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

[sheldonison]

....I am somewhat puzzled by slog_b(sexp_b(z)).

For clarity b is the base and in particular I am (mainly) intrested in bases larger than exp(1/2).

Also to avoid confusion slog_b(sexp_b(z)) =/= z.

This is the tetration analogue of log_b(exp_b(z)) =/= z.

Hey Tommy,
That is interesting actually. So, for \( \log(\exp(z))=z+2n\pi i \) Is that what you have in mind? If so, there is a direct analog since tetration itself is Pseudo periodic, except, of course, as you approach the real axis, the Pseudo periodic part breaks down into approximately what I would call the theta mapping. The theta mapping is exactly 1-periodic.

Consider the very similar 1-cyclic case, \( \theta(z)=\text{Abel}_b(\text{sexp}_b(z)) \), where the Abel function is from the Schroeder function fixed point. As imag(z) increases, theta goes to a constant. Of course, any solution of theta(z)+Period would be valid. And of course, there is a nasty singularity at the real axis, that is mathematically equivalent to the Kneser Riemann mapping. Before I go digging up an old plot of the 1-cyclic mapping for abel(sexp), is that close to what you are looking for?

Another very interesting function is slog_a(sexp_b(z)), where a and b are different tetration bases, where both bases are greater than eta. Also, slog(2sinh(z)), is related.
- Sheldon