
k
1991 k
k 1991 k
k
D
C
.1/
.1/
k
1991 k
k
kD0
kD0
!
!
995
995
X
X
1990 k
k 1991 k
C
:
.1/
.1/k
D
k
k1
kD0
kD0
995
X
k
P
Now, for n 2 N, define Sn D k0 .1/k nk
by setting nk
D 0 for k >
k
k
Pn2
n k, so that S0 D S1 D 1. Use the columns’ theorem to get mD0
Sm D
P
P
n2
k n1k
.1/
or,
which
is
the
same,
S
D
1
S
.
D
1
S
n
n
mD0 m From
k0
kC1
this last relation, prove that SnC1 D Sn Sn1 for every n 2 ZC , and compute
554
B Hints and Solutions
S2 D 0, S3 D S4 D 1, S5 D 0, S6 D S7 D 1. Finally, show that Sm D Sn
whenever 6 divides m n, and observe that
995
X
1991
1991 k
.1/
1991 k
k
kD0
k
!
D S1991 S1989
D S5 S3 D 0 .1/ D 1:
Section 5.1
2. First of all, note that
3
b2 .b a/
a2 .a b/
b
a3
a2 C
b2 D
C
b
a
b
a
2
3
a
a b3
a2 D .a b/
D .a b/
b
ab
D
1
.a b/2 .a2 C ab C b2 /:
ab
Therefore, it suffices to show that a2 C ab C b2 0. To this end, write
b 2 3b2
> 0:
C
a2 C ab C b2 D a C
2
4
Of course, equality holds if and only if a D b.
3. For x ¤ a; b; c, to solve the given equation is the same as to solve the second
degree equation
.x b/.x c/ C .x a/.x c/ C .x a/.x b/ D 0:
Now, observe that the discriminant of such an equation equals
4.a C b C c/2 12.ab C ac C bc/ D 4Œ.a2 C b2 C c2 / .ab C ac C bc/;
which is a strictly positive number.
4. Write the left hand side as a.a C b C c/ C bc. Then, apply inequality (5.2).
5. Firstly, show that a3 C b3 .a C b/ab. Then, deduce that a3 Cb13 Cabc c
abc.aCbCc/ , getting analogous inequalities for the other two summands of the
left hand side. Finally, add the three inequalities thus obtained.
6. Apply Ravi’s transformation to conclude that is sufficient to prove that .y C z/
.x C z/.x C y/ 8xyz, for every positive real numbers x, y and z. To this end,
use inequality (5.2) three times.
B Hints and Solutions
555
7. Apply Ravi’s transformation to rewrite the left hand side as
1
2
z
x
z
y
x
y
C C C C C
:
x
x
y
y
z
z
Then, use the inequality between the arithmetic and geometric means for six
numbers.
8. By adequately grouping in pairs the summands at the left hand side and
applying inequality (5.7), between the arithmetic and harmonic means, we get
aCc
cCa
1
4
1
C
D .a C c/
C
.a C c/ aCb
cCd
aCb
cCd
.a C b/ C .c C d/
and
dCb
1
4
bCd
1
C
D .b C d/
C
.b C d/ :
bCc
dCa
bCc
aCd
.b C c/ C .a C d/
Now, it suffices to add the two inequalities above.
9. The inequality we wish to prove is equivalent to
1 C x C x2 C C x2n1 C x2n .2n C 1/xn
or, which is the same, to
.1 C x2n / C .x C x2n1 / C C .xn1 C xnC1 / 2nxn :
In order to prove this last one, it suffices to apply inequality (5.1) n times,
adding the n inequalities thus obtained.
10. It follows from (5.4) that
p
4
3a4 C b4 D a4 C a4 C a4 C b4 4 a4 a4 a4 b4 D 4ja3bj 4a3 b:
11. We’ve already shown that a2 C b2 C c2 ab C bc C ca, for every positive reals
a, b and c. Multiply both sides of this inequality by a C b C c to get the desired
inequality.
12. Since .ab C bc C ca/2 D .ab/2 C .bc/2 C .ca/2 C 2abc.a C b C c/, it suffices
to show that
.ab/2 C .bc/2 C .ca/2 abc.a C b C c/:
Making x D ab, y D bc and z D ca in (5.3), we get
.ab/2 C .bc/2 C .ca/2 D x2 C y2 C z2
xy C yz C zx
D abc.a C b C c/:
556
B Hints and Solutions
13. Apply inequality (5.3) once more.
14. Expand the right hand side and use the algebraic identity (2.5). After performing the obvious cancellations, apply the inequality of Example 2.3 three times.
Then, use inequality (5.5).
15. Let S D a4 .1 C b4 / C b4 .1 C c4 / C c4 .1 C a4 /. Then, apply the inequality
between the arithmetic and geometric means for two and three numbers, to get
S a4 2b2 C b4 2c2 C c4 2a2
p
3 3 .2a4 b2 /.2b4 c2 /.2c4 a2 /
p
3
D 6 a 6 b 6 c6
D 6a2 b2 c2 :
16. Apply the inequality between the arithmetic and geometric means.
17. Change ai by ai and, then, apply the inequality between the arithmetic and
geometric means.
18. For item (a), it follows from the inequality between the arithmetic and
geometric means that
nŠ D 1 2 n 1C2CCn
n
n
D
nC1
2
n
:
For item (b), argue as in (a), applying the inequality between the arithmetic and
geometric means to the numbers 12 , 22 , . . . , n2 and, then, using the result of
Example 4.18.
19. For 1 k m 1, apply the inequality between the arithmetic and geometric
means to x C k D x C 1„ C ƒ‚
C…
1. Then, multiply the m 1 inequalities thus
k
obtained.
P
20. Since njD1 .S xj / D .n 1/S, the inequality in the statement of the problem
is equivalent to
10
0
n
n
X
X
@ .S xj /A @
jD1
jD1
1
1 A
n2 :
S xj
Then, it suffices to apply inequality (5.7).
21. For item (a), just expand the product of the left hand side. For (b), apart from
the hint given in the statement of the problem, apply the result of Problem 8,
page 104 (with k D 2), to conclude that the sum at the right hand side of item
(a) has exactly 12 n.n 1/ summands.
B Hints and Solutions
557
22. First of all, note that
k1 k2 : : : kn
1
1
1
C
CC
k1
k2
kn
D p1 C p2 C C pn ;
with pi D k1 : : : ki1b
ki kiC1 : : : kn , the hat over ki indicating that the product
contains all of k1 , k2 , . . . , kn , with the exception of ki . Then, we can write
a k2
a kn
p1 ak11 C p2 ak22 C C pn aknn
ak11
C 2 CC n D
:
k1
k2
kn
k1 k2 : : : kn
Now, expand the summands of the numerator in such a way that each aki i appears
pi times and apply the inequality between the arithmetic and geometric means
to get
1
Œ.ak1 C C ak11 / C C .aknn C C aknn / „
ƒ‚
…
k1 k2 : : : kn „1
ƒ‚
…
p1 vezes
.a1 /k1 /p1 : : : .an /kn /
D
n
Y
pi ki
k k :::kn
ai 1 2
pn times
pn k
1
1 k2 :::kn
D a1 a2 : : : an :
iD1
23. Make a0 D 0 and apply (5.7) to get
1
j2
1
CC
:
aj aj1
a1 a0
aj
Then, add the above inequalities for 1 j n and group equal terms to get
the desired inequality. Finally, conclude that equality holds if and only if the
sequence .ak /k1 is an AP.
24. Substituting a, b and c respectively by 6x6 , 6y6 and 6z6 , show that it suffices to
prove that
7x12 C 12x6 y6 C 7y6 z6 C 9y12 C 9x6 z6 2x3 y9 C 6x9 y3 C 6x2 y8 z2 C 12x5 y5 z2 C 6x4 y4 z4 C 6xy7 z4 C 6x8 y2 z2 :
To this end, write the left hand side expression as a sum of seven other
expressions, chosen in such a way that, when applying the inequality between
the arithmetic and geometric means to each one of them, we get the seven
summands of the right hand side. For instance:
2x6 z6 C 2x6 z6 C 2x12 C 2y12 C 2y12 C 2x6 y6 12x5 y5 z2 :
558
B Hints and Solutions
Section 5.2
p
p
2
2
2
2
1. By
p Cauchy’ inequality, we have 12144D 3x C 4y 3 C 4 x C y D
5 x2 C y2 . Therefore, x2 C y2 25 , with equality holding if and only if
y
x
36
3 D 4 . However, since 3x C 4y D 12, equality holds if and only if x D 25 and
48
y D 25 .
2. Adapt the discussion of the case n D 3, presented in the text.
3. Making b1 D b2 D D bn D 1 in Cauchy’s inequality, we get
a1 C a2 C C an q
p
a21 C a22 C C a2n n;
with equality if and only if there exists a nonzero real number such that a11 D
a2
D D a1n D , i.e., if and only if a1 , a2 , . . . , an are all equal. Dividing by
1
n both sides of the above inequality, we get the decided inequality.
4. Apply the inequality between the quadratic and arithmetic means (see last
problem) to the numerator of the fraction at each summand above. Then, add
the results thus obtained.
5. We want to prove that
0
1
n
X
1A
@
xj A @
n2 ;
x
jD1
jD1 j
n
X
10
p
for all positive reals x1 , x2 , . . . , xn . To this end, make aj D xj , bj D p1xj and
apply Cauchy’s inequality.
6. Multiply both
C b C c C dpand applyp
Cauchy’s inequality for n D 4,
psides by ap
with a1 D a, a2 D b, a3 D c, a4 D d and b1 D p1a , b2 D p1b ,
b3 D
2
p
c
p4 . Alternatively, apply (5.7), writing c
d
d
d
16
1
1
1
1
C
C d=4
C d=4
.
4
4 and d D d=4 C
pd=4
, b4 D
D 2c C 2c ,
4
c
D
1
1
c=2 C c=2 ,
d D d4 C d4 C
p
p
7. For the left hand inequality, let S D 2x C 3 C 2y C 3 C 2z C 3. Applying
the inequality between the arithmetic and geometric means for three numbers,
we get
p
S 3 6 .2x C 3/.2y C 3/.2z C 3/
p
D 3 6 8xyz C 12.xy C xz C yz/ C 18.x C y C z/ C 27
p
p
3
6
> 3 18 3 C 27 D 3 9:
For the right hand inequality, apply the inequality between thepquadratic and
arithmetic
means forpthree numbers (cf. Problem 3), with a1 D 2x C 3, a2 D
p
2y C 3 and a3 D 2z C 3, and use the fact that x C y C z D 3.
