626 Higher Engineering Mathematics
Hence the Fourier series is:f(θ)=θ^2 =π^2
3− 4(
cosθ−1
22cos2θ+1
32cos3θ−1
42cos 4 θ+1
52cos5θ−···)Problem 5. For the Fourier series of Problem 4,
letθ=πand show that∑∞
n= 11
n^2=π^2
6Whenθ=π,f(θ )=π^2 (see Fig. 68.3). Hence from the
Fourier series:π^2 =π^2
3− 4(
cosπ−1
22cos2π+1
32cos3π−1
42cos4π+1
52cos5π−···)i.e.π^2 −π^2
3=− 4(
− 1 −1
22−1
32−1
42−1
52−···)2 π^2
3= 4(
1 +1
22+1
32+1
42+1
52+···)i.e.
2 π^2
( 3 )( 4 )= 1 +
1
22+
1
32+
1
42+
1
52+···i.e.π^2
6=1
12+1
22+1
32+1
42+1
52+···Hence∑∞n= 11
n^2=π^2
6Now try the following exerciseExercise 230 Further problems on Fourier
cosine and Fourier sine series- Determine the Fourier series for the function
defined by:
f(x)=⎧
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎩− 1 , −π<x<−π
2
1 , −π
2<x<π
2
− 1 ,π
2<x<πwhich is periodic outside of this range of
period 2π.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
f(x)=4
π(
cosx−1
3cos3x+1
5cos5x−1
7cos7x+···)⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦- Obtain the Fourier series of the function
defined by:
f(t)={
t+π, −π<t< 0t−π, 0 <t<πwhich is periodic of period 2π. Sketch the
given function.
⎡
⎢
⎢
⎢
⎣f(t)=− 2 (sint+^12 sin2t+^13 sin3t
+^14 sin4t+···)⎤
⎥
⎥
⎥
⎦- Determine the Fourier series defined by
f(x)={
1 −x, −π<x< 01 +x, 0 <x<π
which is periodic of period 2π.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
f(x)=π
2+ 1−4
π(
cosx+1
32cos3x+1
52cos5x+···)⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦- In the Fourier series of Problem 3, letx= 0
and deduce a series forπ^2 /8.
[
π^2
8
= 1 +1
32+1
52+1
72+···]68.3 Half-range Fourier series
(a) When a function is defined over the range say 0
toπinstead of from 0 to 2πit may be expanded
in a series of sine terms only or of cosine terms
only. The series produced is called ahalf-range
Fourier series.