Fourier Series
Part 2: Convergence Issues
So far we have proved the following theorem:
∞
Theorem: If 𝑓(𝑥) = 𝑎0 + ∑∞
𝑛=1 𝑎𝑛 cos(𝑛𝑥) + ∑𝑛=1 𝑏𝑛 sin(𝑛𝑥) for 𝑥 ∈ [−𝜋, 𝜋] then
𝑎0 =
1 𝜋
∫ 𝑓(𝑥)𝑑𝑥 ,
2𝜋 −𝜋
𝑎𝑛 =
1 𝜋
∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥 ,
𝜋 −𝜋
𝑏𝑛 =
1 𝜋
∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥
𝜋 −𝜋
In the previous segment we used these formulas to compute a number of Fourier series. In this segment we want to
switch to a more theoretical discussion and we want to figure out the convergence behavior of a Fourier series. There
are two parts to this question:
1. Find out for which functions a Fourier series converges
2. Determine for which functions a Fourier series converges to the original function
Let’s check the first question first (of course, or I would not have called question 1): Let’s start with some function 𝑓(𝑥):

if 𝑎𝑛 =

if 𝑏𝑛 =
1 𝜋
∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥 , does ∑∞
𝑛=1 𝑎𝑛 cos(𝑛𝑥) converge
𝜋 −𝜋
1 𝜋
∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥 , does ∑∞
𝑛=1 𝑏𝑛 sin(𝑛𝑥) converge
𝜋 −𝜋
The series are, on one level, just some infinite series, so we could apply the various convergence tests we know. Let’s
start with the divergence test:
Proposition 1: If 𝑓(𝑥) is a 𝐶 1 ([−𝜋, 𝜋]) function (i.e. f is differentiable and f’ is continuous), then the divergence test fails.
That means, of course, that lim 𝑎𝑛 cos(𝑛𝑥) = 0 and lim 𝑏𝑛 sin(𝑛𝑥) = 0, and tells us nothing about convergence of
𝑛→∞
𝑛→∞
the series.
1
𝜋
To show this, let’s use integration by parts on the coefficients 𝑎𝑛 = 𝜋 ∫−𝜋 𝑓(𝑥)cos(𝑛𝑥) 𝑑𝑥. Let 𝑢(𝑥) = f(x) and
1
𝑣 ′ (𝑥) = cos(𝑛𝑥). Then 𝑢′ (𝑥) = 𝑓′(𝑥) and 𝑣(𝑥) = 𝑛 sin(𝑛𝑥) so that:
𝜋
∫ 𝑓(𝑥)cos(𝑛𝑥) 𝑑𝑥 =
−𝜋
𝜋
1
1 𝜋
1 𝜋
𝑓(𝑥) sin(𝑛𝑥)| − ∫ 𝑓 ′ (𝑥) sin(𝑛𝑥) 𝑑𝑥 = 0 − ∫ 𝑓 ′ (𝑥) sin(𝑛𝑥) 𝑑𝑥
𝑛
𝑛 −𝜋
𝑛 −𝜋
−𝜋
Since 𝑓′ is continuous on a closed, bounded interval, it is bounded by, say, 𝑀. But then
|𝑎𝑛 | ≤
11
2𝑀
𝑀 2𝜋 =
𝑛𝜋
𝑛
so that lim 𝑎𝑛 = 0. But then lim 𝑎𝑛 cos(𝑛𝑥) = 0 for all x because |cos(𝑛 𝑥)| ≤ 1, so that the divergence test
𝑛→∞
𝑛→∞
fails.
The proof for the other series is similar – but not quite: make sure to try it!
Based on this proof it is now easy to show that:
Proposition 2: Suppose 𝑓 is a 𝐶 2 [−𝜋, 𝜋] function (i.e. 𝑓 is twice differentiable and 𝑓′′ is continuous). Then the Fourier
∞
series 𝑎0 + ∑∞
𝑛=1 𝑎𝑛 cos(𝑛𝑥) + ∑𝑛=1 𝑏𝑛 sin(𝑛𝑥) with 𝑎0 =
1 𝜋
∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥
𝜋 −𝜋
1 𝜋
∫ 𝑓(𝑥)𝑑𝑥 ,
2𝜋 −𝜋
𝑎𝑛 =
1
𝜋
𝜋
∫−𝜋 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥 , and 𝑏𝑛 =
converges uniformly.
Note that this theorem will be easy to prove but it is not satisfying: for one thing, it does not show whether the series
converges to the original function, and for another we know from our examples that the Fourier series for a function 𝑓
seems to converge (to the original function) for functions that are not even continuous, while this theorem assumes 𝑓 is
twice continuously differentiable. But, it is a start, so let’s go prove it.
