Zeros of the deformed exponential function

Let $f(x)=\sum_{n=0}^{\infty}\frac{1}{n!}q^{n(n-1)/2}x^n$ ($0<q<1$) 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 $n\ge1$, as $k\to \infty$, \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 $A_{i}=\sum_{m=1}^{\infty}m^i\sigma(m)q^m$ 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.