AngouriMath
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SolveByTakingASquareFactorOutOfARoot(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A square root of a polynomial with a repeated factor, the factor taken out of the
root:sqrt(9 + 3x - 5x^2 + x^3) is sqrt((x - 3)^2 (x + 1)) , which is
|x - 3| sqrt(x + 1) , and the modulus is sgn(x - 3) (x - 3) -- a constant
sign on each side of the root, carried through the integration as a symbol and
written back as the sign. Timofeev's1/sqrt(9 + 3x - 5x^2 + x^3) is
sgn(x - 3) ln(...) that way, where the radical of a cubic was elliptic to
every rule that read it.
root:
sign on each side of the root, carried through the integration as a symbol and
written back as the sign. Timofeev's
every rule that read it.
Remarks
Square roots only: an odd root of a power is the power of the root on the reals
with no sign to keep. The sign is a symbol to the integration, so a symbol's
square is one and its odd powers the symbol, and the answer holds wherever the
factor is not zero, where the integrand is singular anyway. At the top only: a
substitution's own variable is one it knows the sign of.
https://github.com/asc-community/AngouriMath/issues/718
with no sign to keep. The sign is a symbol to the integration, so a symbol's
square is one and its odd powers the symbol, and the answer holds wherever the
factor is not zero, where the integrand is singular anyway. At the top only: a
substitution's own variable is one it knows the sign of.
https://github.com/asc-community/AngouriMath/issues/718
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