B Hints and Solutions
559
8. Apply the inequality
the quadratic and arithmetic means (cf. Probq between
n
. Then, use the result of item (a) of Corollary 4.25.
lem 3), with ak D
k
9. By Cauchy’s inequality, we have
y2
z2
x2
C
C
y2
z2
x2
z2
x2
y2
C
C
z2
x2
y2
y z
z x
x y
C C y z
z x
x y
2
z
x
y
C C
D
:
x
y
z
2
10. Since
a2 b2k C b2k
b2k
a2k
D k
D .ak bk / C
;
ak C bk
ak C bk
ak C bk
we get
n
X
kD1
n
n
n
X
X
X
a2k
b2k
b2k
D
.ak bk / C
D
:
ak C bk
a C bk
a C bk
kD1
kD1 k
kD1 k
Hence,
n
X
kD1
X a2 C b2
1X
1X
a2k
k
k
ak ,
.ak C bk /;
ak C bk
2 kD1
a C bk
2 kD1
kD1 k
n
n
2
n
2
Cy
xCy
and it suffices to show that xxCy
2 , for all positive reals x and y. This
follows immediately from the inequality between the quadratic and arithmetic
means (cf. Problem 3).
11. Start by letting anC1 D a1 . Now, for 1 j n, use Cauchy’s inequality to
write
aj
p
.1 C aj /2 D 1 1 C ajC1 p
ajC1
!2
.1 C ajC1 / 1 C
a2j
ajC1
!
;
so that
1C
a2j
ajC1
.1 C aj /2
:
1 C ajC1
Then, multiply the n inequalities obtained in this way to reach the desired one.
12. Perform Ravi’s transformation (according to the discussion that precedes the
statement of Problem 6, page 120) to show that the given inequality is
equivalent to
560
B Hints and Solutions
p
p
p
p
p
p
2x C 2y C 2z y C z C x C z C x C y;
for positive reals x, y and z. Then, use the inequality
between the quadratic and
q
p
p
yC z
, with equality if
arithmetic means (cf. Problem 3) to show that yCz
2
q2
q
p p
p p
xC y
xC z
and only if y D z; analogously, get xCz
and xCy
, and
2 2
2 2
add these three inequalities to get the desired one.
13. Use Cauchy’s inequality to get
n
X
!2
a k b k ck
n
X
.ak bk /2
kD1
!
kD1
n
X
!
c2k
:
kD1
and
n
X
v
! n
!
u n
X
u X
2 2
4
4
t
ak bk ak
bk
kD1
kD1
kD1
Now, check that the inequality between the quadratic and arithmetic means (cf.
Problem 3) gives
n
X
v
!
u
n
u X
2
4
t
ck n
ck :
kD1
kD1
Finally, it suffices to combine both inequalities above.
14. Make x D 1a , y D 1b , z D 1c and, then, apply Cauchy’s inequality to get
y2
z2
x2
C
C
..y C z/ C .x C z/ C .x C y//
yCz
xCz
xCy
.x C y C z/2 :
Finally, use the inequality between the arithmetic and geometric means to
conclude the proof.
Section 5.3
1. Start by observing that
n
1 nC1
1 C nC1
1
1
1
D
1
C
:
n
nC1
.n C 1/2
1 C 1n
Then, apply Bernoulli’s inequality.
B Hints and Solutions
561
m
2. Write am D mm 1 C am
and apply Bernoulli’s inequality. Do the same with
m
an and, then, add the results.
3. The desired inequality is equivalent to
b6
c6
a6
C
C
ab C ac C bc:
b 2 c2
a 2 c2
a2 b2
Suppose, without loss of generality, that a b c. Then, a6 b6 c6 and
1
a21c2 a21b2 . Applying in succession Chebyshev’s inequality and the
b2 c2
inequality between the arithmetic and geometric means, we get
b6
c6
1 6
1
1
a6
1
6
6
C 2 2 C 2 2 .a C b C c / 2 2 C 2 2 C 2 2
b 2 c2
a c
a b
3
b c
a c
a b
1
1
1
C 2 2C 2 2
a 2 b 2 c2
b 2 c2
a c
a b
D a 2 C b 2 C c2 :
Finally, since we already know that a2 C b2 C c2 ab C ac C bd, there is
nothing left to do.
4. Write x41 C x42 C C x41994 D x31 x1 C x32 x2 C C x31994 x1994 and, then,
apply Chebyshev’s inequality to get
x41 C x42 C C x41994 1
.x3 C x32 C C x31994 /.x1 C x2 C C x1994 /:
1994 1
Now, use the equations of the system.
5. Apply Chebyshev’s inequality to each summand of the left hand side.
6. Let’s show that, for every n > 1, one has
n
X
iD1
i a i b i n
X
iD1
!
i a i
n
X
!
i b i
0:
iD1
To this end, start by observing that the left hand side equals
n
X
iD1
i .1 C C b
i C C n /ai bi n
X
i j a i b j ;
(B.2)
i;jD1
i¤j
where b
i indicates that the summand i is omitted. Then, notice that, to each
pair .i; j/ with i < j, the sum
i j ai bi C i j aj bj i j ai bj i j aj bi D i j .ai aj /.bi bj /
562
B Hints and Solutions
appears in (B.2) exactly once, which means that
n
X
i a i b i iD1
n
X
!
n
X
i a i
iD1
!
i bi i j .ai aj /.bi bj / 0:
iD1
Equality holds when all of the ai ’s or all of the bi ’s are equal. Finally, the usual
Chebyshev’s inequality corresponds to the particular case in which i D 1n , for
1 i n.
7. Apply Chebyshev’s inequality to conclude that the expression at the left hand
side is greater than or equal to
1 3
1
1
1
1
3
3
3
.a C b C c C d /
C
C
C
:
4
bCcCd
aCcCd
aCbCd
aCbCc
Then, apply the result of Corollary 5.20, together with (5.7), to conclude that
the last expression above is greater than or equal to
aCbCcCd
4
3
.a C b C c C d/2
42
D
:
3.a C b C c C d/
12
Finally, observe that the condition given in the statement of the problem is
equivalent to .aCc/.bCd/ D 1, and apply the inequality between the arithmetic
and geometric means.
8. By symmetry we can suppose, without loss of generality, that a b c. This
1
1
1
way, we get bCc
aCc
aCb
. Applying Chebyshev’s inequality and (5.7), we
obtain
bn
cn
1
1
1
an
1
C
C
C
C
.an C bn C cn /
bCc
aCc
aCb
3 bCc
aCc
aCb
3
.an C bn C cn /:
2.a C b C c/
Also by Chebyshev’s inequality, it follows that
a n C b n C cn 1
.a C b C c/.an1 C bn1 C cn1 /;
3
which completes the proof.
9. Suppose, with no loss of generalidade, that x y z. Use Chebyshev’s
inequality to conclude that the expression in the statement of the problem is
greater than or equal to
x3 C y3 C z3
3
.x C 1/ C .y C 1/ C .z C 1/
:
.x C 1/.y C 1/.z C 1/
B Hints and Solutions
563
Apply Chebyshev’s inequality to the first factor and the inequality between the
arithmetic and geometric means to the second factor to conclude that the last
3t3
1
expression above is greater than or equal to .tC1/
2 , where t D 3 .x C y C z/.
Finally, use again the inequality between the arithmetic and geometric means
3t3
3s3
to conclude that t 1, and prove that t s ) .tC1/
2 .sC1/2 .
10. In order to get the first inequality, apply Chebyshev’s inequality to the first
sum, followed by the result of Corollary 5.8. For the second inequality, use the
inequality between the arithmetic and geometric means.
11. Suppose, with no loss of generalidade, that a1 a2 a2n . Arrange the
given numbers in n pairs .b1 ; c1 /, .b2 ; c2 /, . . . , .bn ; cn /, with b1 b2 bn
and bj cj , for each j. We wish to maximize
S D b 1 c1 C b 2 c2 C C b n cn :
To this end, we will show that c1 c2 cn . Indeed, if the sequence .ci /
isn’t nondecreasing, there will exist indices i > j such that ci cj . In this case,
we change the positions of ci and cj in S, after which the new sum will be
S0 D S bi ci bj cj C bi cj C bj ci D S C .bi bj /.cj ci / S:
Hence, the sum will be maximal when c1 c2 cn . Finally, observe
that, for all indices i < j, we have bi ci cj . Suppose that, for some pair of
indices i < j, we would have bj ci . Change ci by bj , so that the new sum is
S00 D S bi ci bj cj C bi bj C ci cj D S C .bi cj /.bj ci / S:
Therefore, we must have bi ci bj cj whenever i < j. In general, we must
have
b 1 c1 b 2 c2 b n cn ;
so that the maximal sum is a1 a2 C a3 a4 C C a2n1 a2n .
12. Adapt, to the present case, the idea of the proof of rearrangement’s inequality.
13. For item (a), it suffices to see that
a.x C y a/ xy D .ax xy/ C a.y a/ D x.y a/ C a.y a/
D .y a/.a x/ 0;
for y a x. In what concerns (b), we want to prove that, given an integer
n > 1 and positive reals x1 ; : : : ; xn , one always has
AM .x1 ; : : : ; xn / GM .x1 ; : : : ; xn /;
564
B Hints and Solutions
where AM and GM denote the arithmetic and geometric means, respectively.
To this end we can suppose, without loss of generality, that x1 xn , with
x1 < xn . If a D AM .x1 ; : : : ; xn /, then x1 < a < xn . Similarly to the proof of
item (a), we obtain a.x1 C xn a/ > x1 xn , so that
AM .x1 C xn a; x2 ; : : : ; xn1 ; a/ D AM .x1 ; x2 ; : : : ; xn1 ; xn /
and
GM .x1 C xn a; x2 ; : : : ; xn1 ; a/ > GM .x1 ; x2 ; : : : ; xn1 ; xn /:
If the numbers x1 C xn a; x2 ; : : : ; xn1 ; a are all equal, we will have
AM .x1 C xn a; x2 ; : : : ; xn1 ; a/ D GM .x1 C xn a; x2 ; : : : ; xn1 ; a/
and, therefore
AM .x1 ; x2 ; : : : ; xn1 ; a/ > GM .x1 ; x2 ; : : : ; xn1 ; a/:
Otherwise, we order those n numbers as y1 yn , with y1 < yn . Since
AM .y1 ; : : : ; yn / D a, it follows that y1 < a < yn . Now, changing y1 per y1 C
yn a and yn per a we obtain, as above,
AM .y1 C yn a; y2 ; : : : ; yn1 ; a/ D AM .y1 ; y2 ; : : : ; yn1 ; yn /
and
GM .y1 C yn a; y2 ; : : : ; yn1 ; a/ > GM .y1 ; y2 ; : : : ; yn1 ; yn /:
Among the numbers y1 C yn a; y2 ; : : : ; yn1 ; a we now have at least two which
are equal to a. If the numbers y1 C yn a; y2 ; : : : ; yn1 ; a are all equal, we will
have
AM .y1 C yn a; y2 ; : : : ; yn1 ; a/ D GM .y1 C yn a; y2 ; : : : ; yn1 ; a/
and, hence,
AM .y1 ; y2 ; : : : ; yn1 ; yn / D GM .y1 ; y2 ; : : : ; yn1 ; yn /:
Otherwise, we operate a third exchange of numbers, as described above.
