Talk:Poisson summation formula

From Wikipedia, the free encyclopedia

[edit] How to prove a summation formula?

As indicated in the application, PSF can help to prove

\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}

Anyone knows details of it? I tried myself, but could not figure it out. Malaybear 11:08, 23 August 2007 (UTC)

Hi Malaybear, and welcome to Wikipedia!
I'm not sure how to prove it myself, but let's think it through together! :) Willow 11:19, 23 August 2007 (UTC)

[edit] Trial proof of summation formula

Here are the basic definitions from the article. The summation equals

S(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=-\infty}^{\infty} f(t + n T) = \frac{1}{T} \sum_{m=-\infty}^{\infty} \tilde{F}(m \omega_0) \ e^{i m \omega_0 t}

where the continuous Fourier transform is defined as

\tilde{F}(\omega) \ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^{\infty} f(t) \ e^{-i\omega t} dt

and the fundamental frequency ω0 is

\omega_0 \ \stackrel{\mathrm{def}}{=}\ \frac{2\pi}{T}.

This suggests that we try setting

S_{0} = \sum_{n=-\infty}^{\infty} \frac{1}{b^{2} + n^{2}} = \frac{1}{b^{2}} + 2\sum_{n=1}^{\infty} \frac{1}{b^{2} + n^{2}}

and set b=0 at the end. We can take S0 as a limit

S_{0} = \lim_{t \rightarrow 0} S(t) = \lim_{t \rightarrow 0} \sum_{n=-\infty}^{\infty} \frac{1}{b^{2} + \left(t + nT \right)^{2}}

where the period T=1 and ω0=2π. Therefore we have

f(t) = \frac{1}{b^{2} + t^{2}}

which we may write as

f(t) = \frac{1}{\left( t + ib\right)\left( t - ib\right)} = \left( \frac{1}{2ib} \right) \left[ \frac{1}{t - ib} - \frac{1}{t + ib}\right]

Using the residue theorem, the Fourier transform of this function is something like

\tilde{F}(\omega) = \left(\frac{\pi}{b} \right) e^{-b \left| \omega \right|}

I'm not totally sure if this is right, but let's try it and see what happens; you can always put the correct constant in later. Plugging into the Poisson summation formula gives

S_{0} = \lim_{t \rightarrow 0} \frac{1}{T} \sum_{m=-\infty}^{\infty} \tilde{F}(m \omega_0) \ e^{i m \omega_0 t} = \sum_{m=-\infty}^{\infty} \tilde{F}(m \omega_0)

which is a geometric series

S_{0} = \tilde{F}(0) + 2 \sum_{m=1}^{\infty} \tilde{F}(m \omega_0) = \left(\frac{\pi}{b} \right) \left[ 1 + 2 \sum_{m=1}^{\infty} e^{-b 2 \pi m} \right]

This equals

S_{0} = \left(\frac{\pi}{b} \right) \left[ -1 + 2 \sum_{m=0}^{\infty} e^{-b 2 \pi m} \right] = \left(\frac{\pi}{b} \right) \left[ -1 + \frac{2}{1 - e^{-2\pi b}} \right]

Willow 13:20, 23 August 2007 (UTC)

Define T0 as the last term, and expand it for small b
T_{0}= \frac{2}{1 - e^{-2\pi b}} = \frac{2}{1-(1-2\pi b+2(\pi b)^2-\frac{4}{3}(\pi b)^3) + \cdots}
which can be written as
T_{0} = \frac{1}{\pi b} \left[ 1+\pi b-\frac{2}{3}(\pi b)^2+(\pi b)^2 + \cdots \right] = \frac{1}{\pi b} \left[ 1+\pi b+\frac{1}{3}(\pi b)^2 + \cdots \right] = \frac{1}{\pi b} + 1 + \frac{\pi b}{3} + \cdots
Submitting this into original equation
S_{0} = \frac{\pi}{b}\left[ -1 + \frac{1}{\pi b} + 1 + \frac{\pi b}{3} + \cdots \right] =\frac{1}{b^2}+\frac{(\pi)^2}{3} + \cdots
Comparing to the previous equation
S_{0} = \sum_{n=-\infty}^{\infty} \frac{1}{b^{2} + n^{2}} = \frac{1}{b^{2}} + 2\sum_{n=1}^{\infty} \frac{1}{b^{2} + n^{2}}
and taking the limit b goes to zero finishes the proof.
2\sum_{n=1}^{\infty} \frac{1}{b^{2} + n^{2}} = \frac{(\pi)^2}{3} + \cdots
—The preceding unsigned comment was added by Malaybear (talk • contribs) 14:22, August 23, 2007 (UTC).

Brilliant, Malaybear — well done! The power of working together clears up both of our confusions! :) Willow 16:47, 23 August 2007 (UTC)