Proof: We already know that
𝜋
1 𝜋
∫ 𝑓(𝑥)cos(𝑛𝑥) 𝑑𝑥 = − ∫ 𝑓 ′ (𝑥) sin(𝑛𝑥) 𝑑𝑥
𝑛 −𝜋
−𝜋
Integrate by parts again, with 𝑢(𝑥) = 𝑓 ′ (𝑥), 𝑣 ′ (𝑥) = sin(𝑛𝑥), and, consequently, 𝑢′ (𝑥) = 𝑓′′(𝑥), 𝑣(𝑥) =
1
𝑛
− cos(𝑛𝑥):
𝜋
𝜋
1 𝜋
1
1
1 𝜋
∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥 = − ∫ 𝑓 ′ (𝑥) sin(𝑛𝑥) 𝑑𝑥 = − (− 𝑓 ′ (𝑥) cos(𝑛𝑥)| + ∫ 𝑓 ′′ (𝑥) cos(𝑛𝑥) 𝑑𝑥) =
𝑛 −𝜋
𝑛
𝑛
𝑛 −𝜋
−𝜋
−𝜋
𝜋
1
1
= 2 (𝑓 ′ (𝜋) − 𝑓 ′ (−𝜋)) cos(𝑛 𝜋) − 2 ∫ 𝑓 ′′ (𝑥) cos(𝑛𝑥) 𝑑𝑥
𝑛
𝑛 −𝜋
Therefore:
𝐾
|𝑎𝑛 | ≤ 2
𝑛
𝐾
for some constant K, so that ∑∞
𝑛=1 𝑎𝑛 cos(𝑛𝑥) converges by the comparision test with ∑ 𝑛2 , a convergent p-series
with 𝑝 = 2, and the convergence is uniform by Weierstrass Convergence theorem.
The proof that ∑∞
𝑛=1 𝑏𝑛 sin(𝑛𝑥) converges is similar. But then the original Fourier series must converge.
The above theorem is nice and easy to prove, but using Abel’s convergence test for series we can do slightly better:
Proposition 3: If 𝑓(𝑥) is a 𝐶 1 ([−𝜋, 𝜋]) function (i.e. f is differentiable and f’ is continuous), then the Fourier series for 𝑓
converges.
Apply Abel’s convergence test (see IRA 4.2 for details). The details are left as an exercise.
But we can do a little better still. Here is a ‘sharp’ result for the convergence of a Fourier series. The proof is a little
detailed so we are going to skip it.
Theorem: (Fourier Series)
Suppose 𝑓 is a piecewise 𝐶 1 [−𝜋, 𝜋] function, i.e. 𝑓 consists of finitely many parts, each of which is continuously
differentiable, and all discontinuities are jump discontinuities, then the Fourier series of 𝑓 converges to 𝑓(𝑥) at
all points where 𝑓 is continuous, and to 1/2(𝑓(𝑥 + ) − 𝑓(𝑥 − )) at each jump.
Example: Check the convergence behavior of the Fourier series of 𝑓(𝑥) = {
𝜋, 0 < 𝑥 ≤ 𝜋
0, −𝜋 ≤ 𝑥 ≤ 0
Pay special attention to the limit of the Fourier series at the jump at 𝑥 = 0
𝜋
2
In the previous segment we worked out the Fourier series for this function to be 𝐹(𝑥) = 2 + ∑∞
𝑛=0 2𝑛+1 sin((2𝑛 + 1) 𝑥).
𝜋
1
1
𝜋
Clearly 𝐹(0) = 2 = 2 (𝑓(0+ ) − 𝑓(0− )) = 2 (𝜋 − 0) = 2 , so that the series does converge at least at the jump as
advertised. And it seems to converge to 𝑓(𝑥) elsewhere as well, judging by several graphs of the N-th partial Fourier
series with 𝑁 = 3, 𝑁 = 10, and 𝑁 = 40, respectively.
3
2
1
3
3
3
2
2
2
1
1
1
1
2
3
3
2
1
1
2
3
3
2
1
1
2
3
Note how the Fourier series ‘overshoots’ near the jump but then quickly dampens down to hug the original function
closely. That behavior is typical for Fourier series and has a number of interesting consequences in the physics of signal
transmission.
Anyway, we have achieved our goal: the Fourier Theorem above tells us for which functions the Fourier Series converges
back to the original function.
We do have one pesky limitation left: each of the function we considered so far was defined on the interval [−𝜋, 𝜋].
While we can extend the function by periodicity to all real numbers and the Fourier series would converge to that
extended periodic function for all 𝑥, what about functions defined on an arbitray interval [a, b]? We will check that out
in our next segment.
Exercises:
1. Prove the second part of Proposition 1
2. Prove the second part of Proposition 2
1
3. Prove Proposition 3. Hint: Prove and then use the fact that | ∑𝑘𝑛=1 sin(𝑛𝑥) | ≤ |sin(𝑥/2)| together with Abel’s
convergence test.
𝜋
4
4. Use the Fourier series for 𝑓(𝑥) = 𝑥 on [−𝜋, 𝜋] to show that = ∑∞
𝑛=0
(−1)𝑛
2𝑛+1
(we already knew that by looking at
the Taylor series for 𝑎𝑟𝑐𝑡𝑎𝑛(𝑥) but our reasoning is different this time around)
5. Use the Fourier series for 𝑓(𝑥) = |𝑥| on [−𝜋, 𝜋] to show that
𝜋2
8
1
= ∑∞
𝑛=0 (2𝑛+1)2
1
6. Show that ∑∞
𝑛=1 𝑛2 =
𝜋2
6
1, 𝑥 < 0
7. Check the Fourier Theorem for 𝑓(𝑥) = {
, especially at the jump 𝑥 = 0. Verify the convergence behavior
𝑥, 𝑥 ≥ 0
graphically, using Mathematica.