Observe that this algorithm ends after a finite number of steps, and when we
reach this point all numbers in our list will be equal to a. In particular, the corresponding arithmetic and geometric means of this las set of (n equal) numbers
B Hints and Solutions
565
will be equal. However, since each performed operation preserves arithmetic
means and increases geometric means, it follows that, at the beginning, we had
AM .x1 ; x2 ; : : : ; xn1 ; a/ > GM .x1 ; x2 ; : : : ; xn1 ; a/:
14. Make a proof by induction on n > 1. For the induction step, it suffices to prove
that
! a2nC1
a2n
a2
1C
1 C n .1 C anC1 /
1C
anC1
a1
a1
or, equivalently, that
a2nC1 1
1
a2
C a2n anC1 C n :
a1
anC1
a1
Finally, observe that such an inequality is true by an immediate application of
the rearrangement’s inequality.
15. (By Prof. Emanuel Carneiro) Let’s make induction on n, leaving the initial case
n D 3 as an exercise. More precisely, let’s show that, in order to maximize the
left hand side expression, one of the xi ’s must be equal to 0, and that this allows
us to invoke the induction hypothesis. To this end, let
E.x1 ; x2 ; : : : ; xn1 ; xn / D x21 x2 C x22 x3 C x23 x4 C C x2n x1 :
and suppose that the expression attains its greatest value2 for some sequence
.x1 ; x2 ; : : : ; xn / such that none of the xi ’s is equal to 0. Substituting xn1 by 0
and x1 by x1 C xn1 , we obtain a new expression, which is less than or equal to
the original one, i.e., which is such that
0 E.x1 C xn1 ; x2 ; : : : ; 0; xn / E.x1 ; x2 ; : : : ; xn1 ; xn /
D 2x1 x2 xn1 C x2 x2n1 C xn1 x2n x2n2 xn1 x2n1 xn :
Upon dividing this last expression by xn1 , we get
2x1 x2 C x2 xn1 C x2n x2n2 xn1 xn 0:
Analogously, performing the exchange operations
xi1 7! 0 and xiC1 7! xiC1 C xi1 ;
Here, we’re tacitly assuming that there exists a sequence .x1 ; x2 ; : : : ; xn / such that x1 C x2 C C
xn D 1 and E.x1 ; x2 ; : : : ; xn / is maximal. Although this could be rigorously proved, such a proof is
beyond the scope of these notes, so that we shall not present it here.
2
566
B Hints and Solutions
for 1 i n (with x0 D xn and xnC1 D x1 ), we get the inequalities
2xiC1 xiC2 C xiC2 xi1 C x2i x2i2 xi1 xi 0;
for 1 i n. Adding all of these, we arrive at the inequality
n
X
xi .xiC1 C xiC3 / 0;
iD1
which is an absurd. Therefore, if .x1 ; x2 ; : : : ; xn / maximizes E.x1 ; x2 ; : : : ; xn /,
then at least one of the xi ’s must be equal to 0. Suppose, with no loss of
generality, that xn D 0. Then,
Emax D E.x1 ; x2 ; : : : ; xn1 ; 0/
D x21 x2 C x22 x3 C x23 x4 C C x2n2 xn1
x21 x2 C x22 x3 C x23 x4 C C x2n2 xn1 C x2n1 x1
4
;
27
where, in the last inequality, we applied the induction hypothesis.
17. Make i D baii , for 1 j n, and conclude that
b1 C C bn a1 C C an , a1 .1 1/ C a2.2 1/ C C an .n 1/ 0:
Then, apply Abel’s inequality to get
n
X
aj .j 1/ a1 min f1 1; 1 C 2 2; : : : ; 1 C C n ng:
jD1
Finally, use the condition given in the statement of the problem, together with
the inequality between the arithmetic and geometric means to show that 1 C
C k k, for 1 k n.
Section 6.1
1. We must have x 1 0 and 3 x > 0, so that x 2 Œ1; 3/.
q
p
p
p
p
p
2. We must have x 0, 3 x 0, 32 3 x 0 and 12 32 3 x.
The maximal domain of f is the set formed by the intersection of the solution
sets of these inequalities.
B Hints and Solutions
567
2 12 j
3. For item (a), just see that f .1/ D f 11 D j112 C1
D 0, f .10/ D f 10
D
2
1
2
24 j102 12 j
j22 32 j
99
5
2
2
D 101 and f 36 D f 3 D 22 C32 D 13 . For item (b), if ja b j D 55
102 C12
and a2 C b2 D 73, then .a; b/ D .8; 3/ or .3; 8/, so that f 38 D f 83 D 55
.
73
If ja2 b2 j D 32 and a2 C b2 D 257, then 2a2 D 289 or 2a2 D 225 and,
hence, a … N. However, setting a2 b2 D 32k and a2 C b2 D 257k, with
k 2 N, we get 2a2 D 289k and 2b2 D 225k, so that k D 2 gives a D 17 and
b D 15 (analogously,
by
b2 a2 D 32k, we obtain a D 15 and b D 17).
15 17letting
32
Therefore, f 17 D f 15 D 257 . Finally, if a2 b2 D 101k and a2 Cb2 D 89k,
for some k 2 N, then a2 D 95k and b2 D 6k, so that .ab/2 D 570k2 ; if
b2 a2 D 101k and a2 C b2 D 89k, we conclude, analogously, that .ab/2 D
570k2 . In any case we reach ap
contradiction, for, .ab/2 > 0 > 570k2 . p
5. Start by making x D 1 and y D 2 in the given relation, to compute f .1C 2/.
6. One has to prove that f .akC1 / D f .ak / C f .r/, for every k 1. To this end, let
x D ak and y D r, so that x C y D akC1 .
7. Adapt, to the present case, the hint given to the previous problem.
8. In the case of gf , notice that one has to shrink the common domain X of f and g,
to avoid those x 2 X for which g.x/ D 0.
9. The image is the set Z of integers.
10. Start by showing that fxg 2 Œ0; 1/, for every x 2 R. Then, show that the image
of fg is precisely the interval Œ0; 1/.
11. If an denotes the n-th natural number which is not a perfect square, then there
exist positive integers s and t, such that 1 s 2t and an D t2 Cs; in particular,
t2 is the greatest perfect square which is less than or equal to an . Since there are
exactly an integers from 1 to an , exactly t of which are perfect squares (these
being 12 , 22 , . . . , t2 ), it follows that an is the .an t/-th non perfect square.
Hence, n D an t D t2 C s t, so that
r
tD
nsC
1
1
C :
4
2
Now, since s > 0, we have
r
2
an D t C s D n C t D n C
nsC
p
1
1
1
C <nC nC :
4
2
2
On
p the1 other hand, by using the fact that s 2t, we get, as above, an > n C
n 2 . Therefore,
p
p
1
1
nC nC
1 < an < n C n C ;
2
2
and, thus, an D bn C
p
n C 12 c.
568
B Hints and Solutions
12. For the first part of item (a), make x D y D 0; for the second, make y D x.
For item (b), use (a). For item (c), consider initially the case of m 2 N, by
induction; then, use (a) to get the case m < 0. Finally, apply the result of (c) to
get (d) and the results of (c) and (d) to get (e).
Section 6.2
1. Show that it is equal to .0; C1/.
2. Start by recalling that x C 1x 2, for every real x > 0, with equality if and only
if x D 1. Then, use this fact to conclude that the image is R n .2; 2/.
3. Let’s analyse the case in which f increases in .1; a \ I and decreases in
Œa; C1/ \ I (the other case is completely analogous). For x0 2 I such that x0 <
a, we have f .x0 / > f .a/, so that x0 is not a minimum point of f . Accordingly,
for x0 2 I such that x0 > a, we have f .x0 / > f .a/ and, as before, x0 is not a
minimum point of f . Hence, a is the only minimum point of f .
4. For y0 2 Y, take x0 2 X such that f .x0 / D y0 . Then, by the definition of the
function f C c, we have .f C c/.x0 / D f .x0 / C c D y0 C c, so that y0 C c 2
Im .f C c/. Since y0 C c is a generic element of Y C c, it follows that Y C c Im .f C c/. Now, if y1 2 Im .f C c/, then, by definition, there exists x1 2 X such
that .f Cc/.x1 / D y1 or, which is the same, f .x1 /Cc D y1 . Then, f .x1 / D y1 c,
and hence y1 c 2 Im .f / D Y. This way, y1 D .y1 c/ C c 2 Y C c and, since
this is valid for every y1 2 Im .f C c/, we get Im .f C c/ Y C c. Therefore,
Im .f C c/ D Y C c.
5. Adapt, to the present case, the hint given for the previous problem.
6. Let’s consider the case a > 0, the case a < 0 being totally analogous. For
b 2
y 2 R, we have f .x/ D y if and only if x C 2a
4a2 D ay or, which is the
same,
b 2
y
4ay C xC
D C 2 D
:
2a
a
4a
4a2
b 2
Since x C 2a
0, the last equation above has a solution if and only if 4ay C
0. However, since a > 0, such a condition is equivalent to y 4a
.
Hence, there exists x 2 R for which f .x/ D y if and only if y 4a , so that
Im .f / D Π4a
; C1/.
7. Use the canonical form f (see the statement of the previous problem) to get, in
the case a > 0,
B Hints and Solutions
569
"
2
#
b 2
<0, xC
2 <0
2a
4a
p
p
ˇ
ˇ p
2
ˇ
bˇ
b
b
< 2 , ˇˇx C ˇˇ <
, xC
,
<xC
<
2a
2a
2a
2a
2a
2a
4a
p
p
b C b ,
<x<
:
2a
2a
f .x/ < 0 , a
b
xC
2a
2
4a
8. In order to show that > 0, use the canonical form of f to get
"
#
b 2
x0 C
af .x0 / D a
2 ;
2a
4a
2
b 2
b 2
4a2 < 0. Now, since x0 C 2a
so that af .x0 / < 0 if and only if x0 C 2a
b 2
0, we must have 0 x0 C 2a < 4a2 and, hence, > 0. For the second part,
use the result of the previous problem.
9. If a rectangle of perimeter 2p has dimensions x and y, then x C y D p. Show
that its area depends on x according to the quadratic function f .x/ D x2 C px,
and apply the result of Proposition 6.25.
10. Letting l be the width and h be the height of the truck (measured in meters),
it is immediate that the load volume of each truck is 18lhm3 . Hence, we have
to maximize the product lh, subjected to the condition that the trucks can enter
the tunnel using the correct lanes. Since the cross section of the trucks are l h
rectangles, the common length d of their diagonals
p should satisfy d 5m.
However, by Pythagoras’ Theorem, we have d D l2 C h2 and, hence,
p
p
p
lh D l d2 l2 l 25 l2 D 25l2 l4 :
Setting x D l2 , we conclude that it suffices to maximize the quadratic function
f .x/ D 25x x2 , with 0 < x < 5.
11. We must find the smallest positive real number a such that f .x/ a, for every
x 2 R; equivalently, this is the same as finding the smallest positive real number
a such that ax2 5x C .a C 1/ 0, for every x 2 R.
