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  "body": "\n## Definition\n\nLet $q \\in \\C$ with $0 \u003c \\abs{q} \u003c 1$, and let $[n]_q! = \\prod_{k=1}^{n} \\frac{1 - q^k}{1 - q}$ be the $q$-factorial, a [[q-analog|GFSDMJHHVA]] of $n!$.\nThe *$q$-exponential* is the [[power series|0MZB06ST2B]]\n$$e_q(z) = \\sum_{n=0}^{\\infty} \\frac{z^n}{[n]_q!},$$\nfor $z \\in \\C$ with $\\abs{z} \u003c \\frac{1}{\\abs{1 - q}}$.\nThe second $q$-exponential is\n$$E_q(z) = \\sum_{n=0}^{\\infty} \\frac{q^{\\binom{n}{2}} z^n}{[n]_q!}, \\qquad z \\in \\C.$$\n\n## Remarks\n\nSince $\\frac{z^n}{[n]_q!} = \\frac{((1 - q) z)^n}{(q; q)_n}$, the [[q-binomial theorem|VVT27RB46B]] 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|Y1X8ZZQ7VM]].\nThe ratio test gives the radius of convergence $\\frac{1}{\\abs{1-q}}$ of $e_q$, since $[n+1]_q \\to \\frac{1}{1 - q}$.\nWith $D_q f(z) = \\frac{f(qz) - f(z)}{(q - 1) z}$, one has $D_q z^n = [n]_q z^{n-1}$, hence $D_q e_q(z) = e_q(z)$.\n\nExample: the products show $e_q(z) E_q(-z) = \\frac{((1-q)z; q)_\\infty}{((1-q)z; q)_\\infty} = 1$.\nCheck on the first terms: $e_q(z) E_q(-z) = (1 + z + \\cdots)(1 - z + \\cdots) = 1 + 0 \\cdot z + \\cdots$.\n\nNon-example: $e_q(x) e_q(y) \\ne e_q(x + y)$ in general.\nThe 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 $q \\ne 1$.\n\nSome authors use $\\sum_n z^n / (q; q)_n$ instead, which is $e_q\\left(\\frac{z}{1-q}\\right)$.\n",
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