Fourier series for periodic functions of period 2π 613
From Section 66.3(i):
a 0 =
1
2π
π
−π
f (x) dx
=
1
2π
0
−π
−k dx +
π
0
k dx
=
1
2π
{[−kx]
0
−π + [kx]
π
0 } = 0
[a 0 is in fact the mean value of the waveform over a
complete period of 2π and this could have been deduced
on sight from Fig. 66.3.]
From Section 66.3(i):
a n =
1
π
π
−π
f (x) cos nx dx
=
1
π
0
−π
−k cos nx dx +
π
0
k cos nx dx
=
1
π
−k sin nx
n
0
−π
+
k sin nx
n
π
0
= 0
Hence a 1 , a 2 , a 3 , ... are all zero (since sin 0 =
sin(−nπ)= sin nπ = 0), and therefore no cosine terms
will appear in the Fourier series.
From Section 66.3(i):
b n =
1
π
π
−π
f (x) sin nx dx
=
1
π
0
−π
−k sin nx dx +
π
0
k sin nx dx
=
1
π
k cos nx
n
0
−π
+
−k cos nx
n
π
0
When n is odd:
b n =
k
π
1
n
−
−
1
n
+
−
−
1
n
−
−
1
n
=
k
π
2
n
+
2
n
=
4k
nπ
Hence b 1 =
4k
π
, b 3 =
4k
3π
, b 5 =
4k
5π
, and so on.
When n is even:
b n =
k
π
1
n
−
1
n
+
−
1
n
−
−
1
n
= 0
Hence, from equation (1), the Fourier series for the
function shown in Fig. 66.3 is given by:
f (x) = a 0 +
∞
n=1
(a n cos nx + b n sin nx)
= 0 +
∞
n=1
(0 + b n sin nx)
i.e.
f (x) =
4k
π
sin x +
4k
3π
sin 3x +
4k
5π
sin 5x + · · ·
i.e.
f (x) =
4k
π
sin x +
1
3
sin 3x +
1
5
sin 5x + · · ·
Problem 2. For the Fourier series of Problem 1
let k = π. Show by plotting the first three partial
sums of this Fourier series that as the series is added
together term by term the result approximates more
and more closely to the function it represents.
If k = π in the Fourier series of Problem 1 then:
f (x) = 4(sin x +
1
3 sin 3x +
1
5 sin 5x + · · ·)
4 sin x is termed the first partial sum of the
Fourier series of f (x), (4sinx +
4
3 sin 3x) is termed
the second partial sum of the Fourier series, and
(4 sin x +
4
3 sin 3x +
4
5 sin 5x) is termed the third partial
sum, and so on.
Let P 1 = 4 sin x,
P 2 =
4 sin x +
4
3 sin 3x
and P 3 =
4 sin x +
4
3 sin 3x +
4
5 sin 5x
.
Graphs of P 1 , P 2 and P 3 , obtained by drawing up
tables of values, and adding waveforms, are shown in
Figs. 66.4(a) to (c) and they show that the series is
convergent, i.e. continually approximating towards a
definite limit as more and more partial sums are taken,
and in the limit will have the sum f (x) = π.
Even with just three partial sums, the waveform is starting to approach the rectangular wave the Fourier series
is representing.
Problem 3. If in the Fourier series of Problem 1,
k = 1, deduce a series for
π
4
at the point x =
π
2
If k = 1 in the Fourier series of Problem 1:
f (x) =
4
π
sin x +
1
3
sin 3x +
1
5
sin 5x + · · ·