12. For even n, say n D 2k, using the triangle inequality we get
f .x/ D .jx ˛1 j C C jx ˛k j/ C .j˛kC1 xj C C j˛2k xj/
k
2k
2k
k
X
X
X
X
.x ˛j / C
.˛j x/ D
˛j ˛j :
jD1
jDkC1
jDkC1
jD1
P
Pk
Now, show that every x 2 .˛k ; ˛kC1 / satisfies f .x/ D 2k
jDkC1 ˛j jD1 ˛j .
Finally, the case of an odd n can be dealt with in an analogous way.
570
B Hints and Solutions
13. For item (a), apply the inequality between the arithmetic and geometric means
to the denominator of the defining formula for f .x/.For item (b), write f .x/ D
2 C x25x
and, then, proceed as in (a). Finally, for item (c), apply the inequality
C1
between the arithmetic and geometric means to get
1
3x3 .1 x3 /3
3
4
1 3x3 C .1 x3 / C .1 x3 / C .1 x3 /
:
3
4
x3 .1 x3 /3 D
.xC10/.xC2/
9
D ..xC1/C9/..xC1/C1/
D .x C 1/ C xC1
xC1
xC1
3
3
a
a
a
x
x
3
x D 2x C 2x ; for (c), write x D 2 C 2 ; finally, for
14. For item (a), write
C 10; for
(d), write
item (b), write
6x D 3x C 3x.
15. Start by writing x4 C y4 D .x2 C y2 /2 2.xy/2 .2 C xy/2 2.xy/2 , and look
at this last expression as a quadratic function of z D xy.
16. First of all, prove that the function f W R ! R, given by f .x/ D x3 C 2x, is
increasing. Then, observe that the equations of the given system can be written
as y D f .x/, z D f .y/, t D f .z/ and x D f .t/. Finally, suppose that x y and
use fact that f is increasing to conclude that x y z t x.
17. For fixed m; n 2 N, let Ik D Œ.kmnC1/n; .kmnC1/m. Since ..kC1/mnC1/n <
2 Cnm
.kmn C 1/m for k > mn
, there exists k0 2 N such that
mn.mn/
[
Ik D Œ.k0 mn C 1/n; C1/:
kk0
Since f has infinitely many strangulation points, there exist natural numbers
k k0 and p such that p is a strangulation point of f and p 2 Ik . Then, f ..kmn C
1/n/ < f .p/ < f ..kmn C 1/m/. However, kmn C 1 is relatively prime to m and
n, so that f .kmn C 1/ C f .n/ < f .kmn C 1/ C f .m/ and, hence, f .n/ < f .m/.
Section 6.3
1. For every z 2 R, we have jzj > z if and only if z < 0. Hence, we want to find
the set of x 2 R for which g.f .x// < 0. Make y D f .x/, and apply the result of
Problem 7, page 161, to conclude that 12 < y < 12 . Now, recall that y D f .x/
to show that 12 < x 72 < 12 .
2. Start by observing that .f ı g/.x/ D f .g.x// D 2g.x/ C 7 and, hence, that
2g.x/ C 7 D x2 2x C 3.
2
3. Make g.x/ D y to obtain x D yC3
2 and, then, f .y/ D f .g.x// D 2x 4x C 1 D
2
C 1.
4 yC3
2 yC3
2
2
B Hints and Solutions
571
4. Since f ı g and g ı g are functions from R to itself, we have f ı g D g ı g ,
.f ı g/.x/ D .g ı f /.x/ , f .g.x// D g.f .x//. Now, it suffices to compute
f .g.x// D .ac/x C .ad C b/, g.f .x// D .ac/x C .bc C d/ and compare the
results.
5. For x; f .x/ ¤ b, we have
f .f .x// D
7.
8.
9.
10.
11.
12.
13.
14.
f .x/ C a
D
f .x/ C b
xCa
xCb
xCa
xCb
Ca
Cb
D
.a C 1/x C .ab C a/
:
.b C 1/x C .a C b2 /
From this, conclude that we must have .aC1/xC.abCa/
D x for every x 2 R,
.bC1/xC.aCb2 /
except for at most four values of x. Hence, except for such values of x, we have
.a C 1/x C .ab C a/ D .b C 1/x2 C .a C b2 /x. Finish by showing that b D 1.
Let’s analyse the case in which f and g are increasing, leaving the analysis of
the other two cases to the reader. Given x1 < x2 in I, it follows from the fact
that f is increasing that f .x1 / < f .x2 /. Now, since f .x1 /; f .x2 / 2 J and g is also
increasing, we have g.f .x1 // < g.f .x2 //.
Make x D vu in both given relations.
Verify that F is injective by showing that F.x1 / D F.x2 / ) x1 D x2 . Show that
the surjectivity of F reduces to its very definition.
Start by assuming that f D g C h, with g; h W X ! R such that g is even and
h is odd. Then, use the definitions of even and odd function to conclude that
g.x/ D 12 .f .x/ C f .x// and h.x/ 21 .f .x/ f .x//.
Initially, show that f .1/ D 0. Then, make a D 1 and b D 1 to compute f .1/.
Finally, make b D a.
It suffices to compute .f ı f /.x/ D f .f .x// D f .f .x// D f .f .x// D
.f ı f /.x/, where, in the second and third equalities, we used the fact that f is
odd.
Take f .x/ D g.x/ for x 0 and f .x/ D x for x 0.
Start by writing
4x2
4.x2 1/
4
CkD
C
Ck
xC1
xC1
xC1
1
4
Ck D 4 xC1C
C .k 8/:
D 4.x 1/ C
xC1
xC1
f .x/ D
1
Hence, letting g.x/ D x C 1 C xC1
, we have f .x/ D 4g.x/ C .k 8/.
Now, Problem 2, together with the fact that g is an odd function, implies that
Im .g/ D Rn.2; 2/, so that (by the result of Problem 5) Im .4g/ D Rn.8; 8/.
Finally, Problem 4 guarantees that Im .f / D R n .k 16; k/, and we must have
k 16 D L and k D L.
15. For item (a), start by showing that every x 2 R can be written as
x D kp C ˛, with k 2 Z and ˛ 2 Œ0; p/. Then, make induction on k to
572
B Hints and Solutions
show
that f .x/ D f .x kp/, for every k 2 Z. For item (b), first observe that
p
g x C jaj
D f .ax ˙ p/ D f .ax/ D g.x/, for every x 2 R; subsequently, note
that if p0 > 0 satisfies g.xCp0 / D g.x/ for every x 2 R, then f .axCap0 / D f .ax/
for every x 2 R. Finally, use this fact to deduce that jajp0 D jap0 j p.
14. In order to show that f is odd, start by observing that f .20 C x/ D f .10 C
.10 C x// D f .10 .10 C x// D f .x/ and, analogously, f .20 x/ D f .x/.
Then, use the fact that f .20 C x/ D f .20 x/. For the second part, compute
f .40 C x/ D f .20 C .20 C x// D f .20 C x/ D f q
.x/.
17. Making g.x/ D f .x/ C 12 , show that g.x C a/ D 14 g.x/2 and, hence, that
g.x C 2a/ D g.x/, for every x 2 R.
18. If X D fx 2 ZI x > 100g, then Im.f / D f .X/ [ f .Z n X/. On the one hand,
f .X/ D fy 2 ZI y > 90g. Now, if 90 x 100, then 101 x C 11 111 and
f .x/ D f .f .x C 11// D f .x C 1/. Therefore, f .90/ D f .91/ D D f .100/ D
f .f .111// D f .101/ D 91. Use strong induction to show that f .90x/ D 91 for
every integer x 1. For the induction step, if f .90 x/ D 91 for every x < n,
observe that, for x D n, one has f .90 n/ D f .f .101 n// D f .91/ D 91.
Finally, conclude that Im.f / D f91; 92; 93; : : :g.
19. From f .f .n// D 4n C 1, we get f .4n C 1/ D f .f .f .n/// D 4f .n/ C 1. On the
other hand, it’s easy to prove by induction, with the aid of items (a) and (b), that
f .2k / D 2kC1 1. Consequently, again by item (c), we obtain f .2kC1 1/ D
f .f .2k // D 4 2k C 1. Due to these facts, and since 1993 D 4 498 C 1,
we successively compute f .1993/ D 4f .498/ C 1, f .498/ D 2f .249/ C 1,
f .249/ D f .4 62 C 1/ D 4f .62/ C 1, f .62/ D 2f .31/ C 1 and f .31/ D
f .25 1/ D 4 24 C 1 D 65.
20. Write N D A1 [ A2 [ A3 [ : : :, with A1 , A2 , . . . pairwise disjoint infinite sets.
Then, for each n 2 N, set f .x/ D n whenever x 2 An .
21. For item (a), use the assumptions on A and B to construct an injection from
A B to N N. For item (b), take f W Z N ! Q such that f .m; n/ D mn . The
first part of (c) now follows from (b) and Lemma 6.43. As for the second part,
let g W Q ! N be defined by letting g.r/ D minfn 2 NI f .n/ D rg.
22. For item (a), let B1 DA1 and, once we have defined B1 ; : : : ; Bk1 , for some
natural k, let Bk D Ak n [k1
jD1 Aj . For (b), since Bk Ak , we conclude that Bk is
S
finite or countably infinite. If fk W Bk ! N is injective, let f W k1 Bk ! N N
be given by f .x/ D .fk .x/; k/ if x 2 Bk and verify that f is injective. Finally, (c)
follows from (b) as a much easier particular case, arguing by contradiction and
taking A1 D I and A2 D Q.
23. Suppose, for the sake of contradiction, that F D fA1 ; A2 ; A3 ; : : :g. Define A D
fx1 ; x2 ; x3 ; : : :g N by choosing x1 … A1 , then x2 … A2 such that x2 > x1 C 1,
then x3 … A3 such that x3 > x2 C 1 etc. Since xk … Ak , we have A ¤ Ak . On the
other hand since x1 < x2 < x3 < , we conclude that A is infinite and N n A
contains the infinite set fx1 C 1; x2 C 1; x3 C 1; : : :g, hence is also infinite.
B Hints and Solutions
573
Section 6.4
1. For item (a), use Proposition 6.39. For item (b), suppose that g1 ; g2 W Y ! X
satisfy gi ı f D IdX and f ı gi D IdY , for i D 1; 2. Use Proposition 6.34 to get
g1 D g1 ı IdY D g1 ı .f ı g2 / D .g1 ı f / ı g2 D IdX ı g2 D g2 .
2. Use several times the result of Proposition 6.34, together with the fact that f 1 ı
f D IdX , f ı f 1 D IdY , g ı g1 D IdZ and g1 ı g D IdY .
3. Apply the result of Example 6.48.
4. See the hint to the next problem.
a
5. For the first part, note that the equation axCb
cxCd D y has, for every y ¤ c , the
single solution x D dyCb
.
cya
6. For the first part, it suffices to see that, for 0 x1 < x2 , we have x2 x1 > 0;
hence, item (a) of Problem 18, page 26, gives
xn2 xn1 D .x2 x1 /.x1n1 C x1n2 x2 C C x2n1 / > 0;
7.
8.
9.
10.
