Commit 00d0cb6f authored by Raymond Toy's avatar Raymond Toy
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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.
parent fe15b0a3

src/code/irrat.lisp

+40 −0
Original line number Diff line number Diff line
@@ -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))