12/03/2015, 09:25 PM
(This post was last modified: 12/03/2015, 09:26 PM by sheldonison.)

(12/02/2015, 12:28 AM)tommy1729 Wrote: Too avoid confusion

R and x real and

R >= 0

X >= 0

A = analytic continuation with respect to x.

L = limit with respect to n going to oo.

Then 2sinh method for base e :

A L A ln^[n]( 2sinh^[R](exp^[n](x)) )

So then call this function TommySexp(R,x). The work I have done is with TommySexp(R,0) which is a base "e" sexp(z=R) function. While my computation algorithm differs, it gives the same results at the real axis. The resulting conjectures is that all of the Taylor series coefficients of TommySexp(R,0) converge, but the radius of convergence appears to changes in abrupt steps, with the radius of convergence getting arbitrarily small. The math is tricky, so I don't regard this as a completed proof, but its more than enough to convince me that the TommySexp is very likely not defined in the complex plane; and is nowhere analytic.

However, there is another usage, TommySexp(0.5,x) which is the half iterate. No calculations as to the radius of convergence have been done, and I don't think anything has been proven. However, we do know that TommySexp(1,x)=exp(x), so this should be studied more.

- Sheldon