11.
so that f .x1 / < f .x2 / and f is increasing, thus injective. For the second part,
let y 2 Œ0; C1/ be given. Since we are assuming that Im.f / D Œ0; C1/, there
exists x 2 Œ0; C1/ such that y D xn ; then, by the definition of the n-th root of
p
p
y, we get x D n y, so that f 1 .y/ D n y.
and .f Setting f .x/ D 2x, we have f 1 .x/ D 2x , so that .f C f 1 /.x/ D 5x
2
f 1 /.x/ D 3x
.
These
last
two
functions
are
bijections
from
R
to
itself.
2
Suppose that f is increasing (the case of a decreasing f is entirely analogous).
Given y1 < y2 in J, take the elements x1 ; x2 2 I such that f .x1 / D y1 and
f .x2 / D y2 . If x1 D x2 , then y1 D f .x1 / D f .x2 / D y2 , which is not the case; if
x2 < x1 , it follows from the fact that f is increasing that y2 D f .x2 / < f .x1 / D
y1 , which is not the case either. Therefore, x1 < x2 or, which is the same,
f 1 .y1 / < f 1 .y2 /. Since the elements y1 < y2 of J were chosen arbitrarily, this
assures that f 1 is also increasing.
Show that, for fixed a; b 2 R, the
system of equations
x3 D a and x f .y/ D b
p
p
1 3
3
admits the single solution x D a and y D f . a b/.
Start by showing that it suffices to prove that any two functions in G commute
in respect to the operation of composition of functions. To this end, use items
(a) and (b), together with the fact that the only function h 2 G of the form
h.x/ D x C a, for some a 2 R, is h D IdR .
Set g D f 1 and start by showing that what is needed to be proved is equivalent
to the existence of positive integers x < y < z, in arithmetic progression and
such that g.x/ < g.y/ < g.z/; then, show that we can suppose that x D 1. To
what is left to do, fix t 2 R and show that one cannot have g.t/ > g.2t 1/ >
g.4t 3/ > g.8t 7/ > > g.1/.
574
B Hints and Solutions
Section 6.5
1. If there existed x1 < x2 in I such that f .x1 / D g.x1 / and f .x2 / D g.x2 /, we
would have f .x1 / > f .x2 / D g.x2 / > g.x1 / D f .x1 /.
2. Initially, note that x D 4 is a solution. Then, look at both sides of the given
equation as functions from .0; C1/ to R and apply the result of the former
problem.
3. Letting g.x/ D f .x/ f .0/ in the given relation, conclude that it suffices to
consider the case
in
which f .0/ D 0. Under this additional hypothesis, make
y D 0 to get f 2x D f .x/
2 , for every x 2 Q. Then, use the given relation to
conclude that f .x C y/ D f .x/ C f .y/, for all x; y 2 Q, so that (according to the
first part of the solution of Example 6.56) f .x/ D f .1/x, for every x 2 Q.
4. Use the given relation to prove by induction that f .n/ D f .1/, for every n 2 N;
then, consider the case n < 0 by making y D x.
5. Make x D y D z D 0 to get f .0/ D 12 ; then, obtain f .1/ D 12 by an analogous
reasoning. Make y D z D 1 to conclude that f .x/ 12 , for every x 2 R. Finally,
make an analogous substitution to prove that f .x/ 12 , for every x 2 R.
6. In each of the intervals Œn; n C f .n/ and Œf .n/; 2f .n/ there are f .n/ C 1 naturals.
Use the increasing character of f , together with the given relation, to conclude
that f .n C k/ D f .n/ C k, for every 1 k f .n/. Now, fix natural numbers
k and n, with n > k. Use the fact that f .n/ n to show that f .n/ > k and, by
what was done above, that f .n/ D n 1 C f .1/.
7. Set x D a D 0 in (a) to get f .0/ D 0. Then, letting x D 0 in (a), show that
f .a/ D a for every a 2 Z. Now taking x 2 Œ0; 1/, use the fact that f .x/ 2 Z and
f .f .x// D 0 to conclude that f .x/ D 0. Finally, for a general x 2 R, change x
by fxg (the fractional part of x) and make a D bxc in (a) to obtain f .x/ D bxc.
8. Start by computing f .x C f .y C f .0/// in two different ways to get f .0/ D 0.
Then, deduce that f .f .y// D y for every y 2 Z, so that f is bijective. From this
last relation, compute f .f .f .x/// in two different ways to conclude that f is odd.
Change x by f .x/ in the given relation to obtain f .f .x/ C f .y// D f .f .x C y//
and, then, f .x/ C f .y/ D f .x C y/, for all x; y 2 Z. Finally, use induction to
get f .x/ D f .1/x, for every x 2 Z, thus arriving at the contradiction 1 D
f .f .1// D f .1/f .1/ 0.
9. Set k D 0 to conclude that f .1/ D 2; then, use induction to show that f .n/ D
n C 1, for every n 2 N. Now, use the given relation to show that, if f .1/ D a,
then a < 0 or a > 0 lead to contradictions. Finally, make another induction to
conclude that f .n/ D n C 1, for every n 2 N.
10. First of all, show that we can suppose that f has no fixed points. Then, write
A D B [ C, where B and C are disjoint sets such that f .x/ > x for every x 2 B
and f .x/ < x for every x 2 C. If P
B has l elements and a is the common value
of jf .xj / xj j, conclude that 0 D kjD1 .f .xj / xj / D .2l k/a and, hence, that
a D 0, which is a contradiction.
B Hints and Solutions
575
11. Set x D y D 0 in the given relation to obtain f .0/ D 2f .0/f .a/ and, hence,
f .a/ D 12 . Set just y D 0 to get f .x/ D f .a x/ for every x 2 R, so that
f .x/ D f .a C x/, also for every x 2 R. Substitute y D a in the given relation
to get
f .x/ D f .x C a/ D f .x/f .0/ C f .a/f .a x/ D
1
1
f .x/ C f .x/ D f .x/;
2
2
for every x 2 R. Now, let y D x in the given relation to find
1
D f .0/ D f .x/f .a C x/ C f .x/f .a x/
2
D f .x/f .x/ C f .x/f .x/ D 2f .x/2 :
Therefore, if we show that f .x/ 0 for every x 2 R, it will follow from the last
relation above that f .x/ D 12 for every x 2 R. To what is left to do, it suffices to
make y D x in the given relation, to obtain
f .2x/ D f .x/f .a x/ C f .x/f .a x/ D 2f .x/2
12.
13.
14.
15.
2
and, hence, f .x/ D 2f 2x 0, for every x 2 R.
Set x D y D 0 to get f .0/ D 1. Then, use induction to prove that f .nx/ D f .x/n ,
for all x 2 Q and n 2 N; therefore, use this relation to show that f .x/ D f .1/x ,
for every x 2 QC . Now, letting y D x, show that f .x/ D f .x/1 and, hence,
f .x/ D f .1/x for every x 2 Q. Finally, use the fact that the codomain of f is the
set of positive rationals to conclude that f .1/ D 1.
Making x C y D a and x y D b, show that f .a/ C f .b/ D 2f aCb
2 , for all
a; b 2 Œ0; 1. Setting x D y, show that f .2x/ D 2f .x/ for every x 2 Œ0; 1, and
conclude that f .a/ C f .b/ D f .a C b/ for all a; b 2 Œ0; 1. From this point, show
that f .x/ D x for every x 2 Œ0; 1 \ Q. To what is left to do, reason as in the
passage from Q to R in Example 6.56.
For item (a), start by making m D 0 in the given relation to get f .f .n// D
f .0/2 C n; then, conclude that f is a bijection. Take k 2 Z for which f .k/ D 0,
and let l D f .0/. We then have l D f .0/ D f .f .k// D f .0/2 C k D l2 C k,
whereas, letting m D k and n D 0 in the stated relation, we get k2 C l D k;
hence, k D l D 0. In what concerns (b), make m D 1 and n D 0 in the given
relation to obtain f .1/ D 1; then, deduce that f .f .n/ C 1/ D n C 1 and, hence,
that f .n/ D n for every nonnegative integer n. Finally, extend the arguments
above to the negative integers, thus showing that f .n/ D n for every n 2 Z.
Let x D y D 1 to conclude that f .1/ D 1. Then, successively show that
f .f .xy// D x2 y2 f .xy/ D f .f .x/f .y//, f .xy/ D f .x/f .y/ and f .x2 / D f .x/2 ,
for every x; y 2 N. Now, if f .x/ < x2 for some x 2 N, write f .x/3 D f .x3 / >
f .xf .x// D x2 f .x/2 to arrive at f .x/ > x2 , which is a contradiction; analogously,
show that one cannot have f .x/ > x2 for some x 2 N, so that f .x/ D x2 for
every x 2 N.
576
B Hints and Solutions
16. Write x 2 in place of x e 2 in place of y in (a) to conclude that f .x/ D 0 for
x 2. Now, let x D y D 0 in (a) to get, from (b), that f .0/ D 1. Finally, for
2
0 < x < 2, write 2 x in place of y in (a) to obtain f .x/ 2x
; on the other
2
2
hand, show that f x C f .x/ D f .2/f .x/ D 0 and, hence, that f .x/ 2x
for
each one of those values of x.
17. Firstly, notice from (b) that f has at most one fixed point in each one of the
intervals .1; 0/ and .0; C1/. If there exists a fixed point x0 of f such that
1 < x0 < 0, let x D y D x0 in (a) to conclude that x20 C 2x0 is another
fixed point of f and, hence, that x20 C 2x0 D x0 , which is a contradiction. Argue
analogously to conclude that f has no fixed points in the interval .0; C1/.
x
Finally, set x D y in (a) to conclude that f .x/ D xC1
for every x 2 S.
18. Start by observing that, if such an f there exist, then f .Fk / D FkC1 for every k 2
N, where .Fn /n1 is the Fibonacci sequence. Then, use Zeckendorf’s theorem
(cf. Problem 30, page 98) to show that one possibility for f is to have
f .Fi1 C Fi2 C C Fij / D Fi1 C1 C Fi2 C1 C C Fij C1 ;
for natural numbers j and 1 i1 < i2 < < ij .
19. If f is such a function, set y D 0 in the given relation to get f .z/ D .c C 1/z for
every z 2 Im.f /, where c D f .0/. Then, use this fact to deduce that cf .x C y/ D
f .x/f .y/ xy for all x; y 2 R. In order to show that c D 0, suppose that c ¤ 0
and conclude that 0 … Im.f /. Now, make x D c and y D c in the last relation
above to obtain f .c/f .c/ D 0 and, then, arrive at a contradiction. Therefore,
we have got f .x/f .y/ D xy, for all x; y 2 R. In order to finish, take y0 2 Im.f /
for which f .y0 / ¤ 0; then, f .y0 / D .c C 1/y0 D y0 and, hence, f .x/y0 D xy0 ,
for every x 2 R.
