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A Fourier series arising in perturbation theory in quantum mechanics is

$$\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} \, .$$

where $n$ is an odd positive integer and $n$ runs through all odd positive integers other than $n$. (The numbers are odd so that the Fourier terms are zero at $\pm a$.)

I have no idea what kind of function produces this series. Is it familiar to anyone?

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3 Answers 3

Taking $n=1$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{1-m^2} = \frac{1}{4}(\pi(1-2x)\sin(\pi x) - \cos(\pi x))$$

Taking $n=3$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{9-m^2} = \frac{1}{36}(3\pi(1-2x)\sin(3\pi x) - \cos(3\pi x))$$

Following this my hunch would be for $a= \frac{1}{2}$

$$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{n^2-m^2} = \frac{1}{4n^2}(n\pi(1-2x)\sin(n\pi x) - \cos(n\pi x))$$

Playing around a bit more with wolfram alpha, my new guess is

For $x \in [0,a]$,

$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$

and For $x \in (-a,0]$,

$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(-1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$

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This, or something similar, is the Fourier series for the fundamental solution for the wave operator (calling time "y") on a product of two circles: $(D_x^2-D_y^2)u=\delta$ (with periodic delta having Fourier expansion with all coefficients 1). The Fourier series for $\delta$ converges in the $-1-\epsilon$ Sobolev space (the point being to legitimize these manipulations!). Since the wave operator is not elliptic, we have no a-priori assurance that the solution $u$ is in any better Sobolev space than is $\delta$, but it may be so by accident.

The manipulations suggested in the other answers are legitimizable as reflecting limits taken in negatively-indexed Sobolev spaces.

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Not by itself, but something similar. The fundamental solution of the wave operator should not depend on only odd values of frequency, and the fundamental solution should be a Fourier series in both $x$ and $y$ variables, so it should be a double summation. For a fixed $n$, one may think that this is more similar to the solution of the reduced wave equation; but we still have the problem with the frequency support. –  Willie Wong Jun 26 '11 at 0:20

You can find a differential equation for the expression by differentiating it twice. Just to get rid of the annoying factors, let's define $z={\pi x}/{2a}$. Then

$$\frac{d^2}{dz^2}\sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz) = \sum_{\substack{m \; odd \\ m\neq n}} \frac{-m^2}{n^2 - m^2} \cos (mz) = $$ $$= \sum_{\substack{m\; odd \\ m\neq n}} \cos (mz) - n^2 \sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz)\; .$$

Or

$$f''(z) + n^2 f(z) = \sum_{\substack{m \; odd \\ m\neq n}} \cos (mz) = -\cos (nz) \; .$$

The sum of cosines should be easy to perform (replace the cosines by complex exponentials, use the geometric summation formula and take the real part), and solving the differential equation is also not difficult.

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Your sum of cosines doesn't converge, as you can see by checking $z=0$. I haven't yet solved the differential equation; it's possible that $f$ might be better behaved, but seems unlikely. –  Jacob Greenfield Feb 20 '11 at 22:06
    
That doesn't matter, just manipulate it formally. I just checked that Sivaram's guess satisfies the equation. –  Raskolnikov Feb 20 '11 at 22:15
    
I suppose you got that $$\sum_{\substack{m \; odd \\ m\neq n}} \cos (mz) = -\cos(nz)$$ from WolframAlpha summing the whole thing (including $m$) to zero? No offense, but even I can tell something's wrong there. The series doesn't converge, so Mathematica returns nonsense. I'll check Sivaram's solution as well. –  Jacob Greenfield Feb 20 '11 at 22:41
    
No, I just replaced the cosines by complex exponentials, used the geometric summation formula and then took the real part. If you're that worried about convergence, it's easy to fix that. Just multiply each term of your original sum by $\theta^m$ with $\theta<1$. Then work everything out as I did. Convergence should then be no problem. In the end result, take $\theta \to 1$ where convergence again will be no problem. –  Raskolnikov Feb 20 '11 at 23:13
    
@Jakob: Don't worry about convergence of $\sum_{m \; odd} \cos (mz)$. Just wave your hands and say, "It's a delta function at zero." –  GEdgar Jun 26 '11 at 0:23

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