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The following I saw as an exercise in a text on modular forms. I am lacking the understanding for why the following is true, but it is nevertheless astonishing:

Why is the numerical value of

$$x:=\sum_{n=1}^{\infty} \exp(-(n/10)^2)$$

so ridiculously close to the value of

$$y:=5\sqrt\pi-\frac12$$

while (and this is the truly surprising aspect) not being equal to it?

And by "ridiculously close" I mean that $x$ and $y$ agree up to the 427-th digit after the decimal point:

\begin{align} x=8. &3622692545275801364908374167057259139877472806\\ &1193564106903894926455642295516090687475328369\\ &2723327081134118121412853331180764328622113012\\ &6254685480139353423101884932655256142496258651\\ &4475413114466047689633981400087319507675739860\\ &2583500950926170092927234872474563201569608877\\ &6295310820270966625045319920380686673873757671\\ &6833994894682925918204397725582580869380029533\\ &6967158956664049274231240924510273274260978066\\ &257808237337\color{red}{62} \end{align}

They first disagree in the red digits.

Question: Can someone explain this in "simple" words? It does not have to be elementary, but I want to understand what machinery has to interact to arrive at this. And how would one even come up with such an example, or similar ones?

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    $\begingroup$ +1. I can see why it would be close (because $\frac{1}{10}(x+1/2)$ is a triangular approximation of $\int_0^{+\infty}e^{-t^2}dt=\frac{\sqrt{\pi}}{2}$) - but it is supposed to undershoot the actual integral in the range where $t\to e^{-t^2}$ is concave ($[0,\frac{1}{\sqrt{2}}]$) and overshoot it where it is convex ($[\frac{1}{\sqrt{2}},+\infty)$) and it is amazing that those two errors almost exactly cancel each other. $\endgroup$
    – user700480
    Commented Nov 7, 2020 at 9:31
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    $\begingroup$ That's because of Poisson summation formula (en.wikipedia.org/wiki/Poisson_summation_formula): the Fourier transform of $\displaystyle e^{-\frac{x^2}{100}}$ is decaying very fast. $\endgroup$
    – user436658
    Commented Nov 7, 2020 at 9:41
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    $\begingroup$ On a related note, Patterns That Eventually Fail by John Baez (with a little help from Greg Egan and Peter Borwein). $\endgroup$
    – PM 2Ring
    Commented Nov 8, 2020 at 12:06
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    $\begingroup$ @PM2Ring This is an awesome read! Thank you. $\endgroup$
    – M. Winter
    Commented Nov 8, 2020 at 13:03
  • $\begingroup$ Glad you like it. :) Wikipedia has a short article on Almost Integers, and I'm sure there's a collection of similar stuff on the XKCD forum thread connected to this comic: m.xkcd.com/217 Sadly, the forum is offline, but I guess you can find the thread on the Wayback Machine. $\endgroup$
    – PM 2Ring
    Commented Nov 8, 2020 at 13:11

1 Answer 1

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Define $$\psi(q)\equiv-\frac12+\frac12\vartheta_3(0,e^{-\pi q})=\sum_{n\ge1}e^{-\pi n^2q}$$ where $\vartheta_3(z,q)$ is Jacobi's third theta function. An application of the Poisson summation formula leads to the identity $$\vartheta_3(0,e^{-\pi/q})=\sqrt q\,\vartheta_3\left(0,e^{-\pi q}\right)$$ whose proof is outlined in Jacobi (1828). Substituting $q:=(\pi k^2)^{-1}$ yields $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac12\vartheta_3(0,e^{-1/k^2})=-\frac12+\frac12\cdot\sqrt{\pi k^2}\vartheta_3(0,e^{-\pi^2k^2})$$ so that $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac k2\sqrt\pi(1+2\psi(\pi k^2))=-\frac12+\frac k2\sqrt\pi+k\sqrt\pi\sum_{n\ge1}e^{-(\pi kn)^2}.$$ When $k=10$ the error has an order of magnitude of $$\log_{10}\left(10\sqrt\pi e^{-100\pi^2}\right)=1+\log_{10}\sqrt\pi-100\pi^2\log_{10}e=\bf-427$$ which corresponds to your observation.

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  • $\begingroup$ While I had already upvoted this, I missed the real gem which was the paper of Jacobi. The way he writes and the concluding paragraph is truly inspiring. By any chance do you have any more such papers by Jacobi related to elliptic / theta which are translated in English? $\endgroup$
    – Paramanand Singh
    Commented Nov 25, 2020 at 9:02

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