Fourier Series
These summary notes should be used in conjunction with, and should not be a replacement
for, your lecture notes. You should be familiar with the following definitions.
• A function f is periodic with period 2L if f (x) = f (x + 2L) for all x. For example sin and
cos are both periodic with period 2π.
• If f is any function defined in the interval (−L, L] or [−L, L) then the 2L-periodic extension of f , denoted F , is defined by
{
f (x)
if x ∈ (−L, L] or x ∈ [−L, L)
F (x) =
F (x + 2L) otherwise
• f is an odd function if f (−x) = −f (x) for all x.
• f is an even function if f (−x) = f (x) for all x.
•

 0∫
f (x) dx =
 2
−n
∫
if f is odd
n
n
f (x) dx
if f is even
0
1. Full-range Fourier Series
The Full-range Fourier series representation of f on (−L, L) is
∞ [
( nπx )
( nπx )]
∑
f (x) ∼ a0 +
an cos
+ bn sin
,
L
L
n=1
where
∫ L
1
a0 =
f (x) dx,
2L∫ −L
( nπx )
1 L
an =
f (x) cos
dx (n = 1, 2, 3, . . .),
L ∫ −L
L
( nπx )
1 L
dx (n = 1, 2, 3, . . .).
bn =
f (x) sin
L −L
L
If f is 2L-periodic then the above series is a representation of f . If f is only defined on (−L, L)
then the series is actually a representation of F , the 2L-periodic extension of f .
Calculation of a Fourier series boils down to computing the coefficients a0 , an and bn , and a
firm grasp of integration by parts is required to complete these calculations successfully.
1
Example 1 For the function defined by f (x) = x2 on the interval −2 < x < 2 sketch the
4-period extension of f and determine the Fourier coefficients.
Sketch f on the interval (−2, 2) and then repeatedly shift this to the right and to the left to
obtain the graph of the 4-period extension F .
Figure 1: The 4-period extension of f
Let’s now compute the Fourier coefficients. Note that L = 2 in this case.
[
]2
∫
1 2 2
1 1 3
1
4
a0 =
x dx =
x
= (8 − (−8)) = .
4 −2
4 3
12
3
−2
an
1
=
2
1
=
2
=
=
=
2
x2 cos
( nπx )
−2
([
2 2
x sin
nπ
dx
2
( nπx )]2
2
4
−
nπ
−2
∫
2
x sin
−2
( nπx )
2
)
dx
))
([
∫ 2
( nπx )]2
( nπx )
8
8
4
2
2
sin(nπ) −
sin(−nπ) −
− x cos
+
cos
dx
nπ
nπ
nπ
nπ
2
nπ −2
2
−2
))
(
(
1
4
4 [ ( nπx ) ]2
4
4
cos(nπ) −
cos(−nπ) + 2 2 sin
dx
0−0−
−
2
nπ
nπ
nπ
nπ
2
−2
))
(
(
1
4
4
4
4
(−1)n + 2 2 (sin(nπ) − sin(−nπ))
−
− (−1)n −
2
nπ
nπ
nπ
nπ
)
(
2
4
8
−
− (−1)n + 2 2 (0 − 0)
nπ
nπ
nπ
16
(−1)n .
2
2
nπ
1
=
2
=
∫
(
Note that sin(nπ) = 0 for all n ∈ Z and cos(nπ) = (−1)n for all n ∈ Z.
2
bn
1
=
2
∫
2
x2 sin
−2
( nπx )
2
dx
( )
Before we dive into the integration observe that x2 defines an even function while sin nπx
2
defines an odd function. Since the product of an even and an odd function is odd, it follows
that bn = 0 for all n. This simple observation saves us from having to perform a rather laborious
integration.
The Fourier series is given by
( nπx ) 4 16 ∑ (−1)n
( nπx )
4 ∑ 16
n
f (x) ∼ +
(−1)
cos
=
+
cos
.
3 n=1 n2 π 2
2
3 π 2 n=1 n2
2
∞
∞
Here are the graphs of the (truncated) series for N = 2, N = 11 and N = 101 terms. Notice
that as N increases the series converges to the extended function.
Figure 2: N = 2
Figure 3: N = 11
3
Figure 4: N = 101
Example 2 For the function defined by
{
x − 2 −2 < x < 0
f (x) =
x+2
0<x<2
sketch the 4-period extension of f and determine the Fourier coefficients.
Figure 5: The extended function
a0
∫
1 2
f (x) dx
=
4 −2
(∫ 0
)
∫ 2
1
=
(x − 2) dx +
(x + 2) dx
4
−2
0
([
]0
[ 2
]2 )
1
x
x2
=
− 2x
+
+ 2x
4
2
2
−2
0
1
([0 − 6] + [6 − 0])
4
= 0
=
4
an =
=
=
=
∫
( nπx )
1 2
f (x) cos
dx
2 −2
2
(∫ 0
∫ 2
( nπx )
( nπx ) )
1
dx +
dx
(x − 2) cos
(x + 2) cos
2
2
2
−2
0
(∫ 0
∫ 0
∫ 2
∫ 2
( nπx )
( nπx )
( nπx )
( nπx ) )
1
x cos
dx − 2
cos
dx +
x cos
dx + 2
cos
dx
2
2
2
2
2
−2
−2
0
0
(∫
)
[
[
( nπx )
( nπx )]0
( nπx )]2
2
1
2
2
x cos
+2
dx − 2
sin
sin
2
2
nπ
2
nπ
2
−2
−2
0
1
(0 − 2 × 0 + 2 × 0)
2
= 0
=
Note that
x cos
( nπx )
2
defines an odd function so the value of the integral between −2 and 2 is zero.
bn =
=
=
=
=
=
=
=
=
=
∫
( nπx )
1 2
f (x) sin
dx
2 −2
2
(∫ 0
∫ 2
( nπx )
( nπx ) )
1
dx +
dx
(x − 2) sin
(x + 2) sin
2
2
2
−2
0
(∫ 0
∫ 0
∫ 2
∫ 2
( nπx )
( nπx )
( nπx )
( nπx ) )
1
dx − 2
dx +
dx + 2
dx
x sin
sin
x sin
sin
2
2
2
2
2
−2
−2
0
0
(∫
)
[
[
( nπx )]0
( nπx )]2
( nπx )
2
1
2
2
dx − 2 −
cos
+2 −
cos
x sin
2
2
nπ
2
nπ
2
−2
−2
0
(
( ∫ 2
)
(
))
)
(
2
1
nπx
2
2
2
n
n
dx − 2 −
2
x sin
+
(−1) + 2 − (−1) +
2
2
nπ nπ
nπ
nπ
0
( ([
)
)
]2
∫ 2
)
(
)
(
nπx
4
1
2
nπx
2
4
4
4
n
n
cos
dx +
2 − x cos
+
−
(−1) −
(−1) +
2
nπ
2
nπ 0
2
nπ nπ
nπ
nπ
0
)
)
( (
[
]
( nπx ) 2
1
4
2
2
8
8
n
2 − (−1) +
sin
+
−
(−1)n
2
nπ
nπ nπ
2
nπ nπ
0
)
(
1
8
8
8
−
(−1)n
− (−1)n +
2
nπ
nπ nπ
4
8
−
(−1)n
nπ nπ
4
(1 − 2(−1)n )
nπ
Note that
x sin
defines an even function so
∫ 2
x sin
−2
( nπx )
2
( nπx )
2
∫
2
dx = 2
x sin
0
5
( nπx )
2
dx.
The Fourier series is given by
∞
( nπx )
∑
4
n
(1 − 2(−1) ) sin
.
f (x) ∼
nπ
2
n=1
Here are the graphs of the series for 5, 50 and 100 terms.
Figure 6: 5 terms
Figure 7: 50 terms
Figure 8: 100 terms
6
2. Half-range Fourier Series
If a function f is defined on the interval (0, L) then the 2L-periodic extension of f can be defined
to be an odd function or an even function. This then gives rise to the Half-range Fourier sine
series for f and the Half-range Fourier cosine series for f , respectively.
Sine Series: In this case the extended function F is defined to be an odd 2L-periodic extension
of f , i.e.

