(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.
(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:
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.)
To be honest this scenario is not "real life" to me. So even if you could prove it, it's not applicable to most cases (e.g. one of the fixed points is repelling and we don't talk about entire functions.)
The typical scenario is that we have some real analytic function with two adjacent fixed points (one attracting one repelling) with multiplier > 0, we know the regular iteration at each point asking whether both iterations are equal, or equivalently one iteration can be continued beyond the other fixed point. Similar (because the limit case of both hyperbolic fixed points get united) scenario: we have one parabolic fixed point with multiplier 1 and want to know whether the regular iteration is analytic there.
So the answer that was given by Karlin& McGregor (hyperbolic case) is: If the function is analytic on C except a closed countable set of isolated singularities, then it can not have such two fixed points except for LFTs. Similar (and earlier) result of Baker (parabolic case): If the function is meromorphic it can not have such fixed point except for LFTs. (And I even found a result of Liverpool that extends the set of functions similar to the one in the hyperbolic case: meromorphic single valued function with the exception of at most countable isolated singularities - I attach the paper to this post. It's theorem 3).
These statements still leave a lot of room for other kinds of functions where two fixed points can have the same regular iteration.
Also not nice in your assumptions is that you hide an aspect of the function that has nothing to do with continuous iteration, but comes already from integer iteration: Whether a singularity would come arbitrarily close to one fixed point. Your assumption looks already as if it only allows entire functions (in the case of repelling fixed point) - without even utilizing continuous iteration. Also "local iteration" seems not a very suitable term anymore, because you don't talk about a vicinity of a point, but a full fledged domain.
Your assumptions would imho be more reasonable if you wanted to take the functional limit of infinite iteration. There I would understand these assumptions.
I mean great if you intuitively know these properties from your extended study of iteration (theory), but I don't - totally don't - see it. So I would need a very small stepped proof. On the other hand you should rather invest your efforts to more fruitful enterprises than proving this statement!
Which I've yet to see a counter example for (so long as we do away with meromorphic functions), which all the evidence further points to.
This would mean, the only time you have a local iteration, is when it is the Schroder iteration about an attracting fixed point (or if you flip it on its head, A schroder iteration about a repelling fixed point). I've yet to see any counter evidence. And additionally, every Schroder iteration is conjugate to this scenario. So I don't share your point of view that this isn't in the real world, I think it very much is.
Also, it's much more important to the stuff MphLee and I talked about. Where we are talking about conjugating between Semi-group, and semi-group actions. I agree it doesn't help with iterating random functions--but foundationally I think it's very important. It would additionally show we can have a Canonical map between two function \(f,g\) which are locally iterable, and additionally share the same multiplier, that being \(\phi(z) = \Psi_f^{-1}(\Psi_g(z))\) which satisfies \(\phi g = f \phi\). Where \(\Psi\) is the Schroder map of the respective function.
But, then again, many of our mathematical preferences diverge, bo. So to each their own, lol.
08/21/2022, 02:32 PM (This post was last modified: 08/21/2022, 02:36 PM by bo198214.)
(08/21/2022, 01:29 AM)JmsNxn Wrote: Well the idea I have for local iteration, is that it is only possible if the function being iterated is conjugate to:
This, was never the problematic point. Problematic was this domain is also the domain for all (integer!) iterates. Which excludes all functions that have a singularity in the attractive basin around the fixed point.
Also Schröder iteration does not have this assumption, so it would not lead to local iteration in most cases. The convergence radius of the iterates can depend on t.
08/21/2022, 03:56 PM (This post was last modified: 08/21/2022, 04:09 PM by bo198214.)
Until now we were doing all our constructions \(f^{\circ t}(x)=\alpha^{-1}(t+\alpha(x))\) in the parabolic case with \(\alpha\) having a pole (i.e. a rather simple singularity) as \(\tan(x)\) has at \(\frac{\pi}{2}\) or \(\cot(x)\) has at 0.
