# Which theta function is $\theta(x;q) = (x;q)(q/x;q)$?

The physics paper I am reading very non-chalantly defines the theta function as

$$\theta(x;q) = (x;q)(q/x;q) \hspace{0.5in} \tilde{\theta}(x;q) = x^{-1/2}(x;q)(q/x;q)$$

where they are using the $q$-Pochammer symbol. This notation is a bit compressed it reads: $$\theta(x;q) = \prod_{i=0}^\infty( 1 - xq^i) \prod_{i=0}^\infty( 1 - (q/x)q^i)$$

There are different versions of the Jacobi Triple Product Identity floating around. For example:

$$\prod_{n > 0} (1 + q^{n-\frac{1}{2}}z)(1 + q^{n-\frac{1}{2}}z^{-1}) = \left( \sum_{l \in \mathbb{Z}} q^{\frac{l^2}{2} }z^l \right) \prod_{n > 0} \frac{1}{1-q^n}$$

However this doesn't seem to match up with the combination of q-Pochhammer symbols listed above.

How to expand $\theta(x;q)$ as an infinite series and recognize the infinite product? Wolfram Alpha was inconclusive.

• Setting $x=-q^{1/2}z$ gets you further, but the Euler factor $(q; q)$ is then still missing; and the original theta functions tend to use $q$ (and $z$) squared. If only ratios of theta functions with the same $q$ are of interest, this shortcut works, but it overloads the notion of theta function even further. By the way, here are some helpful formulae. – ccorn Jun 4 '15 at 21:07

A fairly standard form of one theta function is: $\theta_4(u,q):=\sum_{n=-\infty}^\infty (-1)^n \cos(2nu)\;q^{n^2},$ but also $\theta_4(u,q^{1/2})=(q;q)(x;q)(q/x;q)\;$ where $\;x=q^{1/2}e^{2iu}.$ Thus $\;\theta_4(u,q^{1/2})/(q;q)=(x;q)(q/x;q).\;$ I am not sure where the $x^{-1/2}$ comes from in $\tilde{\theta}(x;q)$.