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A resonance whose sign is symbolic is written in j = -a/b in place of i - #1864
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Rafael-SOWNet merged 2 commits intoOct 13, 2026
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Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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 13, 2026 02:47
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Part of #718. Reopens #1852's decline with an answer.
e^(x/n) sin(x sqrt(-1/n^2))is resonant:1/n + i sqrt(-1/n^2)is zero whenn > 0, and1/n - i sqrt(-1/n^2)is zero whenn < 0. #1852 declined it because neither sign could be chosen. The rule for a polynomial times an exponential times a trigonometric function now writes Euler's identity inj = -a/binstead ofiwhen it cannot show that either term vanishes. Hereais the exponential's rate andbis the frequency. Where the case is resonant at all,j^2 = -1holds whichever signnhas, anda + j bis zero by construction, so one answer holds on both sides. Ifbevaluates to zero, the case is still declined.A logarithm's substitution produces the same integrand, so Rubi's 4.7.5 rows such as
x^m sin(a + ln(k x^n) sqrt(-(1 + m)^2/n^2))^pare answered too.Sizes, measured on master fb527ae, on 2.5.0 and on this branch:
exp(x/n)*sin(x*sqrt((-1)/n^2))exp(x/n)*sin(a + x*sqrt((-1)/n^2))sin(a + ln(k*x^n)*sqrt((-1)/n^2))x^m*sin(a + ln(k*x^n)*sqrt(-(1 + m)^2/n^2))x^m*cos(a + ln(k*x^n)*sqrt(-1/4*(1 + m)^2/n^2))^2x^2*sin(a + ln(k*x^n)*sqrt((-1)/n^2))^3x^m*sin(a + ln(k*x^2)*sqrt(-1/36*(1 + m)^2))^3The test differentiates each of these back at
n = 3and atn = -3.Rubi corpus, compared with master fb527ae:
The allocation gate passed. The suite passed: 15,293 passed, 0 failed.
The #1852 entry in BREAKING-CHANGES.md said the resonance "is declined now". It is amended to the answer, and so is the 4.7.5 entry's sentence that left a symbolic
balone.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura