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Reduce a polynomial times a high half-odd power of a quadratic with numbers for coefficients - #1913
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Rafael-SOWNet merged 3 commits intoOct 16, 2026
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…umbers for coefficients The reduction to R Q^(k + 1/2) + K/sqrt(Q) took a numerator of degree 8 and a power of 9/2 at most, and solved its system as expressions. With numbers for every coefficient the system is now solved in exact rationals, kept in lowest terms, to degree 24 and a power of 25/2, and a coefficient written as an expression that evaluates to a rational is that rational rather than a name. Welz's 1/((3 - 2 x + x^2)^(11/2) (1 + x + 2 x^2)^5) and its ^(21/2), ^10 twin were declined: the pieces over the first quadratic were refused at the bound, and solved as expressions their values came to half a million nodes. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…f-a-quadratic-beside-a-power-of-another
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Part of #718.
Welz's
1/((3 - 2 x + x^2)^(11/2) (1 + x + 2 x^2)^5)and its^(21/2),^10twin were declined, though1/((3 - 2 x + x^2)^(9/2) (1 + x + 2 x^2)^4)is answered.SolveARationalFunctionBesideTheRootOfAQuadraticsplits them into pieces. A piece over the first quadratic is a polynomial times its half-odd power.SolveAPolynomialTimesAnOddHalfPowerOfAQuadraticreduces it toR Q^(k + 1/2) + K/sqrt(Q)by one linear solve, and three things stopped that:-2written as a product, was named for the solve as though it were a symbol. The system then ran as an elimination over expressions, and its values came to 552,490 nodes against a bound of 250,000, so it was refused. A coefficient free of symbols that evaluates to a rational is now that rational.Checks
AHighHalfPowerOfAQuadraticBesideAPowerOfAnotherIntegralTest: 3 cases.Corpus, measured against master 90c9d3e (this branch then merged 1e79b0e):
Both solve on this branch re-run alone (0.07 s and 0.2 s) and neither on master. No losses and no slower rows. Gate passed; suite 15,467 passed, 0 failed.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura