628 Higher Engineering Mathematics
Problem 7. Find the half-range Fourier sine
series to represent the functionf(x)= 3 xin the
range 0≤x≤π.From para. (c), for a half-range sine series:f(x)=∑∞n= 1bnsinnxWhenf(x)= 3 x,bn=2
π∫π0f(x)sinnxdx=2
π∫π03 xsinnxdx=6
π[
−xcosnx
n+sinnx
n^2]π0by parts=6
π[(
−πcosnπ
n+sinnπ
n^2)
−( 0 + 0 )]=−6
ncosnπWhennis odd,bn=6
nHenceb 1 =6
1,b 3 =6
3,b 5 =6
5andsoon.Whennis even,bn=−6
nHenceb 2 =−6
2,b 4 =−6
4,b 6 =−6
6and so on.Hence the half-range Fourier sine series is given by:f(x)= 3 x= 6(
sinx−1
2sin2x+1
3sin3x−1
4sin4x+1
5sin5x−···)Problem 8. Expandf(x)=cosxas a half-range
Fourier sine series in the range 0≤x≤π, and sketch
the function within and outside of the given range.When a half-range sine series is required then an
odd function is implied, i.e. a function symmetrical
about the origin. A graph ofy=cosx is shown in
Fig. 68.6 in the range 0 toπ.Forcosxto be symmetrical
about the origin the function is as shown by the broken
lines in Fig. 68.6 outside of the given range.
From para. (c), for a half-range Fourier sine series:f(x)=∑∞n= 1bnsinnxdx2 210  2 x1 y^5 cos xf(x)Figure 68.6bn=2
π∫π0f(x)sinnxdx=2
π∫π0cosxsinnxdx=2
π∫π01
2[sin(x+nx)−sin(x−nx)]dx=1
π[
−cos[x( 1 +n)]
( 1 +n)+cos[x( 1 −n)]
( 1 −n)]π0=1
π[(
−cos[π( 1 +n)]
( 1 +n)+cos[π( 1 −n)]
( 1 −n))−(
−cos0
( 1 +n)+cos0
( 1 −n))]Whennis odd,bn=1
π[(
− 1
( 1 +n)+1
( 1 −n))−(
− 1
( 1 +n)+1
( 1 −n))]
= 0Whennis even,bn=1
π[(
1
( 1 +n)−1
( 1 −n))−(
− 1
( 1 +n)+1
( 1 −n))]=1
π(
2
( 1 +n)−2
( 1 −n))=1
π(
2 ( 1 −n)− 2 ( 1 +n)
1 −n^2)=1
π(
− 4 n
1 −n^2)
=4 n
π(n^2 − 1 )Henceb 2 =8
3 π,b 4 =16
15 π,b 6 =24
35 πandsoon.