AngouriMath

Symbolics for .NET


Table of contents

AngouriMath

Navigation
Home Quick start Wiki Almanac Research What's new Try Use Angouri.org GitHub Report a bug Donate Contacts

← Back to list of members

SolveALinearOverALinearBesideTheRootOfAQuarticByTheSquareOfTheNumerator​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

A linear over a linear beside the reciprocal of the square root of a cubic or a quartic,
L/(M y) with y = sqrt(P), where gamma P - mu M W = L^4 for
W = 4 L' P - L P' and constants gamma and mu: its integral is
(2 mu/c) artanh(L^2/(c y)) for c^2 = gamma, which is
-(2 mu/s) atan(L^2/(s y)) for s^2 = -gamma. Welz's
(1 + x - sqrt(3))/((1 + x + sqrt(3)) sqrt(x^4 + 4 sqrt(3) x^2 - 4)).

Remarks

u = L^2/(c y) has u' = L W/(2 c P y) and 1 - u^2 = (gamma P - L^4)/(gamma P),
so u'/(1 - u^2) = c L W/(2 y (gamma P - L^4)), which is c L/(2 mu M y) where
the coefficients agree. That uses no more about the root than y^2 = P, so it holds
on either branch, and for c = i s the inverse hyperbolic tangent is
-i atan(L^2/(s y)). The argument holds the square of L because the integrand
vanishes where L does, as the derivative of a function of L^2/y does. The
identity is five coefficients linear in gamma and mu.
 The ansatz over logarithms of A - B y takes its places at rational poles, and
declined Welz's, whose pole is -1 - sqrt(3); the rule for a cubic binomial reads
no quartic, and is asked first. The answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4862 pages online