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Consider the following function $$ \Pi(t) = \begin{cases} 1 \ \ \text{if} \ |t| < 1/2 \\ 0 \ \ \text{if} \ |t| \geq 1/2 \end{cases} $$

This is not a periodic function. Why can we define the fourier coefficients of $\Pi(t)$ as $$c_n = \frac{1}{T} \int_{-T/2}^{T/2} e^{-2 \pi int/T} \Pi(t) \ dt $$

$$ = \frac{1}{T} \int_{-1/2}^{1/2} e^{-2 \pi int/T} \Pi(t) \ dt $$

if the function is not periodic? I thought fourier coefficients were only defined for periodic functions?

fourierguy
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  • You are right. In order to compute the Fourier coefficients of a given function, you need to introduce a periodic extension of it (if it is not already periodic). The coefficients you compute are those of the periodic extension. In your case the function is already defined on whole $\mathbb R$ and it is not periodic. So the problem makes no sense to me. – Avitus Jun 25 '13 at 15:52

2 Answers2

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You've used the variable $T$ in both expressions without telling us what it represents. If we interpret $T$ as the period in the Fourier expansion, and compute

$$ \sum_{n=-\infty}^\infty c_ne^{2\pi int/T} $$

with $c_n$ defined as in your question, we get the periodic function which agrees with your function on the interval $[-T/2,T/2]$ and is periodic with period $2$. So the sum would look like some sort of square wave (unless $T<1/2$ in which case it would be the constant function $1$).

In principle, there is nothing that says you can't define the Fourier coefficients for arbitrary (not necessarily periodic) functions, but the Fourier coefficients are only meaningful if you specify the period you want to sum over.

In this case, you've only integrated from $-T/2$ to $T/2$, so the only information you've used about the function $\Pi$ is the values it takes on the interval $[-T/2,T/2]$. So the Fourier coefficients you get do not give you the function $\Pi$ when you sum over them, but give you the unique function that:

  • agrees with $\Pi$ on the interval $[-T/2,T/2]$
  • is periodic with period $T$ (I'm assuming that's what $T$ is).

You could change the behaviour of $\Pi$ outside $[-T/2,T/2]$ without changing either of the integrals in your question, so, as far as the Fourier series is concerned, you might as well have been integrating over the periodic function rather than the non-periodic $\Pi$.

Incidentally, your second integral in the question is wrong. It should read:

$$ c_n=\int_{-1/2}^{1/2}e^{-2\pi int}\Pi(Tt)dt $$

I'd edit, but it's more useful for you to see where the error is in your question. I'm also not convinced this integral is actually easier to solve, either.

John Gowers
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  • I think I am right. All I did was change the limits of integration. The length is the same. – fourierguy Jun 25 '13 at 16:20
  • I have to agree with Donkey_2009. $T$ wasn't mentioned anywhere until it popped up in those expressions. It's not necessarily a matter of whether you are right or wrong, but of what you are communicating. – bob.sacamento Jun 25 '13 at 16:43
  • $T$ is the period – fourierguy Jun 25 '13 at 16:54
  • @fourierguy - sorry, I thought you were making a substitution. The integral is, in fact right (though you should remove the $\Pi(t)$ from it, since it's just equal to $1$ on that range anyway). – John Gowers Jun 25 '13 at 17:05
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Fourier coefficients are initially explained and taught as having applying to periodic functions, since they involve $\sin()$ and $\cos()$ functions, and since they find great utility among periodic functions. But the process of taking the integral

$\displaystyle \int \exp(ikx) f(x) dx $

has no inherent limit in period length. It can be taken for periods as long as you like, i.e. periods approaching infinity, and can therefore be applied to non-periodic functions. It's just a matter of whether 1) the integral converges and 2) whether the numbers that pop out are useful somehow, and they often times are.

bob.sacamento
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