Apropos "fix"point: are the fractional iterations from there "fix" as well? Gottfried Ultimate Fellow Posts: 889 Threads: 130 Joined: Aug 2007 07/10/2022, 10:44 AM (This post was last modified: 07/10/2022, 01:04 PM by Gottfried.) This is a tiny question which bothered me from time to time, but I took it always as somehow given. Now by the current discussions this comes up again.  Practically this has a relevance, when I thought to ask my brother, who has a 3-D-printer, for a 3 D model of fractional iteration in the complex numbers, where the height $$h \in \mathbb R$$ gives the 3rd dimension, and the integer heights are visualized simply like dots at the floors of the stages in a house. The fractional iterates from one point to another follow then some "noodle" from a point in first floor to the according point in second floor.  (Our chosen interpolation-method define the form of the "noodles".)  The noodles from a fixpoint in the first floor to the same point in the second floor - how is it shaped? Straight, vertical? I remembered I've never consciously seen a discussion of that, I gave it a chance, that it might be simply taken as an axiom. But perhaps, with the obvious informations around from Milnor or Devaney: they might have derived this, or at least might have explicitely mentioned this. (Note, that the "noodles" connecting periodic points have/must have an "exploding", chaotically divergent, form - although I've as well not seen this discussed in the material I have/had made available for me.)             So in this sense: are the fractional iterates of a fixpoint "fix" as well/ are they all identically the same coordinate?      And     is this taken so-to-say axiomatically, or is it a consequence of something? Gottfried Gottfried Helms, Kassel MphLee Long Time Fellow Posts: 369 Threads: 28 Joined: May 2013 07/10/2022, 04:42 PM (This post was last modified: 07/10/2022, 04:44 PM by MphLee.) I'd say that there is nothing algebraically that forces $$f^t(p)=p$$ for every $$t\in\mathbb R$$ when $$p$$ is a fixed point of $$f^1$$. If I understand what you are asking, you are asking if the following conjecture is a theorem Conjecture: let $$f:X\to X$$ be a function and $$f(p)=p$$ a fixed point. Then, given a way to non-integer iterating $$f$$ using a superfucntion, i.e. a function $$\varphi(t,x)$$ s.t. $$\varphi(0,x)=x$$ and $$\varphi(t+1,x)=f(\varphi(t,x))$$, we have that the orbit of the fixedpoint $$p$$ is constant $$\varphi(t,p)=p$$ I'd say there is nothing forcing this, at least algebraically. First of all remember that we can build the superfunction piecewise, a la Andrew Robbins, obtaining k-differentiability on the endpoints. We can define $$\Phi(n):=p$$, the fixed point for every integer $$n\in\mathbb N$$, then define $$\Phi(t)$$ for $$t\in [0,1]$$ to be an arbitrary loop in $$\gamma:[0,1]\to X$$ that connects $$p$$ to itself. You can then evaluate $$\Phi(x):=f^{{\rm floor}(x)}(\gamma (\{x\}))$$. Note that the constant solution $$\forall x.\,K(x)=p$$ is a solution of the superfunction equation. Maybe only adding constraints can force the orbit, the noodle, that connects the fixed point to itself at different iteration height, to be constant/straight line. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Gottfried Ultimate Fellow Posts: 889 Threads: 130 Joined: Aug 2007 07/10/2022, 05:35 PM (This post was last modified: 07/10/2022, 05:45 PM by Gottfried.) MphLee - ah, thanks. Your formulation (the conjecture) sounds good - maybe I should even incorporate it into my question later. Bythe time of thinking about it I got the vague idea, that at least for attracting fixpoints it should be impossible for any iteration - be it as fractionally small as we want - that it could "swim against the attraction field" - so to say... Surely a very vague imagination and possibly a circular thinking (or a knotted "noodlolar" one ) I'll take some time to think about this further... Gottfried Gottfried Helms, Kassel MphLee Long Time Fellow Posts: 369 Threads: 28 Joined: May 2013 07/10/2022, 11:28 PM (This post was last modified: 07/10/2022, 11:30 PM by