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I'd like to simplify the following expression

$$\sum_{n=1}^\infty \cos\left(\frac{n\pi x}{L}\right)e^{-a\left(\frac{n\pi}{L}\right)^2}-(-1)^{-bn\left(\frac{n\pi}{L}\right)^2}\cos\left(\frac{n\pi x}{L}\right)c^{-b\left(\frac{n\pi}{L}\right)^2}$$

Are there identities I could use to do this, something along the lines of $$\frac{2}{L}\sum_{n=1}^\infty \frac{L}{n\pi}\sin\left(\frac{n\pi x}{L}\right)=\frac{L-x}{L}$$ and $$\frac{2}{L}\sum_{n=1}^\infty \frac{L}{n\pi}(-1)^n \sin\left(\frac{n\pi x}{L}\right)=\frac{-x}{L}$$

Thomas Andrews
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N A
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  • You’ll have to define $(-1)^\alpha$ for real $\alpha.$ – Thomas Andrews Aug 29 '21 at 02:23
  • Or maybe you meant $(-1)^n$ in the first formula. – Thomas Andrews Aug 29 '21 at 02:27
  • Is there any reason to believe there is a closed formula for this? Initially, just looking at $x=0,$ it becomes related to the non-standard function $$\sum z^{n^2},$$ which has no closed formula in terms of elementary functions. – Thomas Andrews Aug 29 '21 at 02:31
  • I'm not quite clear what you mean for your first comment - a, b, c, L are all positive constants. – N A Aug 29 '21 at 03:35
  • Also, no, I'm not sure there is a closed formula. The equations are derived from a system which describes groundwater flow, d^2/dh^2 = S/T*dh/dt. When the boundaries are given as specified head, Fourier sine transforms can be used, and for this case the closed sine formulas are employed as part of the solution. I'm trying to use specified flux boundaries instead though, which needs Fourier cosine transforms. The solution has gotten quite complex; I was wondering if anyone might have ideas of a way to solve. When I sum manually everything zeros out. – N A Aug 29 '21 at 03:47
  • To clarify: You have $$(-1)^{-bn(\pi n/L)^2}$$ in your series. The exponent is not an integer. – Thomas Andrews Aug 29 '21 at 13:46

1 Answers1

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The first one is a Theta function. (The second one may also be, modulo a typo.)

See https://en.wikipedia.org/wiki/Theta_function

In particular, $\vartheta(z; t) =1+2\sum_{n=1}^{\infty}e^{i\pi t n^2}\cos(2\pi n z) $.

marty cohen
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