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  "body": "\n## Statement\n\nLet $R$ be a commutative ring, let $h \\in R$, and let $p \\in R[x]$ be a [[polynomial|0Q9EZS90GY]].\nLet $n \\in \\N$ with $\\deg p \u003c n$, where the [[degree|0DPF6WRZW6]] of the zero polynomial is $-\\infty$.\nThen the [[$n$-th forward difference|E38801KVJ1]] of the polynomial function $x \\mapsto p(x)$ on $R$ with step $h$ is zero:\n$$\\Delta_h^n p(x) = 0 \\quad \\text{for all } x \\in R.$$\n\n## Remarks\n\nThe key step is that $\\Delta_h$ lowers the degree: for $m \\ge 1$, $(x + h)^m - x^m = \\sum_{j\u003cm} \\binom{m}{j} h^{m-j} x^j$ has degree less than $m$, and $\\Delta_h$ sends constants to $0$.\nInduction on $n$ then gives the claim.\nThe same computation shows more: if $p$ has degree $d$ and leading coefficient $a$, then $\\Delta_h^d p$ is the constant $d!\\, a\\, h^d$.\nExample: for $p(x) = x^3$ and $h = 1$, the third differences of $0, 1, 8, 27, 64$ are all $6 = 3!$, and the fourth differences are $0$.\nOver $\\R$ there is a converse: a function $f \\colon \\Z \\to \\R$ with $\\Delta^n f = 0$ agrees on $\\Z$ with a polynomial of degree less than $n$.\n\n## Explanations\n\n### L1 human\n\nWrite the values of a polynomial at equally spaced points and take differences again and again.\nFor a polynomial of degree $d$, the $d$-th row is constant and every later row is zero.\nThis is the discrete form of the fact that the $(d+1)$-th derivative of a polynomial of degree $d$ is zero.\n\n### L2 human\n\nThe fact gives a practical test: if a table of values has constant $d$-th differences, the data fit a polynomial of degree $d$.\nWith the [[expansion of iterated differences in shifts|24D1EGHP4G]] it yields identities such as $\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} k^m = 0$ for $0 \\le m \u003c n$.\nCombined with [[Newton's forward difference formula|DDHCMKHD2T]], it shows that a polynomial of degree $d$ is determined by its values at $d + 1$ equally spaced points.\nIt is then given by a finite [[Newton series|Y2HXSEA2EE]].\n",
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