626 Higher Engineering Mathematics
Hence the Fourier series is:
f (θ) = θ
2
=
π 2
3
− 4
cos θ −
1
2 2 cos 2θ +
1
3 2 cos 3θ
−
1
4 2 cos 4θ +
1
5 2 cos 5θ −· · ·
Problem 5. For the Fourier series of Problem 4,
let θ = π and show that
∞
n=1
1
n 2 =
π 2
6
When θ = π, f (θ) = π 2 (see Fig. 68.3). Hence from the
Fourier series:
π
2
=
π 2
3
− 4
cos π −
1
2 2 cos 2π +
1
3 2 cos3π
−
1
4 2 cos 4π +
1
5 2 cos 5π − · · ·
i.e.
π
2
−
π 2
3
= −4
−1 −
1
2 2 −
1
3 2 −
1
4 2 −
1
5 2 − · · ·
2π 2
3
= 4
1 +
1
2 2 +
1
3 2 +
1
4 2 +
1
5 2 + · · ·
i.e.
2π
2
(3)(4)
= 1 +
1
2 2 +
1
3 2 +
1
4 2 +
1
5 2 + · · ·
i.e.
π 2
6
=
1
1 2 +
1
2 2 +
1
3 2 +
1
4 2 +
1
5 2 + · · ·
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
1. 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
cos x −
1
3
cos 3x
+
1
5
cos 5x
−
1
7
cos 7x + · · ·
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
2. 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(sin t +
1
2 sin 2t
+
1
3 sin 3t
+
1
4 sin 4t + · · ·)
⎤
⎥
⎥
⎥
⎦
3. 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
cos x +
1
3 2 cos 3x
+
1
5 2 cos 5x + · · ·
⎤
⎥
⎥
⎥
⎥
⎥
⎦
4. In the Fourier series of Problem 3, let x = 0
and deduce a series for π 2 /8.
π 2
8
= 1 +
1
3 2 +
1
5 2 +
1
7 2 + · · ·
68.3 Half-range Fourier series
(a) When a function is defined over the range say 0
to π instead of from 0 to 2π it may be expanded
in a series of sine terms only or of cosine terms
only. The series produced is called a half-range
Fourier series.