Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In particular,can an oscillatory function with some decay term ( i.e e^(-t)*cos(kt) have a fourier series representation? All the articles I read said that the function has to be periodic,but this one is not. Thanks

share|improve this question

3 Answers 3

No. Since $$ f(x) = \sum_{n=0}^\infty \left(a_n\cos\left(2\pi n\frac{x}{T}\right) + b_n\sin\left(2\pi n\frac{x}{T}\right)\right) $$ is always $T$-periodic (due to $\sin$ and $\cos$ being $2\pi$-periodic), a fourier series can only faithfully represent periodic functions.

You can still apply the fourier transform to non-periodic functions, but instead of coefficients $a_n$,$b_n$, you get continous function $\mathbb{R}\to\mathbb{C}$.

share|improve this answer
    
Thank you for your answer –  Karim Jonson Oct 24 '12 at 0:41

To actually get a Fourier Series, you must have a periodic function. However, a non-periodic function can be considered to haven an infinite period. Then your sum over sin/cos terms with integer frequencies becomes an integral over sin/cos terms with continuous frequencies, and your coefficients become a function.

ie. $ \sum_{-\infty}^{\infty }c_{n}e^{\frac{in\pi x}{L}} \rightarrow \int_{-\infty}^{\infty}F(w)e^{iwx}dw $

Here, n*pi/L is the frequency in the periodic case, and w is the frequency in the non-periodic case. In both cases, we sum/integrate over all frequencies.

The function that would give you your coefficients in the periodic case ($c_{n}$) and which sits under the integral as the continuous analog of the coefficients in the non-periodic case (F(w)) is called the Fourier Transform.

So for a periodic function, the Fourier Transform returns discrete amplitudes, with which you can build a series. For a non-periodic (infinite periodic) function, the Fourier Transform returns a complex-valued function, which is used under the integral to weight the different frequency (w) contributions in an exactly analogous way to the coefficients in the periodic case. However, you can no longer really call it a Fourier series.

One last point: if you are only interested in a particular region of a non-periodic function, then you can construct a new function that is periodic and looks like your region of interest over a single period. Then you can find a Fourier series, but it will only be applicable to the region of the original function that you singled out.

share|improve this answer
    
Thanks for the explanations. Yes,I forgot to specify. I wasn`t interested on the whole R,but on a bounded domain. Can you explain more,or from where I can get more references,about your last statement ( i.e. a bounded region). Can it be applied to R^2 for a scalar function? Say I have defined the function on [0,a] x [0,b]. ( so bassicaly the domain is bounded) –  Karim Jonson Oct 24 '12 at 0:40
1  
Note that if you pretend that $f$ is periodic with period $T$ when it isn't,and if $f(0) \neq f(T)$, then you're essentially expanding a non-continuous function into a fourier-series. Even if $f$ is otherwise well-behaved enough to give you pointwise convergence of the fourier series, you won't get it at $f(0$, then! You'll see things like Gibb's phenomenon at $t=0$. So be carefull when you do this. –  fgp Oct 24 '12 at 1:06

It has to be periodic - suppose you may represent a function $f:\Bbb{R}\rightarrow\Bbb{R}$ as a Fourier series:

$$f(x)=\sum_{-\infty}^{\infty}{c_{n}e^{inx}}.$$

Then, as $e^{inx}$ is $2\pi$-periodic, so is $f$.

share|improve this answer
    
Thank you very much for your answers. I appreciate it. –  Karim Jonson Oct 24 '12 at 0:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.