Even and odd functions and half-range Fourier series 627
22
2
0
A
x
B
f (x)
f(x)5x
2
Figure 68.4
(b) If a half-range cosine series is required for the
function f (x) = x in the range 0 to π then an
even periodic function is required. In Figure 68.4,
f (x) = x is shown plotted from x = 0 to x = π.
Since an even function is symmetrical about the
f (x) axis the line AB is constructed as shown.
If the triangular waveform produced is assumed
to be periodic of period 2π outside of this range
then the waveform is as shown in Fig. 68.4. When
a half-range cosine series is required then the
Fourier coefficients a 0 and a n are calculated as
in Section 68.2(a), i.e.
f (x) = a 0 +
∞
n=1
a n cos nx
where a 0 =
1
π
π
0
f (x)dx
and
a n =
2
π
π
0
f (x)cos nx dx
(c) If a half-range sine series is required for the function f (x) = x in the range 0 to π then an odd periodic function is required. In Figure 68.5, f (x) = x
is shown plotted from x = 0 to x = π. Since an
odd function is symmetrical about the origin the
line CD is constructed as shown. If the sawtooth
waveform produced is assumed to be periodic of
period 2π outside of this range, then the waveform
is as shown in Fig. 68.5. When a half-range sine
series is required then the Fourier coefficient b n is
22
D
C
f (x)
f (x)5x
2
3 x
0
2
2
Figure 68.5
calculated as in Section 68.2(b), i.e.
f (x) =
∞
n=1
b n sin nx
where b n =
2
π
π
0
f (x)sin nx dx
Problem 6. Determine the half-range Fourier
cosine series to represent the function f (x) = 3x in
the range 0 ≤ x ≤ π.
From para. (b), for a half-range cosine series:
f (x) = a 0 +
∞
n=1
a n cos nx
When f (x) = 3x,
a 0 =
1
π
π
0
f (x)dx =
1
π
π
0
3x dx
=
3
π
x 2
2
π
0
=
3π
2
a n =
2
π
π
0
f (x) cos nx dx
=
2
π
π
0
3x cos nx dx
=
6
π
x sin nx
n
+
cos nx
n 2
π
0
by parts
=
6
π
π sin nπ
n
+
cos nπ
n 2
−
0 +
cos 0
n 2
=
6
π
0 +
cos nπ
n 2 −
cos0
n 2
=
6
πn 2 (cos nπ − 1)
When n is even, a n = 0
When n is odd, a n =
6
πn 2 (−1 −1) =
−12
πn 2
Hence a 1 =
−12
π
, a 3 =
−12
π3 2 , a 5 =
−12
π5 2 , and so on.
Hence the half-range Fourier cosine series is given by:
f (x) = 3x =
3π
2
−
12
π
cos x +
1
3 2 cos 3x
+
1
5 2 cos 5x + · · ·
22
2
0
A
x
B
f (x)
f(x)5x
2
Figure 68.4
(b) If a half-range cosine series is required for the
function f (x) = x in the range 0 to π then an
even periodic function is required. In Figure 68.4,
f (x) = x is shown plotted from x = 0 to x = π.
Since an even function is symmetrical about the
f (x) axis the line AB is constructed as shown.
If the triangular waveform produced is assumed
to be periodic of period 2π outside of this range
then the waveform is as shown in Fig. 68.4. When
a half-range cosine series is required then the
Fourier coefficients a 0 and a n are calculated as
in Section 68.2(a), i.e.
f (x) = a 0 +
∞
n=1
a n cos nx
where a 0 =
1
π
π
0
f (x)dx
and
a n =
2
π
π
0
f (x)cos nx dx
(c) If a half-range sine series is required for the function f (x) = x in the range 0 to π then an odd periodic function is required. In Figure 68.5, f (x) = x
is shown plotted from x = 0 to x = π. Since an
odd function is symmetrical about the origin the
line CD is constructed as shown. If the sawtooth
waveform produced is assumed to be periodic of
period 2π outside of this range, then the waveform
is as shown in Fig. 68.5. When a half-range sine
series is required then the Fourier coefficient b n is
22
D
C
f (x)
f (x)5x
2
3 x
0
2
2
Figure 68.5
calculated as in Section 68.2(b), i.e.
f (x) =
∞
n=1
b n sin nx
where b n =
2
π
π
0
f (x)sin nx dx
Problem 6. Determine the half-range Fourier
cosine series to represent the function f (x) = 3x in
the range 0 ≤ x ≤ π.
From para. (b), for a half-range cosine series:
f (x) = a 0 +
∞
n=1
a n cos nx
When f (x) = 3x,
a 0 =
1
π
π
0
f (x)dx =
1
π
π
0
3x dx
=
3
π
x 2
2
π
0
=
3π
2
a n =
2
π
π
0
f (x) cos nx dx
=
2
π
π
0
3x cos nx dx
=
6
π
x sin nx
n
+
cos nx
n 2
π
0
by parts
=
6
π
π sin nπ
n
+
cos nπ
n 2
−
0 +
cos 0
n 2
=
6
π
0 +
cos nπ
n 2 −
cos0
n 2
=
6
πn 2 (cos nπ − 1)
When n is even, a n = 0
When n is odd, a n =
6
πn 2 (−1 −1) =
−12
πn 2
Hence a 1 =
−12
π
, a 3 =
−12
π3 2 , a 5 =
−12
π5 2 , and so on.
Hence the half-range Fourier cosine series is given by:
f (x) = 3x =
3π
2
−
12
π
cos x +
1
3 2 cos 3x
+
1
5 2 cos 5x + · · ·