618 Higher Engineering Mathematics
x
f(x) f(x) 52 x
22 2 0
22
2
2 3
Figure 67.2
From Section 66.3(i),
a 0 =
1
2 π
∫π
−π
f(x)dx
=
1
2 π
∫π
−π
2 xdx=
2
2 π
[
x^2
2
]π
−π
= 0
an=
1
π
∫π
−π
f(x)cosnxdx=
1
π
∫π
−π
2 xcosnxdx
=
2
π
[
xsinnx
n
−
∫
sinnx
n
dx
]π
−π
by parts(see Chapter 43)
=
2
π
[
xsinnx
n
+
cosnx
n^2
]π
−π
=
2
π
[(
0 +
cosnπ
n^2
)
−
(
0 +
cosn(−π)
n^2
)]
= 0
bn=
1
π
∫π
−π
f(x)sinnxdx=
1
π
∫π
−π
2 xsinnxdx
=
2
π
[
−xcosnx
n
−
∫ (
−cosnx
n
)
dx
]π
−π
by parts
=
2
π
[
−xcosnx
n
+
sinnx
n^2
]π
−π
=
2
π
[(
−πcosnπ
n
+
sinnπ
n^2
)
−
(
−(−π)cosn(−π)
n
+
sinn(−π)
n^2
)]
=
2
π
[
−πcosnπ
n
−
πcos(−nπ)
n
]
=
− 4
n
cosnπ
When n is odd, bn=
4
n
. Thus b 1 =4, b 3 =
4
3
,
b 5 =
4
5
, and so on.
When n is even, bn=
− 4
n
. Thus b 2 =−
4
2
,
b 4 =−
4
4
,b 6 =−
4
6
, and so on.
Thus f(x)= 2 x=4sinx−
4
2
sin2x+
4
3
sin3x
−
4
4
sin4x+
4
5
sin5x−
4
6
sin6x+···
i.e. 2 x= 4
(
sinx−
1
2
sin2x+
1
3
sin3x−
1
4
sin4x
+
1
5
sin5x−
1
6
sin6x+···
)
(1)
for values of f(x)between−π andπ.Forvalues
of f(x)outside the range−πto+πthe sum of the
series is not equal tof(x).
Problem 2. In the Fourier series of Problem 1, by
lettingx=π/2, deduce a series forπ/4.
Whenx=π/2, f(x)=πfrom Fig. 67.2.
Thus, from the Fourier series of equation (1):
2
(π
2
)
= 4
(
sin
π
2
−
1
2
sin
2 π
2
+
1
3
sin
3 π
2
−
1
4
sin
4 π
2
+
1
5
sin
5 π
2
−
1
6
sin
6 π
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 , whenπ<x< 2 π.
ThedefinedfunctionisshowninFig.67.3between0and
2 π. 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
(ancosnx+bnsinnx)