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Search for a root of a small or a large number at the size of one - #2067
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Rafael-SOWNet merged 4 commits intoOct 22, 2026
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FindGoodRoot looks for an exact root. It rounded each candidate with the zero range set to 1e-6 and accepted a rational one whose power was within 1e-6 of the base, both absolutely. A root below 1e-6 became the integer 0, and 0 ^ q is within 1e-6 of a base below it, so sqrt(2/10^20), (3/10^30)^(1/3) and sqrt(i/10^20) were 0. And the rounded candidate was what it returned as the real root when no rational one held: the root of 10^12 + 1/2 was 10^6, and (2 + i/10^8)^(1/2) lost its imaginary part. The base is scaled to the size of one by an exact power of ten and the root scaled back, so sqrt(1/10^20) is 1/10^10 exactly. A candidate is rounded only to look for a rational root, which is accepted where its power is the base at the working precision; a real root is the candidate as computed. The real cube root of a negative base stays the one given: (-2/10^21)^(1/3) is -1.26e-7, as (-8)^(1/3) is -2. Fixes #2065. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…umber-is-searched-at-the-size-of-one
…umber-is-searched-at-the-size-of-one # Conflicts: # BREAKING-CHANGES.md
The real root is the candidate as GetAllRoots computed it at the working precision, where it was made a second time from its parts for every candidate: (-27)^(1/3) allocated 8.0% more than master, sqrt(2) 1.4%. Now both are within 0.1% of master, measured per call. Part of #2065. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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October 29, 2026 00:27
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Fixes #2065.
FindGoodRootlooks for an exact root. It rounded each candidate with the zero range set to1e-6, and it accepted a rational candidate whose power was within1e-6of the base, both absolutely. Two wrong answers followed:1e-6became the integer0, and0 ^ qis within1e-6of any base below it. Sosqrt(2/10^20),(3/10^30)^(1/3)andsqrt(i/10^20)were0.10^12 + 1/2was10^6, and(2 + i/10^8)^(1/2)lost its imaginary part.sqrt(1/10^20)is now1/10^10exactly.(-2/10^21)^(1/3)is-1.26e-7, as(-8)^(1/3)is-2.ARootOfASmallNumberTestchecks values against mpmath, exact roots that stay exact, and the two complex cases. 9 of its 12 cases fail on master. The 3 that pass aresqrt(10^30),(-8)^(1/3)and(27/8)^(2/3), which are unchanged.(1 + i/10^8)^(1/2)is asserted to twelve digits. Its modulus,1 + 5e-17, is read as1within1e-16, and that is what #1944's PR changes.Results:
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura