Does tetration take the right half plane to itself?
#4
(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


Messages In This Thread
RE: Does tetration take the right half plane to itself? - by sheldonison - 05/15/2017, 09:00 PM

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