Even and odd functions and half-range Fourier series 625

2

f(x)

(^0) x
22
2  2  3 
Figure 68.2
The square wave shown in Fig. 68.2 is an odd function
since it is symmetrical about the origin.
Hence, from para. (b), the Fourier series is given by:
f(x)=
∑∞
n= 1
bnsinnx
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
2sinnxdx

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
sin3x+
1
5
sin5x

• 
  

  1
  7
  sin7x+···
  )
  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 isshowninFig.68.3intherange
  −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
  22 ^0
  f()
  f() 5 ^2
  ^2
  2  2  
  Figure 68.3
  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
  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
  −
  2sinnθ
  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.