Rank-Wise Approximations of Hyper-Operations

[Catullus]

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.

Iterated exponentials should become closer and closer to iterated mx+b around all of the the fixed points.

[JmsNxn]

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.

Iterated exponentials should become closer and closer to iterated mx+b around all of the the fixed points.

Yes, that is the Schroder iteration!

Tetration for a global function doesn't necessarily behave like that. That's a uniqueness criterion you haven't defined though. All you've said with this is that:

\[
\text{Tet}(s-k) \approx L + e^{-Lk}\text{Tet}(s)\,\,\text{for}\,\,\Re(s) < -R\,\,\text{for}\,\,R\,\,\text{Large}\\
\]

Yes ABSOLUTELY THAT's TRUE!!!!! There are countably infinite solutions to that though...

[Catullus]

Tetration for a global function doesn't necessarily behave like that. That's a uniqueness criterion you haven't defined though. All you've said with this is that:

\[
\text{Tet}(s-k) \approx L + e^{-Lk}\text{Tet}(s)\,\,\text{for}\,\,\Re(s) < -R\,\,\text{for}\,\,R\,\,\text{Large}\\
\]

Yes ABSOLUTELY THAT's TRUE!!!!! There are countably infinite solutions to that though...

A tetration function must do that about all of the fixed points. You could iterate from any of the fixed points, and continue back to where you want it. Like to calculate 2^^.5 you could use any of the fixed points. Tetration needs to be unique.

[Daniel]

Tetration for a global function doesn't necessarily behave like that. That's a uniqueness criterion you haven't defined though. All you've said with this is that:

\[
\text{Tet}(s-k) \approx L + e^{-Lk}\text{Tet}(s)\,\,\text{for}\,\,\Re(s) < -R\,\,\text{for}\,\,R\,\,\text{Large}\\
\]

Yes ABSOLUTELY THAT's TRUE!!!!! There are countably infinite solutions to that though...

A tetration function must do that about all of the fixed points. You could iterate from any of the fixed points, and continue back to where you want it. Like to calculate 2^^.5 you could use any of the fixed points. Tetration needs to be unique.

Daniel
Using a base a bit larger than 1 produces simple dynamics and fractals with chaos only in small areas. I was able to compute the dynamics from one fixed point that was able to give the location of a neighboring fixed point and it's Lyapunov multiplier. This indicates that the Taylors series of an iterated function can give the position of all the other fixed points and consistent Taylor series.

[JmsNxn]

Yes, that is perfectly possible.

Let's say we write \(b^{z} = \exp_b(z)\) for \(b \approx e\), and we discuss \(L(b)\) as the fixed point as a function of \(\exp_b(z)\).

There is a neighborhood where the iterated exponential is indistinguishable for each \(b\approx e\)--at least, topologically. We are just talking about moving \(L\) while we iterate. At least for the Schroder case--these are isomorphic objects until you hit the boundary of Shell-Thron. In many senses, the behaviour of \(\exp_2(z)\) is indistinguishable topologically from \(\exp(z)\).

What your response suggests, is that you are tracing a path across the fixed point \(L(b)\), and finding a holomorphic solution. There are countably infinite solutions to this though. You are choosing one, which I agree is very natural, but there are many many more--which satisfy the uniqueness you are arguing for.

\[
\exp^{\circ s}_b(z)\,\,\text{for}\,\,|z-L(b)| < \delta\\
\]

While:

\[
\exp^{\circ s}_b(L(b)) = L(b)\\
\]

This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.

There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.

This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.

[Daniel]

Yes, that is perfectly possible.

...

This implicit solution exists uniquely across iterations. But if you ask for a tetration solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.

There are infinite solutions to these equations. This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition.

JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of n-period fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary.

[Catullus]

JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of n-period fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary.

How would n period points, for n ∈ ? How would you use the limit formula on them?

[Daniel]

JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of n-period fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary.

Why would n period points, for n ∈ \mathbb{N}? How would you use the limit formula on them?

Please restate your question. Thanks.

[Catullus]

JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of n-period fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary.

Why would n period points, for n ∈ \mathbb{N}? How would you use the limit formula on them?

Please restate your question. Thanks.

How would you use periodic points to do continuous iterations of exponentials?
The limit formula uses fixed points usually, if not always.

[Catullus]

This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.

There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.

This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.

I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations.
For bases not in the Shell-Thron region, maybe you could analytically continue the hyper-operations there.