Let the q-deformation of the exponential function be defined by

$$ e_q(z)=\sum_{n=0}^\infty{\frac{z^n}{[n]_q!}}. $$

Eq. (1.8) of this paper provides the product representation

$$ e_q(z)=\prod_{k=0}^\infty{\left[1+z q^{-k}\left(1-\frac1q\right)\right]}.\tag{*} $$

This formula shows that the roots of $z\mapsto e_q(-z)$ are given by

$$ z_k=\frac{q^{k+1}}{q-1}. $$


Are there similarly simple formulas for the roots of the derivative(s) of the q-exponential function?

Many thanks.

Edit 1:

Computing the logarithmic derivative of the product (*) we find that the zeros of the derivative of $e_q(-z)$ are the same as the zeros of

$$ \sum_{n=1}^\infty{\frac{1}{z-\frac{q^n}{q-1}}},\tag{**} $$

which can be evaluated in terms of the q-digamma function. Taking $q=2$, which is indeed the case I am most interested in, this becomes $$ \sum_{n=1}^\infty{\frac{1}{z-2^n}}. $$

The numerical value of the first root $z_0\approx2.83643\; 10564\; 48581\; 92032\; 21153\; 61015$ (thank you @AntonioVargas) is not recognized by any of the inverse symbolic calculators.

Edit 2:

Inspired by this paper I was able to compute that, for $|z|<q/(q-1)$, $({}^{**})$ can be written as $$ \frac{1}{z(q-1)}\log_q\left(1-z(q-1)\right), $$ where $$ \log_q(1-z)=-\sum_{n=1}^\infty{\frac{z^n}{[n]_q}} $$ is a series representation of the the q-logarithm.

enter image description here

Unfortunately, the smallest zero of $({}^{**})$ is larger than $q/(q-1)$, so this is not very helpful.

Question 2: Does there exist a similar reformulation of $({^{**}})$ that is valid for $q^k/(q-1)<z<q^{k+1}/(q-1)$ for $k\geqslant 2$?

The answer to this question is 'yes'. More precisely, for each integer $K\geq 1$ and $q>1$ we have $$ \sum_{n=1}^\infty{\frac{1}{z-\frac{q^n}{q-1}}}=\sum_{k=1}^K{\frac{1}{z-\frac{q^k}{q-1}}}+\frac{1}{z(q-1)}\log_q\left(1-z\frac{q-1}{q^K}\right),\quad |z|<\frac{q^{K+1}}{q-1}. $$

In particular, setting $K=1$ and $q=2$, the zero $z_0\in[2,4]$ that I am after satisfies $$ \frac{1}{1-\frac{2}{z_0}}+\log_2\left(1-\frac {z_0}{2}\right)=0. $$

  • 2
    $\begingroup$ At first glance the roots do not appear to be anything nice. Taking $q=2$, for instance, $e_q(z)$ has integer roots, but the smallest root of $e_q'(z)$ doesn't appear to be rational: $$z_0 \approx -2.83643\ 10564\ 48581\ 92032\ 21153\ 61015.$$ $\endgroup$ – Antonio Vargas Jun 19 '14 at 14:32

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