0<x<L
 f (x)
−F (−x)
F (x) =

F (x + 2L)
The coefficients are then a0 = an = 0 for all n = 1, 2, 3, . . . and
∫
( nπx )
2 L
bn =
f (x) sin
dx.
L 0
L
Thus
f (x) ∼
∞
∑
bn sin
( nπx )
n=1
L
.
Cosine Series: In this case the extended function F is defined to be an even 2L-periodic
extension of f , i.e.

0<x<L
 f (x)
F (−x)
F (x) =

F (x + 2L)
The coefficients are then bn = 0 for all n = 1, 2, 3, . . . and
∫
1 L
a0 =
f (x) dx
L 0
∫
( nπx )
2 L
an =
f (x) cos
dx.
L 0
L
Thus
f (x) ∼ a0 +
∞
∑
n=1
an cos
( nπx )
L
.
Example 3 For the function defined by f (x) = x2 on the interval (0, 2) determine the Fourier
sine and cosine series.
Sine series:
bn =
=
=
=
∫
( nπx )
2 2 2
x sin
dx
2 0
2
[
∫ 2
( nπx )]2
( nπx )
4
2 2
+
− x cos
x cos
dx
nπ
2
nπ 0
2
0
)
([
∫ 2
( nπx )]2
( nπx )
2
8
4
2
−
−
cos(nπ) +
x sin
sin
dx
nπ
nπ
nπ
2
nπ 0
2
0
)
(
8
4
4 [ ( nπx )]2
n
− (−1) +
cos
nπ
nπ n2 π 2
2
0
7
8
16
(−1)n + 3 3 (cos(nπ) − cos(0))
nπ
nπ
8
16
16
n
= − (−1) + 3 3 (−1)n − 3 3
nπ
nπ
nπ
16
8
n
n
=
((−1) − 1) −
(−1) .
n3 π 3
nπ
= −
Therefore
)
∞ (
( nπx )
∑
16
8
n
n
((−1)
−
1)
−
(−1)
sin
.
F (x) =
3π3
n
nπ
2
n=1
Cosine series:
a0
an
Hence
1
=
2
∫
2
[
]2
4
1 1 3
x dx =
x
= .
2 3
3
0
2
0
∫
( nπx )
2 2 2
=
x cos
dx
2 0
2
[
∫ 2
( nπx )]2
( nπx )
2 2
4
=
x sin
dx
−
x sin
nπ
2
nπ 0
2
0
)
([
∫ 2
( nπx )]2
( nπx )
4
2
2
dx
= −
− x cos
+
cos
nπ
nπ
2
nπ 0
2
0
(
)
4
4
4 [ ( nπx )]2
= −
−
cos(nπ) − 2 2 sin
nπ
nπ
nπ
2
0
16
=
(−1)n .
n2 π 2
( nπx )
4 16 ∑ (−1)n
.
F (x) = + 2
cos
3 π n=1 n2
2
∞
8