4

Under what circumstances is the Fourier series of a function guaranteed to have a finite number of coefficients?

J. M. ain't a mathematician
  • 76,540
okj
  • 2,569
  • 6
    When the function you want to represent is a finite sum of $\sin$ and $\cos$? What conditions are you looking for? –  Jan 03 '12 at 21:52
  • 4
    Indeed, as Jonas says, on the face of it the answer is simply that this happens if and only if the function is a trigonometric polynomial, i.e., a polynomial in sine and cosine. Are you looking for something deeper than this? If so, what? – Pete L. Clark Jan 03 '12 at 21:55

1 Answers1

8

If it is annihilated by a constant-coefficient differential operator! :)

Not hard to prove, I think, under the assumption that you have a periodic function.

Edit: a "constant coefficient" differential operator is of the form $c_n{d^n\over dx^n}+c_{n-1}{d^{n-1}\over dx^{n-1}} + \ldots + c_1{d\over dx}+c_0$, where the coefficients $c_i$ are constants.

As an earlier comment noted, without the assumption of periodicity, in addition to exponentials and/or sines-and-cosines, also polynomial multiples of exponentials and sine/cosine are annihilated by suitable constant-coefficient differential operators. For example, $xe^x$ is annihilated by $({d\over dx}-1)^2$. Even more simply, $x^n$ is annihilated by ${d^n\over dx^n}$, after all. But under an assumption such as that $f(x)=\sum_k a_k\,e^{ikx}$, polynomial multiples are excluded.

paul garrett
  • 55,317