slog_b(sexp_b(z)) How does it look like ?

[sheldonison]

Im not very confident about how to prove periodicity of functions that are not analytic !
As often " nonanalytic calculus " has less " tools " then " analytic calculus " so I find it harder.

Has the periodicity for the base change been confirmed ? Or is it not ?

I haven't seen or written a rigorous proof of coo nowhere analytic functions, but I am working on a rigorous proof for a related coo nowhere analytic problem, iterated logarithm of x^^n. I still have some gaps, but I know the basic flow of the proof, and I will post it sometime in the next few months.

As to your original question,
For f=b^z, with fixed point L=b^L, \( b^{L+\delta}=L\times(1+\delta\ln(L)), \lambda=\ln(L), \text{period}=\frac{2\pi i}{\ln(\lambda)} \)

So, there if \( \alpha(z) \) is the Abel function for f=b^z, then the
iterated function from the Schroeder equation fixed point can be represented generically as follows
\( \alpha^{-1}(z) = f^z = L + \sum_{n = 1}^{\infty} a_n\exp( z \ln(\lambda)) \)

As \( \Im(z) \) gets aribrarily large, the function can be well approximated by \( \alpha^{-1}(z) \approx L+a_1\exp\(z \ln(\ln(L))) \). As imag(z) increases, \( \text{sexp}_b(z)\approx \alpha^{-1}(z+k+n\text{Period}) \), where we can solve for k easily enough.

The key is that the pseudo period for sexp(z), 4.4+1.05i for base e, is much larger than the period of the 1-cyclic kneser \( \theta \) mapping, so the 1-cyclic mapping converges to the constant k much quicker than sexp(z) converges to L.
\( \text{sexp}_b(z)=\alpha^{-1}(z+ n\text{Period}+\theta(z)) \)

So \( \alpha(\text{sexp}(z)) = z + \theta(z) \)

If the slog(sexp) is well behaved as imaginary(z) gets larger, decaying to a constant +z, than the answer to Tommy's original question is as follows, noting that theta(z) is analytic in the upper half of the complex plane, with singularities at the real axis.
\( \text{slog}(\text{sexp}(z)) \approx z + k + n\text{Period} + \theta (z) \)
- Sheldon