AngouriMath

Navigation

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

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

Method (no overloads)

Summary

x^(n - 1) g(x^n) with a symbolic n, which is g(u)/n under
u = x^n: the power in front is the derivative of the power inside up to the
constant n, and the rest mentions x only as x^n. Rubi's 6.5.2 and
6.6.2 -- (e x)^(n - 1) (a + b sech(c + d x^n))^p -- write the power in front as
(e x)^(n - 1), which is e^(n - 1) x^(n - 1) for a positive e, and
the answer says so. A concrete n is the general substitution's; a symbolic
one was nobody's, and x^(n - 1) e^(x^n) was declined.
https://github.com/asc-community/AngouriMath/issues/718

Remarks

x^(k n - 1) with k half an odd number is the same substitution by half
the power: under u = x^(n/2), x^(k n - 1) dx is (2/n) u^(2k - 1) du and g(x^n) is g(u^2). Beside an exponential of x^n that is a moment
of the Gaussian: Rubi's 2.3, f^(a + b x^n) x^(-1 + 5n/2), and 6.1.3's
x^(-1 + n/2) sinh(a + b x^n).
https://github.com/asc-community/AngouriMath/issues/1501

AngouriMath

Table of contents

AngouriMath

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

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