Posts: 1,616
Threads: 102
Joined: Aug 2007
I just was investigating the subject with linear fractional functions, and I came up with this example:
\(f(z)=\frac{2z}{1+z}\)
This function has exactly two fixed points, one at 0 and one at 1.
With a bit of calculating one can come up with an explicit formula for the iteration of this function:
\(f^{\circ t}(z)=\frac{2^tz}{1+(2^t1)z}\)
As all the iterates of the function are analytic at both fixed points, it must be the regular iteration at both fixed points.
And hence we have an example of a function that can be analytically iterated at 2 fixed points at the same time.
Of course one can say linear fractional functions are somehow the trivial regular iterations (though I am not sure whether I could really give an explicit formula for the general iteration of \(f(z)=\frac{a+bz}{c+dz}\) ). But then for which other functions is that also possible?!
Posts: 1,616
Threads: 102
Joined: Aug 2007
08/02/2022, 10:18 PM
(This post was last modified: 08/02/2022, 10:20 PM by bo198214.
Edit Reason: typo
)
And just as we are at the topic, I also have an example for a function having analytic iteration at a *parabolic* fixed point: \(f(z)=\frac{z}{1+z}\)
I was always wondering when that case would occur.
This function has a parabolic fixed point at 0 and one can write an explicit formula for the regular iteration:
\(f^{\circ t}(z) = \frac{z}{1+tz}\) which is analytic at 0 for all t!
So this makes it a bit clearer in my head:
It seems that a regular iteration at a parabolic fixed point is analytic is as rare as an iteration is analytic at 2 hyperbolic fixed points.
It would even say that they are related:
if and only if the regular iteration at a parabolic fixed point is analytic then the perturbed function (with two (or more?) exploded hyperbolic fixed points is analytic at all fixed points).
And maybe this scenario happens only for fractional linear functions, who knows.
But I really withdraw my conjecture that most parabolic fixed points of \(b^z\) have iterations analytic at the fixed point.
Posts: 1,176
Threads: 123
Joined: Dec 2010
08/03/2022, 02:52 AM
(This post was last modified: 08/03/2022, 08:12 AM by JmsNxn.)
I think this makes sense, particularly because these are meromorphic functions.
Iterating meromorphic functions tend to be a whole nother beast compared to iterating transcendental entire functions. If we want to go this route, we equally have an iteration for \(f^{\circ t}(z) = \lambda^t z\), this is holomorphic at \(\infty\) and at \(0\), where they are both fixed points. By which any conjugation using linear fractional transformation makes another solution. We can also probably map \(\infty \mapsto A\) and \(0 \mapsto B\), then we have a holomorphic iteration at both fixed points.
But! and this is the big but, this is an iteration of a meromorphic function, not an iteration of an entire function.
Things get a lot more complicated with meromorphic functions. I challenge you to find an entire function that can be iterated about two fixed points!
This reminds me of the problem, there are only two kinds of entire fractional iterations:
$$
\begin{align}
f^{\circ t}(z) = z_0 + \lambda^t(zz_0)\\
g^{\circ t}(z) = z+tz_0\\
\end{align}
$$
But as soon as you relax entire into meromorphic, there are a bunch of linear fractional transformation solutions to a meromorphic fractional iteration.
So essentially, because you have added poles to your iteration, this greatly changes the problem. Milnor has nothing to say about this.
It's also important to note that this still isn't a counter example to the proposition. Because this iteration does not have the local expansion:
$$
\begin{align}
f^{\circ t}(z) &= \lambda^t z + O(z^2)\,\,\text{for}\,\,z<\delta\,\,\Re(t) > 0\\
f^{\circ t}(z) &= 1 + \mu^t (z1) + O(z^2)\,\,\text{for}\,\,z1 < \delta\,\,\Re(t) > 0\\
\end{align}
$$
Because there are poles in \(f^{\circ t}\) at precisely:
$$
\begin{align}
(2^t1)z &= 1\\
2^t  1 &= 1/z\\
t &= \log(11/z)/\log(2) + k\pi i/\log(2)\,\,\text{for}\,\,k\in\mathbb{Z}\\
\end{align}
$$
So take \(\Im(z) > 0\) and \(z < \delta\), then there's a pole in t at about \(t \approx K \) for \(K\) arbitrarily large as \(z \to 0\) (choose k = 1). This means the above local expansion is invalid for \(z \approx 0\). I would call this a different kind of iteration than regular, simply because it doesn't have the taylor expansion at 0 that we typically see with regular iteration. But that's for you to decide.
To clarify, what I mean by no analytic iteration at both fixed points, I mean the above expansion is valid. And as I said before, this can't happen because the Schroder coordinate change will become discontinuous somewhere along a path to the other fixed point.
