Higher Engineering Mathematics

666 FOURIER SERIES

Hence

π^2

8

= 1 +

1

32

+

1

52

+

1

72

+···

Problem 5. Deduce the Fourier series for the

functionf(θ)=θ^2 in the range 0 to 2π.

f(θ)=θ^2 is shown in Fig. 70.4 in the range 0 to 2π.
The function is not periodic but is constructed out-
side of this range so that it is periodic of period 2π,
as shown by the broken lines.

− 4 π− 2 π 02 π 4 π

4 π^2

f(θ) = θ^2

f(θ)

θ

Figure 70.4

For a Fourier series:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

a 0 =

1

2 π

∫ 2 π

0

f(θ)dθ=

1

2 π

∫ 2 π

0

θ^2 dθ

=

1

2 π

[

θ^3

3

] 2 π

0

=

1

2 π

[

8 π^3

3

− 0

]

=

4 π^2

3

an=

1

π

∫ 2 π

0

f(θ) cosnθdθ

=

1

π

∫ 2 π

0

θ^2 cosnθdθ

=

1

π

[

θ^2 sinnθ

n

+

2 θcosnθ

n^2

−

2 sinnθ

n^3

] 2 π

0

by parts

=

1

π

[(

0 +

4 πcos 2πn

n^2

− 0

)

−(0)

]

=

4

n^2

cos 2πn=

4

n^2

whenn=1, 2, 3,···

Hencea 1 =

4

12

,a 2 =

4

22

,a 3 =

4

32

and so on

bn=

1

π

∫ 2 π

0

f(θ) sinnθdθ=

1

π

∫ 2 π

0

θ^2 sinnθdθ

=

1

π

[

−θ^2 cosnθ

n

+

2 θsinnθ

n^2

+

2 cosnθ

n^3

] 2 π

0

by parts

=

1

π

[(

− 4 π^2 cos 2πn

n

+ 0 +

2 cos 2πn

n^3

)

−

(

0 + 0 +

2 cos 0

n^3

)]

=

1

π

[

− 4 π^2

n

+

2

n^3

−

2

n^3

]

=

− 4 π

n

Henceb 1 =

− 4 π

1

,b 2 =

− 4 π

2

,b 3 =

− 4 π

3

, and so on.

Thus f(θ)=θ^2

=

4 π^2

3

+

∑∞

n= 1

(

4

n^2

cosnθ−

4 π

n

sinnθ

)

i.e.θ^2 =

4 π^2

3

+ 4

(

cosθ+

1

22

cos 2θ+

1

32

cos 3θ+···

)

− 4 π

(

sinθ+

1

2

sin 2θ+

1

3

sin 3θ+···

)

for values ofθbetween 0 and 2π.

Problem 6. In the Fourier series of Problem 5,

letθ=πand determine a series for

π^2

12

.

Whenθ=π,f(θ)=π^2

Hence π^2 =

4 π^2

3

+ 4

(

cosπ+

1

4

cos 2π

+

1

9

cos 3π+

1

16

cos 4π+···

)

− 4 π

(

sinπ+

1

2

sin 2π

+

1

3

sin 3π+···

)