06/06/2022, 04:38 AM
(06/06/2022, 04:07 AM)Catullus Wrote:(06/06/2022, 04:00 AM)JmsNxn Wrote: But there is only one Schroder iteration of the exponential. There are countable TETRATIONS which satisfy this. I apologize if I talked too loosely.Tetraion is iterated exponentiation.
Using the limit formula on any of the fixed points should produce the same tetration function.
No, that's incorrect.
I just explained why that's incorrect.
A tetration function:
\[
\text{Tet}(s)\\
\]
Is a function such that:
\[
\begin{align}
\text{Tet}(0) = 1\\
\text{Tet}(s+1) = e^{\text{Tet(s)}}\\
\end{align}
\]
As I just showed, there are countable solutions to this equation when you use Schroder's iteration.
To reiterate. The inverse Schroder function about \(L\) of \(e^z\) is an entire function, \(\Psi^{-1}(z)\). This function equals \(1\) countably infinite times (Picard). Therefore, \(\Psi^{-1}(e^{-Ls_0})\) equals \(1\) countably infinite times. Therefore \(\Psi^{-1}(e^{L(s-s_0)})\) is a tetration function. Therefore there are countably infinite tetration functions spawned from the fixed point \(L\) using Schroders iteration.
All I am doing is filtering through preimages.

