18.03 Fourier Series Using Complex Exponentials
Jeremy Orloff
For ease of notation we assume periodic functions of period 2π and work on the
interval [−π, π]. The extension to other periods is easy.
First we’ll give the Fourier theorem and then we’ll motivate and prove it.
Theorem (Fourier):
Then
Suppose f (t) is continuous and periodic with period 2π.
∞
X
1
f (t) =
cn e , where cn =
2π
n=−∞
int
Z
π
f (t)e−int dt
−π
The proof of this uses the
Orthogonality relations:
1
2π
Z
π
int
e
·e
−imt
=
−π
0
1
if m 6= n
if m = n.
The proof of the orthogonality relations is a trivial integration. (And the statement
and proof are much easier than the analogous statements with sines and cosines.)
Proof of the formula for the Fourier coefficients cn :
Z π
Z π
∞
∞
X
X
1
1
int
−in0 t
cn e . then
If f (t) =
f (t) · e
dt =
cn eint · e−in0 t = cn0
2π −π
2π −π
n=−∞
n=−∞
The first equality is just a substitiution of the series for f (t) and the second follows
from the orthogonality relations.
Proof every continuous (period 2π) function equals its Fourier series:
See the note on Fourier completeness for this. Since cos(t) is a sum of complex
exponentials the proof there suffices.
Comments
Z π
1
¯ dt is an inner product on the vector
f (t)g(t)
1. The bilinear form hf, gi =
2π −π
space of periodic functions. The Fourier theorem and orthogonality relations show
the functions {eint } form an orthonormal basis. They are analogous to the standard
vectors i, j, k in three-dimensional space.
2. Also see the note on Fourier completeness for the definition of convolution and the
periodic delta function.
3. The following notation is often used for the Fourier coefficients
Z π
1
ˆ
f (n) =
f (t) · e−int dt.
2π −π
With this notation the Fourier theorem says
∞
X
f (t) =
fˆ(n) eint .
−∞
1
18.03 Fourier Series Using Complex Exponentials
Theorem:
tiplication:
Just like the Laplace transform, Fourier connects convolution and mulf[
∗ g(n) = 2π fˆ(n) · ĝ(n).
Proof: The proof is simple algebra, but it gives some insight into the algebraic
meaning of convolution.
First note that
Z π
Z π
2πeint if n = m
i(m−n)u
in(t−u) imu
int
int
imt
e
du =
e
e du = e
e ∗e =
0
if n 6= m.
−π
−π
Now
(f ∗ g)(t) =
X
=
X
n
fˆ(n)eint ∗
X
ĝ(m)eimt
m
fˆ(n)ĝ(m) eint ∗ eimt
n, m
by the formula just above, most of these terms are 0
X
=
2π fˆ(n)ĝ(n)eint
n
This last equality shows the Fourier coefficients of f ∗ g are 2π fˆ(n)ĝ(n) as claimed.
Note: In the Fourier game the factors of 2π that occur throughout are sometimes
put in other places. For example, some authors define
Z π
1
ˆ
f (n) = √
f (t)e−int dt
2π −π
and then
1 Xˆ
f (n)eint
f (t) = √
2π
2