Uniqueness of fractionally iterated functions
#1
A troubling question that occured to me is if my derivation of the Taylor's series of is correct, due to its generality it must contain all solutions that don't have a super-attracting fixed point. Since all functions except the successor function have finite fixed points, if f(z) is smooth then my approach should be valid. So shouldn't all valid methods give the same results? Yeah, lots of different tetrations and all, but shouldn't they all agree on their common areas? Aren't we all studying parts of the same "elephant"?
Daniel
Reply
#2
(07/04/2022, 02:12 AM)Daniel Wrote: A troubling question that occured to me is if my derivation of the Taylor's series of is correct, due to its generality it must contain all solutions that don't have a super-attracting fixed point. Since all functions except the successor function have finite fixed points, if f(z) is smooth then my approach should be valid. So shouldn't all valid methods give the same results? Yeah, lots of different tetrations and all, but shouldn't they all agree on their common areas? Aren't we all studying parts of the same "elephant"?

Your iteration method is the basis for iterations. But no, they don't agree.

The same fallacy seems to be making its way around this forum, and I keep on having to correct it.

If \(f\) is a holomorphic function, and has two fixed points \(x_0, x_1\). Then the iteration \(f^{\circ s}(z)\) for \(z \approx x_0\) is NOT THE SAME FUNCTION, as the iteration \(f^{\circ s}(z)\) for \(z \approx x_1\). You CANNOT make them one function. It's incorrect. If you iterate \(\sqrt{2}^z\) about \(z\approx 2\), it is NOT THE SAME iteration as iterating \(\sqrt{2}^z\) about \(z\approx 4\).

We can also iterate from periodic points too, and that can be even MORE COMPLICATED. They are not the same elephant.

So, as your iteration method works, it works to construct Schroder's iteration about a fixed point. This is commonly referred to as the standard iteration, or the regular iteration.

The higher order problems are to create an iteration which works globally (and hence for no fixed points--like Kneser). Or to construct super functions, which there are uncountably many.

Your iteration method is precisely the local iteration. And any local iteration looks like your iteration method. The trouble lies in extending iterations to larger domains (excluding other fixed points), and dealing with more exotic constructions.
Reply
#3
(07/04/2022, 11:45 PM)JmsNxn Wrote:
(07/04/2022, 02:12 AM)Daniel Wrote: A troubling question that occured to me is if my derivation of the Taylor's series of is correct, due to its generality it must contain all solutions that don't have a super-attracting fixed point. Since all functions except the successor function have finite fixed points, if f(z) is smooth then my approach should be valid. So shouldn't all valid methods give the same results? Yeah, lots of different tetrations and all, but shouldn't they all agree on their common areas? Aren't we all studying parts of the same "elephant"?

Your iteration method is the basis for iterations. But no, they don't agree.

The same fallacy seems to be making its way around this forum, and I keep on having to correct it.

If \(f\) is a holomorphic function, and has two fixed points \(x_0, x_1\). Then the iteration \(f^{\circ s}(z)\) for \(z \approx x_0\) is NOT THE SAME FUNCTION, as the iteration \(f^{\circ s}(z)\) for \(z \approx x_1\). You CANNOT make them one function. It's incorrect. If you iterate \(\sqrt{2}^z\) about \(z\approx 2\), it is NOT THE SAME iteration as iterating \(\sqrt{2}^z\) about \(z\approx 4\).

So, as your iteration method works, it works to construct Schroder's iteration about a fixed point. This is commonly referred to as the standard iteration, or the regular iteration.

The higher order problems are to create an iteration which works globally (and hence for no fixed points--like Kneser). Or to construct super functions, which there are uncountably many.

Your iteration method is precisely the local iteration. And any local iteration looks like your iteration method. The trouble lies in extending iterations to larger domains (excluding other fixed points), and dealing with more exotic constructions.

My derivation is symbolic, but based on the Lyapunov multiplier. Different fixed points have difference multipliers and lead to different equations. I'm not just talking crap here, I wrote software in the early Nineties that with a fixed point mapped to the origin, computed the position of a neighboring fixed point and its multiplier, which must be different from the multiplier of the first fixed point. It takes a great deal of computer power to push the Taylor's series out far enough that the algorithm can compute through the region of chaos between fixed points.
Daniel
Reply
#4
I understand that, Daniel. I apologize if my response seemed hostile. Wasn't my intention, lol.

All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you  observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.
Reply
#5
(07/05/2022, 12:06 AM)JmsNxn Wrote: I understand that, Daniel. I apologize if my response seemed hostile. Wasn't my intention, lol.

All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you  observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.

Hey, it's all good JmsNxn. I'm always happy to get thoughtful feedback, even if it is not what I have hoped for. Smile

My construction not only encompasses the cases of Abel and Schroeder's functional equations. See my page on generating flows based on the Abel's case. I'd run your test but unfortunately I'm now without Mathematica for the first time in thirty years, so I guess I need to get good at GP-Pari.   Huh
Daniel
Reply
#6
(07/05/2022, 12:17 AM)Daniel Wrote:
(07/05/2022, 12:06 AM)JmsNxn Wrote: I understand that, Daniel. I apologize if my response seemed hostile. Wasn't my intention, lol.

All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you  observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.

