Iteration with two analytic fixed points

[JmsNxn]

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 ...

No, bo! We're good. I just want to make sure we're on the same page. I'm in total agreement. Again, I'm just trying to carve out a difference in types of iterations.

You're completely okay (but you can apply it to \(\sqrt{2}\)).

I'll stop bringing it up further out.

[bo198214]

(but you can apply it to \(\sqrt{2}\)).

Ok, I think that is where we stopped with your proof attempts ... for base \(\sqrt{2}\) one has as domain whole \(\mathbb{C}\).
And we know already by several means (one from Karlin&McGregor, but also because the multipliers are not reciprocal) that it can not have the same regular iteration at both fixed points, but maybe its still interesting to have another proof.
So here you have met all your assumptions, how would *your* proof go in this scanario (function is entire, domain is \(\mathbb{C}\))?

[JmsNxn]

(but you can apply it to \(\sqrt{2}\)).

Ok, I think that is where we stopped with your proof attempts ... for base \(\sqrt{2}\) one has as domain whole \(\mathbb{C}\).
And we know already by several means (one from Karlin&McGregor, but also because the multipliers are not reciprocal) that it can not have the same regular iteration at both fixed points, but maybe its still interesting to have another proof.
So here you have met all your assumptions, how would *your* proof go in this scanario (function is entire, domain is \(\mathbb{C}\))?

There exists an immediate attracting basin \(\mathcal{A}\) about the fixed point \(z=2\). This domain is connected and open. By which:

\[
\begin{align}
f^{\circ n}(z) &\to 2\,\,\text{as}\,\,n\to\infty\\\\
f(z) &= \sqrt{2}^z\\
\end{align}
\]

And \(A\) is defined as the maximal connected domain which satisfies this--where \(2 \in A\) and \(f : A \to A\).

The Schroder function is holomorphic here; By which \(\Psi(A) \to \mathbb{C}\). This function is nonsingular, by which:

\[
\frac{d}{dz} \Psi(z) \neq 0\\
\]

So that locally, near \(z = 2\) we have:

\[
f^{\circ t}(z) = \Psi^{-1}\left(\log(2)^t \Psi(z)\right)\\
\]

For \(|\log(2)^t| < 1\). This function is analytically continuable to:

\[
f^{\circ t}(z) : \mathbb{C}_{\Re(t) > 0} \times A   \to A\\
\]

This can be constructed in a manner of ways. The manner I find myself most comfortable with is with Mellin transforms. We can always express these \(\sqrt{2}^z\) iterates using integrals.

The second way, is the manner that I code in my solutions. These would essentially just be: program in the inverse Schroder, program in the Schroder. Add in a recursive protocol, which pulls us near \(z=2\) for high accuracy in Taylor data, then pull back using the inverse. And this will run smoothly.

From here, we can note that \(A\) is actually simply connected--which is written in Devaney's Chaotic dynamical systems: Attracting basins of geometric fixed points are simply connected. So even on its maximal domain \(f^{\circ t}(z) : A \to A\) is mappable to \(f^{\circ t}(z) : \mathbb{D} \to \mathbb{D}\)--So long as \(f\) is euclidean (transcendental entire), we are okay.

Again, all the crazy behaviour starts happening when \(t \to -\infty\), where the neighbourhoods aren't as nice. And there's no iteration for \(t \in \mathbb{C}\). But there is a local iteration if we fix \(\Re(t) > 0\). Where upon, we are somewhat locally around the fixed point. As you limit \(t \to -\infty\) everything turns to chaos! And yes it can still be holomorphic there. But it may not be a local iteration, in the sense that it's writable as \(f^{\circ t}(z) : \mathcal{H} \times A \to A\).

I don't want to press this too far Bo, I just hope you can understand that a local iteration is a stricter requirement than Regular iteration; but the two ideas pretty much coincide everywhere.

EDIT:

I'd also like to add, that a lot of your uses of regular iterations, especially lately, are Group actions. These are essentially \(\{\mathbb{R}, +\}\) group actions on a function. I am focused on \(\{\mathcal{H}, +\}\) semi-group actions on a function. And they have their disagreements, despite being expandable in the same manner and ultimately just being regular iteration in a different disguise.

Don't get me wrong, everything I'm doing is regular iteration, just with further restrictions

[Daniel]

Sorry, but what are the definitions for local and regular iteration?

