Fourier Series Worksheet - EN212 Engineering Mathematics III

PAPUA NEW GUINEA UNIVERITY OF TECHNOLOGY
SCHOOL OF MATHEMATICS AND COMPUTER SCIENCE
EN212 ENGINEERING MATHEMATICS III
SEMESTER 1 2026
TOPIC 3- FOURIER SERIES
Worksheet: Fourier Series
In this worksheet, there will be some problems related to Fourier series topic
that was introducced in week 7 snd will continue this week of lectures. Students are
encouraged to practice and solve these problems in term of improving their ability
in this topic. This worksheet will not be evaluated and it is just for acquaintance
of students with future quizzes. If you have any question regarding this worksheet,
see your tutors or you can come to my office in MCS216.
Problem 1
Find the Fourier series for f(x) = x2on −π≤x≤π.
Solution.
Problem 2
Find Fourier series for the following function:
f(x) = (0,−π < x ≤0
1,0< x ≤π
Solution.
Problem 3
Find the Fourier series for the following functions:
f(x) = (0,−π < x ≤0
sin(x),0< x ≤π
and use it to find a value for
∞
X
n=1
1
4n2−1.
Date: April 21, 2026.
1

2PAPUA NEW GUINEA UNIVERITY OF TECHNOLOGY SCHOOL OF MATHEMATICS AND COMPUTER SCIENCE EN212 ENGINEERING MATHEMATICS III SEMESTER 1 2026 TOPIC 3- FOURIER SERIES
Solution.
Problem 4
Find the Fourier series for the following function:
f(x) = 
0,−π < x < −π
2
sin(x),−π
2≤x≤π 2
0,π
2< x ≤π
Solution.
Problem 5
Use Fourier series of the following function to find
∞
X
n=1
1
nsin nπ 2:
f(x) = 
0,−π < x < −π
4
1,−π
4≤x≤π 4
0,π
4< x ≤π
Solution.
Problem 6
Use Fourier series to solve the boundary value problem (Modeling a vibrating
string):
ytt = 9yxx,0≤x≤π, t > 0,
y(x, 0) = x(π−x),0< x < π,
yt(x, 0) = 0,0< x < π.
Solution.
Problem 7
Use Fourier series to solve the boundary value problems:
ut=uxx,0<x<∞, t > 0,
ux(0, t)=0, t > 0,
ux(π, t)=0, t > 0,
u(x, 0) = (1,0<x<1
0,1<x<∞
Solution.

PAPUA NEW GUINEA UNIVERITY OF TECHNOLOGYSCHOOL OF MATHEMATICS AND COMPUTER SCIENCEEN212 ENGINEERING MATHEMATICS IIISEMESTER 1 2026TOPIC 3- FOURIER SERIES3
Problem 8
Find the Fourier series of the periodic function f(x),
f(x) = (−k, −π≤x < 0
+k, 0< x ≤π
such that f(x+ 2π) = f(x).
Solution.