08/15/2018, 03:10 AM
(This post was last modified: 08/15/2018, 03:46 AM by sheldonison.)
(08/14/2018, 11:09 PM)Chenjesu Wrote: ... We want to know what raising something to the "1/2" height means, at least according to this conjecture, and z in this case is the height of tetration:
\( \exp_b^{ \frac{1}{2}}(1)=s \exp_{b}(\frac{1}{2})= \underbrace{b^{b^{b^{...}}}}_{\text{\frac{1}{2} times}} \)
How should this result be interpreted to yield a final answer in terms of standard mathematical functions for b=e?
This is a very good question. William Paulsen's paper from earlier this year, "figure 1" would be a good starting place;
You might also want to view my post showing pictures of the Riemann mapping: https://math.eretrandre.org/tetrationfor...p?tid=1172 But this figure is hard to explain; the short explanation of this figure is as follows:
- Start with the Schroeder function for the complex fixed point L~=0.318131505 + 1.33723570I; then e^L=L
- Turn the Schroder function into an Abel function, Abel(exp(z))=Abel(z)+1.
- But this is a complex valued Abel function for the real number line!
- It has singularities at z=0,1,e,e^e....
- Take the Abel function of the real axis ... then you get the repeating function above in the darkened line, along with the requisite singularities, starting with -infinity
- Now wrap the repeating 1-cyclic pattern around the real axis via the mapping \( y \mapsto \exp(2\pi\i z) \)
- Take the Rieman mapping of that circle function so that it maps to the unit circle.
- Put the singularity at z=1. Now, z=-1 corresponds to 0.5, exp(0.5), exp(exp(0.5)) etc. Unwrap it onto the original drawing, and magically you have a real valued tetration of the real axis.
As far as I know, nobody has a rigorously proven computation technique published; Paulsen's paper gives numerical results just like Trapmann's paper, along with heuristic arguments for why it should equal Kneser's solution. There are a few really good programs that can easily calculate sexp(-0.5) for any base, like my pari-gp program, fatou.gp In the future, I expect someone will show how to make one of these computation schemes mathematically rigorous ...
- Sheldon

