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  "body": "\n## Example\n\nBabbage's functional equation asks for maps $f \\colon X \\to X$ with $f(f(x)) = x$ for all $x \\in X$.\nIts solutions are exactly the [[involutions|GP56NBN9RY]] of $X$, and also exactly the [[iterative roots|27C6J3YNZY]] of order $2$ of the identity.\n\nTake $X = \\R \\setminus \\set{-1}$ and $f(x) = \\frac{1 - x}{1 + x}$.\nFirst, $f$ maps $X$ into $X$: the equation $\\frac{1 - x}{1 + x} = -1$ would give $1 - x = -1 - x$, which is impossible.\nSecond, for $x \\in X$,\n$$f(f(x)) = \\frac{1 - \\frac{1-x}{1+x}}{1 + \\frac{1-x}{1+x}} = \\frac{(1 + x) - (1 - x)}{(1 + x) + (1 - x)} = \\frac{2x}{2} = x.$$\nSo $f$ is an involution and solves Babbage's equation.\nOther solutions on suitable sets are $x \\mapsto c - x$, $x \\mapsto 1/x$ and the identity.\n\n## Remarks\n\nOn an interval $I \\subseteq \\R$, every [[continuous|0X3WGX95FE]] solution other than the identity is strictly decreasing.\nIndeed, a continuous involution is injective, hence strictly monotone.\nIf it is increasing and $f(x) \u003e x$ for some $x$, then $x = f(f(x)) \u003e f(x) \u003e x$, a contradiction; the case $f(x) \u003c x$ is the same.\nSo an increasing continuous solution is the identity.\nThe more general equation $f^n(x) = x$ asks for the iterative roots of order $n$ of the identity; for example the rotation $z \\mapsto e^{2\\pi i/n} z$ of $\\C$ is one.\n\n## Explanations\n\n### L1 human\n\nBabbage's equation asks for a rule that, applied twice, brings every point back to where it started.\nMirror reflections, the change of sign $x \\mapsto -x$ and the reciprocal $x \\mapsto 1/x$ are such rules.\nThe map $x \\mapsto \\frac{1 - x}{1 + x}$ is a less obvious one, and a short computation checks it.\n",
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