using sinh(x) ?

[sheldonison]

....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

How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0,
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(0))) \)

This function gives the exact same results for all values of n as the following equivalent equation:
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z+\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))]}} ) \)

Let \( k=\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))-n \), then the following equation is also exactly equivalent:
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{SuperFunction(z+k+n)) \) 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
\( \ln^{[n]}(\operatorname{SuperFunction}(0.5i*\pi/\ln(2)+k+n)) \) 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