672 FOURIER SERIES
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 exercise.
Exercise 242 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
3cos 3x+1
5cos 5x−1
7cos 7x+···)⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦- 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 sin 2t+^13 sin 3t+^14 sin 4t+···)⎤⎥
⎥
⎥
⎦- Determine the Fourier series defined by
f(x)={
1 −x, −π<x< 01 +x,0<x<πwhich is periodic of period 2π.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
f(x)=π
2+ 1−4
π(
cosx+1
32cos 3x+1
52cos 5x+···)⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦- 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+···]71.3 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.(b) If ahalf-range cosine seriesis required for the
functionf(x)=xin the range 0 toπthen aneven
periodic function is required. In Figure 71.4,
f(x)=xis shown plotted fromx=0tox=π.
Since an even function is symmetrical about the
f(x) axis the lineABis constructed as shown. If
the triangular waveform produced is assumed to
be periodic of period 2πoutside of this range
then the waveform is as shown in Fig. 71.4.
When a half-range cosine series is required then
the Fourier coefficientsa 0 andanare calculated− 2 π^0BAf(x)
f(x) = x−ππ 2 ππxFigure 71.4