AngouriMath
SolveAnEvenPolynomialOverASymbolicBiquadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A polynomial in x^2 over a + b x^2 + c x^4 with a symbol in it, by the
two roots inx^2 : with q = sqrt(b^2 - 4 a c) and
r = (-b ± q)/(2c) , (d + e x^2)/(a + b x^2 + c x^4) is
(d + e r_1)/(q (x^2 - r_1)) - (d + e r_2)/(q (x^2 - r_2)) , and
1/(x^2 - r) is atan(x/sqrt(-r))/sqrt(-r) for any complex r . A
higher even degree is divided down first, and an odd numerator is left to the
substitutionu = x^2 . And over x^(2k) times the biquadratic, a power of
x on both sides cancelled first: in w = x^2 the remainder R over
w^k Q is the expansion of R/Q at w = 0 up to w^(k-1) , over
w^k , plus (d + e w)/Q , so the terms in x^(-2j) go by the power
rule and the rest as above. Split inx , the repeated factor x^(2k) beside
the quartic went to the Hermite reduction or to the conjugates of a sum of two squares,
and2 a (1 - b t^2)^4/(t^4 (1 - 2 b t^2 + (a^2 + b^2) t^4)) , which the root of the
quotient makes of Rubi'scot(c + d x)^(5/2)/(a + b tan(c + d x))^(3/2) , was
declined after seconds in one spelling and answered at length in the other.
two roots in
higher even degree is divided down first, and an odd numerator is left to the
substitution
rule and the rest as above. Split in
the quartic went to the Hermite reduction or to the conjugates of a sum of two squares,
and
quotient makes of Rubi's
declined after seconds in one spelling and answered at length in the other.
Remarks
as one; over the rationals it is factored, over symbols it was declined, and Rubi's
quartic files --
even numerators to it:
that is zero as written declines, the square of a quadratic being another shape.
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