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 Iteration with two analytic fixed points tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 08/12/2022, 09:54 PM (08/12/2022, 05:24 PM)bo198214 Wrote: (08/12/2022, 02:14 AM)tommy1729 Wrote: you got the fixpoints derivatives connection right.  but how do you know both regulars agree ? They don't, I made a short sketch why it can not in this particular case (or generally for entire functions) from the paper of Karlin&McGregor just some post ago yes but that limit is the whole neigbourhood of the second fixpoint. I mean analytic from say the open interval between 0 and 1. So when fixpoint 0 and 1 are given  analytic around x belonging to ]0,1[ Im sorry im responsable for some errors miscommunications etc. I lack time and the ideas are coming fast. *** Then again hmm if the regular must be analytic in 0 when taking regular in 0 and similar to the point 1 ... then I guess we have  analytic in  [0,1[ when expanded in 0. and ]0,1] when expanded in 1. and the issue is not analytic continuation. So that seems like 2 different superfunctions yeah.... maybe *** I was thinking about fusing the two methods. But then they are not regular anymore so we loose the semi-group homom probably. hmmm *** im thinking ... Thanks for your efforts anyway. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 08/12/2022, 10:03 PM But I wonder ... What if we place both real fixpoints at -oo and +oo ? Maybe that works ?? regards tommy1729 JmsNxn Ultimate Fellow Posts: 935 Threads: 111 Joined: Dec 2010 08/12/2022, 11:28 PM (08/11/2022, 07:01 AM)bo198214 Wrote: (08/11/2022, 12:01 AM)JmsNxn Wrote: Again bo, not to be the nitpicker that I am. As Milnor would describe it, a polynomial is a map from $$\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}$$. It is not a "euclidean mapping" as Milnor describes, because it's not a transcendental entire function.But James, really, you making up a lot of artificial conditions. Needs to be euclidian, needs to have a vicinity independent of t, etc, etc. This all is not needed! Look at the Karlin&McGregor paper, the class of functions they describe contains all *meromorphic functions*. They use standard regular iteration - the vicinity *depends* on t (and explicitly state that on the 4th line of this paper). And from *there* they conclude that the only functions inside their class of functions that can have the same regular iterations at both fixed points can only be the linear fractional functions. Yes, there was a misread here. I thought you were arguing we could do it for this polynomial, and I was trying to say you can't--and we agree. Also, I'm mostly unconcerned with polynomials, lol; and they behave essentially the same as rational functions; that's all I was getting at. You are absolutely correct though, I think our wires just got crossed. We are on the same page. Exact same page. The vicinity depends on $$t$$ is built in to regular function as well; which is the old timey definition of what it means, I wasn't confused by that. So to keep it on the same page then; can we agree that if $$f^{\circ t}(z) : D \times H \to H$$ for two domains $$D\subset \mathbb{C}$$ and $$H\subset \mathbb{C}$$--where $$D$$ is closed under addition (is a semigroup under $$\{+\}$$); such that: \begin{align} f^{\circ t}(f^{\circ s}(z)) &= f^{\circ t+s}(z)\\ f^{\circ 1}(z) &= f(z)\\ \end{align} Then there can only be one point $$p \in H$$ such that $$f(p) = p$$. I just want to make sure we can agree on that. Which is essentially my statement that there can't be a local iteration about two fixed points. Which has started this whole spiral. (And come to think of it, I think I learned this by way of Sheldon, but I never bothered to look into the proof, as it was just "apparent" to me as to how Schroder iterations behave. We'll have a branching problem of some kind at the other fixed point.) « Next Oldest | Next Newest »

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