# What is the series expansion of reciprocal of theta function $\frac{1}{\theta(z;q)}$?

"The" theta function is an ambiguous concept, but one definition I have found is:

$$\theta(z;q) = (z;q)_\infty(q/z;q)_\infty = \frac{1}{(q;q)_\infty}\sum_{k \in \mathbb{Z}}z^k q^{\binom{k}{2}} \tag{\ast}$$

I found this recently on arXiv but maybe there is a textbook. The second equation is known as the Jacobi triple product:

$$(q;q)_\infty(z;q)_\infty(q/z;q)_\infty = \sum_{k \in \mathbb{Z}}z^k q^{\binom{k}{2}}$$

Therefore it might make better sense to call the theta function this result of this triple product rather than definition ($\ast$) Before I forget $(z;q)$ is an abbreviation for a factorial called the Pochammer symbol

$$(z;q) = \prod_{i=0}^\infty (1 - z q^i)$$

My question is if there is an easy expansion for the reciprocal of the theta function. I certainly could not think of one:

$$\frac{1}{\theta(z;q)} = \frac{1}{(z;q)_\infty} \frac{1}{(q/z;q)_\infty} = (q;q)_\infty \frac{1}{\sum_{k \in \mathbb{Z}}z^k q^{\binom{k}{2}}} \tag{\ast} = \sum \;?\,?\,?$$

• There are some minus signs missing in the triple product. – ccorn Jan 1 '16 at 15:56

Andrews in [1] (equation (2.1)) sights the partial fraction expansion $${\prod_{n\gt0}\frac{(1-q^n)^2}{(1-zq^n)(1-z^{-1}q^{n-1})}}=\sum_{m\ge0}\left(\frac{(-1)^mq^{m(m+1)/2}}{1-z^{-1}q^m}-z\frac{(-1)^mq^{m(m+1)/2}}{1-zq^m}\right)\tag{*}$$ where $1\lt|z|\lt|q|^{-1}$ and $|q|\lt1$. Making the substitution $z{\to}z/q$, one obtains an expression for $1/\theta(z,q)$. The Laurent series expansion of the righthand side of $(*)$ is given in Lemma 1 of the paper and proved in Section 2. To wit one has, $${\prod_{n\gt0}\frac{(1-q^n)^2}{(1-zq^n)(1-z^{-1}q^{n-1})}}=\sum_{\substack{(N,r)\in\mathbb{Z}\\r\ge|N|}} {(-1)^{r+N}z^Nq^{(r^2-N^2)/2+(r+N)/2}}\tag{**}$$.