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  "body": "\n## Definition\n\nLet $U \\subseteq \\C$ be an open set containing $0$, and let $f \\colon U \\to \\C$ be [[holomorphic|0NAHXR1FRH]] with $f(0) = 0$.\nLet $\\lambda = f'(0)$ and assume $0 \u003c \\abs{\\lambda} \u003c 1$.\nLet $V$ be a neighbourhood of $0$ on which the iterates $f^n$ of $f$ are defined and tend to $0$.\nThe *Koenigs function* of $f$ is the function $\\varphi \\colon V \\to \\C$ given by\n$$\\varphi(z) = \\lim_{n \\to \\infty} \\frac{f^n(z)}{\\lambda^n},$$\nwhere $f^n$ is the $n$-th [[iterate|AYVTR2ES5B]] of $f$.\n\n## Remarks\n\nThe limit exists, uniformly near $0$, and $\\varphi$ is holomorphic with $\\varphi(0) = 0$ and $\\varphi'(0) = 1$.\nIt solves [[Schröder's equation|VTJV7JVJEA]] $\\varphi \\circ f = \\lambda \\varphi$, since replacing $z$ by $f(z)$ shifts the index $n$ by one.\nIt is the unique solution with these normalisations, by the [[Koenigs linearization theorem|G02NBNN6CJ]].\n\nExample: $f(z) = z / (2 - z)$ on the unit disc, with $\\lambda = f'(0) = 1/2$.\nFor $\\varphi(z) = z / (1 - z)$ one has $\\varphi(f(z)) = \\frac{z/(2-z)}{(2 - 2z)/(2 - z)} = \\frac{z}{2(1 - z)} = \\frac{1}{2} \\varphi(z)$.\nSo $\\varphi(f^n(z)) = 2^{-n} \\varphi(z)$, and since $f^n(z) \\to 0$ and $\\varphi(w) / w \\to 1$ as $w \\to 0$, the limit of $2^n f^n(z)$ is $\\varphi(z)$.\n\nNon-example: for $f(z) = z^2$ one has $\\lambda = 0$, so the quotient $f^n(z) / \\lambda^n$ is not defined, and $f$ has no Koenigs function.\n\nWhen $f$ is a univalent holomorphic map of the unit disc into itself with $f(0) = 0$ and $0 \u003c \\abs{f'(0)} \u003c 1$, the limit exists on the whole disc and $\\varphi$ is univalent there.\n",
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