# how can I use a fourier expansion of a periodic function to find the sum of a series?

• Dec 17th 2010, 01:10 PM
czechman345
how can I use a fourier expansion of a periodic function to find the sum of a series?
If you can recommend a good online resource for this, or can just explain it on your own that would be great. I'm not sure what math topic this technically falls under, so I apologize if I'm posting it in the wrong area.

• Jan 14th 2011, 01:32 AM
Ackbeet
Off-hand, the only way I know of that you'd be able to do this is if you had an independent way of evaluating the periodic function. So, let's say you have a $T$-periodic function $f(t),$ (so the function repeats itself every $T$ units in time) and you know its expression exactly. Moreover, you've computed its Fourier series expansion thus:

$\displaystyle f(t)=\dfrac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos\left(\frac{2\pi n t}{T}\right)+b_{n}\sin\left(\frac{2\pi n t}{T}\right)\right],$ where

$\displaystyle a_{0}=\frac{1}{T}\int_{T}f(t)\,dt,$

$\displaystyle a_{n}=\frac{1}{T}\int_{T}f(t)\cos\left(\frac{2\pi n t}{T}\right)dt,$ and

$\displaystyle b_{n}=\frac{1}{T}\int_{T}f(t)\sin\left(\frac{2\pi n t}{T}\right)dt.$

The expression

$\displaystyle \int_{T}$

just means "integrate over any period". It turns out it doesn't matter which period you integrate over.

Suppose further that it's possible to substitute a value for $t$ into your Fourier series there to get the exact expression for the series that you're trying to evaluate. Why then, you can evaluate the series by plugging the same value for $t$ into your known expression for $f(t),$ and you've evaluated your series.

All of this depends on one key fact: that you can find a function $f(t)$ whose Fourier series is that intimately related to the series you're trying to evaluate. That might be a tall order. Or there might be a trick for doing that. I don't know.

In any case, this is one way I could see someone using Fourier series to evaluate a series.