05/15/2017, 09:00 PM
(This post was last modified: 05/15/2017, 11:25 PM by sheldonison.)
(05/10/2017, 07:46 PM)JmsNxn Wrote: This question is rather straightforward. Let \( 1 < \alpha < \eta \). We know that there exists a unique bounded tetration function \( F(z) = \alpha \uparrow^2 z \) that is holomorphic for \( \Re(z) > 0 \). But is \( \Re (F(z)) > 0 \)? Does the bounded tetration function take the right half plane to itself?So are you asking if it is a 1-1 mapping from the right half of the plane to itself? Certainly there's a singularity at z=-2.
Lets think about base sqrt(2), with two fixed points, 2 and 4. As \( \Re(z) \) gets arbitrarily large \( \text{sexp}_{\sqrt{2}}(z) \approx 2 \)
At the imaginary axis, sexp(0)=1. At half the period, sexp(period/2) the value would have grown to something like 2.468. In between, imag(z) gets to about +/- 0.7208.
So at the imaginary axis, it will roughly "circle" or loop around the fixed point of 2 an infinite number of times, as follows: This very roughly circular path is what the right half of the plane is mapped to! Of course, it would be more complicated if I repeated this experiment at the singularity at z=-2 and again mapped one imaginary period, but I don't see the plane mapped to itself at z=-2 either. The imaginary period is \( =\frac{-2\pi i}{\ln(\ln(2))} \approx \) 17.143
- Sheldon