q-exponential

Remarks

Since $\frac{z^n}{[n]_q!} = \frac{((1 - q) z)^n}{(q; q)_n}$, the q-binomial theorem gives the products $e_q(z) = \frac{1}{((1 - q) z; q)_\infty}$ and $E_q(z) = (-(1 - q) z; q)_\infty$, with the q-Pochhammer symbol. The ratio test gives the radius of convergence $\frac{1}{\abs{1-q}}$ of $e_q$, since $[n+1]_q \to \frac{1}{1 - q}$. With $D_q f(z) = \frac{f(qz) - f(z)}{(q - 1) z}$, one has $D_q z^n = [n]_q z^{n-1}$, hence $D_q e_q(z) = e_q(z)$.

Example: the products show $e_q(z) E_q(-z) = \frac{((1-q)z; q)_\infty}{((1-q)z; q)_\infty} = 1$. Check on the first terms: $e_q(z) E_q(-z) = (1 + z + \cdots)(1 - z + \cdots) = 1 + 0 \cdot z + \cdots$.

Non-example: $e_q(x) e_q(y) \ne e_q(x + y)$ in general. The coefficient of $xy$ is $1$ on the left, and on the right it is $\frac{2}{[2]_q} = \frac{2}{1 + q}$, which differs from $1$ for $q \ne 1$.

Some authors use $\sum_n z^n / (q; q)_n$ instead, which is $e_q\left(\frac{z}{1-q}\right)$.