Higher Engineering Mathematics

674 FOURIER SERIES

=

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

and so on.

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

sin 2x+

1

3

sin 3x

−

1

4

sin 4x+

1

5

sin 5x−···

)

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=cosxis shown in
Fig. 71.6 in the range 0 toπ. For cosxto be sym-
metrical about the origin the function is as shown
by the broken lines in Fig. 71.6 outside of the given
range.

−π

− 1

0 π

1

2 πx

f(x)

y = cos x

Figure 71.6

From para. (c), for a half-range Fourier sine series:

f(x)=

∑∞

n= 1

bnsinnxdx

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)

)

−

(

−cos 0

(1+n)

+

cos 0

(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

and so on.

Hence the half-range Fourier sine series forf(x)in

the range 0 toπis given by:

f(x)=

8

3 π

sin 2x+

16

15 π

sin 4x

+

24

35 π

sin 6x+···