Using the Fourier Series To Find Some Interesting Sums

Preliminaries

If f(x) is periodic with period 2p and f’(x) exists and is finite for -π<x<π, then f can be written as a Fourier series:

[itex]f(x)=\sum_{n=-\infty}^{\infty}a_{n}e^{inx} [/itex]

where

[itex]a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt [/itex].

We shall also need Parseval’s formula. It says that for such an f we have:

[itex]\parallel f \parallel^{2}=\sum_{n=-\infty}^{\infty}\vert a_{n}\vert^{2} [/itex]

where

[itex]\parallel f \parallel^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\vert f(t) \vert ^{2}dt [/itex]

We shall also use the fact that

[itex]\int_{-\pi}^{\pi}e^{-int}dt=0 [/itex]

and

[itex]e^{-in\pi}=(e^{-i\pi})^{n}=(-1)^{n}=(e^{i\pi})^{n}=e^{in\pi} [/itex]

First Series

Let f(x) be periodic with period 2π where f(x) = x for -π<x<π and f(-π)=f(π)=0.  Let us calculate a_n for n≠0. The easiest way is to use partial integration:

[itex] a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}te^{-int}dt =\frac{(-1)^{n}}{2\pi in}(\pi-(-\pi))-\frac{(-1)^{n}}{2\pi in}\int_{-\pi}^{\pi}e^{-int}dt=\frac{(-1)^{n}}{in}-0[/itex]

For n=0 we have

[itex]a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}tdt =\frac{1}{2\pi}\frac{1}{2}(\pi^{2}-(-\pi)^{2})=0 [/itex]

The only thing left to calculate is

[itex]\parallel f \parallel^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}t^{2}dt=\frac{1}{2\pi}\cdot\frac{1}{3}(\pi^{3}-(-\pi)^{3})=\frac{\pi^{2}}{3} [/itex]

Applying Parseval’s formula, we get

[itex]\frac{\pi^{2}}{3}=\sum_{n=-\infty}^{-1}\frac{1}{n^{2}}+0+\sum_{n=1}^{\infty}\frac{1}{n^{2}} [/itex]

Now [itex](-n)^{2}=n^{2} [/itex], so we can merge the two sums on the right:

[itex]\frac{\pi^{2}}{3}=2\sum_{n=1}^{\infty}\frac{1}{n^{2}} [/itex]

or

[itex] \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}[/itex]

which is our first result.

Second Series

Let f(x) be periodic with period 2π where f(x) = x² for -π≤x≤π. Then

[itex]a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}t^{2}e^{-int}dt=\frac{(-1)^{n}}{2\pi in}(\pi^{2}-(-\pi)^{2})-\frac{(-1)^{n}}{2\pi in}\int_{-\pi}^{\pi}2te^{-int}dt [/itex]

We have already calculated the difficult part of the last integral under “First Series”, therefore it is easy to find

[itex]a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}t^{2}e^{-int}dt=0-\frac{(-1)^{n}}{2\pi in}\int_{-\pi}^{\pi}2te^{-int}dt=\frac{2\cdot(-1)^{2n}}{n^{2}}=\frac{2}{n^{2}} [/itex]

For n=0 we have

[itex]a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}t^{2}dt =\frac{1}{2\pi}\frac{1}{3}(\pi^{3}-(-\pi)^{3})=\frac{\pi^{2}}{3} [/itex]

The only thing left to calculate is

[itex]\parallel f \parallel^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}t^{4}dt=\frac{1}{2\pi}\cdot\frac{1}{5}(\pi^{5}-(-\pi)^{5})=\frac{\pi^{4}}{5} [/itex]

Applying Parseval’s formula, we get

[itex]\frac{\pi^{4}}{5}=\sum_{n=-\infty}^{-1}\frac{4}{n^{4}}+\frac{\pi^{4}}{9}+\sum_{n=1}^{\infty}\frac{4}{n^{4}}[/itex]

Now [itex](-n)^{4}=n^{4} [/itex], so we can merge the two sums on the right:

[itex]\frac{\pi^{4}}{5}-\frac{\pi^{4}}{9}=2\sum_{n=1}^{\infty}\frac{4}{n^{4}} [/itex]

Rearranging gives us

[itex] \sum_{n=1}^{\infty}\frac{1}{n^{4}}=\pi^{4}\frac{9-5}{2\cdot 4\cdot5\cdot 9}=\frac{\pi^{4}}{90}[/itex]

which is our second result.

Third series

Now it would be obvious to use f(x)=x³, but that function is somewhat problematic around ±π. A better function is f(x) = x³-π²x which is 0 for ±π. Therefore: Let f(x) be periodic with period 2π where f(x) = x³-π²x for -π≤x≤π. Then

[itex] a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(t^{3}-\pi^{2}t)e^{-int}dt=\frac{(-1)^{n}}{2\pi in}((\pi^{3}-\pi^{2}\cdot\pi)-((-\pi)^{3}-(-\pi)^{2}\cdot(-\pi))-\frac{(-1)^{n}}{2\pi in}\int_{-\pi}^{\pi}(3t^{2}-\pi^{2})e^{-int}dt[/itex]

[itex]=0-\frac{(-1)^{n}}{2\pi in}\cdot 3\int_{-\pi}^{\pi}t^{2}e^{-int}dt +\frac{(-1)^{n}}{2\pi in}\cdot\pi^{2} \int_{-\pi}^{\pi}e^{-int}dt [/itex]

Now, the last integral is 0 (as mentioned in the preliminaries) and the first is solved in “Second series”. The result is that

[itex]a_{n}=\frac{6}{in^{3}}[/itex]

For n=0 we have

[itex]a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(t^{3}-\pi^{2}t)dt=\vert_{-\pi}^{\pi}(\frac{t^{4}}{4}-\pi^{2}\frac{t^{2}}{2})=0 [/itex]

The only thing left to calculate is

[itex]\parallel f \parallel^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(t^{6}-2 \pi^{2}t^{4}+\pi^{4}t^{2}) dt=\frac{1}{2\pi}(2\cdot\frac{\pi^{7}}{7}-2\cdot 2\pi^{2}\frac{\pi^{5}}{5}+2\cdot \pi^{4}\frac{\pi^{3}}{3})=\frac{8\pi^{6}}{105}[/itex]

Applying Parseval’s formula:

[itex]\frac{8\pi^{6}}{105}=\sum_{n=-\infty}^{-1}\frac{36}{n^{6}}+0+\sum_{n=1}^{\infty}\frac{36}{n^{6}} [/itex]

Now [itex](-n)^{6}=n^{6} [/itex], so we can merge the two sums on the right and rearrange:

[itex]\sum_{n=1}^{\infty}\frac{1}{n^{6}}=\frac{1}{2\cdot 36}\cdot\frac{8\pi^{6}}{105}=\frac{\pi^{6}}{945}[/itex]

which is our third result.

Summing up

All the results in this insight are well-known, but the proofs are usually rather obtuse. I just thought that it would be fun to show that a trivial application of Fourier series theory would give the same answers.

Now I expect everybody has understood how to continue with increasing powers of x in f(x).  Who is going to do the next two?