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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $P_n$ be the [[Legendre polynomial|QFYGJVA996]] of degree $n$.\nIts zeros $x_1, \\ldots, x_n$ are the nodes of the $n$-point [[Gauss–Legendre rule|RGGXDRCJVE]] on $[-1, 1]$.\nLet $E_{n+1}$ be the Stieltjes polynomial: the unique monic polynomial of degree $n + 1$ with\n$$\\int_{-1}^{1} P_n(x) E_{n+1}(x) x^k \\, dx = 0 \\quad \\text{for } k = 0, 1, \\ldots, n.$$\nThen:\n\n1. the zeros $y_1, \\ldots, y_{n+1}$ of $E_{n+1}$ are real, simple, lie in $(-1, 1)$, and interlace with $x_1, \\ldots, x_n$;\n2. the interpolatory rule $K_{2n+1}(f) = \\sum_{i=1}^{n} a_i f(x_i) + \\sum_{j=1}^{n+1} b_j f(y_j)$ on these $2n + 1$ nodes satisfies $K_{2n+1}(p) = \\int_{-1}^{1} p(x) \\, dx$ for every polynomial $p$ of degree at most $3n + 1$.\n\n## Remarks\n\nClause 2 follows from clause 1 by division.\nWrite $p = q \\, P_n E_{n+1} + r$ with $\\deg q \\le n$ and $\\deg r \\le 2n$.\nThe term $q P_n E_{n+1}$ vanishes at all $2n+1$ nodes and has integral $0$ by the defining property of $E_{n+1}$.\nThe remainder $r$ is integrated exactly, because an interpolatory rule with $2n + 1$ nodes is exact up to degree $2n$.\nClause 1 is harder; it was proved by Szegő for the Legendre weight, and it can fail for other weight functions.\nExample: for $n = 1$, $P_1 = x$ and the conditions give $E_2(x) = x^2 - \\tfrac35$, so the Kronrod rule uses the nodes $0, \\pm\\sqrt{3/5}$ and is the $3$-point Gauss–Legendre rule, exact up to degree $5$.\nIn practice the difference $\\abs{K_{2n+1}(f) - G_n(f)}$, where $G_n$ is the Gauss rule, serves as an error estimate that costs only $n + 1$ new function values.\n\n## Explanations\n\n### L1 human\n\nA Gauss rule samples a function at a few clever points and gives a very accurate integral, but it does not tell how accurate.\nThe Kronrod extension adds new points between the old ones, keeps the old samples, and gives a second, more accurate value.\nComparing the two values tells how large the error is.\n\n### L2 human\n\nThe $n$-point Gauss rule has [[degree of exactness|ZYT1JNT1Y0]] $2n - 1$, and a fresh $(2n+1)$-point Gauss rule would have degree $4n + 1$ but it would reuse almost none of the old nodes.\nThe Kronrod rule gives up some degree, $3n + 1$ instead of $4n + 1$, in exchange for reusing all $n$ old nodes.\nPairs such as $G_7$–$K_{15}$ and $G_{10}$–$K_{21}$ are the basis of adaptive integrators, for example in the QUADPACK library.\n",
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