Fourier series for a non-periodic function over range 2π 619
x
f(x) f(x) 5 x
22 2 0
2 3
Figure 67.3
It is more convenient in this case to take the limits from 0
to2πinstead of from−πto+π.ThevalueoftheFourier
coefficientsareunalteredbythischangeoflimits.Hence
a 0 =
1
2 π
∫ 2 π
0
f(x)dx=
1
2 π
[∫π
0
xdx+
∫ 2 π
π
0dx
]
=
1
2 π
[
x^2
2
]π
0
=
1
2 π
(
π^2
2
)
=
π
4
an=
1
π
∫ 2 π
0
f(x)cosnxdx
=
1
π
[∫π
0
xcosnxdx+
∫ 2 π
π
0dx
]
=
1
π
[
xsinnx
n
+
cosnx
n^2
]π
0
(from Problem 1,by parts)
=
1
π
{[
πsinnπ
n
+
cosnπ
n^2
]
−
[
0 +
cos0
n^2
]}
=
1
πn^2
(cosnπ− 1 )
Whennis even,an=0.
Whennis odd,an=
− 2
πn^2
Hencea 1 =
− 2
π
,a 3 =
− 2
32 π
,a 5 =
− 2
52 π
, and so on
bn=
1
π
∫ 2 π
0
f(x)sinnxdx
=
1
π
[∫π
0
xsinnxdx−
∫ 2 π
π
0dx
]
=
1
π
[
−xcosnx
n
+
sinnx
n^2
]π
0
(from Problem 1,by parts)
=
1
π
{[
−πcosnπ
n
+
sinnπ
n^2
]
−
[
0 +
sin0
n^2
]}
=
1
π
[
−πcosnπ
n
]
=
−cosnπ
n
Henceb 1 =−cosπ=1,b 2 =−
1
2
,b 3 =
1
3
, and so on.
Thus the Fourier series is:
f(x)=a 0 +
∑∞
n= 1
(ancosnx+bnsinnx)
i.e. f(x)=
π
4
−
2
π
cosx−
2
32 π
cos3x
−
2
52 π
cos5x−···+sinx
−
1
2
sin2x+
1
3
sin3x−···
i.e. f(x)
=
π
4
−
2
π
(
cosx+
cos3x
32
+
cos5x
52
+···
)
+
(
sinx−
1
2
sin2x+
1
3
sin3x−···
)
Problem 4. For the Fourier series of Problem 3:
(a) what is the sum of the series at the point of
discontinuity (i.e. atx=π)? (b) what is the
amplitude and phase angle of the third harmonic?
and (c) letx=0, and deduce a series forπ^2 /8.
(a) The sum of the Fourier series at the point of dis-
continuity is given by the arithmetic mean of the
two limiting values off(x)asxapproaches the
point of discontinuity from the two sides.
Hence sum of the series atx=πis
π− 0
2
=
π
2
(b) The third harmonic term of the Fourier series is
(
−
2
32 π
cos3x+
1
3
sin3x
)
This may also bewrittenin theformcsin( 3 x+α),
where amplitude,c=
√√
√
√
[(
− 2
32 π
) 2
+
(
1
3
) 2 ]
= 0. 341