626 Higher Engineering Mathematics
Hence the Fourier series is:
f(θ)=θ^2 =
π^2
3
− 4
(
cosθ−
1
22
cos2θ+
1
32
cos3θ
−
1
42
cos 4 θ+
1
52
cos5θ−···
)
Problem 5. For the Fourier series of Problem 4,
letθ=πand show that
∑∞
n= 1
1
n^2
=
π^2
6
Whenθ=π,f(θ )=π^2 (see Fig. 68.3). Hence from the
Fourier series:
π^2 =
π^2
3
− 4
(
cosπ−
1
22
cos2π+
1
32
cos3π
−
1
42
cos4π+
1
52
cos5π−···
)
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= 1
1
n^2
=
π^2
6
Now try the following exercise
Exercise 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<π
which is periodic outside of this range of
period 2π.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
f(x)=
4
π
(
cosx−
1
3
cos3x
+
1
5
cos5x
−
1
7
cos7x+···
)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
- Obtain the Fourier series of the function
defined by:
f(t)=
{
t+π, −π<t< 0
t−π, 0 <t<π
which 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< 0
1 +x, 0 <x<π
which is periodic of period 2π.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
f(x)=
π
2
+ 1
−
4
π
(
cosx+
1
32
cos3x
+
1
52
cos5x+···
)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
- 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.