AngouriMath
SolveByTakingARootOfAPerfectSquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A square root of a perfect square in x is the modulus:
sqrt(x^2) is |x| , which for a real x is sgn(x) x , and
sqrt(a (x + h)^2) is sqrt(a) sgn(x + h) (x + h) for a positive number
a , and a square in a power of x the same way:
sqrt(a^2 + 2 a b x^2 + b^2 x^4) is sqrt(b^2) sgn(x^2 + a/b) (x^2 + a/b) ,
the square Rubi's 1.2.2 and 1.2.3 write. Every rule that reads a root of a quadratic
assumes it is not one:
1/sqrt(1 + csch(x)^2) under u = tanh(x) is sqrt(u^2)/(u^2 - 1) ,
and the table rule forsqrt(a w^2 + b w + c) beside a linear, reached after the
partial fractions and a reciprocal, answered it with a logarithm of zero.
https://github.com/asc-community/AngouriMath/issues/718
the square Rubi's 1.2.2 and 1.2.3 write. Every rule that reads a root of a quadratic
assumes it is not one:
and the table rule for
partial fractions and a reciprocal, answered it with a logarithm of zero.
https://github.com/asc-community/AngouriMath/issues/718
Remarks
At every depth, unlike SolveByTakingASquareFactorOutOfARoot(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), since the
misreading is at depth; and the square root only, not(x^2 + 2x + 1)^(5/2) ,
which that rule writes assgn(x + 1) (x + 1)^5 at the top. Below a
substitution the variable is one whose sign the substitution knows, and a sign
written for it is a factor the rest of the search has to carry:
tanh(x)/(a + b tanh(x)^2)^(5/2) under u = e^(2x) holds
(1 + 2u + u^2)^(5/2) , and with that rewritten the search ran past its budget
where it had answered in twenty seconds. The one half-odd power taken beyond the
square root is of the bare square,(x^2)^(3/2) , which is sgn(x) x^3 :
it is what a root of(b + a u^2)/u^2 -- a + b coth(x)^2 under
u = tanh(x) -- leaves once the quotient is written apart, and nothing else
reads it.
misreading is at depth; and the square root only, not
which that rule writes as
substitution the variable is one whose sign the substitution knows, and a sign
written for it is a factor the rest of the search has to carry:
where it had answered in twenty seconds. The one half-odd power taken beyond the
square root is of the bare square,
it is what a root of
reads it.
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