However according to Ecalle starting with the julia function or logit j one can obtain \(\alpha(x)=\int \frac{{\rm d}x}{j(x)}\), and hence it has to be of the general form:
\[ c\log(x) + a_{-m}x^{-m} + ... + a_{-1}x^{-1} + a_0 + a_1 x + a_2 x^2 + ... \]
And the criterion that \(f^{\circ t}\) is analytic is only that the logit j converges.
So I was looking what one could construct with a \(\log\) and came up with this scenario:
\begin{align}
\alpha(x) &= \log(x) + \frac{1}{x}\\
\alpha'(x) &= \frac{1}{x} - \frac{1}{x^2}\\
j(x) &= \frac{1}{\alpha'(x)} = \frac{1}{\frac{1}{x} - \frac{1}{x^2}} = \frac{x^2}{x-1}
\end{align}
So the logit j is analytic at 0, which according to Ecalle causes \(f^{\circ t}\) to be analytic at 0 for all t too. How does \(f^{\circ t}\) look like? First we have to find the inverse of \(\alpha\):
\begin{align}
\log(x)+\frac{1}{x} &= y\\
xe^{\frac{1}{x}} &= e^y\\
-\frac{1}{x}e^{-\frac{1}{x}} &=- e^{-y}\\
-\frac{1}{x} &= W(-e^{-y})\\
x&=\frac{-1}{W(-e^{-y})}
\end{align}
Just need to choose the right branch of the LambertW (0 for x>0 and -1 for x<0 in \(f^{\circ t}(x)\)). So we have
\[ f^{\circ t}(x) = \frac{-1}{W(-e^{-(t+\log(x)+\frac{1}{x})})} \]
Nobody would believe that this gourd is analytic at 0, but according to Ecalle it is!
08/22/2022, 03:04 AM (This post was last modified: 08/22/2022, 03:31 AM by JmsNxn.)
(08/21/2022, 02:32 PM)bo198214 Wrote:
(08/21/2022, 01:29 AM)JmsNxn Wrote: Well the idea I have for local iteration, is that it is only possible if the function being iterated is conjugate to:
This, was never the problematic point. Problematic was this domain is also the domain for all (integer!) iterates. Which excludes all functions that have a singularity in the attractive basin around the fixed point.
Also Schröder iteration does not have this assumption, so it would not lead to local iteration in most cases. The convergence radius of the iterates can depend on t.
No no no no.
You've seemed to misunderstand.
There is always a half plane \(\mathcal{H}\) which is closed under additions of elements, where the Schroder iteration is conjugate to:
I'm aware the domain depends on \(t\) as we try to grow it indefinitely. But if we stick to the half plane where \(\mathcal{H} = \{t \in \mathbb{C}\,|\,|\lambda^t| < \delta\}\) for \(\lambda\) the multiplier, then we are stuck in a neighborhood of the fixed point (hence local iteration). Which is then conjugate to this scenario.
Choose \(|z| < \epsilon\) such that \(|\Psi(z)| < \rho\) and then \(|\Psi^{-1}(q)| < \epsilon\) when \(|q| < \rho\). Which this construction is always possible, so long as we shrink \(\epsilon, \rho\), and our multiplier \(|\lambda| < 1\). Then we have that: For \(\mathcal{H} = \{t \in \mathbb{C}\,|\,|\lambda^t| < 1\}\):
Where \(\mathcal{N} = \{z \in \mathbb{C}\,|\, |z| < \epsilon\}\).
This is where the "local iteration" terminology is inspired from. Which, as you can see, is biholomorphic to the above uses of the Unit disk.
What I want to say, is that this completely describes every possible local iteration of \(f : \mathbb{C} \to \mathbb{C}\). Which, iconically avoids the LFT conundrums. Such that we have an equivalence of statements: The Schroder iteration of a euclidean function \(f\) = The local iteration of a euclidean function \(f\). So that the equivalence is reflexive. Whereby, if we can identify that this is a local iteration of some kind, that it must be a Schroder iteration.
