618 Higher Engineering Mathematics
xf(x) f(x) 52 x22 2 0
22 2 2 3 Figure 67.2From Section 66.3(i),a 0 =1
2 π∫π−πf(x)dx=
1
2 π∫π−π2 xdx=
2
2 π[
x^2
2]π−π= 0an=1
π∫π−πf(x)cosnxdx=1
π∫π−π2 xcosnxdx=2
π[
xsinnx
n−∫
sinnx
ndx]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)]
= 0bn=1
π∫π−πf(x)sinnxdx=1
π∫π−π2 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
ncosnπWhen 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
2sin2x+4
3sin3x−4
4sin4x+4
5sin5x−4
6sin6x+···i.e. 2 x= 4(
sinx−1
2sin2x+1
3sin3x−1
4sin4x+1
5sin5x−1
6sin6x+···)
(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(
sinπ
2−1
2sin2 π
2+1
3sin3 π
2−1
4sin4 π
2+1
5sin5 π
2−1
6sin6 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)