MphLee.) Really nice Gottfried... swimming against the attraction field... interesting. We need an expert opinion on this. I'm sure is some trivial proposition one can find in any book on dynamics... but I'm too ignorant. I'm sure James can settle this in 1 second. To think about it... if the orbit crosses the basin of attraction of the fixed point... maybe it is not necessary that from that moment each successive point of the orbit has a distance from the fixed point that is monotone decreasing. Conjecture 2: let $$f:X\to X$$, $$X$$ a metric space, $$p\in X$$ attracting and $$I\subseteq X$$ its basin of attraction. Let $$\chi:[0,\infty)\to X$$ be a $$x_0$$-initialized superfucntion that is continuous, i.e. $$\chi(0)=x_0$$ and $$\chi(t+1)=f(\chi(t))$$ if exists a $$r\in [0,\infty)$$ s.t. $$\chi({r})\in I$$ is in the basin of attraction of $$p$$ then $$D(t):=d(\chi(t+r),p)$$ is monotone decreasing. I.e. for $$r exp(1/2). And probably by analytic continuation to all bases larger than eta ( e^(1/e) ). but for bases =< exp(1/2) we get issues with multiple fixpoints or derivatives at the fixpoints. THAT is also the reason why I proposed alternative similar methods. Since those alternatives agree on bases larger than exp(1/2) by the uniqueness criterion , they must be the 2sinh method extended to lower bases !! I hope that is clear to everyone. Regards tommy1729 ps wiki will not accept the 2sinh method in the tetration section , even when it does mention non-C^oo solutions. meh ! MphLee Long Time Fellow Posts: 369 Threads: 28 Joined: May 2013 07/16/2022, 04:22 PM (This post was last modified: 07/17/2022, 10:50 PM by MphLee.) Typos and errors fixed. I hope everyone forgives me, but I need to put all of this on solid, formal grounds. I need to prove every step. Now I'm doing it for James' post... but I've done only for the first half of it. Quote:So, this is actually a defining property of the standard Schroeder iteration. But it's a little difficult to fully flesh out why. Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them. If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p$$, then the iteration is expressible via Shroeder iteration (provided that $$|f'(p)| \neq 0,1$$).  For convenience, assume that $$|f'(p)| < 1$$. You can actually prove this pretty fast. Assume that $$f^t(x)$$ is a super function in $$t$$ about a fixed point $$p$$, and $$x$$ is in the neighborhood of $$p$$. Assume that $$f^t(p) = p$$.  Well then: $$\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\$$ So that: $$\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\$$ And we know that there must be some 1-periodic function $$\theta(t)$$, such that: $$f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\$$ In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroeder iteration. Now let's add one more constraint, let's say that $$f^{t}(f^{s}(x)) = f^{t+s}(x)$$. Well then, we have that $$\theta$$ must be constant. By which, we are guaranteed that it's a Schroeder iteration... So to clarify. Any fractional iteration with a constant noodle is a Schroeder iteration. But there are plenty of superfunctions of $$f$$ which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law). I'm trying to scan the deep meaning behind your moves... what is happening in the background. I'd like a feedback so I can go on to the next step and then to Tommy's observations. When you ask for superfunction about a fixed point $$p$$ you, in fact, are actually taking a complex object. You are not just taking A superfunction but an entire family of superfunctions $${\varphi}_{x\in U}:\mathbb R^+\to X$$, superfunctions of $$f:X\to X$$, parametrized by a small $$f$$-stable neighborhood of the fixed point $$p\in U\subseteq X$$. In other words you asking for a $$\varphi:\mathbb R\times U\to U$$, i.e. a family $$\{\varphi_x\}_{x\in U}:\mathbb R^+\to X$$ of superfunctions of $$f$$ restricted to the neighborhood of the fixed point $$p$$ (that should be the local part about it). Notice that, as recent posts by me, James and Tommy are making clear... being a superfunction is something weaker than being