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  "body": "\n## Statement\n\nLet $M$ be an additive commutative monoid, let $G$ be an additive abelian group, let $h \\in M$, and let $f \\colon M \\to G$ be a function.\nLet $\\Delta_h^n f$ be the [[$n$-th forward difference|E38801KVJ1]] of $f$ with step $h$.\nThen for every $n \\in \\N$ and every $x \\in M$,\n$$\\Delta_h^n f(x) = \\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} f(x + kh),$$\nwhere $\\binom{n}{k}$ is the [[binomial coefficient|092GEF38YD]].\n\n## Remarks\n\nWrite $\\Delta_h = E_h - I$, where $E_h$ is the [[shift operator|91QMF4R1CY]] and $I$ is the identity.\nThe operators $E_h$ and $I$ commute, so the [[binomial theorem|0EHWPWGHB2]] gives $(E_h - I)^n = \\sum_k \\binom{n}{k} (-1)^{n-k} E_h^k$, and $E_h^k f(x) = f(x + kh)$.\nA direct proof is an induction on $n$ with Pascal's rule.\nExample: for $n = 3$ the formula reads $\\Delta_h^3 f(x) = f(x + 3h) - 3f(x + 2h) + 3f(x + h) - f(x)$.\nWith $f(x) = x^3$, $h = 1$ and $x = 0$ this gives $27 - 24 + 3 - 0 = 6 = 3!$.\n\n## Explanations\n\n### L1 human\n\nThe $n$-th difference of a function can be computed in one step from $n + 1$ equally spaced values.\nThe weights are the binomial coefficients with alternating signs, as in a row of Pascal's triangle.\nFor example, the second difference is $f(x + 2h) - 2f(x + h) + f(x)$.\n\n### L2 human\n\nThe formula is the inverse of [[Newton's forward difference formula|DDHCMKHD2T]], which writes shifted values in terms of differences.\nTogether the two formulas say that the sequences $(f(x + kh))_k$ and $(\\Delta_h^k f(x))_k$ are each other's binomial transform.\nThe second difference $f(x + h) - 2f(x) + f(x - h)$, divided by $h^2$, is the standard finite difference approximation of $f''(x)$.\nCombined with the fact that [[differences of polynomials vanish beyond the degree|EFYGR426DF]], it gives identities.\nAn example is $\\sum_{k=0}^n (-1)^k \\binom{n}{k} k^m = 0$ for $0 \\le m \u003c n$.\n",
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