Rank-Wise Approximations of Hyper-Operations

[JmsNxn]

I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points.

What if the different Schröder iterations flowed together nicely?

That would be quite the function!

I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points.

The only way I can interpret "flowed together nicely" is Kneser. Kneser makes sure \(L\) and it's complex conjugate \(L^*\) flow together properly. Though, they are never allowed to produce an iteration holomorphic in the neighborhood of both points.

And I mean this absolutely, you can't have holomorphy in \(s\) on some domain, and holomorphy in \(z\) on some domain which includes \(L,L^*\)--the closest you get is Kneser.

[Catullus]

1. \(\forall x\:\tau(x,1)=1\)
   
2. \(\forall y\in\Bbb{C-N}\:\tau(x,-y)=x\uparrow\tau(x,-y-1)\)
   
3. For a given k, using \(\tau\) to do \(\text{sexp(slog(}x)+k)\) produces the same iteration of exponentials, (Or principled logarithms, if k is negative, and if possible,) for any branch of slog, if slog happens to branch.
   
4. If \(\tau(x,y)\) approaches any of the fixed points or n-cycles of the function \(x\uparrow k\), it will approach continuously iterating \(m*y+b\).
   
5. \(\tau\) is mostly holomorphic.
   

Does \(\tau\) exist?
If so, is it unique?

[Daniel]

I've shared this before here I believe.  I solved the symbolic Taylor's series for \( f^n(z) \) based on a fixed point. The series also gave the position of the next fixed point and it's Lyapunov multiplier.

[JmsNxn]

I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points.

What if the different Schröder iterations flowed together nicely?

That would be quite the function!

I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points.

The only way I can interpret "flowed together nicely" is Kneser. Kneser makes sure \(L\) and it's complex conjugate \(L^*\) flow together properly. Though, they are never allowed to produce an iteration holomorphic in the neighborhood of both points.

And I mean this absolutely, you can't have holomorphy in \(s\) on some domain, and holomorphy in \(z\) on some domain which includes \(L,L^*\)--the closest you get is Kneser.

1.) \( \tau \)(x,y) = x. If y = 1.
2.) \( \tau \)(x,y+1) = x^\( \tau \)(x,y). For -y-1 ∉ \( \mathbb{N} \).
3.) For a given k, using \( \tau \) to do sexp(slog(x)+k) produces the same iteration of exponentials. (Or principled logarithms if k is negitave.) For any branch of slog. (If slog happens to branch.)
4.) If \( \tau \)(x,y) approaches any of the fixed points or n-cycles of the function f(k) = x^k, it will approach continuously iterating y*m+b.
5.) \( \tau \)(x,y) is mostly holomorphic.
Does \( \tau \)(x,y) exist?
If so is it unique?

The key you've added here is is \(\tau\) mostly holomorphic. Well, it isn't holomorphic at the fixed points/ periodic points. Then Kneser satisfies your solution.

Think of it this way:

\[
\exp^{\circ s}(z) = F(s,z)\\
\]

Then \(F(s,z)\) can't be holomorphic on intersecting domains for different fixed points. We cannot construct something near multiple fixed points.

Think \(\sqrt{2}\). Let's try:

\[
\exp^{\circ s}_{\sqrt{2}}(4+h)\\
\]

And:

\[
\exp^{\circ s}_{\sqrt{2}}(4-h)\\
\]

Despite looking like similar functions, there is no way to make the transition between the two functions to be holomorphic at \(4\) (\(h=0\)). This CANNOT BE DONE. These are two different analytic functions, and at \(4\) there is a singularity or the value \(4\). You cannot paste the two together.

[Catullus]

Could similar uniqueness criterion work for higher hyper-operations?

[JmsNxn]

Could similar uniqueness criterion work for pentation and higher hyper-operations?

They work for the iteration of any function, so yes.

[Catullus]

Could similar uniqueness criterion work for pentation and higher hyper-operations?

They work for the iteration of any function, so yes.

So how do pentation and higher hyper-operations behave? What about factorial iterations? What about using this to iterate functions in the Fast-growing hierarchy? Like doing f3(.5). Or f4(.5).

[JmsNxn]

Could similar uniqueness criterion work for pentation and higher hyper-operations?

They work for the iteration of any function, so yes.

So how do pentation and higher hyper-operations behave?

Let's start with pentation, and consider \(\sqrt{2}\). For, \(\sqrt{2}\uparrow^3 z\) we have a fixed point \(\omega_2\) such that:

\[
\sqrt{2}\uparrow^3 \infty = \omega_2
\]

This is \(\infty\) in the sense of the closing point of \(\Re(z) > 0\). If we choose tetration as the Schroder iteration about \(2\):

\[
\sqrt{2} \uparrow \uparrow \omega_2 = \omega_2\\
\]

While additionally, this is a geometrically attracting fixed point. Therefore we just rinse and repeat the construction of tetration.

---

If we talk about the repelling fixed points (discussions of \(4\), and the subsequent iterations)--it gets very chaotic. But the answer to your question is that it continues to hold.

[Catullus]

Could similar uniqueness criterion work for pentation and higher hyper-operations?

They work for the iteration of any function, so yes.

If you did that to do tetration, pentation, et cetera. How can you, when the hyperoperations converge as the rank becomes larger and larger, approximate the convergence?
2[1]2+1/1 = 5. 2[2](2+1/2) = 5. 2[3](2+1/3) ~ 5.040. 2[4](2+1/4) ~ 5.051. Do they converge? If so what do they converge to? How can you approximate the convergence in a way, such that it becomes better and better as the rank of the hyperoperation increases?

[Catullus]

If they work for the iteration of any function then, then they would work for the iteration of the super function function. Then you could apply the super function operation to addition a non natural number of times, giving you non natural number rank hyper-operations.
What would happen if you did that?