Using the Fourier Series To Find Some Interesting Sums - Comments

[Svein]

Svein submitted a new PF Insights post

Using the Fourier Series To Find Some Interesting Sums

Continue reading the Original PF Insights Post.

 

[micromass]

Svein submitted a new PF Insights post

Using the Fourier Series To Find Some Interesting Sums

Continue reading the Original PF Insights Post.

Nice article. I think a follow up could consist of finding a general formula for
$$\sum \frac{1}{n^{2p}}$$
You can do that with Fourier series too!

 

[Svein]

Nice article. I think a follow up could consist of finding a general formula for
$$\sum \frac{1}{n^{2p}}$$
You can do that with Fourier series too!

Yes - but I thought it best to keep it reasonably simple the first time through.

 

[stevendaryl]

Very nice. I have done similar tricks for evaluating summations, but I didn't know the trick of using Parseval’s formula. My favorite trick is using representations of the Dirac delta function, and that's how I would evaluate $\sum \frac{1}{n^2}$. However, it's a lot more convoluted.

Start with the representation for the delta function:

$\delta(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_n cos(nx)$

Now, integrate both sides from $-x$ to $+x$.

$sign(x) = \frac{x}{\pi} + \frac{2}{\pi} \sum_n \frac{1}{n} sin(nx)$

(where $sign(x) = \pm 1$ depending on whether $x>0$ or $x < 0$)

Integrate again, this time from $0$ to $x$:

$|x| = \frac{x^2}{2\pi} + \frac{2}{\pi} \sum_n \frac{1}{n^2} (1 - cos(nx))$

Using a trig identity, $1-cos(nx) = 2 sin^2(\frac{n}{2} x)$. So we have:

$|x| = \frac{x^2}{2\pi} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} x)$

Now, we set $x = \pi$ to get the identity:

$\pi = \frac{\pi}{2} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}$

So:

$\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}$

Drat! The sum on the right side is only over odd values of $n$, because $sin^2(\frac{n}{2} \pi) = 0$ when $n$ is even. But all is not lost. We can reason as follows:

$\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \sum_{even\ n} \frac{1}{n^2}$

but $\sum_{even\ n} \frac{1}{n^2} = \sum_{n} \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n} \frac{1}{n^2}$. So we have:

$\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \frac{1}{4}\sum_{n} \frac{1}{n^2}$

So $\sum_{odd\ n} \frac{1}{n^2} = \frac{3}{4} \sum_n \frac{1}{n^2}$. Putting this back into our result, we have:

$\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \frac{3}{4} \sum_{n} \frac{1}{n^2}$

So $\sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$

Using Parseval is a lot simpler.

 

[Greg Bernhardt]

Nice Insight @Svein!