using sinh(x) ?

[bo198214]

\( \operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x))) \)

Ok, first let us verify that it is indeed an iteration of exp, i.e that \( f^z(x)=\text{TommySexp_e}(z,x) \) indeed satisfies:
\( f^{v+w}(x)=f^{v}(f^{w}(x)) \) and \( f^1(x)=\exp(x) \).

neglecting some rules of properly evaluating limits we get
\( f^v(f^w(x))=\lim_{n\to\infty} \ln^{[n]} (2\sinh^{[u]}(\exp^{[n]}(\ln^{[n]}(2\sinh^{[v]}(\exp^{[n]}(x))))))=\lim_{n\to\infty} \ln^{[n]}(2\sinh^{u+v}(\exp^{[n]}(x))=f^{u+v}(x) \).

and \( f^1(x)=\lim_{n\to\infty} \ln^{[n]}(2\sinh(\exp_^{[n]}(x))=\exp(x) \)
because towards infinity \( 2\sinh \) gets arbitrarily close to \( \exp \).

Basically thats the iteration equivalent of the Abel function Lévy proposes:
\( \beta(x) = \lim_{n\to\infty} \alpha(\exp^{[n]}(x)) - \alpha(\exp^{[n]}(x_0)) \)
where \( \alpha \) is the Abel function of \( 2\sinh \) (or in Lévy's case \( \exp(x)-1 \)).

The superfunction \( \sigma \) is then (the inverse of \( \beta \)):
\( \lim_{n\to\infty} \alpha(\exp^{[n]}(x)) - \alpha(\exp^{[n]}(x_0))=y \)
\( \sigma(y)=x=\lim_{n\to\infty} \log^{[n]}(\alpha^{-1}(y+\alpha(\exp^{[n]}(x_0)))) \)
\( \sigma(y)=\lim_{n\to\infty} \log^{[n]}(2\sinh^{[y]}(\exp^{[n]}(x_0)))) \)
which is the same as Tommy's superfunction.

We can do something similar with not only \( 2\sinh \) or \( \exp(x)-1 \) but with any function that does not deviate too much from exp at infinity (i.e. all functions \( h \) such that \( \log^{[n]}(h(\exp^{[n]}(x)))\to \exp(x) \)).

as for the ROC i assume a plot for increasing n says more than a thousand words.

plot n = 1 -> 100. z = 1/2

\( \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x))) \)

I think you were not attentive when proposing to compute \( \exp^{[100]} \), already \( \exp^{[6]}(0) \) can not be computed in even sage's multiple precision arithmetic.