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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $g:R \rightarrow R$ be continuous and $ 2\pi$-periodic, let $m \in N$. How many solution in class of $m$-times continuously differentiable $2\pi$-periodic functions has equation $$f^{(m)}=g ?$$

Edit. Obviously, if $f$ is a solution in this class and $C$ is a constant then $f+C$ is also a solution. Are there another solutions?

share|cite|improve this question
More generally if $f$ is a solution and $p$ a polynomial of degree less than $m$ then $f + p$ is a solution, and conversely if $f_1, f_2$ are two solutions then $f_1 - f_2$ is a polynomial of degree less than $m$. So once you find a single solution (e.g. by the method below) you have all solutions. – Qiaochu Yuan May 3 '12 at 15:55
up vote 2 down vote accepted

If $g$ is $2\pi$-periodic then you can use Fourier series and write $$ g(x)=\sum_{k=-\infty}^\infty g_ke^{i k x} $$ and so also $$ f(x) = \sum_{k=-\infty}^\infty f_ke^{i k x}. $$ By direct substitution you get $$ \sum_{k=-\infty}^\infty (ik)^m f_k e^{i k x}=\sum_{k=-\infty}^\infty g_k e^{i k x} $$ then, all the solutions that can be cast in the form $$ (ik)^mf_k=g_k $$ are periodic solutions of the given equation.

share|cite|improve this answer

Your Answer


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.