618 Higher Engineering Mathematics
x
f (x)
f (x) 5 2x
0
2␲
22␲
22␲
␲
2␲
2␲
3␲
Figure 67.2
From Section 66.3(i),
a 0 =
1
2π
π
−π
f (x) dx
=
1
2π
π
−π
2x dx =
2
2π
x 2
2
π
−π
= 0
a n =
1
π
π
−π
f (x) cos nx dx =
1
π
π
−π
2x cos nx dx
=
2
π
x sin nx
n
−
sin nx
n
dx
π
−π
by parts (see Chapter 43)
=
2
π
x sin nx
n
+
cos nx
n 2
π
−π
=
2
π
0 +
cos nπ
n 2
−
0 +
cos n(−π)
n 2
= 0
b n =
1
π
π
−π
f (x) sin nx dx =
1
π
π
−π
2x sin nx dx
=
2
π
−x cos nx
n
−
−cos nx
n
dx
π
−π
by parts
=
2
π
−x cos nx
n
+
sin nx
n 2
π
−π
=
2
π
−π cos nπ
n
+
sin nπ
n 2
−
−(−π)cos n(−π)
n
+
sin n(−π)
n 2
=
2
π
−π cos nπ
n
−
π cos(−nπ)
n
=
−4
n
cos nπ
When n is odd, b n =
4
n
. Thus b 1 = 4, b 3 =
4
3
,
b 5 =
4
5
, and so on.
When n is even, b n =
−4
n
. Thus b 2 =−
4
2
,
b 4 =−
4
4
, b 6 =−
4
6
, and so on.
Thus f (x) = 2x = 4 sin x −
4
2
sin 2x +
4
3
sin 3x
−
4
4
sin 4x +
4
5
sin 5x −
4
6
sin 6x + · · ·
i.e. 2x = 4
sin x −
1
2
sin 2x +
1
3
sin 3x−
1
4
sin 4x
+
1
5
sin 5x −
1
6
sin 6x + · · ·
(1)
for values of f (x) between −π and π. For values
of f (x) outside the range −π to +π the sum of the
series is not equal to f (x).
Problem 2. In the Fourier series of Problem 1, by
letting x = π/2, deduce a series for π/4.
When x = π/2, f (x) = π from Fig. 67.2.
Thus, from the Fourier series of equation (1):
2
π
2
= 4
sin
π
2
−
1
2
sin
2π
2
+
1
3
sin
3π
2
−
1
4
sin
4π
2
+
1
5
sin
5π
2
−
1
6
sin
6π
2
+ · · ·
π = 4
1 − 0 −
1
3
− 0 +
1
5
− 0 −
1
7
− · · ·
i.e.
π
4
= 1 −
1
3
+
1
5
−
1
7
+· · ·
Problem 3. Obtain a Fourier series for the
function defined by:
f (x) =
x, when 0 < x < π
0, when π < x < 2π.
The defined function is shown in Fig. 67.3 between 0 and
2π. The function is constructed outside of this range so
that it is periodic of period 2π, as shown by the broken
line in Fig. 67.3.
For a Fourier series:
f (x) = a 0 +
∞
n=1
(a n cos nx + b n sin nx)