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SolveAFunctionOfTheRootsOfTheSecantPlusAndMinusOneInTheirDifference(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A function of p = sqrt(sec(u) + 1) and q = sqrt(sec(u) - 1), u = a + b x,
in their difference w = p - q: p + q = 2/w, so p = (w^2 + 2)/(2 w),
q = (2 - w^2)/(2 w) and sec(u) = (w^4 + 4)/(4 w^2), and
dx = -8 K w dw/(b (w^4 + 4)) for K = cot(u) p q, which is 1 or -1.
Charlwood's sqrt(sqrt(1 + sec(x)) - sqrt(sec(x) - 1)) is -8 w^(3/2)/(w^4 + 4) in w.

Remarks

w' = -b sec(u) tan(u) w/(2 p q) and K tan(u) = p q, so (K/b) w' is
-sec(u) w/2, which is -(w^4 + 4)/(8 w); K is constant wherever it is
continuous, its square being cot(u)^2 (sec(u)^2 - 1). Each step uses no more about
the roots than p^2 = sec(u) + 1 and q^2 = sec(u) - 1, so the answer holds on
every branch of either. Where the integrand is real, K changes sign only at the
multiples of 2 pi, where w = sqrt(2): the integral in w is taken from
there, so that the answer is continuous through them. Declined before, the roots of the two sums
being no substitution any rule tries. The answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

Table of contents

AngouriMath

SolveAFunctionOfTheRootsOfTheSecantPlusAndMinusOneInTheirDifference​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

SolveAFunctionOfTheRootsOfTheSecantPlusAndMinusOneInTheirDifference(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)