an iteration, like being an interpolation is weaker than being a superfunction. In other words we just ask for the family $$\varphi_x$$ to satisfy the mild conditions $$\varphi_x(t+1)=f(\varphi_x(t))\quad \quad \varphi_x(0)=x$$ No other coherence conditions, like the semigroup property... a property that is very strong. Note that what I write as $$\varphi_x(t)$$ we could write as $$\varphi(t,x)$$ and James denotes it as $$\varphi_x(t)=f^t(x)$$. I use $$\varphi_x$$ notation to emphasize that it is a FAMILY OF POSSIBLE SOLUTIONS of the equation $$\varphi\circ S=f\circ \varphi$$. We can think of this as an assignment $$\varphi_{-}:U\to [S,f]$$ where $$[S,f]$$ is the space of all the superfunctions of $$f$$... Now... we want to know if $$\varphi_p(t)$$ is a constant path or... if it loops around $$p\in X$$ infinitely many times. The answer James gives is... it is constant if the family $$\varphi_x$$ satisfies the stronger condition of being an iteration: i.e. $$\varphi_{\varphi_x(t)}(s)=\varphi(s,\varphi(t,x))=\varphi(s+t,x)=\varphi_x(s+t)$$ The reasoning, subject to multipliers conditions to ensure convergence of Schroeder functions, seems to begin in the following way. Given a family of superfunctions of $$f$$ locally about a fixed point $$p\in U$$, i.e. given a family of superfunctions $$\varphi_{x\in U}$$ and given a function $$\Psi\in [f,{\rm mul}_\lambda]$$, a solution of the Schroeder eq. of $$f$$ we define another family $$A_x=\Psi\circ \varphi_x$$ this time this family lies in the space $$[S,{\rm mul}_\lambda]$$ that is, each $$A_x$$ is a superfunction of multiplication $${\rm mul}_\lambda$$ by $$\lambda$$. $$A_x(t+1)=\lambda A_x(t)$$ this is what James denotes by $$A_x(t)=\Psi(f^t(x))$$. Remember that we have a family of functions in the space  $$[S,{\rm mul}_\lambda]$$ and remember, in this space there is a special solution: exponentiation $$\exp_\lambda \in [S,{\rm mul}_\lambda]$$. What James does now is measuring the difference between the $$A_x$$ and $$\exp_\lambda$$. They differ by a periodic function. $$\theta_x(t)\lambda^t=A_x(t)$$ If the periodic function is constant $$\theta_x(t)=K$$ then $$K\lambda^t=\Psi(\varphi_x(t))$$ and $$\varphi_x(t)=\Psi^{-1}(K\lambda^t)$$ is purely obtained by the Schroeder function. James calls it the Schroeder iteration. At this level of detail each $$\theta_x$$ can be a different periodic function. What I managed to prove is that... if we ask the family $$\phi_x$$ tho satisfy the additional condition $$\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)$$, weaker than being an iteration but implied by it we obtain that if $$x\sim_f y$$ then $$\theta_x=\theta_y$$ Proposition. let $$\varphi_{-}:U\to [S;f|_U]$$ be a family of superfunctions of $$f|_U$$ for $$U\subseteq X$$ an $$f$$ stable subset, let $$\Psi \in [f, \lambda ]$$ and define the family of 1-periodic function $$\theta_x(t)=\frac{\Psi(\varphi_x(t))}{\lambda^t}$$ we prove thatif for each $$x\in U$$ we have $$\varphi_x(0)=x$$ (a section); This condition is not needed for each $$t$$ $$\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)$$ (weakly iterative) then for $$x,y\in U$$ are $$f$$ connected, i.e. $$f^n(x)=f^m(y)$$ for some $$m,n\in\mathbb N$$, $$\theta_x(t)=\theta_y(t)$$ Proof. assume $$\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)$$. We prove that $$\theta_x=\theta_{f^n(x)}$$. Compute $$\theta_{f^n(x)}(t)=\frac{\Psi(\varphi_{f^n(x)}(t))}{\lambda^t}=\frac{\Psi(\varphi_{x}(t+n))}{\lambda^t}=\theta_x(t+n)$$ by assumption. $$\theta_{f^n(x)}(t)=\theta_{x}(t)$$ by periodicity. If $$f^n(x)=f^m(y)$$ then $$\theta_x(t)=\theta_y(t)$$ by transitivity of the identity. $$\square$$ At this point I can't wait to translate the rest of the argument... Maybe $$\phi_x$$ satisfies semigroup, is an iteration, iff $$\theta_x$$ is constant. If this was the case, as Tommy seemed to suggest elsewhere... this is something like a criterion to select among the space of superfunctions the special ones... something like an uniqueness criterion. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »

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