Raymond Toy pushed to branch rtoy-issue-78-unneded-code-code-in-complex-acos at cmucl / cmucl

Commits: 00d0cb6f by Raymond Toy at 2020-05-24T10:17:44-07:00 Derive the values on the branch cuts for acosh

Conclusion: on the branch cuts, acosh is continuous with quadrant I for 9 <= x < 1 and with quadrant II for x < -1. This, of course, is consistent with Kahan's coutner-clowkwise continuity principla.

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1 changed file:

- src/code/irrat.lisp

Changes:

===================================== src/code/irrat.lisp ===================================== @@ -1726,6 +1726,46 @@ Z may be any number, but the result is always a complex." (asinh (imagpart (* (conjugate sqrt-1+z) sqrt-1-z))))))))

+;; acosh(z) = 2*log(sqrt((x+1)/2) + sqrt((x-1)/2)) +;; +;; For z = x, 0 <= x < 1 +;; acosh(z) = 2*log(sqrt((x+1)/2) + sqrt((x-1)/2)) +;; = 2*log(sqrt((x+1)/2) + i*sqrt((1-x)/2)) +;; = 2*(log(1) + i*arg(sqrt((x+1)/2) + i*sqrt((1-x)/2))) +;; = 2*i*atan(sqrt((1-x)/2), sqrt((x+1)/2)) +;; = 2*i*atan(sqrt((1-x)/(1+x))) +;; +;; For z = -x, x > 1 +;; acosh(z) = 2*log(sqrt((1-x)/2) + sqrt((-x-1)/2)) +;; = 2*log((i*sqrt((x-1)/2) + 0 + i*sqrt((1+x)/2)) + 0) +;; = 2*log(i*(sqrt((x-1)/2) + sqrt((1+x)/2)) + 0) +;; = 2*(log(sqrt((x-1)/2) + sqrt((1+x)/2)) + i*arg(sqrt((x-1)/2) + sqrt((1+x)/2)) + 0) +;; = 2*(log(sqrt((x-1)/2) + sqrt((1+x)/2)) + i*pi/2) +;; = 2*log(x+sqrt(x+1)*sqrt(x-1)) + i*pi +;; +;; For z = x + i0, 0 <= x < 1 +;; acosh(z) = 2*log(sqrt((1+x)/2+i0) + sqrt((x-1)/2+i0)) +;; = 2*log(sqrt((1+x)/2)+i0 + i*sqrt((1-x)/2) + 0) +;; = 2*log(sqrt((1+x)/2) + i*sqrt((1-x)/2)) +;; = 2*(log(1) + i*arg(sqrt((1+x)/2) + i*sqrt((1-x)/2)) +;; = 0 + 2*i*atan(sqrt((1-x)/2)/sqrt((1+x)/2)) + +;; = 0 + 2*i*atan(sqrt((1-x)/(1+x)) +;; +;; This is the same value we got for acosh(x), 0 <= x < 1. Hence, +;; acosh is continuous with quadrant I on the branch cut for 0 <= x < +;; 1. +;; +;; Finally, for z = -x + i0, x > 1 +;; acosh(z) = 2*log(sqrt((1-x)/2+i0) + sqrt((-x-1)/2+i0)) +;; = 2*log(i*sqrt((x-1)/2) + 0 + i*sqrt((1+x)/2 + i0)) +;; = 2*log(i*(sqrt((x-1)/2) + sqrt((1+x)/2)) + 0) +;; = 2*(log(sqrt((x-1)/2) + sqrt((1+x)/2)) + i*arg(0, (sqrt((x-1)/2) + sqrt((1+x)/2))) +;; = 2*(log(sqrt((x-1)/2) + sqrt((1+x)/2)) + i*pi/2) +;; = 2*log(sqrt((x-1)/2) + sqrt((1+x)/2)) + i*pi +;; +;; We see that this is the same expression for acosh(z), z < -1. +;; Hence, acosh(z) is continuous with quadrant II on the branch cut x +;; < -1. (defun complex-acosh (z) "Compute acosh z = 2 * log(sqrt((z+1)/2) + sqrt((z-1)/2))

View it on GitLab: https://gitlab.common-lisp.net/cmucl/cmucl/-/commit/00d0cb6f5f931b01000bef17...