# Asymptotics of q-Pochhammer symbol

In one of the papers I'm reading there is an asymptotic formula for q-Pochhammer symbol it is written as: $(x,q)_n \propto e^{\frac{1}{\hbar}(Li_2(x)-Li_2(xq^n))}$ where $q=e^{\hbar}$. How can I derive this formula and how do I generalize it to for example $(q^2,q^2)_n$ ? I'm interested in the limit $\hbar \rightarrow 0$, $n \rightarrow \infty$ where $q^n$ is fixed if it isn't clear.

## 1 Answer

The $q$-Pochhammer symbol $(x;q)_n=\prod_{k=0}^{n-1}(1-xq^k)$ can more generally be written as

$$(x;q)_n = \frac{(x;q)_\infty}{(q^n x;q)_\infty}$$

as you can check for yourself. The "infinite" $q$-Pochhammer symbol can be expanded as

$$(x;q)_\infty= \exp \bigg[ \frac 1 \hbar \sum_{k=0}^\infty \frac{B_k \hbar^k}{k!} \text{Li}_{2-k}(x)\bigg], \quad |x|<1$$

where we have set $q=e^\hbar$. This expansion is a small exercise which you can complete yourself: start by expanding $\log (x;q)_\infty$ in powers of $x$ using Taylor series and then some results about the Lambert series should finish the proof.

These two results prove your expression to leading order in $\hbar$. Applying it for $(q^2;q^2)_n$ is then straight-forward, but the limit $\hbar \to 0$ with $\hbar n$ constant is trickier. I hope this helps.

PS. May I ask where you read this?

• could you please provide a reference for the expansion in terms of polygamma – Mohammad Al Jamal Feb 26 '18 at 21:57