Rank-Wise Approximations of Hyper-Operations
#10
(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.
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Messages In This Thread
RE: Rank-Wise Approximation of hyper operations - by JmsNxn - 06/06/2022, 04:38 AM

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