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  "body": "\n## Definition\n\nLet $f$ be an [[entire function|0419794R0T]] that is not constant.\nFor $r \u003e 0$, let $M(r) = \\max_{\\abs{z} = r} \\abs{f(z)}$.\nThe *order* of $f$ is\n$$\\rho = \\limsup_{r \\to \\infty} \\frac{\\log \\log M(r)}{\\log r} \\in [0, \\infty],$$\nwhere $\\limsup$ is the [[limit superior|0P8BA7T7S5]].\nBy convention, constant functions have order $0$.\n\n## Remarks\n\nEquivalently, $\\rho$ is the infimum of the numbers $\\lambda \\ge 0$ such that $M(r) \u003c \\exp(r^\\lambda)$ for all sufficiently large $r$.\nExample: for $f(z) = e^{z^2}$, $M(r) = e^{r^2}$, so $\\log \\log M(r)/\\log r = 2$ and the order is $2$.\nNon-example for a bound: $e^{z^2}$ does not satisfy $M(r) \u003c \\exp(r^{\\lambda})$ for large $r$ when $\\lambda \u003c 2$, so its order is not at most $1$.\nFor a nonconstant polynomial, $\\log M(r) \\sim d \\log r$ with $d$ the degree, so the order is $0$.\nIn terms of the Taylor coefficients of $f(z) = \\sum_n a_n z^n$, the order is $\\limsup_{n \\to \\infty} \\dfrac{n \\log n}{\\log(1/\\abs{a_n})}$, where terms with $a_n = 0$ are omitted.\nIf $0 \u003c \\rho \u003c \\infty$, the *type* of $f$ is $\\sigma = \\limsup_{r \\to \\infty} r^{-\\rho} \\log M(r)$; functions of order $1$ and finite type, or of order less than $1$, are those of [[exponential type|9P8D8YAHQS]].\n",
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