So this function is not holomorphic for \(\Re(t) > A\) for some \(A \in \mathbb{R}\) about \(z =0\) and about \(z=1\). I guess I should've clarified that's what I mean by no holomorphic iteration about two fixed points.
Also, I think this is a very important distinction we should be considering. Is it okay to let the regular iteration have poles? Is it still regular iteration if there is a pole. If the answer is yes, I suggest that we make a restriction to types of regular iteration, one without poles and one with poles.
For example, using your parabolic iteration, there are poles in \(t\) at precisely:
$$
t = 1/z\\
$$
Is that still allowed as a regular iteration? I've never seen the term regular iteration, which allows for poles, in either variable, but if so, then yes I'll withdraw the claim that holomorphic about two points, and change it to, a regular iteration in a half plane can't contain two fixed points. And additionally, recall that there are no poles in \(z\) or poles in \(t\). Noticeably, this function \(f^{\circ t}(z)\) is holomorphic on \(\mathbb{C}^2 /\{t = 1/z\}\) which is a very different type of domain than what you will see with entire functions. If we let \(z\) get arbitrarily large here, then the function is getting more and more volatile near \(t = 0\)and similarly if you take very large iterates \(t >T\) then we shrink the domain of holomorphy about \(z=0\). A feature we'll never really see with entire functions.
To me, iteration about a fixed point, means that we have a fixed neighborhood \(N\) and a fixed domain \(D\) in \(\mathbb{C}\) such that this object is holomorphic; and D is closed under addition of elements; \(\{D,+\}\) is a semigroup. Notably, this parabolic iteration is \(\mathbb{R}\) analytic in \(t\), but isn't holomorphic for any strip \(\Im(t) < \delta\) and \(z < \delta\). You have to shrink \(\delta\) as \(t\) grows, and thereby you are making a more dynamic kind of iteration.
It doesn't have the form: \(f^{\circ t}(z) : D \times N \to N\); where we are mapping a semi group, but it does have the form \(f^{\circ t}(z) : \mathbb{R} \times N \to N\). This is then a real iteration in my book, and not a complex iteration.
Because yes, there are singularities with \(\sqrt{2}^z\)'s regular iteration about \(2\), but they are not poles. They are logarithmic singularities/ branching problems, and additionally locally about \(z \approx 2\) it's holomorphic in a half plane (which is a semigroup under \(+\)). Which is a standard feature of Schroder iteration. Poles require us to think of these things meromorphically, and that can change a lot of the rules of how we manipulate these things.
I think this is a very important problem that we'll have to clarify our language on. Neither of these iterations are holomorphic on a domain \(D \subset \mathbb{C}\) that is a semigroup under \(+\), while additionally \(\mathbb{R}^+\subset D\); while \(z \in N\), where \(N\) is a neighborhood about the fixed point \(p\); \(N = \{zp < \delta\}\). That's the exception of these iterations, which breaks the law that you can't iterate about two fixed points; you aren't mapping a semigroup in the complex plane; you're mapping a semi group on \(\mathbb{R}\), or you are mapping a semigroup in \(\widehat{\mathbb{C}}\). Which is a very important distinction.
I think the really technical statement is: On a closed Riemann Surface (\widehat{\mathbb{C}}\)) we can iterate about multiple fixed points. On open Riemann surfaces (for example the open \(\mathbb{C}\))we cannot iterate about two fixed points. And specifically in this case. This is how Milnor would handle this "counterexample".
Posts: 1,616
Threads: 102
Joined: Aug 2007
08/03/2022, 08:12 AM
(This post was last modified: 08/03/2022, 08:17 AM by bo198214.)
No, no, no James, you can not wind that easily out of the situation with statements like "can you even call it regular iteration" ...
Regular iteration is not something designed for entire function or so (OMG its meromorphic  I die).
It is a general method for a function analytic at a fixed point (i.e. in a (possibly small) vicinity of the fixed point) and this is what I provided.
The given constructions are as proper regular iterations as it can get.
(08/03/2022, 02:52 AM)JmsNxn Wrote: It's also important to note that this still isn't a counter example to the proposition. Because this iteration does not have the local expansion:
$$
\begin{align}
f^{\circ t}(z) &= \lambda^t z + O(z^2)\,\,\text{for}\,\,z<\delta\,\,\Re(t) > 0\\
f^{\circ t}(z) &= 1 + \mu^t (z1) + O(z^2)\,\,\text{for}\,\,z1 < \delta\,\,\Re(t) > 0\\
\end{align}
$$
Because there are poles in \(f^{\circ t}\) at precisely:
$$
\begin{align}
(2^t1)z &= 1\\
2^t  1 &= 1/z\\
t &= \log(11/z)/\log(2) + k\pi i/\log(2)\,\,\text{for}\,\,k\in\mathbb{Z}\\
\end{align}
$$ Only because there are poles in t does not mean that it lacks the given expansion. Put it in your favourite computer algebra tool and look at the coefficients, these are:
$$\frac{2^t z}{1+(2^t1)z} = 2^t z (2^t  1)2^t z^2 + (2^t  1)^22^t z^3 (2^t  1)^32^t z^4 + \dots$$
$$\frac{z}{1+tz}=z tz^2 + t^2z^3 t^3z^4 + \dots$$
Remember regular iteration was always a local property. We take the powerseries expansion at the fixed point, then there can only be one formal powerseries parametrized by t hat satisfies \(f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}\). Then we prove that this powerseries has a radius of convergence for all t (hyperbolic case), which it has i.e. my hyperbolic series converges for \(z<\left\frac{1}{2^t1}\right\).
