Fourier series over any range 635

Whennis even,an= 0

a 1 =

− 8

π^2

, a 3 =

− 8

π^232

, a 5 =

− 8

π^252

and so on.

Hence the half-range Fourier cosine series for f(x)in
the range 0 to 2 is given by:

f(x)= 1 −

8

π^2

[

cos

(πx

2

)

+

1

32

cos

(

3 πx

2

)

+

1

52

cos

(

5 πx

2

)

+ ···

]

Problem 5. Find the half-range Fourier sine

series for the functionf(x)=xin the range

0 ≤x≤2. Sketch the function within and outside of

the given range.

A half-range Fourier sine series indicates an odd func-
tion. Thus the graph of f(x)=xintherange0to2is
shown in Fig. 69.5 and is extended outside of this range
so as to be symmetrical about the origin, as shown by
the broken lines.

f (x )

f (x ) 5 x

(^2422) x
2
0 42 6
22
Figure 69.5
From para. (c), for a half-range sine series:
f(x)=
∑∞
n= 1
bnsin
(nπx
L
)
bn=
2
L
∫ L
0
f(x)sin
(nπx
L
)
dx

2
2
∫ 2
0
xsin
(nπx
L
)
dx

⎡
⎢
⎣
−xcos
(nπx
2
)
(nπ
2
) +
sin
(nπx
2
)
(nπ
2
) 2
⎤
⎥
⎦
2
0

⎡
⎢
⎣
⎛
⎜
⎝
−2cosnπ
(nπ
2
) +
sinnπ
(nπ
2
) 2
⎞
⎟
⎠−
⎛
⎜
⎝^0 +
sin0
(nπ
2
) 2
⎞
⎟
⎠
⎤
⎥
⎦

−2cosnπ
nπ
2

− 4
nπ
cosnπ
Henceb 1 =
− 4
π
(− 1 )=
4
π
b 2 =
− 4
2 π
( 1 )=
− 4
2 π
b 3 =
− 4
3 π
(− 1 )=
4
3 π
and so on.
Thus the half-range Fourier sine series in the range 0 to
2 is given by:
f(x)=
4
π
[
sin
(πx
2
)
−
1
2
sin
(
2 πx
2
)

• 1
  3
  sin
  (
  3 πx
  2
  )
  −
  1
  4
  sin
  (
  4 πx
  2
  )
  +···
  ]
  Now try the following exercise
  Exercise 233 Further problems on
  half-range Fourier series over rangeL

1. Determine the half-range Fourier cosine series
   for the function f(x)=x in the range
   0 ≤x≤3. Sketch the function within and out-
   side of the given range.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

f(x)=

3

2

−

12

π^2

{

cos

(

πx

3

)

+

1

32

cos

(

3 πx

3

)

+

1

52

cos

(

5 πx

3

)

+···

}

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

2. Find the half-range Fourier sine series
   for the function f(x)=x in the range
   0 ≤x≤3. Sketch the function within and out-
   side of the given range.
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
   f(x)=

6

π

(

sin

(πx

3

)

−

1

2

sin

(

2 πx

3

)

+

1

3

sin

(

3 πx

3

)

−

1

4

sin

(

4 πx

3

)

+···

)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