Uniqueness of fractionally iterated functions Daniel Long Time Fellow Posts: 279 Threads: 94 Joined: Aug 2007 07/04/2022, 02:12 AM (This post was last modified: 07/04/2022, 09:02 AM by Daniel.) A troubling question that occured to me is if my derivation of the Taylor's series of $f^n(z)$ 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 JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/04/2022, 11:45 PM (This post was last modified: 07/04/2022, 11:47 PM by JmsNxn.) (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 $f^n(z)$ 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. Daniel Long Time Fellow Posts: 279 Threads: 94 Joined: Aug 2007 07/05/2022, 12:01 AM (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 $f^n(z)$ 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 JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/05/2022, 12:06 AM 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. Daniel Long Time Fellow Posts: 279 Threads: 94 Joined: Aug 2007 07/05/2022, 12:17 AM (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. 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.   Daniel JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/05/2022, 12:53 AM (This post was last modified: 07/05/2022, 12:54 AM by JmsNxn.) (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. 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.   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. Gottfried Ultimate Fellow Posts: 901 Threads: 130 Joined: Aug 2007 07/05/2022, 01:18 AM (This post was last modified: 07/05/2022, 01:22 AM by Gottfried.) (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 JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/05/2022, 01:21 AM (This post was last modified: 07/05/2022, 01:24 AM by JmsNxn.) (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. « Next Oldest | Next Newest »

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