AngouriMath

Navigation

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

SolveByReducingThePolynomialOverALinearFactor(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A polynomial numerator over a denominator with a power of a linear in it beside a
factor that is not a polynomial -- P(x)/((a + b x)^k S(x)) -- with the
polynomial written in powers of the linear at its root: what the power divides goes
over S alone, and each lower power over its own power of the linear.

Remarks

P(x) = p_0 + p_1 (a + b x) + ... + (a + b x)^k Q(x) is a polynomial identity,
its coefficients the Taylor coefficients of P at -a/b over the powers
of b, so the integrand is Q/S + sum p_j /((a + b x)^(k-j) S) exactly and
everywhere. (A + B x + C x^2 + D x^3)/((a + b x) sqrt(c + d x)) was answered
through the substitution u = sqrt(c + d x) term by term, a cubic in
u^2 - c over a symbolic quadratic in u each time, in a hundred kilobytes
of piecewise that did not evaluate within the corpus's budget; reduced, it is a
quadratic over the root, three powers, and one p_0/((a + b x) sqrt(c + d x)).
Only beside a factor that is not a polynomial in x: a quotient of polynomials
is the partial fractions' and they do this and more.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

SolveByReducingThePolynomialOverALinearFactor​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

SolveByReducingThePolynomialOverALinearFactor(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)