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I want to compute the residue integral for the $q$-theta function, and derive its properties. First, I'll briefly explain the definition


\begin{align} & \theta(a;q)=(a;q)(q/a;q)=\prod_{i=0}^{\infty}(1-aq^i)(1-a^{-1}q^{i+1}), \qquad (q;q) =\prod_{i=0}^{\infty}(1-q^{i+1}) \end{align} where $(a;q)=\prod_{i=0}^{\infty}(1-aq^i).$

Now I'd like to check the following properties:

$\begin{align} \theta(qz; q)&=-\frac{1}{z}\theta(z;q),\\ \theta(z^{-1};q)&=\theta(qz;q)=-\frac{1}{z}\theta(z;q), \end{align}$

and I am not sure about the last property which follows:

$\theta'(1;q)=-(q;q)^2.$

Is it true?


I tried to manipulate the equation from the definition, but even the first property does not seem easy... If you know about this function, please give me hint or explicit calculation.

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  • $\begingroup$ one can state a general shifting property of the Theta function, that is $\theta(q^n z, q)=\theta(z,q)(-\frac{\sqrt{q}}{z})^nq^{-n^2/2}$ $\endgroup$ – GGG Feb 10 '17 at 9:13
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The function you are studying comes up a lot in $q$-series. The first property of $(z;q)$, the $q$-Pochhammer symbol, is $(zq;q) = 1/(1-z)(z;q)$ and also $(q/z;q)=z/(z-1)(1/z;q)$. Using this, it is easy to deduce the first property of $q$-theta. First, using the definition of $q$-theta gives $$\theta(qz;q) = (qz;q)(1/z;q) = 1/(1-z)(z;q)(z-1)/z(q/z;q) = -\theta(z;q)/z.$$ Second, using the definition of $q$-theta again gives $$\theta(1/z;q)=(1/z;q)(zq;q)=-(1-z)/z(q/z;q)(zq;q)=-(q/z;q)(z;q)/z=-\theta(z;q)/z.$$ The third property is true, and one way to prove it is to express $q$-theta in terms of Ramanujan theta $$\theta(z;q)(q;q)=f(-z,-q/z)= -(z-1)+(z^2-1/z)q-(z^3-1/z^2)q^3+(z^4-1/z^3)q^6+\dots$$ and thus $d/dz$ and evaluating at $z=1$ both sides gives $$\theta'(1;q)(q;q)=-1+3q-5q^3+7q^6-9q^{10}+\dots=-(q;q)^3$$ where the right right is a well known result and thus proving $\theta'(1;q)=-(q;q)^2$.

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