Hey, it's all good JmsNxn. I'm always happy to get thoughtful feedback, even if it is not what I have hoped for. Smile

My construction not only encompasses the cases of Abel and Schroeder's functional equations. See my page on generating flows based on the Abel's case. I'd run your test but unfortunately I'm now without Mathematica for the first time in thirty years, so I guess I need to get good at GP-Pari.   Huh

Lol, ya!

Pari-gp is the ultimate tool. I only use mathematica to plot complicated graphs. Even that is difficult.

You can always sail the high seas and yarr me matey, acquire Mathematica by more nefarious means.

All I'm saying, is that mathematically, you cannot have an iteration holomorphic at \(2\) and at \(4\). This is detailed beautifully in Trappmann's and Kouznetsov's paper on iterated exponentials. It deals specifically with \(\sqrt{2}\). They define exactly 4 iteration types. and sadly, those are the only ones about \(2,4\). And all of them cannot be holomorphic at both fixed points. It's just a cruel joke that nature plays on us.

https://www.researchgate.net/profile/Hen...ion_detail


The only solution to this is to do something extravagant and fancy with super functions, then you can get holomorphy on a larger domain, but still, it is not holomorphic at both fixed points.
Reply
#7
(07/04/2022, 11:45 PM)JmsNxn Wrote: The same fallacy seems to be making its way around this forum, and I keep on having to correct it.

If \(f\) is a holomorphic function, and has two fixed points \(x_0, x_1\). Then the iteration \(f^{\circ s}(z)\) for \(z \approx x_0\) is NOT THE SAME FUNCTION, as the iteration \(f^{\circ s}(z)\) for \(z \approx x_1\). You CANNOT make them one function. It's incorrect. If you iterate \(\sqrt{2}^z\) about \(z\approx 2\), it is NOT THE SAME iteration as iterating \(\sqrt{2}^z\) about \(z\approx 4\).

We can also iterate from periodic points too, and that can be even MORE COMPLICATED. They are not the same elephant.
Daniel - If I got James right the very old thread "Bummer!" should be enlightening, where Henryk  noticed that problem first time.

Gottfried
Gottfried Helms, Kassel
Reply
#8
(07/05/2022, 01:18 AM)Gottfried Wrote:
(07/04/2022, 11:45 PM)JmsNxn Wrote: The same fallacy seems to be making its way around this forum, and I keep on having to correct it.

If \(f\) is a holomorphic function, and has two fixed points \(x_0, x_1\). Then the iteration \(f^{\circ s}(z)\) for \(z \approx x_0\) is NOT THE SAME FUNCTION, as the iteration \(f^{\circ s}(z)\) for \(z \approx x_1\). You CANNOT make them one function. It's incorrect. If you iterate \(\sqrt{2}^z\) about \(z\approx 2\), it is NOT THE SAME iteration as iterating \(\sqrt{2}^z\) about \(z\approx 4\).

We can also iterate from periodic points too, and that can be even MORE COMPLICATED. They are not the same elephant.
Daniel - If I got James right the very old thread "Bummer!" should be enlightening, where Henryk  noticed that problem first time.

Gottfried

Ya what is it, the same values up to like 1E-13, but then they disagree.

It's intrinsic to the iteration, if you follow Milnor, he even explains this dilemma. You cannot pass from an iteration in the fatou set and be holomorphic once we hit the Julia set. \(4\) is in the Julia set, and the iteration about \(2\) is in the fatou set. To construct the iteration about \(4\) we use it as in the fatou set of \(\log\), then \(2\) is in the Julia set. It dates back very far. Milnor honestly covers so much in this forum, and so much is rediscovered on this forum.

John Milnor is truly a god among men. His complex dynamics is unmatched, even though it dates to the 70s or what ever.
Reply


Possibly Related Threads…
Thread Author Replies Views Last Post
  [MSE] iterated sin using Besselfunction 1st kind Gottfried 0 87 10/30/2022, 12:38 PM
Last Post: Gottfried
  Bessel functions and the iteration of \(e^z -1 \) JmsNxn 8 597 09/09/2022, 02:37 AM
Last Post: tommy1729
  The iterational paradise of fractional linear functions bo198214 7 471 08/07/2022, 04:41 PM
Last Post: bo198214
  Universal uniqueness criterion? bo198214 57 111,927 06/28/2022, 12:00 AM
Last Post: JmsNxn
Question Iterated Hyperbolic Sine and Iterated Natural Logarithm Catullus 2 566 06/11/2022, 11:58 AM
Last Post: tommy1729
  A question concerning uniqueness JmsNxn 4 10,000 06/10/2022, 08:45 AM
Last Post: Catullus
  The weird connection between Elliptic Functions and The Shell-Thron region JmsNxn 1 707 04/28/2022, 12:45 PM
Last Post: MphLee
  Using a family of asymptotic tetration functions... JmsNxn 15 7,119 08/06/2021, 01:47 AM
Last Post: JmsNxn
  Generalized Kneser superfunction trick (the iterated limit definition) MphLee 25 13,011 05/26/2021, 11:55 PM
Last Post: MphLee
  [Exercise] A deal of Uniqueness-critrion:Gamma-functionas iteration Gottfried 6 7,485 03/19/2021, 01:25 PM
Last Post: tommy1729



Users browsing this thread: 1 Guest(s)