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.

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

share|improve this question
5  
When the function you want to represent is a finite sum of $\sin$ and $\cos$? What conditions are you looking for? –  Jonas Teuwen Jan 3 '12 at 21:52
3  
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 3 '12 at 21:55

1 Answer 1

up vote 7 down vote accepted

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.

share|improve this answer
    
@ Paul Garrett: Could you expound on this answer? I'm not familiar with the term "constant-coefficient differential operator". –  okj Jan 3 '12 at 22:09

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.