4
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In this question the summation goes from $-\infty$ to $\infty$ and the answer has a pretty "good" closed form.

Now I came across the sum $\sum_{n=1}^\infty q^{-n^2} z^n$ where $|z|<1$ and I don't know how to crack it.

Please help me!

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    $\begingroup$ That can be expressed in terms of the Jacobi theta function: en.m.wikipedia.org/wiki/Theta_function#Auxiliary_functions. That’s about as close to a closed form as you can get $\endgroup$ – Jacob Nov 9 '18 at 3:57
  • $\begingroup$ Jakobi functions work well when the lower bound of summation is $-\infty$. Instead I'm interested in sum starting from 1 $\endgroup$ – Mikalai Parshutsich Nov 9 '18 at 4:01
  • $\begingroup$ The sum can be expressed in terms of basic hypergeometric functions, concretely as $_0\phi_1 \left[0;\frac{1}{q},\frac{z}{q}\right] -1$. I don't know if a simpler form exists. $\endgroup$ – pregunton Nov 16 '18 at 19:53
  • $\begingroup$ @pregunton I've read some articles about basic hypergeometric functions but I didn't understand what the general form of $_0 \phi_1$ looks like. Could you express it as a sum? $\endgroup$ – Mikalai Parshutsich Nov 20 '18 at 11:53
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    $\begingroup$ Sorry, I made a mistake, what I meant is $_1\phi_2$. We have $-1+{}_1\phi_2\left[ \begin{matrix} 1/q\\0\;\;0 \end{matrix}\:;\frac{1}{q},\frac{z}{q}\right] = -1+\sum_{n=0}^\infty\frac{(1/q;1/q)_n}{(0,0,1/q;1/q)_n}\left((-1)^n \frac{1}{q^{n \choose 2}} \right)^{1+2-1}\left(\frac{z}{q}\right)^n$. The first fraction equals 1 because the $q$-shifted factorials cancel (the definition of $q$-shifted factorial is given in the article I linked), and simplifying the remaining terms we end up with $-1+\sum_{n=0}^\infty q^{-n^2}z^n = \sum_{n=1}^\infty q^{-n^2}z^n$. $\endgroup$ – pregunton Nov 20 '18 at 13:14

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