AngouriMath
SolveAGaussianInAReciprocal(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
An exponential of a quadratic in the reciprocal of a linear, beside a whole power of
the linear:L^m F^(A/L^2 + B/L + C) with L = c + d x . Under u = 1/L ,
dx = -du/(d u^2) , and it is -(1/d) u^(-m - 2) F^(A u^2 + B u + C) : the
Gaussian beside a power, which the table's moments answer, written back with
u = 1/L . Rubi's 2.3, f^(a + b/x^2) x^m and F^(a + b/(c + d x)^2) (c + d x)^m .
https://github.com/asc-community/AngouriMath/issues/1501
the linear:
Gaussian beside a power, which the table's moments answer, written back with
https://github.com/asc-community/AngouriMath/issues/1501
Remarks
The exponent is rewritten into u a node at a time -- b/L^k is b u^k --
rather than by substitutingx = (1/u - c)/d and asking the simplifier to see
that1/(1/u)^2 is u^2 .
rather than by substituting
that
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