This is the standard setup of regular iteration I remember a theorem of Écalle that states  in the parabolic case  that this formal iteration powerseries has a positive radius of convergence for all t if and only if the iterative logarithm series has a positive radius of convergence.
The iterative logarithm or Julia function \(j\) is a function that satisfies \(j\circ f = f' \cdot j\) and it has properties similar to the normal logarithm, i.e. \(\text{logit}[f^{\circ t}] = t\cdot\text{logit}[f]\) (and one can construct the Abel function via \(\int \frac{1}{j} \)).
We can verify Écalles theorem with our parabolic example:
$$\text{logit}\left[\frac{z}{1+z}\right] =  z^2$$ (do the math yourself)
And voilá this "series" has infinite radius of convergence.
This brings me to an idea of reverse engineering, just looking at simple expression for the logit and then obtaining a function with parabolic fixed point that also has positive radius of convergence for all t.
But alas I am anyways already above my time budget for tetration  not good for my health *sigh*
Quote:To me, iteration about a fixed point, means that we have a fixed neighborhood \(N\) and a fixed domain \(D\) in \(\mathbb{C}\) such that this object is holomorphic; and D is closed under addition of elements; \(\{D,+\}\) is a semigroup. Notably, this parabolic iteration is \(\mathbb{R}\) analytic in \(t\), but isn't holomorphic for any strip \(\Im(t) < \delta\) and \(z < \delta\). You have to shrink \(\delta\) as \(t\) grows, and thereby you are making a more dynamic kind of iteration.
...
But honestly James this is a lot of explaining why a counter example does not work. Better would be that these cases come up while developing the proof, because I can not see the role of this demand that we have a fixed \(D\times N\) in a proof, on the other hand I see your point that this would be a nice property to have.
Posts: 1,176
Threads: 123
Joined: Dec 2010
Again, bo. I agree with everything. But you summarize my disagreement in the statement:
$$
z < \frac{1}{2^t 1}\\
$$
That's a real iteration. Sure, then that's the regular iteration.
Then, LOCAL ITERATION, as I described is an entirely different idea.
WE MUST BE ABLE TO FIX \(\delta\) in
$$
z < \delta\\
$$
What you have produced is a regular iteration, that IS NOT A LOCAL ITERATION.
So you have admitted that when I say local iteration it is not a regular iteration. And the definitions intersect, but are not the same.
There can be no local iteration about two fixed points. And you are trying to prove something I absolutely agree with. But our definitions do not align. And your statement that a local iteration is regular iteration is false.
Posts: 1,616
Threads: 102
Joined: Aug 2007
08/03/2022, 10:38 AM
(This post was last modified: 08/03/2022, 10:42 AM by bo198214.)
So what.
For me there was the conjecture in the space that an iteration can not be holomorphic at two (hyperbolic) fixed points at the same time. Which would boil down to if I have a regular iteration at one point then there is no continuation to the other fixed point. (And it would be totally enough for me to have this statement on the real line only!)
And I simply gave a counter example to that.
I though it was clear from my last sentence that I understood what you would like to have  a vicinity of the fixed point that is independent of t.
Good, if that helps you with the proof. But I neither saw a proof nor even a fixed conjecture.
Ps: I never claimed equality of regular iteration with what you call local iteration. I did not even mention "local iteration" in my post.
Posts: 247
Threads: 79
Joined: Aug 2007
(08/03/2022, 10:38 AM)bo198214 Wrote: So what.
For me there was the conjecture in the space that an iteration can not be holomorphic at two (hyperbolic) fixed points at the same time. Which would boil down to if I have a regular iteration at one point then there is no continuation to the other fixed point. (And it would be totally enough for me to have this statement on the real line only!)
And I simply gave a counter example to that.
I though it was clear from my last sentence that I understood what you would like to have  a vicinity of the fixed point that is independent of t.
Good, if that helps you with the proof. But I neither saw a proof nor even a fixed conjecture.
