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  "body": "\n## Definition\n\nLet $X$ be a set, let $f \\colon X \\to X$ be a [[function|0EPAT51DXQ]], and let $n \\ge 1$ be an integer.\nAn *iterative root of order $n$* of $f$ is a function $g \\colon X \\to X$ such that\n$$g^n = f,$$\nwhere $g^n = g \\circ \\cdots \\circ g$ is the $n$-th [[iterate|AYVTR2ES5B]] of $g$.\n\n## Remarks\n\nExample: on $X = \\R$, the map $g(x) = x + 1$ is an iterative root of order $3$ of $f(x) = x + 3$.\nCheck: $g^2(x) = x + 2$ and $g^3(x) = x + 3 = f(x)$.\n\nNon-example: $g(x) = 3x$ is not an iterative root of order $3$ of $f(x) = x + 3$, since $g^3(x) = 27x \\ne x + 3$.\n\nThe roots of order $2$ are the [[functional square roots|E7890PHMGP]].\nRoots are rarely unique: both $2x$ and $-2x$ are iterative roots of order $2$ of $4x$.\nSome maps have no continuous root of a given order.\nIf $g \\colon \\R \\to \\R$ were continuous with $g^2(x) = -x$, then $g$ would be injective, hence strictly monotone.\nThen $g^2$ would be increasing, but $x \\mapsto -x$ is decreasing.\nFor odd $n$ the same map has the root $g(x) = -x$, since $(-1)^n = -1$.\n",
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