Zeros of the deformed exponential function

Let $f(x)={\sum }_{n=0}^{\mathrm{\infty }}\frac{1}{n!}{q}^{n(n-1)\mathrm{/}2}{x}^n$ ($0<q<1$) be the deformed exponential function. It is known that the zeros of $f(x)$ are real and form a negative decreasing sequence $(x_k)$ ($k\ge 1$). We investigate the complete asymptotic expansion for $x_k$ and prove that for any $n\ge 1$, as $k\to \mathrm{\infty }$, \begin{align*} x_k=-kq^{1-k}\Big(1+\sum_{i=1}^{n}C_i(q)k^{-1-i}+o(k^{-1-n})\Big), \end{align*} where $C_i(q)$ are some $q$ series which can be determined recursively. We show that each $C_i(q)\in \mathbb{Q}[A_0,A_1,A_2]$, where $A_i={\sum }_{m=1}^{\mathrm{\infty }}{m}^i\sigma (m)q^m$ and $\sigma (m)$ denotes the sum of positive divisors of $m$. When writing $C_i$ as a polynomial in $A_0,A_1$ and $A_2$, we find explicit formulas for the coefficients of the linear terms by using Bernoulli numbers. Moreover, we also prove that $C_{i}(q)\in $\mathbb{Q}$[E_2,E_4,E_6]$,where$E_2$,$E_4$and$E_6$ are the classical Eisenstein series of weight 2, 4 and 6, respectively.