Ps: I never claimed equality of regular iteration with what you call local iteration. I did not even mention "local iteration" in my post.
As I have mentioned before, I did a numerical experiment that from a fixed point my method of iteration was able to show where the next fixed point was and what it's Lyapunov multiplier was. If it is important I could try and recreate my experiment.
Daniel
Posts: 1,616
Threads: 102
Joined: Aug 2007
(08/03/2022, 12:33 PM)Daniel Wrote: As I have mentioned before, I did a numerical experiment that from a fixed point my method of iteration was able to show where the next fixed point was and what it's Lyapunov multiplier was. If it is important I could try and recreate my experiment. Yes, Daniel, you mentioned that several times, but until you concretely show what you did it remains quite vague for me what it could be.
So why not showing us (in a separate thread), what you did exactly!
Posts: 247
Threads: 79
Joined: Aug 2007
(08/03/2022, 02:23 PM)bo198214 Wrote: (08/03/2022, 12:33 PM)Daniel Wrote: As I have mentioned before, I did a numerical experiment that from a fixed point my method of iteration was able to show where the next fixed point was and what it's Lyapunov multiplier was. If it is important I could try and recreate my experiment. Yes, Daniel, you mentioned that several times, but until you concretely show what you did it remains quite vague for me what it could be.
So why not showing us (in a separate thread), what you did exactly!
Good idea. Now that I have Mathematica again I should be able to make a notebook where I can demonstrate my point.
Daniel
Posts: 1,176
Threads: 123
Joined: Dec 2010
08/04/2022, 05:20 PM
(This post was last modified: 08/04/2022, 05:51 PM by JmsNxn.)
Alright, I feel this is the thread where I should explain why there can't be an iteration about two fixed pointsit comes with a few caveats.
To begin, let me define what I refer to as a local iteration, which is a strengthening of regular iteration (so it is more restrictive). Let's assume that \(D\subset \mathbb{C}\) is a domain that is closed under addition of elements. So that \(\{D,+\}\) is a semigroup.
Let's also assume that \(E\) is a SIMPLY CONNECTED DOMAIN (this is the missing criteria which allows for the double fixed point iteration that Bo has found). Let's assume that \(f : E \to E\) and then a local iteration for \(t \in D\) and \(z \in E\), is a holomorphic function:
$$
f^{\circ t}(z) : D \times E \to E\\
$$
Where \(f^{\circ 1}(z) = f(z)\) and this function satisfies \(f^{\circ t_0}(f^{\circ t_1}(z)) = f^{\circ t_0 + t_1}(z)\).
To begin, since \(E\) is simply connected, it is safe to assume that \(E = \mathbb{D}\) the unit disk (conjugate with a riemann mapping). A mapping from the unit disk to itself always has a fixed point, and so again, without loss of generality, we can assume one of the fixed points is \(0\). By schwarz's lemma, this means that \(f'(0) \le 1\) and that \(f(z) \le z\). Where \(f(z) = z\) only when \(f'(0) = 1\). This means that \(f(z) = e^{i\theta}z\) for some \(\theta \in [0,2\pi)\).
So, if \(f'(0) < 1\), then \(f(z) < z\) and there are no more fixed points other than \(0\) by construction. So all that's left is if \(f\) has a neutral fixed point at \(0\). Wlog we can assume that \(f'(0) = 1\). Well then \(f(z) = z\), which is only fixed points, and is therefore the trivial iteration. And can be discarded. Or you can just observe that \(e^{i\theta}z\) has no other fixed points than zero.
So there you have it, this is what I mean when I say you can't have an iteration about two fixed points. Bo's solution explicitly doesn't satisfy these conditions. In fact, you can make infinitely many of these iterations about 2 fixed points.
Let \(h^{\circ t}(z) = \lambda^t z\) for some \(\lambda \in \mathbb{C}/\{0\}\). Take a linear fractional transformation which maps \(\infty \to A\) and \(0 \to B\). Let's call it \(\mu\). Then:
$$
g^{\circ t}(z) = \mu(h^{\circ t}(\mu^{1}(z)))\\
$$
Is a holomorphic fractional iteration (it will be regular), about both fixed points \(A\) and \(B\).
This clears up what I was looking for, and what I mean by no iteration about two fixed points; I specifically mean no local iteration as I've defined above. Again, I think the main issue we were having, Bo, was semantics.
Hope that makes sense
EDIT: Additionally I should add, that for the case \(b^z\), so long as the fixed point isn't parabolic, then the immediate basin of attraction of any fixed point (or a fixed point of \(\log_b\)), is simply connected. This can be found in Introduction to Chaotic Dynamical Systems by Devaney. By which, the maximal domain of the iteration is this basin, as the Julia set is the boundary of the basin. Additionally any orbit of the julia set gets arbitrarily close to infinity, in a singular manner.
