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Write a biquadratic's terms by the sign of each root where both roots are real
Over a + b x^2 + c x^4, 1/(x^2 - r) was written atan(x/sqrt(-r))/sqrt(-r)
for every root r. Where the discriminant is negative the roots are
conjugate and the two arctangents sum to a real function, but where it is
a positive number times even powers and the coefficients are real, both
roots are real, and for a positive root the arctangent's argument is
imaginary: on its cut wherever x^2 > r, where its value is decided by the
sign of a rounding residue. There each term is now written by the sign of
its root, as the quadratic rule writes 1/(x^2 - r): the arctangent for a
negative root and ln((x - sqrt(r))/(x + sqrt(r)))/(2 sqrt(r)) for a
positive one. The discriminant's square root is taken by halving its
powers -- any root serves, the other swapping the two -- since
sqrt(4 b^2) is 2 |b|, past which no root's sign can be read:
cosh(x)^2/(a + b sech(x)) has -1 for a root with 2 b, and came to 6,909
characters with 2 |b|, 1,329 now. Where the roots' product a/c is a
positive number times even powers too, the two share a sign, that of
their sum, and one piecewise on it holds both terms. Over coefficients
with the imaginary unit in them the roots are not real whatever the
discriminant, and the arctangents are kept: Rubi's 1.1.1.2
1/((a - i a x)^(3/4) (a + i a x)^(1/4)) was answered NaN by a piecewise
none of whose arms held.
Rubi's 7.4.2 e^(3 acoth(a x)) sqrt(c - c/(a x))/x is
u^4/((a - u^2)(2 a - u^2)) under u = sqrt((a x + 1)/x), discriminant a^2.
The rule that writes each linear once checks its arms in interval
arithmetic, and between 0 and 1/a both arctangents were on their cuts, so
it declined and the answer that remained had no value there.
Part of #718.
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura1 parent fc5d732 commit 5dc25a6
3 files changed
Lines changed: 175 additions & 5 deletions
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