🚀 go-pugleaf

On Wed, 23 Nov 1994 JohnHanna@aol.com wrote:

> For an even more startling result, try graphing
>
>                 y = abs(sin^-1(x))    in Zdecm
>
> Can anyone explain why this function is continuous but not differentiable at
> x={-1,1}?

here's my response, which is nowhere near as clear as I had intended, but
I hope interested readers can decipher it.

As noted previously, for complex numbers x+iy,

sin(x+iy) = sin(x)cosh(y)+icos(x)sinh(y).

This is periodic with period 2Pi, in more than the real
sense:  sin(z+2Pi)=sin(z) for all complex numbers z.

In the complex plane, then, the line Re(z)=Pi/2  or  z=Pi/2+iy has
sin(z) = cosh(y).

So to get values of sine from -1 to 1, do the usual thing.  To get
values greater than 1, let x=Pi/2 and let y decrease from 0.  To
get values less than 1, let x=-Pi/2 and let y increase from 0 (in
both cases, the y's could increase OR decrease, but this is the
scheme TI seems to have chosen, and it also appears to be fairly
standard).  It helps keep arcsine an odd function.

Thus the following figure in the complex plane maps, under the sine
function, onto the real axis:

         |
         |               (The corners are at -Pi/2 and Pi/2; the
         |                horizontal segment thus lies on the real
          ------          axis).
                |
                |
                |

Anyway, this is the image of the real line under the arcsin, as
implemented on the TI-85.  I think this answers why abs((sin^-1)(x))
is continuous but not differentiable as graphed on the TI-85.

Then the inverse sine does two different things, depending on the
range:  We get the usual arcsin for inputs in [-1,1]. For
(sin^-1)(w) with w>1, we must solve sin(Pi/2-iy)=cosh(y)=w for y,
then Pi/2 - iy is our output.

So finally, what do you see when you graph abs(sin^-1(x)) on the
TI-85?  on [-1,1] we see |arcsin(x)| with the "regular" real-valued
arcsin.  on [1,infinity) we see sqrt((Pi/2)^2+((cosh^-1)(x))^2).
On (-infinity,-1] we see sqrt((Pi/2)^2+(cosh^-1)(-x))^2).

Closing note:  A universal definition for the inverse sine is

(sin^-1)(z) = -iLog(iz + (1-z^2)^(1/2) ).

This is of course multi-valued; if you take the principal branch
of both the Log and the root, I believe you get what TI has for
Re(z)>=0.  For Re(z)<0, use (sin^-1)(-z) = -(sin^-1)(z) (this last is a
guess).

Regards,

Mark Janeba			(503)370-6123
Dept. of Mathematics		e-mail:  mjaneba@willamette.edu
Willamette University		"Speaking for myself, not the university"
Salem, Oregon 97301-3922 USA