MA 222
Using symmetries to simplify Fourier series
K. Rotz
Example 1. Let f be the periodic function which is given on one period by
 t
−
if −2 < t < 0
f (t) = t 2

if 0 ≤ t < 2.
2
Find the Fourier series of f .
The graph of f is shown to the right. The period is 4 = 2p,
so p = 2. Notice that it is an even function since it has symmetry
around the y-axis. Therefore, we’re in case 2 of section 4 in the
handout, so we can automatically conclude the following:
1
All of the bn coefficients are equal to zero since f is even.
Now let’s turn our attention to computing the a0 and an terms.
By Fourier’s theorem,
Z
Z
1 p
1 2
a0 =
f (t) dt =
f (t) dt.
p −p
2 −2
−2
2
−1
Because f is even, we can just compute the integral over [0, 2] and
double it:
Z
2 2
a0 =
f (t) dt
2 0
Z 2
t
dt
=
0 2
= 1.
For general n, we have
1
an =
p
Z
p
nπt
1
f (t) cos
dt =
p
2
−p
Z
2
f (t) cos
−2
nπt
dt.
2
Again using the fact that f is even, the integrand is an even function, so we can compute the integral
from t = 0 to t = 2 and double it:
Z
2 2
nπt
an =
f (t) cos
dt
2 0
2
Z 2
1
nπt
=
t cos
dt.
2
0 2
1
MA 222
Using symmetries to simplify Fourier series
K. Rotz
To compute this integral, it’s necessary to use integration by parts: let u = 12 t, dv = cos nπt
dt. Then
2
2
du = 12 dt and v = nπ
sin nπt
.
Continuing,
2
Z 2
1
nπt 2
nπt
1
an =
t sin
sin
dt.
−
nπ
2 0 nπ 0
2
When plugging the bounds into the first expression, we get zero in each case, because sin nπ = 0 for
integer values of n. Integrating the second term,
Z 2
nπt
2
nπt 2
1
sin
dt = 2 2 (cos
an = −
)
nπ 0
2
nπ
2 0
2
= 2 2 (cos(nπ) − 1).
nπ
For even values of n, cos(nπ) = 1, and of odd values, cos(nπ) = −1, so we can write cos(nπ) = (−1)n .
Therefore
Z 2
nπt
4
nπt 2
1
sin
dt = 2 2 (− cos
an =
)
nπ 0
2
nπ
2 0
2
= 2 2 ((−1)n − 1).
nπ
n
If n is even, (−1) = 1, and if n is odd, (−1) = −1. So

2


if n is even
 2 2 (1 − 1)
nπ
an =


 2 (−1 − 1) if n is odd
2 2
n π

if n is even
0
=

 −4
if n is odd.
n2 π 2
In particular, the first few values of an are obtained by plugging in specific values of n.
a1 = −
4
π2
a2 = 0
a3 = −
4
9π 2
a4 = 0
a5 = −
4
25π 2
The full Fourier series of f (t) is therefore
a0
πt
3πt
5πt
f (t) =
+ a1 cos
+ a3 cos
+ a5 cos
2
2
2
2
4
4
1
4
πt
3πt
5πt
= − 2 cos
− 2 cos
−
cos
− ··· .
2 π
2
9π
2
25π 2
2
2
···
MA 222
Using symmetries to simplify Fourier series
K. Rotz
Example 2. Let f be the periodic function which is given on one period by
(
−2 if −3 < t < 0
f (t) =
2
if 0 ≤ t < 3.
Find the Fourier series of f .
2
−3
3
First, note that the interval over which f is defined is (−3, 3).
Therefore its period is 2p = 6, so p = 3. The graph of f (t)
over one period is given to the left. Notice that it is an odd
function, because when it is flipped around the x-axis and then
again around the y-axis, the graph is unchanged. This puts us in
case 1 of section 4 of the handout. We can automatically conclude
that
a0 = 0 and an = 0 for n ≥ 1.
−2
Half of the work is done already, and not a single integral needed
to be computed!
On the other hand, according to case 1, the simplified formula
for the bn ’s is given by
Z
nπt
2 p
f (t) sin
dt
bn =
p 0
p
Z
nπt
2 3
f (t) sin
=
dt
3 0
3
Z
4 3
nπt
=
sin
dt
3 0
3
since f = 2 on the interval [0, 3]. Integrating (which involves the substitution u =
3
nπt 3
4
−
bn =
cos
3
nπ
3
0
4
=
(− cos nπ + 1)
nπ

0
if n is even
=
8

if n is odd.
nπ
3
nπt
),
3
MA 222
Using symmetries to simplify Fourier series
K. Rotz
Plugging in some small values of n, we get
b1 =
8
π
b2 = 0
b3 =
8
3π
b4 = 0
b5 =
8
5π
Now dumping these into Fourier’s formula, we arrive at the final answer:
πt
3πt
5πt
f (t) = b1 sin
+ b3 sin
+ b5 sin
+ ···
3
3
3
8
πt
8
3πt
8
5πt
= sin
+
sin
+
sin
+ ···
π
3
3π
3
5π
3
4
···