20. From (a), notice that f .x C k/ D f .x/ C k, for every k 2 N. Now, take m; n 2 N
2
3 n3
and use the fact that .n Cm/
2 N to compute
n3
f
.n3 C m/3 n3
m3
C
n3
n3
in two different ways, thus obtaining the equality
.a C n2 /3 D a3 C .n6 C 3n3 m C 3m2 /;
where a D f
m
n
. Then, conclude that f .x/ D x, for every x 2 QC .
B Hints and Solutions
577
Section 6.6
1. For the second part, show that the x-coordinates of the intersection points of the
graphs are the solutions
of the equation f .x/ D x. In our case, the only such
p
5C 5
solution is x D 2 .
2. If jf .x/j M for every x 2 I and .x0 ; y0 / 2 Gf , then jy0 j D jf .x0 /j M, so that
M y0 M. Hence, .x0 ; y0 / belongs to the horizontal strip of the cartesian
plane bounded by the parallel lines y D M and y D M.
3. Since the bisector of the odd quadrants of the cartesian plane is the set of points
.x; y/ for which x D y, we conclude that x0 2 I is a fixed point of f if and only if
f .x0 / D x0 , i.e., if and only if .x0 ; x0 / 2 Gf . Therefore, the fixed points of f are
exactly the abscissas of the points where the graph of f intersects that bisector.
In Fig. B.1, if f is increasing for x < x1 and decreasing for x > x3 , its fixed
points are x1 , x2 and x3 , which, in turn, are the abscissas of the points A, B and
C, respectively.
4. Such a point .x; y/ satisfies x 2 I and y D f .x/ D g.x/. Therefore, it suffices to
solve, for x 2 I, the equation f .x/ D g.x/.
5. Let’s prove item (a), the proof of (b) being totally analogous. If f is even, then
f .x/ D f .x/ for every x 2 I. Hence, for x 2 I, we have
.x; y/ 2 Gf , y D f .x/ , y D f .x/ , .x; y/ 2 Gf ;
However, since the points .x; y/ and .x; y/ are symmetric with respect to the
vertical axis, we conclude that the same is true of Gf .
6. Functions f2 and f4 are always nonnegative and vanish only at x D 0. Also,
it is clear that they are even functions, so that (by the result of the previous
problem) their graphs are symmetric with respect to the vertical axis of the
cartesian system. On the other hand, as jxj increases to C1, it’s evident that
y
C
B
x1
x2
A
Fig. B.1 Fixed points of a function f
x3
x
578
B Hints and Solutions
y
x4
O
x2
x
Fig. B.2 Graphs of f2 and f4
the values of f2 and f4 become bigger and bigger, eventually surpassing any
predefined value. Finally,
jxj < 1 ) x4 < x2 ; jxj > 1 ) x4 > x2 and jxj D 1 ) x4 D x2 D 1;
justifying the fact that the graph of f4 is situated below (resp. above) the graph
of f2 in the interval .1; 1/ (resp. outside the interval Œ1; 1). See Fig. B.2.
Functions f3 and f5 are positive for x > 0, negative for x < 0 and vanish
at x D 0. Also, since they are odd functions, their graphs are symmetric with
respect to the origin of the cartesian plane. Now, as jxj increases, it is clear that
the values of jf3 j and jf5 j become bigger and bigger, eventually surpassing any
predefined value. Finally,
jxj < 1 ) jx5 j < jx3 j; jxj > 1 ) jx5 j > jx3 j and jxj D 1 ) jx5 j D jx3 j;
which justifies the fact that, in the interval .1; 1/ (resp. outside the interval
Œ1; 1), the graph of f5 is closer to (resp. further away from) the horizontal axis
than the graph of f3 . See Fig. B.3.
7. Note that f is the inverse of the function g W R ! R such that g.x/ D x3 , for
every x 2 R. Then, apply the result of Proposition 6.63.
8. Verify that the graph of the integer part function is the union of the sets Œn; n C
1/ fng, when n varies in Z.
9. For item (a), start by observing that, for x 2 Œ0; p/, the periodicity of f
guarantees that f .x C kp/ D f .x/, for every k 2 Z. Hence, .x; y/ 2 Gf , y D
f .x/ , y D f .xS
C kp/ , .x C kp; y/ 2 Gf . Then, letting F D Gf \ .Œ0; p/ R/,
we have Gf D k2Z .F C k/, where F C k denotes the translation of F, of k
units, in the direction of the vertical axis. For item (b), note that fxg D x for
x 2 Œ0; 1/. Since fg is periodic of period 1, it follows from (a) that Fig. B.4
sketches the graph of fg.
B Hints and Solutions
579
x3
y
x5
x
O
Fig. B.3 Graphs of f3 and f5
y
...
−4
...
−3
−2
−1
0
1
2
3
x
Fig. B.4 Graph of x 7! fxg
10. For the item (a), it suffices to show that .x; y/ is on the graph of f if and only if
.x a; y/ is on the graph of g. Indeed, if .x; y/ belongs to the graph of f , then
f .x/ D y and, hence, g.x a/ D f ..x a/ C a/ D y, so that .x a; y/ belongs
to the graph of g; moreover, the converse statement can be established in an
analogous way. For item (b), it suffices to show that .x; y/ is on the graph of f
if and only if .x; y C a/ is on the graph of g. Actually, if .x; y/ belongs to the
graph of f , then f .x/ D y and, hence, g.x/ D f .x/ C a D y C a, so that .x; y C a/
belongs to the graph of g; as in the previous case, the converse statement can
be proved analogously. The other items can be similarly dealt with.
11. At a first glance, it may seem that we cannot use the results of the previous
problem, for, the domain of the function under scrutiny is not the set R of real
numbers and its expression doesn’t match any of the ones considered there.
Nevertheless, since
580
B Hints and Solutions
y
(0, 2)
(−1, 0)
O
Fig. B.5 Graph of x 7!
x
2x
xC1
2x C 2 2
2
2x
D
D 2
;
xC1
xC1
xC1
we can reason in a way analogous to that of the previous problem, thus
sketching the graph of f in the following way: firstly, we sketch the graph of
x 7! 1x ; then, we translate it one unit to the left, thus obtaining the graph of
1
x 7! xC1
; secondly, we stretch the previous graph in the vertical direction, by
2
a factor 2, thus obtaining the graph of x 7! xC1
; reflecting the result along
2
the horizontal axis, we obtain the graph of x 7! xC1
; finally, if we vertically
2x
translate the reflected graph two units above, we obtain the graph x 7! xC1
. The
final result is shown in Fig. B.5.
12. For the first part, show that, with the graph of f at our disposal, we can get the
graph of g by reflecting, along the horizontal axis, the portion of the graph of
f situated below such axis. For items (a) and (c), apply this procedure to the
1
graphs of the functions f .x/ D x2 4, f .x/ D xC1
and f .x/ D x2 jx C 2j C 2,
1
respectively. Also notice that, according to Problem 10, the graph of f .x/ D xC1
1
can be obtained by translating the graph of x 7! x one unit to the left. In what
concerns the graph of f .x/ D x2 jx C 2j C 2, take a separate look at the cases
x 2 and x > 2, observing that jx C 2j is respectively equal to x 2 and
x C 2.
13. Apply a (clockwise) rotation of 4 radians to the original cartesian system (this
is equivalent to a counterclockwise rotation, of 4 radians, of the hyperbola of
B Hints and Solutions
581
equation x2 y2 D 4). By standard analytic geometry (see, for instance, Chapter
2
2
p
p
p
6 of [4]), we obtain the curve of equation xy
xCy
D 2 2 or, which
2
2
is the same, xy D 1.
Section 6.7
1. For the first part of item (a), let A D .1; 0/ and P be the point on the
_
trigonometric circle such that AP D x (of course, in
the counterclockwise
direction). Since sin x is the ordinate of P, if x1 ; x2 2 2 ; 2 , with x1 < x2 ,
we clearly
have sin x1 < sin x2 , so that the restriction of the sine function to the
interval 2 ; 2 is increasing and, as such, injective. Since this restriction has
image Œ1; 1, the arcsin function is well defined and, by Problem 8, page 176,
is also increasing. The first part of item (b) can be dealt with in a similar way,
observing that cos x is the abscissa
of P. Finally, the first part of item (c) follows
from the fact that, for x 2 2 ; 2 , the real number tan x is the ordinate of the
!
intersection point of the half-line OA with thevertical
line passing through A.
2. For item (a), observe that cot x D tan x C 2 and use the results of
Problema 10, page 194. For items (b) to (d), use the result of Proposition 6.63.
3. For items (a) and (b), we refer the reader to the discussion of Example 6.68,
more precisely to Eq. (6.15); as for item (c), we suggest the reader to recall
items (a) and (e) of Problem 10, page 194.
p
4. Argue as in Example 6.68 to get f .x/ D a2 C b2 cos.x ˛/, where cos ˛ D
p a
and sin ˛ D p 2b 2 . Now, show that f ha period 2
.
2
2
jj
a Cb
a Cb
5. For item (a), start by using trigonometric identities to get f .x/ D 2 sin2 x C
2
2 sin x C 1. Then, observe that the quadratic
1 attains
y 7! 2y C 2yC
function
1
1
its maximum value at y D 2 , increases on 1; 2 and decreases on 12 ; 1 . Item
(b) follows from the expression obtained above for f .x/, together with the fact
that the sine function has period 2. Finally, for item (c), sketch the graphs of
x 7! 2 sin x and x 7! cos.2x/ on the interval Œ0; 2 and in a single cartesian
coordinate system; then, add corresponding ordinates of the points of these two
graphs to get the desired sketch of the graph of f .
6. Use the result of Problem 4, page 193.p
7. Perform the change of variable x D 5 cos , with 0 . Then, use
the discussion of Example 6.68 to conclude that f .x/ D 25 cos. ˛/, with ˛
satisfying cos ˛ D 35 and sin ˛ D 45 .
8. Suppose, by the sake of contradiction, that f is periodic, of period > 0. Then,
we must have f . / D f .0/ D 2. Use this equality, together with the fact that
the cosine of every real number is at most 1, to get cos D cos.˛ / D 1.
Conclude that there must exist (nonzero) integers k and l for which D 2k
and ˛ D 2l, and arrive at a contradiction.
9. Initially, show that the equality f .x C 3/ D f .x/ implies the equality
582
B Hints and Solutions
15
5x
C
.1/ sin
n
n
n
5x
D sin
;
n
whenever nx ¤ 2 C k, for every k 2 Z. Then, if n is odd, for instance, show
cos 15
D 0, for every x 2 R such
C 15
that the last equality implies sin 5x
n
2n
2n
that nx ¤ 2 C k, for every k 2 Z. Finally, conclude that n divides 15.
10. Suppose that the given function is periodic of period p > 0. Compute f .x C p/
with the aid of the standard trigonometric identities for the sine of the sum of
two arcs and, then, take a careful look at the equality f .x C p/ D f .x/.
Section 7.1
n
1
1
1. For item (a), observe that p
D 1=x
D x. For (b), argue in an analogous
n
1=x
way. Item (c) now follows immediately from the results of Problems
p
p6 and 8,
page 176. Finally, for item (d), Problem 6, page 176, gives m a > n a if and
only if an > am ; the rest follows from Corollary 1.3.
