Good morning! I'm having trouble with this problem...
how can we find the exact value of
$$\sum_{n=0}^\infty e^{-n^2}.$$
Thank you in advance to anyone who can help.
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4Simply there is no closed form of this sum. – Crostul Mar 02 '18 at 18:07
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2look up jacobi theta functions – tired Mar 02 '18 at 18:09
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Wolfram Alpha says this here $$\frac{1}{2} \left(1+\vartheta _3\left(0,\frac{1}{e}\right)\right)$$ – Dr. Sonnhard Graubner Mar 02 '18 at 18:14
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Are you sure the exponent is squared? If not, then the exact value is easy to find. – Clayton Mar 02 '18 at 18:17
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@Clayton, If the exponent non squared it's clear at all that is a progression geometric series such that convergent – zeraoulia rafik Mar 02 '18 at 18:28
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@zera: I’m not sure what you said, but it sounds like you agree with me. If so, then you understand why I ask if the OP made a typo. – Clayton Mar 02 '18 at 18:42
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If you had $e^{-n^2\pi\sqrt{r}}, r\in\mathbb {Q}^{+} $ in place of $e^{-n^2}$ then you could get closed form of the sum. See https://math.stackexchange.com/a/2596065/72031 – Paramanand Singh Mar 02 '18 at 20:37
1 Answers
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Hint: This may helping you and assured that no closed form exist only it is a special case for $\theta$ and $\psi$ functions
The sum :$\displaystyle\sum_{n=0}^\infty e^{-n^2}$ is a particular case of $\psi$ function at $\displaystyle x=\frac 1 \pi$ which is defined as :$\displaystyle\psi(x)=\sum_{n=0}^\infty e^{-n^2\pi x}.$ , For more information try to check (Edwards 2001, p. 15)and satisfies the functional equation :$\displaystyle\frac{1+2\psi(x)}{1+2\psi(x^{-1})}=\frac {1}{\sqrt{x}}$ , In your case you have :$\displaystyle \frac{1+2\psi(\pi)}{1+2\psi({\pi}^{-1})}=\sqrt{\pi}$ , and for more information about this you can see :(Jacobi 1828; Riemann 1859; Edwards 2001, p. 15) .
Jaideep Khare
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zeraoulia rafik
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