[JmsNxn]

Sorry, but what are the definitions for local and regular iteration?

As I've defined Local iteration, is a little sketchy; but it is essentially the idea that the iterate \(f^{\circ t}(z)\) can be split into:

\[
f^{\circ t}(z) : D \times A \to A\\
\]

For some fixed domains \(t \in D\) and \(z\in A\)--where \(t_0,t_1 \in D\) means \(t_0+t_1 \in D\). This only ever happens "locally" or when we have a Schroder function. It can't happen anywhere else; and then, it only happens in a half plane and near the fixed point.

Regular iteration is much more general, and means to solve the Taylor expansion similar to what you've done, Daniel. We are just solving a problem with the coefficients of the \(N\)'th order polynomial. Additionally Regular iteration handles pretty much everything about a fixed point. To remind you, much of your Matrix equations are actually just producing regular iteration.

Local iteration is a term I am arguing for; in that it means that we can split the domains like I wrote above. And they are still "regular iterations" but they are specific kinds of regular iterations.

Hope that helps

[Daniel]

Sorry, but what are the definitions for local and regular iteration?

As I've defined Local iteration, is a little sketchy; but it is essentially the idea that the iterate \(f^{\circ t}(z)\) can be split into:

\[
f^{\circ t}(z) : D \times A \to A\\
\]

For some fixed domains \(t \in D\) and \(z\in A\). This only ever happens "locally" or what we have a Schroder function. It can't happen anywhere else; and then, it only happens in a half plane and near the fixed point.

Regular iteration is much more general, and means to solve the Taylor expansion similar to what you've done, Daniel. We are just solving a problem with the coefficients of the \(N\)'th order polynomial. Additionally Regular iteration handles pretty much everything about a fixed point. To remind you, much of your Matrix equations are actually just producing regular iteration.

Local iteration is a term I am arguing for; in that it means that we can split the domains like I wrote above. And they are still "regular iterations" but they are specific kinds of regular iterations.

Hope that helps

Awesome, thanks for the info.

[bo198214]

There exists an immediate attracting basin \(\mathcal{A}\) about the fixed point \(z=2\). This domain is connected and open.

I am not seeing how this proves that both fixed points can not have the same iteration. Your approach looks a bit like:
We take the attracting basin continue the iteration to this domain and then show that it can not have a repelling fixed point.

But that was clear from the beginning already, the attractive basin can not contain any other fixed points by definition.
What would be needed is to show that there is no way to continue the iteration beyond the attractive basin, i.e. to the repelling fixed point.

I don't want to press this too far Bo, I just hope you can understand that a local iteration is a stricter requirement than Regular iteration; but the two ideas pretty much coincide everywhere.

I thought I emphasized that too much already, and now you ask me whether I understand that? (That it is regular iteration is just by be analytic at the fixed point, that's how regular iteration is defined.)

[JmsNxn]

There exists an immediate attracting basin \(\mathcal{A}\) about the fixed point \(z=2\). This domain is connected and open.

I am not seeing how this proves that both fixed points can not have the same iteration. Your approach looks a bit like:
We take the attracting basin continue the iteration to this domain and then show that it can not have a repelling fixed point.

\(\mathcal{A}\) is a maximal domain of this iteration (it equates to a boundary of a function whose maximal domain is \(f : \mathbb{D} \to \mathbb{D}\)). So in \(\sqrt{2}\)'s case, the value \(4\) is on the boundary of \(\mathcal{A}\)--and this function is not holomorphic anywhere on the boundary of \(A\). Thereby we cannot have that \(f^{\circ t}(z)\) is holomorphic near \(4\). Because this is the maximal domain. The only draw back to this statement, is that we are not using LFT's; which are meromorphic except for \(\lambda z + c\)--and these are mappable to \(\lambda z\). And then, this iteration has the entirety of \(\mathbb{C}\) as their \(A\). This is the only example of such happening. Where it is the only example of additionally having a fixed point at \(\infty\) which is repelling.



Trying to say that we can iterate about two fixed points in the manner I am talking is just flat out wrong. And you continue to doubt it. You can iterate about two fixed points, but never in a semi-group manner. And especially! you don't even come close for entire transcendental functions (mappings of \(\mathbb{C} \to \mathbb{C}\) with an essential singularity at \(z = \infty\)).



So for you to ask that a function: \(f : \mathbb{C} \to \mathbb{C}\) has a "local iteration" representation about two fixed points is nonsense. And, frankly, I'm a little tired of having to continuously argue this point. You have iterated LFTs and Meromorphic functions--which are of an entirely different mathematical class. For entire functions; even polynomial ones--you can't have a SEMI-GROUP ACTION holomorphic at two fixed points.

We can find half roots, and third roots, which are holomorphic on larger domains than \(A\); but the thing is, they are not holomorphic for \(\Re(t) > 0\).

I hope to not burn bridges. But I'll leave it alone now. I don't want to talk about this anymore.



Every function \(b^z\) for \(1 < b < \eta\) follows this rule. The main repelling fixed point is on the boundary of \(A\), and within \(A\) is the left half plane, and a good amount of the right half plane.



And the function \(f^{\circ t}(z)\) has its MAXIMAL DOMAIN on \(A\) for \(\Re(t) > 0\). By which--even asking that it can be reconcilable with another fixed point is nonsense. \(4 \not\in A\).



ADDITIONALLY: \(A\) is always simply connected; and therefore mappable to \(\mathbb{D}\). So that all of your regular iterations, are actually actions of the semigroup \(\mathbb{C}_{\Re(t) > 0}\) (or some equivalent halfplane) on \(f: \mathbb{D} \to \mathbb{D}\) with \(f(0) = 0\). The only time your regular iterations stray from this is with parabolic iteration. Whereby you are mapping \(f: \mathbb{D} \to \mathbb{D}\) but \(f(1) = 1\)--so the fixed point is on the boundary.



I'm sorry bo, but for you to say I am the one repeating myself isn't correct. You are making similar claims over and over; and I'm trying to reexplain something that is taught heavily in complex dynamics. We choose the dynamics of our functions like we choose our fixed points. There's no having your cake and eating it too.

I have nothing but respect for you, but I'm just very frustrated right now. Everyone of your holomorphic solutions satisfies the rule \(\mathbb{C}_{\Re(t) > 0}\) acts on \(f\) near the first fixed point; and \(\mathbb{C}_{\Re(t) < 0}\) acts on the other point--and that is absolutely no coincidence. \(\mathbb{C}_{\Re(t) > 0}\) cannot act on \(f\) about both fixed points.

I'm fine being an asshole here--But much of what you've been arguing has been wrong since my original statements; in the sense that you've found contradicitons. I am trying to classify the types of regular iterations. And one of the most important classes is: SEMIGROUP ACTION. It's pretty much all everyone cares about.

[bo198214]

So this seems to be a dead end, it is as frustrating for me.
Ok, try it in a different way. You say:

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 \(f(p) = p\).

But for me this has nothing to do with non-integer iteration. Your proposition would already be valid for just (positive) integer iteration, isn't it? (I rephrase your proposition to see what I mean)

if \(f^{\circ n}(z) : \mathbb{N} \times H \to H\) for a domain \(H\subset \mathbb{C}\):

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

I didn't think deeply about it. It could be wrong. If you have a counter-example please put it out.

But if this is true, then the preconditions of your statement are so restrictive that just no function with two fixed points can fit.
And then you can not call it a statement about non-integer iteration.

I'm sorry bo, but for you to say I am the one repeating myself isn't correct.

This is already the second time that you claim I said something that I didn't say (and also didnt mean).
Please be more careful what you allege people have said.
I don't know how one can misread this:

I thought I emphasized that too much already, and now you ask me whether I understand that?

[JmsNxn]

So this seems to be a dead end, it is as frustrating for me.
Ok, try it in a different way. You say:

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 \(f(p) = p\).

But for me this has nothing to do with non-integer iteration. Your proposition would already be valid for just (positive) integer iteration, isn't it? (I rephrase your proposition to see what I mean)

if \(f^{\circ n}(z) : \mathbb{N} \times H \to H\) for a domain \(H\subset \mathbb{C}\):

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

I didn't think deeply about it. It could be wrong. If you have a counter-example please put it out.

But if this is true, then the preconditions of your statement are so restrictive that just no function with two fixed points can fit.
And then you can not call it a statement about non-integer iteration.

I'm sorry bo, but for you to say I am the one repeating myself isn't correct.

This is already the second time that you claim I said something that I didn't say (and also didnt mean).
Please be more careful what you allege people have said.
I don't know how one can misread this:

I thought I emphasized that too much already, and now you ask me whether I understand that?

Ya I apologize bo. I was in a bad mood when I made that post and it was late and I was grumpy and tired, lol.

The only difference, which I would say that is different from what you are saying; is that yes you are right if \(H\) is simply connected (and mappable to \(\mathbb{D}\)) you are correct. It can be reduced to the action \(\mathbb{N} \times H \to H\). If \(H\) is not simply connected this isn't necessary. But If you require that the function is a semi-group action; this will force only one fixed point, for an arbitrary domain \(H\). And then we're pretty much back to your scenario of \(\mathbb{N}\)-actions. I see that as a strength though (which means we are reducing the fractional iteration to the natural action).

Another point, is that these iterations are always periodic if the fixed point is on the interior: (It's mapping a s.c. domain to a s.c. domain and has a fixed point on its interior--and therefore is geometric). Therefore since it's periodic, you can make "disks" about the fixed point by drawing Jordan curves like:

\[
\partial\mathcal{B}(z) = \{f^{\circ it}(z)\,|\, 0 \le t < \ell\}\\
\]

(where \(i \ell\) is the period of \(f^{\circ t}\)).

Which the interior of this jordan curve, has the fixed point \(p\), and additionally is invariant under maps. The interior looks like:

\[
\mathcal{B}(z) = \{w \in \mathbb{C}\,|\, |w - p| < |f^{\circ it}(z) - p|\}\\
\]

where \(f : \mathcal{B} \to \mathcal{B}\) for arbitrary \(z\).

This crucially uses the semi-group action--which only happens about a fixed point if its contracting (\(\mathcal{B}\) in this case, is a simply connected domain). To ask that you have \(\mathcal{B}(z) \cap \mathcal{B}(z') = \emptyset\) can't happen, unless we are no longer univalent in the strip \(0 < \Im(t) < \ell\) & \(\Re(t) > 0\), but every Schroder iteration is univalent here.

The only exception to this rule, would be if the semi-group action fails; we cannot construct \(B(z)\) because for some \(t\), the value \(f^{\circ t}(z)\) diverges, or somewhere within its interior. This would equate to your observations involving a fixed point with mult \(\lambda\) and nearby fixed point with mult \(1/\lambda\).

Where about the second fixed point we have a semigroup action for \(\Re(t) < 0\), and the first fixed point still has the semi group action for \(\Re(t)> 0\); And additionally these will have different domains in \(z\). We would have an \(A^+\) and an \(A^-\) attracting basins for the separate cases. Then if \(A^+\) and \(A^-\) intersect non trivially; we must have that the Schroder functions about both don't agree (otherwise they would create a non permissable function \(f^{\circ t}(z) : \mathbb{C} \times A^+ \cap A^- \to A^+ \cup A^-\)--they're not the same function. Because any entire function can at most omit a single point. Here we  crucially assume that \(f\) is euclidean; by which there are infinite periodic points, where therefore \(A^+ \cup A^- \subset \mathbb{C}\), so there can't be an entire function which sends there--\(A^+ \cup A^-\) omits more than one point.

Which is again seen by your iterations about two fixed points; in that as we let \(t \to -\infty\) or we grow \(t\)'s domain, we start entering in \(z\)'s domain being dependent on \(t\)'s domain.

---

But it seems we have hit a stand still. I rather understand my definition is very restrictive. But you get A LOT of cool shit for free when you are this restrictive, which is why I like it. And why I like to see if iterations can be reduced to this case. It can extend to parabolic iterations too, but there also it is very restrictive. Just swap \(f(0) = 0\) in \(f: \mathbb{D} \to \mathbb{D}\) to \(f(1) = 1\).

Either way, it seems it's a large divergence of our interests. As this is needed when we start talking about group actions on arbitrary integrals, rather than integral equations of the form \(y '(x) = p(x)g(y(x))\) (which generates a semi-group). So it interests me. Where as you just want to have a god damned iteration--which I can respect.

---

Again, I'd like to apologize for my tone in the last post; it was a little rude of me. Being in a bad mood is no excuse to be rude.