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
- 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
)
+···
}
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
- 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
)
+···
)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