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

SquareRootLogarithmAnsatz

← Back to list of classes

Description

Summary

A rational function of x and one square root y = sqrt(P(x)) of a quadratic,
a cubic or a quartic, integrated by an ansatz over logarithms of A - B y and
arctangents of A/(B y) for polynomials A and B, and a rational
part in y.

Remarks

The curve y^2 = P(x) is elliptic for a squarefree cubic or quartic P, so no
substitution makes these rational; the ones that are elementary at all -- Welz's
(1 + x)/((x - 2) sqrt(1 + x^3)), which is -(2/3) atanh((1 + x)^2/(3 sqrt(1 + x^3))),
Bronstein's x/sqrt(x^4 + 10x^2 - 96x - 71) -- are integrated by logarithms of
A - B y. The norm (A - By)(A + By) = A^2 - B^2 P is where such a logarithm
has its poles, so the A and B that can occur are the ones for which
A - By vanishes at the places the integrand has poles: at a rational pole
a with P(a) a square, A/B agrees with a branch of y to some
order there, which is a Padé approximant of the branch's series -- Welz's
(1 + x)^2/3 is the second-order Taylor polynomial of sqrt(1 + x^3) at
2, where the norm (1 + x)(2 - x)^3 has its triple root -- and at infinity
likewise for a quartic with a square leading coefficient, which is Bronstein's. Where
P(a) is the negative of a square the branch is imaginary, the conjugate pair of
logarithms is one arctangent of A/(B y), and the norm is A^2 + B^2 P.
Each Padé approximant at each place, for a few orders, is a candidate; the identity
sum c_k g_k' = f is one in Q(x)[y]/(y^2 - P), brought over one denominator
and matched coefficient by coefficient, and solved exactly over the rationals.
 An irreducible quadratic factor of the denominator is a pair of conjugate places, at
which the branch begins with a line alpha + beta x over the rationals or a
Q(sqrt(d)) of its own, lifted to any power of the factor; what such a candidate
contributes to the system is the conjugate difference of its logarithms over
sqrt(d), which is rational, so that candidates over different fields sit in one
system over the rationals. For a quadratic P the curve is rational and Euler's
substitutions answer in principle, but through a rational function whose residues can
lie in a field of degree four -- Timofeev's (3 + x)/((1 + x^2) sqrt(1 + x + x^2)) -- and here it is a logarithm and an arctangent of (1 +- x)/(sqrt(2) y). A
rational pole where P is neither a square nor the negative of one wants the field
of its square root, and is left for now. The answer is checked against the integrand at
sampled points before it is returned.
https://github.com/asc-community/AngouriMath/issues/718

Members

























Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4378 pages online