Posts: 1,214
Threads: 126
Joined: Dec 2010
(06/10/2022, 11:08 PM)Catullus Wrote: (06/10/2022, 08:41 PM)JmsNxn Wrote: 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?
(06/10/2022, 08:41 PM)JmsNxn Wrote: 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.
Posts: 213
Threads: 47
Joined: Jun 2022
06/11/2022, 03:01 AM
(This post was last modified: 12/19/2022, 04:38 AM by Catullus.)
 \(\forall x\:\tau(x,1)=1\)
 \(\forall y\in\Bbb{CN}\:\tau(x,y)=x\uparrow\tau(x,y1)\)
 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.
 If \(\tau(x,y)\) approaches any of the fixed points or ncycles of the function \(x\uparrow k\), it will approach continuously iterating \(m*y+b\).
 \(\tau\) is mostly holomorphic.
Does \(\tau\) exist?
If so, is it unique?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Posts: 281
Threads: 95
Joined: Aug 2007
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.
Daniel
Posts: 1,214
Threads: 126
Joined: Dec 2010
06/11/2022, 03:11 AM
(This post was last modified: 06/11/2022, 03:12 AM by JmsNxn.)
(06/11/2022, 03:01 AM)Catullus Wrote: (06/10/2022, 11:16 PM)JmsNxn Wrote: (06/10/2022, 11:08 PM)Catullus Wrote: (06/10/2022, 08:41 PM)JmsNxn Wrote: 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?
(06/10/2022, 08:41 PM)JmsNxn Wrote: 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 y1 ∉ \( \mathbb{N} \).
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 ncycles 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:
\[
\exp^{\circ s}_{\sqrt{2}}(4h)\\
\]
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.
Posts: 213
Threads: 47
Joined: Jun 2022
06/11/2022, 03:19 AM
(This post was last modified: 07/19/2022, 09:31 AM by Catullus.)
Could similar uniqueness criterion work for higher hyperoperations?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Posts: 1,214
Threads: 126
Joined: Dec 2010
(06/11/2022, 03:19 AM)Catullus Wrote: Could similar uniqueness criterion work for pentation and higher hyperoperations?
They work for the iteration of any function, so yes.
Posts: 213
Threads: 47
Joined: Jun 2022
06/11/2022, 03:23 AM
(This post was last modified: 06/11/2022, 03:26 AM by Catullus.)
(06/11/2022, 03:22 AM)JmsNxn Wrote: (06/11/2022, 03:19 AM)Catullus Wrote: Could similar uniqueness criterion work for pentation and higher hyperoperations?
They work for the iteration of any function, so yes. So how do pentation and higher hyperoperations behave? What about factorial iterations? What about using this to iterate functions in the Fastgrowing hierarchy? Like doing f3(.5). Or f4(.5).
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Posts: 1,214
Threads: 126
Joined: Dec 2010
(06/11/2022, 03:23 AM)Catullus Wrote: (06/11/2022, 03:22 AM)JmsNxn Wrote: (06/11/2022, 03:19 AM)Catullus Wrote: Could similar uniqueness criterion work for pentation and higher hyperoperations?
They work for the iteration of any function, so yes. So how do pentation and higher hyperoperations 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.
Posts: 213
Threads: 47
Joined: Jun 2022
06/11/2022, 06:12 AM
(This post was last modified: 07/09/2022, 07:06 AM by Catullus.)
(06/11/2022, 03:22 AM)JmsNxn Wrote: (06/11/2022, 03:19 AM)Catullus Wrote: Could similar uniqueness criterion work for pentation and higher hyperoperations?
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?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Posts: 213
Threads: 47
Joined: Jun 2022
06/26/2022, 08:41 AM
(This post was last modified: 07/09/2022, 07:16 AM by 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 hyperoperations.
What would happen if you did that?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
