AngouriMath
SolveAPowerOverABinomialInAPowerOfASineOverItsRoots(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
The reciprocal of a + b s^n , s a sine or a cosine of a linear and a symbol in
a or b , over the roots of the binomial: with q = 2 for an even n and 1 for an odd one, m = n/q and r_k the m -th roots of
-a/b , 1/(a + b s^n) = (1/(m a)) sum r_k/(r_k - s^q) , and each term is an
arctangent oftan(y) or of tan(y/2) .
arctangent of
Remarks
Each term is written F(K u)/K with only K^2 fixed, so its derivative is the
same whichever rootK is: (atan(K tan(y))/K)' = 1/(cos(y)^2 + K^2 sin(y)^2) .
A root that makesK^2 zero or infinite -- 1/(1 + sin(y)^5) has -1 -- is
another shape, and declines.n up to 4 , and numbers for a and
b , are answered by the rules before it. Rubi's 4.1.7 and 4.2.7
1/(a + b sin(x)^5) , 1/(a - b cos(x)^6) and the like were declined.
https://github.com/asc-community/AngouriMath/issues/718
same whichever root
A root that makes
another shape, and declines.
https://github.com/asc-community/AngouriMath/issues/718
Summary
SolveAReciprocalOfABinomialInAPowerOfASineOverItsRoots(AngouriMath.Entity,AngouriMath.Entity.Variable) asked first where a power of
the function stands above or below the binomial: the rules before it in the dispatcher spent
the budget on `sinh(y)^4/(a + b sinh(y)^3)`, and the bare reciprocal they answer is left to them.
the function stands above or below the binomial: the rules before it in the dispatcher spent
the budget on `sinh(y)^4/(a + b sinh(y)^3)`, and the bare reciprocal they answer is left to them.
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