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  "body": "\n## Definition\n\nLet $f \\colon [-1, 1] \\to \\C$ be Lebesgue integrable, and let $P_n$ denote the [[Legendre polynomial|QFYGJVA996]] of degree $n$.\nThe *Legendre transform* of $f$ is the sequence $(\\tilde f(n))_{n \\ge 0}$ given by\n$$\\tilde f(n) = \\int_{-1}^{1} P_n(x)\\, f(x)\\, dx, \\qquad n = 0, 1, 2, \\dots$$\n\n## Remarks\n\nThe integrals converge because $\\abs{P_n(x)} \\le 1$ on $[-1, 1]$.\nLet $f \\in L^2([-1, 1])$, the [[$L^2$ space|SASP14MA1Z]].\nThen the inverse transform is the [[Fourier–Legendre series|RK2G9HN63E]]\n$$f(x) = \\sum_{n=0}^{\\infty} \\frac{2n + 1}{2}\\, \\tilde f(n)\\, P_n(x),$$\nwith convergence in $L^2([-1, 1])$; this follows from $\\int_{-1}^{1} P_m P_n\\, dx = \\frac{2}{2n+1} \\delta_{mn}$ and the completeness of the Legendre polynomials.\nIf $f$ is twice continuously differentiable on $[-1, 1]$, integration by parts gives $\\int_{-1}^{1} P_n(x)\\, \\frac{d}{dx}\\big[(1 - x^2) f'(x)\\big]\\, dx = -n(n+1)\\, \\tilde f(n)$, the rule that makes the transform useful for Legendre's equation.\n\nExample: $f(x) = x^2$.\nSince $x^2 = \\frac{1}{3} P_0(x) + \\frac{2}{3} P_2(x)$, orthogonality gives $\\tilde f(0) = \\frac{1}{3} \\cdot 2 = \\frac{2}{3}$, $\\tilde f(2) = \\frac{2}{3} \\cdot \\frac{2}{5} = \\frac{4}{15}$, and $\\tilde f(n) = 0$ otherwise.\nDirectly, $\\tilde f(0) = \\int_{-1}^{1} x^2\\, dx = \\frac{2}{3}$.\n\nNon-example: $f(x) = \\frac{1}{1 - x}$ is not integrable on $[-1, 1]$, and $\\tilde f(0) = \\int_{-1}^{1} \\frac{dx}{1 - x}$ diverges.\n",
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