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  "body": "\n## Definition\n\nLet $X$ be a set and let $f \\colon X \\to X$ be a [[function|0EPAT51DXQ]].\nA *functional square root* of $f$ is a function $g \\colon X \\to X$ such that\n$$g(g(x)) = f(x) \\quad \\text{for all } x \\in X.$$\nSo it is an [[iterative root|27C6J3YNZY]] of order $2$ of $f$.\n\n## Remarks\n\nExample: on $X = (0, \\infty)$, let $f(x) = \\frac{x}{1 + 2x}$ and $g(x) = \\frac{x}{1 + x}$.\nCheck: $g(g(x)) = \\frac{x/(1+x)}{1 + x/(1+x)} = \\frac{x}{1 + 2x} = f(x)$.\n\nNon-example: $g(x) = \\frac{x}{2}$ is not a functional square root of $f(x) = \\frac{x}{2}$, since $g(g(x)) = \\frac{x}{4} \\ne \\frac{x}{2}$ for $x \\ne 0$.\n\nCommon notations are $g = f^{1/2}$ and $g = f^{[1/2]}$; the notation hides the fact that roots are not unique.\nFor instance $x \\mapsto \\pi - x$ and $x \\mapsto x$ are both functional square roots of the identity of $\\R$.\n",
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