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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\C$ be given by a power series $f(x) = \\sum_{n \\ge 0} c_n x^n$ that converges for every $x \\in \\R$; for example, $f$ may be a polynomial.\nFor $h \\in \\R$, let $T_h f$ be the shifted function $(T_h f)(x) = f(x + h)$, and let $D = d/dx$.\nThen for all $x, h \\in \\R$\n$$(T_h f)(x) = \\sum_{n=0}^\\infty \\frac{h^n}{n!} \\, (D^n f)(x).$$\nIn operator notation, $T_h = e^{hD}$ on such functions.\n\n## Remarks\n\nThe right side is the [[Taylor series|0Z49YH6ZWY]] of $f$ at $x$, evaluated at the increment $h$.\nA power series that converges on all of $\\R$ is an [[analytic function|081WKEDK6C]] whose Taylor series at every point converges to it on all of $\\R$, which gives the identity.\nFor a polynomial of degree $d$ the sum stops at $n = d$, and the identity is exact algebra.\n\nExample: $f(x) = x^2$.\nThen $f + h f' + \\frac{h^2}{2} f'' = x^2 + 2hx + h^2 = (x + h)^2$.\n\nThe hypothesis of analyticity is needed.\nFor the [[smooth non-analytic function|09RTYYHXKA]] $f(x) = e^{-1/x^2}$ for $x \\ne 0$, $f(0) = 0$, all derivatives vanish at $0$, so the right side at $x = 0$ is $0$, while $f(h) \u003e 0$ for $h \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nKnowing all derivatives of a nice function at one point determines its values nearby.\nThe formula packages this as \"shifting by $h$ equals the exponential of $h$ times differentiation\".\nIt is Taylor's formula written as an identity between operators.\n\n### L2 human\n\nSince $e^{hD} e^{kD} = e^{(h+k)D}$, the identity explains why shifts compose by adding the steps.\nIn the calculus of finite differences, the forward difference $\\Delta = T_1 - \\id$ becomes $e^D - 1$, and on polynomials this gives formal identities such as $D = \\log(1 + \\Delta) = \\Delta - \\Delta^2/2 + \\Delta^3/3 - \\cdots$, a finite sum on each polynomial.\nIn quantum mechanics the same identity says that the momentum operator generates translations.\n",
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