EVEN AND ODD FUNCTIONS AND HALF-RANGE FOURIER SERIES 671

L

The function is defined by:

f(x)=

{

−2, when−π<x< 0

2, when 0 <x<π

From para. (b), bn=

2

π

∫π

0

f(x) sinnxdx

=

2

π

∫π

0

2 sinnxdx

=

4

π

[

−cosnx

n

]π

0

=

4

π

[(

−cosnπ

n

)

−

(

−

1

n

)]

=

4

πn

(1−cosnπ)

Whennis even, bn=0.

Whennis odd, bn=

4

πn

(1−(−1))=

8

πn

Hence b 1 =

8

π

,b 3 =

8

3 π

,b 5 =

8

5 π

,

and so on

Hence the Fourier series is:

f(x)=

8

π

(

sinx+

1

3

sin 3x+

1

5

sin 5x

+

1

7

sin 7x+···

)

Problem 4. Determine the Fourier series for

the functionf(θ)=θ^2 in the range−π<θ<π.

The function has a period of 2π.

A graph off(θ)=θ^2 is shown in Fig. 71.3 in the range
−πtoπwith period 2π. The function is symmetrical
about thef(θ) axis and is thus an even function. Thus
a Fourier cosine series will result of the form:

f(θ)=a 0 +

∑∞

n= 1

ancosnθ

From para. (a),

a 0 =

1

π

∫π

0

f(θ)dθ=

1

π

∫π

0

θ^2 dθ

=

1

π

[

θ^3

3

]π

0

=

π^2

3

− 2 π −π (^02) π θ
π^2
f(θ)
π
f(θ) = θ^2
Figure 71.3
and an=
2
π
∫π
0
f(θ) cosnθdθ

2
π
∫π
0
θ^2 cosnθdθ

2
π
[
θ^2 sinnθ
n

• 
  

  2 θcosnθ
  n^2
  −
  2 sinnθ
  n^3
  ]π
  0
  by parts

  
  

  2
  π
  [(
  0 +
  2 πcosnπ
  n^2
  − 0
  )
  −(0)
  ]

  
  

  4
  n^2
  cosnπ
  When n is odd, an=
  − 4
  n^2

  
  

. Hence a 1 =

− 4

12

,

a 3 =

− 4

32

,a 5 =

− 4

52

, and so on.

Whennis even,an=

4

n^2

. Hencea 2 =

4

22

,a 4 =

4

42

,

and so on.

Hence the Fourier series is:

f(θ)=θ^2 =

π^2

3

− 4

(

cosθ−

1

22

cos 2θ+

1

32

cos 3θ

−

1

42

cos 4 θ+

1

52

cos 5θ−···

)

Problem 5. For the Fourier series of Problem 4,

letθ=πand show that

∑∞

n= 1

1

n^2

=

π^2

6

Whenθ=π,f(θ)=π^2 (see Fig. 71.3). Hence from

the Fourier series:

π^2 =

π^2

3

− 4

(

cosπ−

1

22

cos 2π+

1

32

cos 3π

−

1

42

cos 4π+

1

52

cos 5π−···

)