iterating z + theta(z) ? [2022] tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 07/01/2022, 09:52 PM ok this might be a repost , a silly question , an ignored idea or a mistake but anyways  We know fractional iterations do not agree on fixpoints usually ( or always ) Lets say we have fixpoint at 0 and around 0 the function behaves like a polynomial a(z). Now lets say we have another fixpoint at 1. Now clearly - using fixpoint formula alone and assuming they must merge from one fix to another ; 1) the derivatives of the fixpoints must match. 2) without a fixpoint in the way of the 2 others that ' misbehaves ' or ' blocks ' the path. Now even better than 1)  1b) around every fixpoint the function behaves like the polynomial a(z). --- Now if we want to expand the function easily it works best if the polynomial a(z) is of high degree. But if the region around 0 is copied to the region around 1 , the same should happen to the region around 1 ; mapping to the region around 2. This implies that 2 should be fix too. In fact this implies that our function should be somewhat periodic and have fixpoints at all integers and no other real fixpoints. Taking this one step further : the function should be entire. and the iterations should be somewhat periodic themselves. --- This brings us to f(z) = z + sin(2 pi z)/5 clearly f(f(z)) = z + sin(2 pi z)/5 + sin(2 pi  (  z + sin(2 pi z)/5 ) )/5 what is also of the form z + periodic. ( funny thing z + theta(z) occurs here often )  So how do the iterates of f(z)  =  z + sin(2 pi z)/5 behave ?? Is the half-iterate of the same type ( z + periodic ) ? Do the fixpoints sometimes match for noninteger iterates ? So I wonder about the general idea  f^[t](z) for real t > 0. and ofcourse f(z) is just a random case. crazy idea i guess. regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,179 Threads: 123 Joined: Dec 2010 07/01/2022, 10:03 PM (This post was last modified: 07/01/2022, 10:05 PM by JmsNxn.) Well right off the bat I'm going to say that you can't use a fixed point iteration. If $$f^{\circ s}(z)$$ is a holomorphic function on some domain $$\mathcal{A}$$, there cannot be any periodic or fixed points in $$\mathcal{A}$$ besides one fixed point. You can go one step further: If $$f^{\circ s}(z)$$ is holomorphic at a fixed point $$L$$, it is not holomorphic about any other fixed point/periodic point $$L_2$$ of $$f$$. This is because somewhere in this iteration we either hit a Julia set, or we hit different fatou set. Best case scenario we get something like Kneser's iteration, which is holomorphic everywhere EXCEPT at periodic/fixed points. What this means for your function is pretty simple. The iteration $$f^{\circ s}(z)$$ at $$0$$, has a discontinuity in $$z$$ as it tries to approach $$f^{\circ s}(z)$$ at $$1$$. This is because, even if both are attracting, somewhere $$z$$ will hit the Julia set; remember every basin is separated by a julia set. If one is repelling and one is attracting, then the repelling is in the Julia set. Best case scenario is that they are both repelling, and your holomorphic on deleted disks about the fixed point For an iteration like this, the best way to do it is to talk about upper half planes and lower half planes with a fixed point at $$\infty$$ in both cases, and somehow trying to glue them together on the real line. No fucking clue what that would look like though. tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 07/01/2022, 10:23 PM (07/01/2022, 10:03 PM)JmsNxn Wrote: Well right off the bat I'm going to say that you can't use a fixed point iteration. If $$f^{\circ s}(z)$$ is a holomorphic function on some domain $$\mathcal{A}$$, there cannot be any periodic or fixed points in $$\mathcal{A}$$ besides one fixed point. You can go one step further: If $$f^{\circ s}(z)$$ is holomorphic at a fixed point $$L$$, it is not holomorphic about any other fixed point/periodic point $$L_2$$ of $$f$$. This is because somewhere in this iteration we either hit a Julia set, or we hit different fatou set. Best case scenario we get something like Kneser's iteration, which is holomorphic everywhere EXCEPT at periodic/fixed points. What this means for your function is pretty simple. The iteration $$f^{\circ s}(z)$$ at $$0$$, has a discontinuity in $$z$$ as it tries to approach $$f^{\circ s}(z)$$ at $$1$$. This is because, even if both are attracting, somewhere $$z$$ will hit the Julia set; remember every basin is separated by a julia set. If one is repelling and one is attracting, then the repelling is in the Julia set. Best case scenario is that they are both repelling, and your holomorphic on deleted disks about the fixed point For an iteration like this, the best way to do it is to talk about upper half planes and lower half planes with a fixed point at $$\infty$$ in both cases, and somehow trying to glue them together on the real line. No fucking clue what that would look like though. like 2 black holes tearing up the space between them. well i was poetic ... JmsNxn Ultimate Fellow Posts: 1,179 Threads: 123 Joined: Dec 2010 07/01/2022, 10:35 PM (07/01/2022, 10:23 PM)tommy1729 Wrote: like 2 black holes tearing up the space between them. well i was poetic ... Lmao. It is sort of like two black holes, the orbits become untenable somewhere between the two and cause a short circuit. Yeah, I've encountered this proof by contradiction so many times trying to do interesting iterations. Fixed points really screw up everything, it's god damned frustrating. The central problem, is you can't move from Julia to Fatou, or Fatou to Julia. It just... breaks down. You could create a super function though... Not sure what it would look like. So if you were to fix $$z$$, and only care about $$s$$ this can leave the Julia set/fatou set, but you lose the semi-group property, and only have the super function identity. It kind of quarantines the errors to the abel function (which would allow you to move $$z$$). I wouldn't give up yet though, iterating $$z + \theta(z)$$ with a fixed point at $$\infty$$ in the lower and upper half planes is very viable. but then you'd have to restrict to the upper half plane or lower half plane in $$z$$. That would probably look pretty weird though. Then you may be able to at least have continuous at the fixed points on $$\mathbb{R}$$... I didn't mean to discourage you, just that local fixed point iteration is doomed here. tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 07/04/2022, 01:28 PM Yes only a fixpoint at +oo i. That is a nice idea. Unfortunately transcendental entire functions are not analytic at complex oo , what was the reason for not talking about it earlier. anyway , I considered f(z) = z + exp( 2 * 3.1415926535 i * z ) and plotted f(z),f(f(z)),f(f(f(z))) using Sage if i can thrust Sage they look like this : ( see pics , where the white bottom is 1 iter , the yellow bottom is 2 iter and the red bottom  is 3 iter. further iterates have red bottoms  ) One wonders how the half-iterate , 1/3-iterate etc look like. does it preserve the id(z) + periodic(z) structure ??   I thought I had a book about this , but turns out it was not that. regards tommy1729 Attached Files Thumbnail(s)             JmsNxn Ultimate Fellow Posts: 1,179 Threads: 123 Joined: Dec 2010 07/04/2022, 11:37 PM Hey, tommy. So a transcendental entire function does not have a fixed point at infinity you are right. Due to Picard's theorem. But a transcendental entire function restricted to $$\Im(z) > 0$$, can absolutely have a fixedpoint at $$i\infty$$. We usually just call that infinity though, where from context it's implied that it's $$i\infty$$ because we're in the upper half plane. This is no different than saying that $$e^z$$ has a zero at $$\infty$$ when $$\Re(z) < 0$$. Kind of an abuse of notation, but not really when you think of $$\infty$$ as a boundary value of a simply connected domain on the Riemann sphere. « Next Oldest | Next Newest »

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