imho a core issue

[sheldonison]

one of the most important things imho is the sequence of derivatives of half-iterates of exp(x).

i have been thinking about it for a long time and its about time i ask about this.

there are some partial results but in general the question is quite open and tetration seems to be almost " immune " to standard calculus " tricks ".

in my imagination i always conjecture

tommysexp(tommyslog(x)+1/2) (around x=0) = a0^2 + a1^2 x + a2^2 x^2 + ...

where the squares indicate POSITIVE and the radius is assumed to be at least 1/3.

although the POSITIVE part seems unlikely , i am considering it.
....
tommy1729

The coefficients should ultimately settle into a pattern, determined by the nearest singularity. In the case of tommysexp(tommyslog(z)+0.5), the nearest singularity is only approximate, as increasing the number of iterated logarithms will bring the nearest singularity arbitrarily closer; but I would expect the approximate pattern could hold for dozens, or thousands, or even millions of derivatives, depending on where the half iterate is generated. For example, based on previous calculations I've done, the half iterate of 0.5 would be approximately one, where the nearest singularity would have a radius of approximately 0.46, and the pattern would hold for millions of derivatives.

I did some work iterating logarithms of superexponential of bases other than base "e", and approximating how the taylor series coefficients change as the sequence of functions converges as n increases, \( \text{basechangesexp}(z)=\lim_{n\to\infty} \log^{[n]}\text{cheta}(z+n+k) \), or in the case of tommysexp, \( \text{tommysexp}(z)=\lim_{n\to\infty} \log^{[n]}\text{superfunction2sinh}(z+n+k) \). I focused on cheta, the upper superexponential for base eta=exp(1/e), which is used for the base change sexp function, but it turns out, that iterated logarithms for cheta for n+2 logarithms behaves similar to iterated logarithms for tommysexp for n logarithms. I have some vacation time, so I will try and post the results later.
- Sheldon