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  "body": "\n## Summary\n\nThe $q$-exponential $e_q(z) = \\sum_{n \\ge 0} z^n / [n]_q!$ is the [[q-analog|GFSDMJHHVA]] of the [[exponential function|0J0FCWV32D]].\nIt is the solution of $D_q f = f$, $f(0) = 1$, for the [[q-derivative|PMJT3FZE0Y]] $D_q$, just as $e^z$ solves $f' = f$.\nBy the [[q-binomial theorem|VVT27RB46B]] it is an infinite product, and it tends to $e^z$ as $q \\to 1$.\nA second $q$-exponential $E_q$ satisfies $e_q(z) E_q(-z) = 1$, which replaces the rule $e^z e^{-z} = 1$.\n\n## Explanations\n\n### L1 human\n\nReplace each factorial $n!$ in the series of $e^z$ by its $q$-analogue $[n]_q! = [1]_q [2]_q \\cdots [n]_q$, where $[k]_q = 1 + q + \\cdots + q^{k-1}$.\nThe result is the $q$-exponential.\nIts first terms are $1 + z + \\frac{z^2}{1 + q} + \\cdots$, which become $1 + z + \\frac{z^2}{2} + \\cdots$ at $q = 1$.\nMany familiar rules of the exponential survive in a modified form, but $e_q(x) e_q(y) = e_q(x + y)$ fails in general.\n",
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