Nowhere analytic superexponential convergence

[tommy1729]

well according to Bo it is not real-analytic and it doesnt have a radius , and according to mike it does have a radius.

I was not talking about \( \log^{[n]}(\exp^{[n]}(x)) \) which is for each \( n \) the identity function. Of course it is real-analytic, even entire.
I was talking about tommysexp construction and the basechange.
In each step it has a radius of convergence, which though diminishes to 0 for \( n\to\infty \).

and here you are wrong.

you say \( \log^{[n]}(\exp^{[n]}(x)) \) is entire.

it is not.

i suggest you plot \( \log^{[2]}(\exp^{[2]}(z)) \) MINUS z on the complex plane ; and see ' how entire it actually is '.

in fact even log(exp(z)) is not entire.

hint : log(exp(z)) is periodic and id(z) is not.

regards

tommy1729