Again, this may seem stupid to you Bo, but it is related to Semi-group actions in particular. By which we can consider the semi-group of a half plane ACTING on the function \(f : \mathbb{D} \to \mathbb{D}\). This obviously has no effect of how we iterate, or iteration through numerical evidence, so on and so forth. But it does help a lot with the work Mphlee and I have been trying to put together. Which is far more algebraic stuff; nothing to do with plugging in numbers. I'll shut up now
08/22/2022, 03:09 AM (This post was last modified: 08/22/2022, 03:56 AM by JmsNxn.)
(08/21/2022, 03:56 PM)bo198214 Wrote: Until now we were doing all our constructions \(f^{\circ t}(x)=\alpha^{-1}(t+\alpha(x))\) in the parabolic case with \(\alpha\) having a pole (i.e. a rather simple singularity) as \(\tan(x)\) has at \(\frac{\pi}{2}\) or \(\cot(x)\) has at 0.
However according to Ecalle starting with the julia function or logit j one can obtain \(\alpha(x)=\int \frac{{\rm d}x}{j(x)}\), and hence it has to be of the general form:
\[ c\log(x) + a_{-m}x^{-m} + ... + a_{-1}x^{-1} + a_0 + a_1 x + a_2 x^2 + ... \]
And the criterion that \(f^{\circ t}\) is analytic is only that the logit j converges.
So I was looking what one could construct with a \(\log\) and came up with this scenario:
\begin{align}
\alpha(x) &= \log(x) + \frac{1}{x}\\
\alpha'(x) &= \frac{1}{x} - \frac{1}{x^2}\\
j(x) &= \frac{1}{\alpha'(x)} = \frac{1}{\frac{1}{x} - \frac{1}{x^2}} = \frac{x^2}{x-1}
\end{align}
So the logit j is analytic at 0, which according to Ecalle causes \(f^{\circ t}\) to be analytic at 0 for all t too. How does \(f^{\circ t}\) look like? First we have to find the inverse of \(\alpha\):
\begin{align}
\log(x)+\frac{1}{x} &= y\\
xe^{\frac{1}{x}} &= e^y\\
-\frac{1}{x}e^{-\frac{1}{x}} &=- e^{-y}\\
-\frac{1}{x} &= W(-e^{-y})\\
x&=\frac{-1}{W(-e^{-y})}
\end{align}
Just need to choose the right branch of the LambertW (0 for x>0 and -1 for x<0 in \(f^{\circ t}(x)\)). So we have
\[ f^{\circ t}(x) = \frac{-1}{W(-e^{-(t+\log(x)+\frac{1}{x})})} \]
Nobody would believe that this gourd is analytic at 0, but according to Ecalle it is!
Is this analytic in the domain \(|\Im t| < \delta\) (you can add \(\Re t > 0\), if you like) and \(|x| < \epsilon\)? Both domains fixed? This shouldn't be right, due to the branching of LambertW?
I just want to tie it in, so that you can see that it's not a "semi-group action in the complex plane" but is a holomorphic iteration. It's only a semi-group action on \(\mathbb{R}\), but trying to extend further from that, we encounter divergence (of the semi-group action).
I don't doubt this post is 100% correct. I'm just trying to shamelessly get my point across, lol
Where \(\mathbb{F} = \{(t,z) \in \mathbb{C}^2\,|\,f^{\circ t}(z) \,\,\text{is holomorphic}\}\). What I am asking of local iteration, is that this set is separable:
So another way of looking at this, is if the Domain of holomorphy, is separable in both variables. This lets us refer to semi-group actions, and additionally cleans up a lot of the differential theory when finding \(\text{logit}\).
(08/22/2022, 03:09 AM)JmsNxn Wrote: Is this analytic in the domain \(|\Im t| < \delta\) (you can add \(\Re t > 0\), if you like) and \(|x| < \epsilon\)? Both domains fixed?
The claim was that \(x\mapsto f^{\circ t}(x)\) is analytic for all t at 0. I didn't say anything about fixing the domain of analyticity. (Like I said several times this is so restrictive that you can not work with it in most cases. I think you can not even apply it to our beloved base sqrt 2.)
Does that answer your concerns or did you rather claim that it is not analytic even if we don't fix the domain?
Then I would have to investigate further ...
