06/29/2010, 03:18 AM
(This post was last modified: 06/29/2010, 05:05 PM by sheldonison.)

(06/28/2010, 11:23 PM)tommy1729 Wrote:How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0,(06/27/2010, 03:31 AM)sheldonison Wrote: ....This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases.

- Sheldon

no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula.

regards

tommy1729

This function gives the exact same results for all values of n as the following equivalent equation:

Let , then the following equation is also exactly equivalent:

As n goes to infinity, k converges quickly to the approximate value, k=0.067838366 (for n=0: k=-0.0734181, for n=1: k=0.0663658, for n>=2: k=0.067838366)

Next, this equation can be used to compute TommySexp at z=0.5i*pi/ln(2), in the region where the attracting fixed point is, for increasing values of n

as n increases. My guess is that everywhere in the region around the i=0.5*pi/ln(2) value, where the Superfunction converges to the attracting fixed point, TommySexp will converge to the repelling fixed point of e^z. with all derivatives going to zero.

For example, consider TommySexp for n=0, n=2, and n=10 at i0.5pi/ln(2). For n=0, real(x)=0, img(f)=i0.905, and for real(x)=1,img(f)=i1.572. This is a well defined analytic function. The function flattens out for n=2, and by the time n=10, it is converging towards the fixed point of "e"=0.318+i1.337 (hence my guess that all derivatives go to zero). Notice how the graph of TommySexp, for n=0, n=2 and n=10 flattens out and converges to the fixed point of "e" as n increases.

- Sheldon