2. For item (a) suppose, by the sake of contradiction, that a > 0. Then, by the
Archimedian property of N, we can choose n 2 N such that n > 1a . This is
equivalent to a < 1n , which is an absurd. For item (b), choose n 2 N such that
n > cb
a .
3. For item (b), let r1 ; : : : ; rn 2 .a; b/ \ Q, with r1 ; : : : ; rn pairwise distinct. If
r D minfr1 ; : : : ; rn g, then r 2 .a; b/ \ Q, and item (a) gives a < aCr
2 < r.
Finally, argue similarly, using the second part of (a), to get infinitely many
irrational numbers in the interval .a; b/.
4. For item (a), suppose that we already have the truth of the result for a 0. If
b 0, then b 0 and, by item (a), there existe r 2 Q and ˛ 2 R n Q such
that r; ˛ 2 .b; a/. Hence, r; ˛ 2 .a; b/, with r 2 Q and ˛ 2 R n Q.
If a 0 < b, show that it suffices to apply, to the interval .0; b/, the result we
assumed to be known. For item (b),puse the Archimedian property of natural
numbers to get n 2 N such that n > ba2 . Finally, in what concerns (c), starts by
using the Archimedian
property to guarantee the existence of m 2 N for which
p
m 2
m
> b); continue by showing that, if m is the least possible
n > b (resp. n
p
5.
6.
7.
8.
2
natural number fullfilling this requirement, then m > 1 and .m1/
2 .a; b/.
n
Adapt the proof of items (b) and (c) of the previous problem to the present case.
Use the result of Example 7.2, together with Theorem 7.4.
Use the result of item (b) of Problemp
4. p
1 p
1
p D p
For a D k2 and b D k2 C 1, we have j a bj D pjabj
< 2k
.
2
2
aC b
k C1C k
Since every element of X is positive, it follows that inf.X/ D 0.
9. Observe that, for k 2 N, every number of each of the forms 31k and 1 31k belong
to C. Now, use the fact that 31k < 1k , for every k 2 N.
B Hints and Solutions
583
10. Adapt, to the present case, the proof of Proposition 7.6.
11. Use the result of Proposition 7.6, together with the result of Problem 10, to
1
establish the existence of xn 2 X and yn 2 Y such that ˛ 2n
< xn < ˛ and
1
˛ < yn < ˛ C 2n .
12. For item (a), show that, if ˛ (resp. ˇ) is an upper bound for X (resp. for cX),
then c˛ (resp. ˇc ) is an upper bound for cX (resp. for X). The other cases can be
dealt with similarly.
13. For item (a), start by showing that, if ˛ and ˇ are upper bounds for X and
Y, respectively, then ˛ C ˇ is an upper bound for X C Y; then, conclude that
sup.X C Y/ sup X C sup Y. Show next that, if sup.X C Y/ sup X < sup Y,
then Y would have an upper bound less than sup Y.
15. The given conditions imply that Rn has vertices .0; 0/, .an ; 0/, .0; bn / and
.an ; bn /, for some integers an and bn . Passing to a subsequence, if necessary,
we can assume without loss of generality that an ; bn > 0. Show that one can
assume, also without loss of generality, that a1 a2 a3 . Then, show
that one cannot have b1 > b2 > b3 > .
16. Since jf .x/j 1 for every x 2 R, we can let L D supfjf .x/jI x 2 Rg. Suppose
that supfjg.x/jI x 2 Rg > 1 and apply triangle inequality to the relation given
in the statement of the problem to reach a contradiction.
Section 7.2
1. Show that, if a > b, then we would have an > aCb
2 > bn for every sufficiently
large n.
p
p
aj
p < p1 jan aj; then, use the convergence of
2. Write j an aj D pjaa nC
a
a
n
.an /n1 to a.
3. For n > 16a2 , show that
jajn
nŠ
<
4. Firstly, note that
s
n
jajn
n
. n2 / 2
<
2jaj
p
n
n
<
1
.
2n
p
1
. n n/k
nk
!
<1
D
jajn
jaj
jaj
1
as n ! C1. Hence, fixed a real number ˛ such that jaj
< ˛ < 1, there exists
q
nk
n nk
n
n0 2 N for which n > n0 ) jajn < ˛, or jajn < ˛ . However, since ˛ n ! 0
as n ! C1, the same happens with
˛ > 0, and note that, for n > k,
nk
.
an
Alternatively, let jaj D 1 C ˛, with
!
!
n
X
n j
n
jaj D .1 C ˛/ D
˛ >
˛kC1
j
k
C
1
jD0
n
n
D
n.n 1/ : : : .n k/ kC1
˛ :
.k C 1/Š
584
B Hints and Solutions
Hence, for n > k, we have
.k C 1/Š
nk
nk
<
n
kC1
jaj
˛
n.n 1/ : : : .n k/
D
1
.k C 1/Š
;
1
kC1
˛
n.1 n / : : : .1 nk /
with the last expression above tending to 0 as n ! C1.
5. Write l D .1 tn /l C tn l and use the triangle inequality to get jcn lj .1 tn /jan lj C tn jbn lj. Then, show that jan lj; jbn lj < implies
jcn lj < .
6. First of all, note that jbn lj maxfjan lj; jcn ljg, for every n 2 N. Hence,
given > 0 and taking n0 2 N such that n > n0 ) jan lj; jcn lj < , we
n
have jbn lj < forpn > n0 , so that bn ! l.
n n
7. For item (a), write n2 C1 D pnC1 p1 < p1n . For (b), write
n
n
p
a C bn
an C b
n2 C an C b n D p
D q
:
n2 C an C b C n
1 C an C nb2 C 1
For item (c), observe that, if an D
Problem 18, page 26,
qn
ann 1
<
:
ann1 C ann2 C C an C 1
n
q
p
Finally, for (d), write n an C bn D a n 1 C . ba /n and use the result of the
previous item.
Adapt, to the present case, the reasoning that led to the proof of Proposition 7.18.
For the first part, write q D 1 C ˛, with ˛ > 0, and note that an D .1 C ˛/n 1 C n˛, so that an > M for n > M1
.
˛
Let b D 1q , so that b > 1. Use the first part of the previous problem to get k 2 N
n
such that bk > a. Then, show that
for every sufficiently large n and,
a2n > kn
n 2n
from this, conclude that a q D bk b2n1kn ! 0 when n ! C1.
2nk
First, show that jam an j m2mk
2 Ck2 C n2 Ck2 ; then, make k ! C1.
p
Observe that anC1 D 1 C an , for every n 1. Then, successively conclude
that a2nC1 a2n D an an1 and that .an /n1 is increasing. Next, note that
1CanC1
n
a2nC1 D 1 C an implies anC1 D 1Ca
< 2, and apply BolzanoanC1 <
anC1
Weierstrass theorem to guarantee the existence
p of l D limn!C1 an . Now, make
n ! C1 in the recurrence relation anC1 D 1 C an and use the result of item
(a) of
pProblem 2, together with item (b) of Proposition 7.18, to get the equation
l D 1 C l.
0 < an 1 D
8.
9.
10.
11.
12.
p
n
1 C qn , then an > 1 and, by item (a) of
B Hints and Solutions
585
13. Setting ˛ D minfa1 ; a2 g and ˇ D maxfa1 ; a2 g, conclude that .an /n1 is
bounded (to this end, separately consider the cases 0 < ˛ ˇ 4, 4 ˛ ˇ
and 0 < ˛ 4 ˇ). Now, if a1 a2 a3 or a1 a2 a3 ,
p
p
write akC2 akC1 D akC1 ak1 to conclude that the given sequence is
monotonic, thus convergent; make k ! C1 in the recurrence relation satisfied
k
by the sequence to conclude that ak ! 4. If a1 a3 a2 or a1 a3 a2 ,
conclude, as above, that the subsequences of the terms of even and odd indices
k
k
are convergent, say, a2k1 ! c and a2k
d; then, makepk ! p
C1 in the
p ! p
given recurrence to conclude that d D c C d and c D c C d, so that
c D d D 4.
14. Use the fact that tn D t0 t1 tn1 D 0 to write
p
p
p
p
ak D t0 . k k C n/ C t1 . k C 1 k C n/ C p
p
C tn1 . k C n 1 k C n/
t1
t0
p
p
D p
p
kC1C kCn
kC kCn
tn1
p
:
p
kCn1C kCn
Then, make k ! C1.
15. Perform the trigonometric substitution xn D 2 cos yn , with yn 2 Œ0; 2 , and use
some Trigonometry to conclude that ynC1 2yn , for every n 1. Then, show
y
that yn nC2k
2kC1
for every k 1, so that yn D 0 for every n 1.
2k
16. First of all, use the given condition to show that j.anC1 an / .akC1 ak /j < 2n .
Then, make n ! C1 to conclude that there exists l D limn!C1 .anC1 an /,
and show that akC1 ak D l, for every k 2 N.
2k
17. Use the definition of an to conclude that ak1 C 1 D kC1
ak , for every integer
k 2. Then, show that
ak1 ak <
2.k C 1/2
.ak akC1 /
.k C 1/.k C 2/
and conclude that the sequence .an /n1 is decreasing for n > 2. Finally, make
2k
k ! C1 in the equality ak1 C 1 D kC1
ak to conclude that an ! 1 when
n ! C1.
18. Successively show that a2n D k C an1 and a2nC1 a2n D an an1 ; then,
2
conclude that
p .an /n1 increases. Now, use the fact that an < k C an to get
1
an < 2 .1C 4k C 1/ for every n 1, so that the given sequence is convergent.
p
n
Make n ! C1 in a2n D k C an1 to find out an ! 12 .1 C 4k C 1/. From
here, items (b) and (c) are relatively easy.
586
B Hints and Solutions
19. First of all, use the inequality between p
the arithmetic and geometric means of
two numbers to conclude that akC1 a, for every k 2 N. Now, make k !
C1 in the given
precurrence relation to conclude that, if .an /n1 converges, then
its limit equals a. To what is left to do, use the triangle inequality to obtain
ˇ
ˇ
1ˇ
a
a ˇˇ
jakC1 ak j D ˇˇak ak1 C
2
ak ak1 ˇ
ˇ
ˇ
ˇ
1
a ˇˇ
D jak ak1 j ˇˇ1 2
ak ak1 ˇ
1
jak ak1 j;
2
for each k > 2. Finally, establish the convergence of .an /n1 from the result of
Example 7.28.
20. Let a1 , a2 , . . . , an be the initial terms of the AP’s, and say that the i-th AP is
that of initial term ai and common ratio di . Choose n 2 N greater than all of
the ai ’s, and let ki be the number of terms of the i-th AP belonging to the set
f1; 2; : : : ; ng. Then, on the one hand, k1 C k2 C C km D n; on the other,
ai C .ki 1/di n < ai C ki di . Use these relations to show that
S
Sm X 1
<
< 1C ;
n
d
n
iD1 i
m
1C
where S D
P
m ai
iD1 di
. Then, make n ! C1.
