Weierstrass Theorem & Bernstein Polynomials

Weierstrass Theorem and Bernstein
Polynomials
Nicole M. Samuels
Abstract
Let f: [0,1] →Rbe continuous over [0,1], we show that fcan be
approximated by a specialized polynomial, called a Bernstein Polynomial
of the Nth degree, denoted by BN(f, x) where N= 1,2, .... Furthermore,
we present Voronovskaya’s Theorem regarding the error of approximation
and provide an estimate in terms of the modulus of continuity.
Contents
1 Weierstrass theorem 1
2 Voronovskaya’s Theorem and an approximation for the distance
of Bernstein polynomials and functions bounded on [0,1] 8
References 19
3 Plagiarism declaration 20
1 Weierstrass theorem
We define the set of polynomials of degree at most nby
Pn={pn(x)|pn(x) = a0+a1x+... +akxk, k ∈N, ak∈Rwhere k≤n}.
Definition 1.1 (Continuity [7]).We say a function is continuous at a point
x in its domain S if the following is true: given any ϵ > 0, there exists δ > 0
such that if x′∈Sand |x−x′|< δ then |f(x)−f(x′)|< ϵ.
A function is continuous over its domain S if it is continuous at every point
in its domain.
Consider some interval [a, b] where a, b ∈R. We denote the set of continuous
functions over [a, b] by C[a, b].
Definition 1.2. [4] The uniform norm of a function fover [a, b]is defined
by
∥f∥= max
a≤x≤b|f(x)|,
1

for any f∈C[a, b].
Theorem 1.3 (Weierstrass Theorem [4]).Let f(x)be a function in C[a, b].
Then there exists a polynomial pn(x)∈Pnfor a chosen ϵ > 0such that
∥f(x)−pn(x)∥< ϵ.
Definition 1.4 (Binomial Coefficient).Consider some n, r ∈Nwhere 0≤r≤
n, we define binomial coefficient by
n
r=n!
r!(n−r)!.
Theorem 1.5 (Binomial Formula).Let x, y ∈Rand n ∈N. Then we have
(x+y)n=
n
X
k=0 n
kxn−kyk=
n
X
k=0 n
kxkyn−k.(1)
This formula is called the Binomial Formula.
Definition 1.6 (Bernstein Polynomials).[2] Let f(x)be defined over [0,1]. For
N= 1,2,3, ... we define a Nth Bernstein polynomial,
BN(f, x) =
N
X
n=0
fn
NN
nxn(1 −x)N−n.
We can also say that this is the Bernstein Polynomial of order Nof the function
f(x).
Example 1.7. Let f(t) = 1. We seek the Bernstein polynomial associated
with 1
BN(1, x) =
N
X
n=0 N
nxn(1 −x)N−n.
We find that by the Binomial Formula (1),
BN(1, x) = ((1 −x) + x)N≡1.
Example 1.8. Let f(t) = t. We have that
n
NN
n=N−1 n−1,(2)
since
n
NN
n=n
NN! n!(N−n)!=(N−1)! (n−1)!(N−n)!=N−1
n−1.
2

This is true for 1≤n≤N. Using Equation (2), the Bernstein polynomial of
function tis
BN(t, x) =
N
X
n=0
n
NN
nxn(1 −x)N−n
=x
N
X
n=1 N−1
n−1xn−1(1 −x)(N−1)−(n−1).
Let s=n−1, by the Binomial Formula (1) we get
BN(t, x) = x
N−1
X
s=0 N−1
sxs(1 −x)(N−1)−s≡x.
Example 1.9. Let f(t) = t2. We find the Bernstein polynomial associated with
BN(t2, x). Using the result in Example 1.8, we observe that
BN(t2, x)−x
N
=
N
X
n=0
n2
N2N
nxn(1 −x)N−n−
N
X
n=0
n
N2N
nxn(1 −x)N−n.
⇒BN(t2, x)−x
N
=N−1
N
N
X
n=2
n(n−1)
N(N−1) N
nxn(1 −x)N−n,
since n2−n
N2=n(n−1) N2=1 N
n(n−1)
N=N−1 N
n(n−1)
N(N−1).
Observe that
n(n−1)
N(N−1) N
n
=n(n−1)
N(N−1) N! n!(N−n)!
=n(n−1)
N(N−1) N(N−1)(N−2)! n(n−1)(n−2)!((N−2) −(n−2))!
=N−2
n−2.
This gives the result
BN(t2, x)−x
N
=N−1
Nx2
N
X
n=2 N−2
n−2xn−2(1 −x)(N−2)−(n−2).
3

Setting s=n−2, we get
BN(t2, x)−x
N=N−1 Nx2
N−2
X
s=0 N−2
sxs(1 −x)(N−2)−s,
which by the Binomial Formula (1) is just
BN(t2, x)−x
N=N−1 Nx2⇒BN(t2, x) = N−1 Nx2+x N.
Lemma 1.10. [2]
N
X
n=0 n
N−x2N
nxn(1 −x)N−n=x(1 −x)
N.
Proof. We expand the left-hand side and get
N
X
n=0 n
N−x2N
nxn(1 −x)N−n
=
N
X
n=0 n
N2−2n Nx+x2N
nxn(1 −x)N−n
=
N
X
n=0 n
N2N
nxn(1 −x)N−n−2x
N
X
n=0 n
NN
nxn(1 −x)N−n
+x2
N
X
n=0 n
NN
nxn(1 −x)N−n.
If we let f(x) = x2, g(x) = xand h(x)=1,we get that
N
X
n=0 n
N−x2N
nxn(1 −x)N−n
=BN(f, x)−2xBN(g, x) + x2BN(h, x).
This gives, by Examples 1.7-1.9, the following result
BN(f, x)−2xBN(g, x) + x2BN(h, x)
=N−1
Nx2+x N−2x2+x2
=x
N−x2 N=x(1 −x) N.
We now present the proof proposed by Bernstein regarding the Weierstrass
Theorem.
4

Bernstein’s proof of Theorem 1.3. [6] We will restrict ourselves to the interval
x∈[0,1] by considering a linear bijective map F: [0,1] →[a, b] defined by
f(x) = (b−a)x+a. This means that for all f∈C[a, b], f ◦F−1is a continuous
function on [0,1] since Fand F−1are both continuous. Assuming there exists
aPnsuch that Pn→f◦F−1uniformly on [0,1], then this would mean that
Pn◦Fconverges uniformly to fon [a, b].
To start off the proof, we begin with the identity
n
N−x2=n N2−2n Nx+x2.
Multiplying both sides by N
nxn(1 −x)N−nand using Lemma 1.10, we get
n
N−x2N
nxn(1 −x)N−n
=n
N2−2n Nx2+x2N
nxn(1 −x)N−n
⇒
N
X
n=0 n
N−x2N
nxn(1 −x)N−n
=
N
X
n=0 n
N2−2n Nx2+x2N
nxn(1 −x)N−n
=x(1 −x)
N.
(3)
For any fixed x∈[0,1], we will approximate the sum of polynomials over all
nfor which n
Nis not close to x. We choose a number δ > 0 and let Sδdenote
the set of all nsatisfying
n
N−x≥δ.
This implies 1
δ2n N−x2≥1.(4)
So multiplying both sides of Equation (4) by BN(1, x) where n∈Sδ, we get
X
n∈SδN
nxn(1 −x)N−n
≤1
δ2X
n∈Sδn
N−x2N
nxn(1 −x)N−n.
5