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= 1
bnsinnx
Whenf(x)= 3 x,
bn=
2
π
∫π
0
f(x)sinnxdx=
2
π
∫π
0
3 xsinnxdx
=
6
π
[
−xcosnx
n
+
sinnx
n^2
]π
0
by parts
=
6
π
[(
−πcosnπ
n
+
sinnπ
n^2
)
−( 0 + 0 )
]
=−
6
n
cosnπ
Whennis odd,bn=
6
n
Henceb 1 =
6
1
,b 3 =
6
3
,b 5 =
6
5
andsoon.
Whennis even,bn=−
6
n
Henceb 2 =−
6
2
,b 4 =−
6
4
,b 6 =−
6
6
and so on.
Hence the half-range Fourier sine series is given by:
f(x)= 3 x= 6
(
sinx−
1
2
sin2x+
1
3
sin3x
−
1
4
sin4x+
1
5
sin5x−···
)
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= 1
bnsinnxdx
2
21
0 2 x
1 y^5 cos x
f(x)
Figure 68.6
bn=
2
π
∫π
0
f(x)sinnxdx
=
2
π
∫π
0
cosxsinnxdx
=
2
π
∫π
0
1
2
[sin(x+nx)−sin(x−nx)]dx
=
1
π
[
−cos[x( 1 +n)]
( 1 +n)
+
cos[x( 1 −n)]
( 1 −n)
]π
0
=
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)
)]
= 0
Whennis 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 π
andsoon.