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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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Mikalai Parshutsich
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    That can be expressed in terms of the Jacobi theta function: https://en.m.wikipedia.org/wiki/Theta_function#Auxiliary_functions. That’s about as close to a closed form as you can get – Jacob Nov 09 '18 at 03:57
  • Jakobi functions work well when the lower bound of summation is $-\infty$. Instead I'm interested in sum starting from 1 – Mikalai Parshutsich Nov 09 '18 at 04:01
  • 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. – pregunton Nov 16 '18 at 19:53
  • @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? – Mikalai Parshutsich Nov 20 '18 at 11:53
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    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 $-1+\sum{n=0}^\infty q^{-n^2}z^n = \sum_{n=1}^\infty q^{-n^2}z^n$. – pregunton Nov 20 '18 at 13:14
  • Since I haven't found enough information about this particular function (it could happen not because of lack of information but just because I looked into wrong sources) I will accept your answer and give all the bounty if you express $_1\phi_2$ in terms of $_r\phi_s$ where $r>s$. Those functions are widely studied... I hope that such expression do exist! – Mikalai Parshutsich Nov 20 '18 at 16:01

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