Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

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


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)


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.
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