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Answer a family with a whole-number parameter - #2205
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sin(x) = 1/2 in x and y was declined. Its families come from the periodic root pi/6 + 2 pi n_1, and the check that each family holds pinned n_1 at 1.37, where it does not, and asked Simplify, which knows nothing about n_1. The non-square solver now pins an n_1 a family brings at a whole number, and decides the residual as forall n_1 in ZZ : residual = 0. The quantifier establishes that n_1 is whole, and a new reader takes a whole number of periods out of a trigonometric function's argument under it: sin and cos lose a whole multiple of 2 pi, tan and cot one of pi, as Complex.sin_add_int_mul_two_pi and Complex.tan_add_int_mul_pi have it. The reader is asked only under a quantifier over whole numbers, and outside one nothing changes. sin(x) - 1/2 in x, y is [[pi/6 + 2 pi n_1, t_1], [pi - pi/6 + 2 pi n_1, t_1]], and forall n in ZZ : sin(x + 2 pi n) = sin(x) is True. Part of #212 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Codecov Report❌ Patch coverage is Additional details and impacted files@@ Coverage Diff @@
## master #2205 +/- ##
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Coverage 86.57% 86.57%
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Files 301 301
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Branches 19367 19410 +43
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- Misses 4874 4878 +4
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Part of #212: the second of the three items left there, families with an integer parameter.
sin(x) = 1/2inxandywas declined. Its families come from the periodic rootpi/6 + 2 pi n_1, and the check that each family holds pinnedn_1at1.37, where it does not hold, and askedSimplify, which knows nothing aboutn_1.n_1that a family brings at a whole number, and decides the residual asforall n_1 in ZZ : residual = 0.QuantifierFactstakes a whole number of periods out of a trigonometric function's argument.sin,cos,secandcsclose a whole multiple of2 pi, andtanandcotone ofpi, asComplex.sin_add_int_mul_two_piandComplex.tan_add_int_mul_pihave it. It also finds the multiple where the argument is gathered into one product,tan(pi * (1/4 + n)).sin(x + pi n)keeps itspi n.sin(x) - 1/2inx, ynull[[pi / 6 + 2 * pi * n_1, t_1], [pi - pi / 6 + 2 * pi * n_1, t_1]]cos(x) - 1inx, ynull[[2 * pi * n_1, t_1]]tan(x) - 1inx, ynull[[pi / 4 + pi * n_1, t_1]]forall n in ZZ : sin(x + 2 pi n) = sin(x)Trueforall n in ZZ : tan(pi/4 + pi n) = 1TrueStill open on #212: the family of
sqrt(y) - x, whose check needssqrt(t^2) = tunder the sector condition, and the parameters' names in the result.Tests:
PeriodByFactsTest(7 decided, 3 not claimed, and nothing assumed outside a quantifier) andAFamilyWithAWholeNumberParameterIsAnswered. The latter checks each family at whole values ofn_1and other values of the rest. Full suite 17838/0, interactive trio, AOT smoke 64/0, all frameworks build.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura