1
c Anthony Peirce.
Introductory lecture notes on Partial Differential Equations - °
Not to be copied, used, or revised without explicit written permission from the copyright owner.
Lecture 13: Full Range Fourier Series
(Compiled 3 March 2014)
In this lecture we consider the Full Range Fourier Series for a given functions defined on an interval [−L, L]. Otside this
interval we see that the Fourier Series represents the periodic extension of the function f (x).
Key Concepts: Full Range Fourier Series; Periodic Extension; Complex Fourier Series.
13.1 Fourier Series
We consider the expansion of the function f (x) of the form
³ nπx ´
³ nπx ´
a0 X
+
an cos
+ bn sin
= S(x)
2
L
L
n=1
∞
f (x) ∼
(13.1)
where
1
an =
L
ZL
f (x) cos
³ nπx ´
L
dx
a0
1
=
2
2L
−L
bn =
1
L
ZL
f (x) dx = average value of f .
−L
ZL
f (x) sin
³ nπx ´
L
(13.2)
dx
−L
Observations:
³ nπ
´
³ nπx ´
³ nπ
´
nπT
2L
(1) Note that cos
(x + T ) = cos
provided
= 2π, T =
and similarly sin
(x + 2L) =
L
L
L
n
L
³ nπx ´
sin
. Thus each of the terms of the Fourier Series S(x) on the RHS of (13.1) is a periodic function having
L
a maximal period 2L. As a result the function S(x) is also periodic.
How does this relate to f (x) which may not be periodic?
The function S(x) represented by the series is known as the periodic extension of f on [−L, L].
(2) If f (or its periodic extension) is discontinuous at a point x0 then S(x) converges to the average value of f
across the discontinuity.
S(x0 ) =
ª
1©
−
f (x+
0 ) + f (x0 )
2
(13.3)
2
Example 13.1
½
f (x) =
1
a0 =
π
an =
=
1
π
1
π
Zπ
−π
Zπ
0
x
−π < x < 0
0≤x≤π
1
f (x) dx =
π
Zπ
x dx =
0
L=π
π
2
1
π
(13.5)
f (x) cos(nx) dx
−π
Zπ
x cos(nx) dx
0


¯π
Zπ


sin(nx) ¯¯
1
1
x
=
−
1.
sin(nx)
dx

π
n ¯0 n
0
¯π ¾
½
¯
1 π sin
1
=
(nπ)
% + 2 cos(nx)¯¯
π
n
n
0
¤
1 £
n
1 2
(−1)n − 1
=
n
2
(−1) − 1 −2 0
πn
2
m = 0, 1, 2, . . .
a2m+1 = −
π(2m + 1)2
Zπ
1
bn =
f (x) sin(nx) dx
π
=
(13.4)
3 4
−2 0
(13.6)
(13.7)
−π
Zπ
x sin(nx) dx
0


¯π
Zπ


cos(nx) ¯¯
1
1
=
−x
1.
cos(nx)
dx
+

π
n ¯0 n
0
¯π ¾
½
¯
1
cos(nπ) 0. cos 0
1
=
−π
+
+ 2 sin(nx)¯¯
π
n
n
n
0
= (−1)n+1 /n
∞
a0 X
f (x) =
+
an cos(nx) + bn sin(nx)
2
n=1
£
¤
∞
∞
X
2 X cos (2m + 1)x
sin(nx)
π
+
(−1)n+1
= −
4
π m=0
(2m + 1)2
n
n=1
(13.8)
(13.9)
Fourier Series
3
10 terms of the Fourier Series
4
3
f(x)
2
1
0
−1
−4
−2
0
x/π
2
4
Figure 1. Truncated Fourier Series approximation to f (x) using 10 terms. Notice the periodic extension of the function that
was sampled on [−π, π] and the oscillations
in theª Fourier Series near the points of discontinuity. Also note that at the point
©
of discontinuity x = π, S(π) = 12 f (π + ) + f (π − )
13.2 It can be useful to shift the interval of integration from [−L, L] to [c, c + 2L]
³ nπx ´
³ nπx ´
Since the periodic extension fe (x) is periodic with period 2L (as are the basis functions cos
and sin
).
L
L
1
an =
L
ZL
f (x) cos
³ nπx ´
L
1
dx =
L
−L
1
bn =
L
ZL
f (x) sin
³ nπx ´
L
1
dx =
L
−L
c+2L
Z
fe (x) cos
c
c+2L
Z
fe (x) sin
c
³ nπx ´
L
³ nπx ´
L
dx
(13.10)
dx.
(13.11)
Example 13.2 Previous Example:
½
f (x) =
0 −π < x < 0
x
0≤x≤π
(13.12)
On [π, 3π]
½
fe (x) =
an
=
=
=
1
π
1
π
1
π
3π
R
0
x − 2π
fe (x) cos(nx) dx
π < x < 2π
2π ≤ x ≤ 3π
π
3π
R
t = x − 2π
x = t + 2π
2π
Rπ
since
(x − 2π) cos(nx) dx
0
t cos(nt) dt.
(13.13)
dx = dt
x = π ⇒ t = −π
x = 3π ⇒ t = π
cos n(t + 2π) = cos t
(13.14)
4
13.3 Complex Form of Fourier Series
³ nπx ´
³ nπx ´
a0 X
+
an cos
+ bn sin
2
L
L
n=1
∞
f (x) =
cos
³ nπx ´
L
=
e i(
Therefore f (x) =
a0
2
=
a0
2
nπx
L
ei(
nπx
L
) − e−i( nπx
L )
; sin
=
2
L
2i
∞
o b n nπx
o
X
nπx
nπx
an n i( nπx
n
+
e L ) + e−i( L ) +
ei( L ) − e−i( L )
2
2i
n=1
µ
¶
µ
¶
∞
X
nπx
a
+
ib
an − ibn
n
n
i( nπx
)
+
e L +
e−i( L )
2
2
n=1
↑
=
³ nπx ´
) + e−i( nπx
L )
↑
c0
∞
X
↑
cn
cn ei(
nπx
L
(13.15)
c−n
)
n=−∞
an − ibn
1
cn =
=
2
2L
ZL
n
³ nπx ´
³ nπx ´o
f (x) cos
− i sin
dx
L
L
(13.16)
f (x)e−i(
(13.17)
−L
1
=
2L
ZL
nπx
L
) dx
b−n = −bn
−L
Therefore
f (x) =
∞
X
cn ei(
nπx
L
)
(13.18)
n=−∞
1
cn =
2L
ZL
f (x)e−i(
nπx
L
) dx.
(13.19)
−L
Example 13.3
½
−1 −π ≤ x < 0
L=π
1
0<x<π


Z0
Zπ

1 
cn =
−
e−inx dx + e−inx dx

2π 
0
−L


¯
¯π
−inx ¯0
e−inx ¯0 
1  e
−π
−
+
=
2π 
(−in)
(−in) 
½
ª
i ©
0
n even
+inπ
−inπ
=
−2 + e
+e
=
(2/iπn) n odd
2πn
f (x) =
(13.20)
(13.21)
(13.22)
(13.23)
Therefore
f (x) =
¡
¢
2
ei (2n+1)x .
πi(2n + 1)
n=−∞
∞
X
(13.24)