Commit fe15b0a3 authored by Raymond Toy's avatar Raymond Toy
Browse files

Derive the values on the branch cuts for atanh 

Conclusion:  atanh is continuous with quadrant Iv for x > 1 and
quadrant II for x < -1.  This, of course, is consistent with Kahan's
coutner-clowkwise continuity principla.
parent 53cf274f

src/code/irrat.lisp

+38 −1
Changes for src/code/irrat.lisp: 38 added lines, 1 removed line.
Original line number Diff line number Diff line
@@ -1449,7 +1449,44 @@ Z may be any number, but the result is always a complex." 
;; +infinity, but the following code returns approx 176 + i*pi/4. The

;; reason for the imaginary part is caused by the fact that arg i*y is

;; never 0 since we have positive and negative zeroes.



;;

;; The branch cut for atanh is on the real axis for x < -1 and x > 1.

;; Let's derive the values on the branch cut.

;;

;; atanh(z) = 1/2*(log(1+z)-log(1-z))

;;

;; For z = x, x > 1:

;;   atanh(x) = 1/2*(log(1+x) - log(1-x))

;;            = 1/2*(log(1+x) - [log(x-1) + i*arg(1-x))

;;            = 1/2*(log(1+x) - (log(x-1) + i*pi))

;;            = 1/2*(log(1+x) - log(x-1) _ i*pi)

;;            = 1/2*log((1+x)/(x-1)) - i*pi/2

;;

;; For z = -x, x > 1

;;   atanh(x) = 1/2*(log(1-x) - log(1+x))

;;            = 1/2*((log(x-1) + i*arg(1-x)) - log(1+x))

;;            = 1/2*((log(x-1) + i*pi) - log(1+x))

;;            = 1/2*(log((x-1)/(x+1)) + i*pi)

;;            = 1/2*log((x-1)/(x+1)) + i*pi/2

;;

;; For z = x - i0, x > 1

;;  atanh(z) = 1/2*(log(1+x - i0) - log(1-x+i0))

;;           = 1/2*(log(1+x) + i*arg(1+x,-0) - (log(x-1) + i*arg(1-x, +0)))

;;           = 1/2*(log(1+x) - i*0 - (log(x-1) + i*pi))

;;           = 1/2*(log(1+x) - log(x-1) - i*pi)

;;           = 1/2*log((1+x)/(x-1)) - i*pi/2

;;

;; This is the same answer we get for atanh(x), x > 1.  Hence, atanh

;; is continuous with quadrant IV along the branch cut x > 1.

;;

;; Similary, for z = -x + i0, x > 1:

;;  atanh(z) = 1/2*(log(1-x + i0) - log(1+x-i0))

;;           = 1/2*((log(x-1) + i*arg(1-x, +0)) - (log(1+x)+i*arg(1+x, -0)))

;;           = 1/2*(log(x-1) + i*pi - (log(1+x) - i0))

;;           = 1/2*(log(x-1) - log(1+x) + i*pi)

;;           = 1/2*log((x-1)/(x+1)) + i*pi/2

;; This is same answer as atanh(x), x < -1.  Thus, atanh is continuous

;; with quadrant II on the branch cut for x < -1

(defun complex-atanh (z)

  "Compute atanh z = (log(1+z) - log(1-z))/2"

  (declare (number z))