q-difference equation

Remarks

Example: let $0 < \abs{q} < 1$ and $e_q(x) = \sum_{k \ge 0} x^k / [k]_q!$ for $\abs{x} < 1/\abs{1 - q}$, where $[k]_q = (1 - q^k)/(1 - q)$ and $[k]_q! = [1]_q \cdots [k]_q$. Since the q-derivative satisfies $D_q x^k = [k]_q x^{k-1}$, one gets $D_q e_q = e_q$. Written out, this is the first-order linear $q$-difference equation $e_q(qx) = \big(1 + (q - 1)x\big) e_q(x)$.

Non-example: $y(x + 1) = x\, y(x)$, satisfied by the gamma function, relates $y$ at $x$ and at the translate $x + 1$. It involves no dilate $q^j x$, so it is an ordinary difference equation and not a $q$-difference equation.

The $q$-derivative $D_q y(x) = \big(y(qx) - y(x)\big)/\big((q - 1)x\big)$ lets one rewrite any $q$-difference equation in terms of $D_q, D_q^2, \dots$, as a $q$-analog of a differential equation. Many authors assume $\abs{q} \ne 1$, or at least that $q$ is not a root of unity.