I know how to expand a function $f(x)$ into a Fourier series with the period $2L$: $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos(n\pi x/L)+\sum_{n=0}^\infty b_n\sin(n\pi x/L),$$ but what if I want to expand $f(x)$ into a series of the type $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos(g(n)\pi x/L)+\sum_{n=0}^\infty b_n\sin(g(n)\pi x/L),$$ where $g(n)$ can be a slightly more complicated function of $n$? Does this even make sense?

To clarify, let me describe a recent scenario in which I've run into problems, what I've tried and where I've failed:

I have a solution the the transient, one-dimensional heat equation on the form $$\psi(x,t) = \sum_{n=0}^\infty a_ne^{-\kappa\lambda_n^2t}\cos(\lambda_nx),$$ where $$\lambda_n = (1+2n)\frac{\pi}{2L},$$ so I guess the period is here $4L$ (note that the period in the present example ($4L$) is different from the definition above ($2L$)). The initial condition is $$\psi(x,0) = \phi(x),$$ where $$\phi(x)=\frac{RL^2}{2k}\left[1-\frac{x^2}{L^2}\right].$$ Thus, I'm faced with the problem of finding the $a_n$ given the condition $$\sum_{n=0}^\infty a_n\cos(\lambda_nx) = \phi(x).$$ I've found that $$\int_{-2L}^{2L}\cos(\lambda_nx)\cos(\lambda_kx)dx = \cases{2L, \quad n = k\\0, \quad n \neq k},$$ so I figured I could just write $$a_k2L = \int_{-2L}^{2L}\phi(x)\cos(\lambda_kx)dx$$ $$\Rightarrow a_k = -\frac{8 L^2 R}{k\pi^2(1 + 2n)^2}.$$ However, inserting this, the initial condition isn't satisfied.

Is my general approach to this problem right? If not, how am I supposed to go about solving it?

  • $\begingroup$ If I understood your question correctly: Müntz's theorem (if you translate it to the language of Fourier series) gives a necessary and sufficient condition on $g(n)$ for the ability to expand any periodic function to a "$g(n)$-Fourier" series. (It needs to have enough different values.) Your example of $\lambda_n$ looks like it should work. en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem $\endgroup$ – Yoni Rozenshein Mar 19 '13 at 17:33
  • $\begingroup$ Why are you integrating on $-2L,2L$? Shouldn't it be on $-L,L$? $\endgroup$ – Alex R. Mar 19 '13 at 17:48
  • $\begingroup$ @YoniRozenshein: Thanks, but I need more than an existence theorem right now :). $\endgroup$ – andreasdr Mar 19 '13 at 18:10
  • $\begingroup$ @Alex: I gave the general definition of a Fourier series at the top (period $2L$). In my example, the period is $4L$, so the $L$ is different (I've added a clarification in the text). $\endgroup$ – andreasdr Mar 19 '13 at 18:11
  • $\begingroup$ It all boils down to what the boundary conditions are. You do not specify those, so I suspect there's something amiss there. $\endgroup$ – Ron Gordon Mar 19 '13 at 18:17

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