21. Suppose such a function does exist, let k > 1 be an integer and xj D kj , for
c
0 j k. Given a real number x 2 Œxj ; xjC1 , show that d.f .x/; f .xj // k˛C1=2
.
c
Letting Cj denote the circle in the plane with center f .xj / and radius k˛C1=2
,
conclude that x 2 Œxj ; xjC1 ) f .x/ 2 Cj . Then, use this fact to show that
P
c2
1 k1
jD0 A.Cj / D k2˛ . Finally, use the fact that k > 1 can be chosen arbitrarily
and ˛ > 0 to reach a contradiction.
22. Start by proving that anC1 an C 1 and, hence, that an n for every n 1.
Now, use telescoping products to show that, for a fixed p 2 N, we have
anCp Y
1
1C p
1<
<
an
a
nCj
jD0
p1
!
and, then,
anCp Y anCjC1 Y
1
1C p
<
<
1<
an
anCj
anCj
jD0
jD0
p1
Continue by showing that
p1
!
:
B Hints and Solutions
587
!
1
1
1
D 1:
1C p
1C p
lim 1 C p
n!C1
an
anC1
anCp1
Now, if y D 1, conclude that what is asked to be proved follows from the limit
above. If y < 1, conclude from the limit above that we can choose np 2 N such
an
that n > np ) anCp
> y. Use the fact that ak k for every k to show that we
can choose a natural number n0 such that
n > n0 and p natural furnish
1
p
1C an0 C1
< y x. Then, note that
an
an
< y x:
p
anCp anCp C anCp
Fix k > n0 ; np , so that
p0 > p for which
> y; establish the existence of a natural number
ak
akCp
ak
ak
y>
:
akCp0
akCp0 C1
Finally, suppose that akCpak C1 x and arrive at a contradiction.
0
22. Let c0 ; c1 ; c2 ; : : : be defined by c0 D 1 and
cn D
an1
1 C an
cn1 ; 8 n 1:
Rewriting such a relation as cn D an1 cn1 an cn , we get the telescoping sum
c1 C c2 C C cn D a 0 a n c n :
(B.3)
cn
On the other hand, the assertion of the problem is equivalent to cn1
< 21=n for
infinitely many values of n 2 N. Arguing by contradiction, suppose that there
exists a natural N such that the opposite inequality is true for every n N.
Then, for n > N we have
cn cN 2
1
1
NC1
C NC2
CC 1n
D C 2.1C 2 CC n / ;
1
1
where C D cN 2.1C 2 CC n / is a positive constant. If 2k1 n < 2k , then
1
n
X
1
jD1
j
1C
1
1
C
2
3
1
C
1
1
CC
4
7
1 C 1 C 1 C C 1 D k;
C
1
2k1
1
CC k
2 1
588
B Hints and Solutions
so that cn C 2k for 2k1 n < 2k . Let r 2 N be such that 2r1 N < 2r ,
and let m > r. Then,
c2r C c2r C1 C C c2m 1 D
D .c2r C C c2rC1 1 / C .c2rC1 C C c2rC2 1 /
C C .c2m1 C C c2m 1 /
C .2r 2r1 C 2rC1 2r2 C C 2m1 2m /
D
C .m r/
;
2
thus showing that the sum of the cn ’s can be taken arbitrarily large. However, (B.3) guarantees that this sum cannot exceed a0 . This contradiction
finishes the proof.
Section 7.3
2. For (a),
the
note that there are more numbers of the form fj˛g than intervals
k of
kC1
form pk ; kC1
p . For (b), take 1 s < t p C 1 such that fs˛g; ft˛g 2 p ; p ,
write fs˛g D s˛ bs˛c,ft˛g D t˛ bt˛c and observe that 0 < jfs˛g ft˛gj < 1p .
3. Use Corollary 7.33 to inductively construct a sequence .ak /k1 such that ak D
mk C nk ˛, for integers mk and nk satisfying item (a), and such that jak lj < 1k ,
for every k 1.
!
4. For item (c), choose A and B in such that O … AB , ˛ 2 R n Q and make
!
!
X D f.m1 C n1 ˛/OA C .m2 C n2 ˛/OBI m1 ; m2 ; n1 ; n2 2 Zg:
For item iii., conclude that there exist distinct lines r and s through O, such that
X \ r; X \ s ¤ ; and X \ r is not dense in r. Show that one can reconstruct X
from X \ r and X \ s, apply Kronecker’s lemma and make d D s.
5. For item (b), write n D ˛n C ˇn and take integer and fractionary parts. For (c),
argue by contradiction,
n o using the fact
o ˛ and ˇ are irrational to show that one
n that
n
1
n
cannot have either ˛ D 1 ˛ or ˇ D 1 ˇ1 .
Section 7.4
1. Use que fact that
1
anC1
D
1
an
1
an C1 ,
for every n 2 N.
Pn1 1
n1
2. Use the result of Problem 5, page 84, to show that kD1
ak akC1 D a1 an , for
every integer n 1. Then, make n ! C1, noting that an D a1 C .n 1/r.
B Hints and Solutions
3. Make Sn D
589
Pn
kD1
2k1
ak
and show that
1
1
1
C 2 C C n1
a
a
a
1
2n 1
2a
1
n DaC
:
a1 a a
an
.a 1/Sn D aSn Sn D a C 2
2n 1
an
Now, use the results of Example 7.12 and of Problem 4, page 218, to conclude
n
a
2
that Sn ! a1
C .a1/
2.
p
p
p
P
P
1
2
> k1 p12k . Since
4. Note that k C k 1 < 2k, so that k1 p p
p1
n
kC k2 1
1
n
for n 1, use the comparison test to show that the given series
diverges.
5. Show that n3 1000n2 > .n 500/3 for every sufficiently large natural n.
Then,
use the comparison test, together with the convergence of the series
P
1
k>500 .k500/3=2 .
6. Let r > 0 be the common ratio of the AP. For item (a), we have
a1
1
1
a1 C.k1/r > 2.k1/r , provided k > r C 1. For item (b), we have
1
ak
D
1
1
1
1
< k
k1 :
D
a2k
a1 C .2k 1/r
.2 1/r
2 r
Now, apply the comparison test.
7. Start by showing that
X1
x2A
x
<
N
N
N
X
1 X 1 X 1
;
2a bD0 3b cD0 5c
aD0
for some natural number N. Then, make N ! C1 and use the formula for the
sum of a geometric series.
P
a
9. Use the comparison test to show that the series j1 10jj converges. Now, letting
x denote its sum, note that, for k n, we have
x
n
X aj
X 9
X
aj
1
D
D n:
j
j
j
10
10
10
10
jD1
jnC1
jnC1
10. For the first part, suppose that, for some x 2 .0; 1/, we have x D 0:a1 a2 a3 : : : D
0:b1 b2 b3 : : :, with an ¤ 0 for infinitely many values
P of nakand, accordingly,
P
a1
bk
bn ¤ 0 for infinitely many values of n. Then, 10
< k1 10
k D
k1 10k P
b1
b1 C1
9
k2 10k D 10 , so that a1 b1 . Analogously, show that a1 b1 , so that
10 C
a1 D b1 . Then, argue similarly to establish, by induction on n, that an D bn ,
for every n 1. In what concerns the second part, define f .x/ D .y; z/, with
590
11.
12.
13.
14.
15.
16.
17.
18.
B Hints and Solutions
y D 0:a1 a3 a5 : : : and z D 0:a2 a4 a6 : : : if x D 0:a1 a2 a3 : : :, with an ¤ 0 for
infinitely many values of n. Then, use the result of the first part and the former
problem to show that f is well defined and is a surjection.
kj
Use the fact that a2k C k12˛ 2ja
, for every k 2 N; then, apply the result of
k˛
Proposition 7.39, together with the comparison test and Proposition 7.48.
p
Use the fact that ak akC1 12 .ak C akC1 / for every k 2 N, together with the
comparison test.
Imposing that Fj ˛ j , for every j k C 1, deduce that FkC2 ˛ kC2 if ˛ C 1 ˛ 2 . Use this a priori estimate
to show that Fn ˛ n for every n 3, where
p
p
p
3
4
˛ D minf 2; 3; 1C2 5 g. Finally, note that ˛ > 1 and apply the comparison
test.
For item (a) (the proof
of item (b) is analogous), let q D lC1
2 .0; 1/ and take
2
p
n
n0 2 N such that an < q for n > n0 or, which is the same, an < qn para
n > n0 . Then, adapt the reasoning used at the proof of the ratio test.
Apply Theorem 7.53.
P
k
Start by observing that the series k0 .1/
is absolutely convergent. Then, let
kŠ
P
P
.1/k
a D k0 kŠ and use the fact that e D k0 kŠ1 , together with Theorem 7.53,
to get ae D 1.
P
P
n
For item (a), let sn D nkD1 ak and tn D nkD1 a'.k/ , so that s1 s2 s3 !
s, where s D supfsn I n 1g, and analogously for .tn /n1 . Given n 2 N, prove
that there exists m 2 N such that sn tm ; then, make n ! C1 to conclude
that s t. Change the roles of the two sequences to get the reverse inequality.
C
For item (b), note that jan j D aC
D 12 jan j C an and a
n C an , so that an P
n D
P
C
1
ja
.
For
(c),
apply
the
result
of
(b)
to
both
j
a
a
and
a
n
n
k1 k
k1 k .
2
By the definition of convergent series, it suffices to show that the sequence
.xn /n1 , given for n 1 by xn D a1 b1 C a2 b2 C C an bn , is convergent. As
we know, this is the same as showing that it is a Cauchy sequence. So, let m
and n be natural numbers, with m > n. Let M > 0 be such that jsk j < M for
every k 1. It follows from Abel’s identity (5.19) (with the roles of ai and bi
exchanged) and from triangle inequality that
ˇ
ˇ m1
ˇ
ˇX
si .bi biC1 / C sm bm sn bn ˇ
jxm xn j D ˇ
iDn
m1
X
jsi j.bi biC1 / C jsm jbm C jsn jbn
iDn
D M.bn bm / C Mbm C Mbn
n
D 2Mbn ! 0:
Hence, .xn /n1 is indeed a Cauchy sequence.
19. In the notations of the former problem, change ak by .1/k and bk by ak .
B Hints and Solutions
591
20. For item (a), observe that 2 sin.a C jh/ sin h2 D cos.a C .j 1/h/ cos.a C jh/.
Now, sum from j D 0 to j D k and transform the result in a product. For item
(b), note
P that it suffices to show that the sequence .sn /n1 , given for n 1 by
sn D nkD1 sin k, is bounded. To this end, apply the result of item (a) to show
that sn D
sin
sin nC1
2
sin 12
n
2
1
.
sin 12
p1 . Now,
n
and, hence, that jsn j use the unit circle to show
OAnC1 , we have sin ˛n D
21. Letting ˛n D An b
P
that sin ˛ < ˛ for every ˛ 2 0; 2 . Finally, use the divergence of k1 p1k to
P
show that k1 ˛k also diverges.
22. Letting ˛n D 2ˇn , it suffices to show tha