2Fourier series

IB Methods

2.2 Convergence of Fourier series

When Fourier first proposed the idea of a Fourier series, people didn’t really

believe in him. How can we be sure that the infinite series actually converges?

It turns out that in many cases, they don’t.

To investigate the convergence of the series, we define the partial Fourier

sum as

S

n

f =

n

X

m=−n

ˆ

f

m

e

imθ

.

The question we want to answer is whether

S

n

f

“converges” to

f

. Here we

have to be careful with what we mean by convergence. As we (might) have seen

in Analysis, there are many ways of defining convergence of functions. If we

have a “norm” on the space of functions, we can define convergence to mean

lim

n→∞

kS

n

f − fk = 0. Our “norm” can be defined as

kS

n

f − fk =

1

2π

Z

π

−π

|S

n

f(θ) −f(θ)|

2

dθ.

However, this doesn’t really work if

S

n

f

and

f

can be arbitrary functions, as in

this does not necessarily have to be a norm. Indeed, if

lim

n→∞

S

n

f

differs from

f

on finitely or countably many points, the integral will still be zero. In particular,

lim S

n

f

and

f

can differ at all rational points, but this definition will say that

S

n

f converges to f.

Hence another possible definition of convergence is to require

lim

n→∞

S

n

f(θ) − f(θ) = 0

for all

θ

. This is known as pointwise convergence. However, this is often too

weak a notion. We can ask for more, and require that the rate of convergent is

independent of θ. This is known as uniform convergence, and is defined by

lim

n→∞

sup

θ

|S

n

f(θ) − f(θ)| = 0.

Of course, with different definitions of convergence, we can get different answers

to whether it converges. Unfortunately, even if we manage to get our definition

of convergence right, it is still difficult to come up with a criterion to decide if a

function has a convergent Fourier Series.

It is not difficult to prove a general criterion for convergence in norm, but

we will not do that, since that is analysis. If interested, one should take the IID

Analysis of Functions course. In this course, instead of trying to come up with

something general, let’s look at an example instead.

Example. Consider the sawtooth function f(θ) = θ.

−3π

−π

π

3π

Note that this function is discontinuous at odd multiples of π.

The Fourier coefficients (for n 6= 0) are

ˆ

f

n

=

1

2π

Z

π

−π

e

−inθ

θ dθ =

−

1

2πin

e

−inθ

θ

π

−π

+

1

2πin

Z

π

−π

e

inθ

dθ =

(−1)

n+1

in

.

We also have

ˆ

f

0

= 0.

Hence we have

θ =

X

n6=0

(−1)

n+1

in

e

inθ

.

It turns out that this series converges to the sawtooth for all

θ 6

= (2

m

+ 1)

π

, i.e.

everywhere that the sawtooth is continuous.

Let’s look explicitly at the case where

θ

=

π

. Each term of the partial Fourier

series is zero. So we can say that the Fourier series converges to 0. This is the

average value of lim

ε→0

f(π ± ε).

This is typical. At an isolated discontinuity, the Fourier series is the average

of the limiting values of the original function as we approach from either side.