AngouriMath

Navigation

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

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

Method (no overloads)

Summary

A whole power of x below -1 beside a power of a binomial B = a + b x^n that is
not whole and a rational function R of x^n with a value at zero, its pole
at zero taken out by the reduction that raises the power of x by n:

int x^m B^p R(x^n) = C x^(m + 1) B^(p + 1) + int x^(m + n) B^p S(x^n),
C = R(0)/(a (m + 1)),   S(t) = (R(t) - C ((m + 1) a + b (m + 1 + n (p + 1)) t))/t,

repeated until the power of x is -1 or more, and the rest asked on its own.

Remarks

The derivative of x^(m + 1) B^(p + 1) is x^m B^p ((m + 1) a + b (m + 1 + n (p + 1)) x^n) on every branch, so the step is exact, and S is a rational function since what
it divides by t vanishes at zero. Rubi's 1.1.3.4 1/(x^3 (1 - x^3)^(2/3) (1 + x^3)) was declined: split over 1 + x^3 its terms are not elementary apart, and the
cube-root ansatz stops short of the ninth degree its denominator has. One step leaves
-(1 - x^3)^(1/3)/(2 (1 + x^3)), which is answered.
Last in the chain, and at the question asked only: asked before the scaling of the
variable and by parts, or below them, it answered six solved rows of Rubi's 1.1.3.4,
1/(x^6 (a + b x^3)^(1/3) (c + d x^3)) among them, in other spellings, three far
shorter and one twice as long. Not for the binomial alone, which Chebyshev's
theorem decides. A factor B left in the rest is taken into the power. The answer
is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

Table of contents

AngouriMath

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

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