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09/03/2022, 07:49 PM
(This post was last modified: 09/03/2022, 07:56 PM by bo198214.)
So I was approaching that from a different angle:
\[ z + z^2 = \lim_{c \to 1} cz+z^2 \]
Because for \(0<c<1\) we have the regular iteration, which allows us to compute \(f^{\circ t}_c(z_0)\) for \(z_0 \) in the basin of attraction of 0.
Just drawing for c=0.5 and c=0.9:
So what happens if we further move \(c\to 1\):
I guess in the limit \(f^{\circ \frac{1}{2}}(x) = 0 = f^{\circ \frac{3}{2}}(x)\) for \(x>0\) (and in consequence for all real \(x\)).
And this is the reason, why the half iterate does not exist.
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09/04/2022, 06:50 AM
(This post was last modified: 09/04/2022, 06:51 AM by bo198214.)
Maybe the previous animation is even a bit misleading, not only \(\lim_{c\to 1}f^{\circ \frac{1}{2}}(x) = 0\), but rather:
\[ \lim_{c\to 1} f^{\circ t}(x) = 0,\quad {\rm for all}\quad t\notin\mathbb{Z} \]
I marked t=0.01 as a sample in this graph:
Compared to \(f'(0)=1\) where the iterates of a point remain in the same petal, for any other \(f'(0)=e^{i\varphi}\), \(0<\varphi<2\pi\) the iterates need to traverse around the fixed point 0. And it seems some get entangled there
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09/08/2022, 04:21 AM
(This post was last modified: 09/08/2022, 07:11 AM by JmsNxn.)
(09/04/2022, 06:50 AM)bo198214 Wrote: Compared to \(f'(0)=1\) where the iterates of a point remain in the same petal, for any other \(f'(0)=e^{i\varphi}\), \(0<\varphi<2\pi\) the iterates need to traverse around the fixed point 0. And it seems some get entangled there
Hey, Bo. I don't have the fancy graphs you just posted. Not even sure how to conjure the graphs you are making using my own code. But there is a Cyclone, a tornado, a star shaped domain, at a neutral fixed point \(f'(0) = e^{i\varphi}\). If \(e^{i\varphi}\) is a unit of the circle, then there is a dimension of \(n\) on the amount of spirals about this \(0\). We usually cancel that out in discussions, where we call it parabolicand depending on the Milnor multiplier/valit of the \(f(0)\), we achieve that many petalsbut the "swirling rate" is based off of \(\varphi\). Nothing more than a rate like \(\varphi  \frac{p}{q} < \frac{1}{q^{1+\epsilon}}\) for \(p,q \in \mathbb{Z}\)which is the basis of siegel disks. If your \(\varphi\) satisfies this construction, you can remap this domain much more favourably.
What you have described above, is a symptom of a domain of holomorphy that is a petal, and it is swirling. By such the domain seems much more insignificant.
So if you fix the petal as real to real, and you vary the swirling, its common that you will hit a flat line of \(y =0\) for all values involved.
You are trying to swirl what needs to be complex, and keeping it real.
You are trying to eat your cake and have it too.
Are we writing \(e^{\pi i t}z + o(z)\) or; are we writing a summation of these weird fibonacci things we've learned? By which, creating a holomorphic solution in \(t\), which is real valued, is asking for trouble. But it gets really complicated. And especially because we will have to add much more terms than LFT'sbut they are related in principle.
Jesus, I hope this makes sense. I've been really busy lately, just trying to get out what I can.
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No, James, this has nothing to do with realvalued iteration, though the curves look similar to the thread " Real Multivalued/Parametrized Iteration Groups".
This is plain old regular (hyperbolic) iteration approaching the parabolic case in the limit.
What I remember from Milnor he does not really consider the proper parabolic case, rather says just take the kfold iterate then you have multiplier 1. Can you perhaps refer to the section where he talks about parabolic case != 1?!
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09/10/2022, 05:38 AM
(This post was last modified: 09/10/2022, 05:49 AM by JmsNxn.)
(09/08/2022, 12:19 PM)bo198214 Wrote: No, James, this has nothing to do with realvalued iteration, though the curves look similar to the thread "Real Multivalued/Parametrized Iteration Groups".
This is plain old regular (hyperbolic) iteration approaching the parabolic case in the limit.
What I remember from Milnor he does not really consider the proper parabolic case, rather says just take the kfold iterate then you have multiplier 1. Can you perhaps refer to the section where he talks about parabolic case != 1?!
I realize this isn't a thread on real valued iterations, but real valued iterations are iterations which map a curve to a curve. And that's more so what I meant. You are right though, I'm being incomplete.
Perhaps here, I have a later edition of the book. But Milnor specifically explains that if \(e^{2\pi i \varphi}\) is the multiplier, and \(\varphi\) is a \(q\)'th root of unity, then the different surfaces are modulo each other upto \(q\); exactly like the action \(z \mapsto e^{2\pi i/q}z\). So yes, he only talks about \(=1\), but he describes how to construct abel functions on every parabolic multiplier, and gives pretty clear language on this. Where there are \(q\) solutions, just like \(j = k \,\,\text{mod}\,\,q\). And then he has a whole section on siegel disks(Section 11)which requires a discussion of irrational numbers which are of the above form as I wrote. In this case, we don't solve for the Abel function, we solve for the inverse Schroder function as \(\lambda = e^{2 \pi i \varphi}\) and \(f(h(w)) = h(\lambda w)\)which does implicitly solve an Abel function.
I'm not sure what you mean by no discussion of the general parabolic case, because Milnor gives a large treatment on how to conjugate every parabolic case to the case \(\lambda = 1\), and for nonparabolic, but "indifferent", he has a lot to say about different ways of describing it. It's very Riemann surface heavy, but he essentially describes the problem, though most of it is hidden in subtle descriptions.
You are correct, he reduces it to \(\lambda = 1\), but he does give a treatment describing what the surface looks like for every parabolic fixed point; and does so a bit off hand, and not to full depth. But there is the above statement, that parabolic fixed points \(e^{2 \pi i j/q}z + o(z)\), where \(j\) is coprime to \(q\), behave like the action \(z \mapsto e^{2\pi i j/q} z\) locally. Additionally, we can construct an abel function about here, and not only that. Expect the swirling I discussed. This then becomes a "permutation of the petals"or something like that. This is hard to explain, but I hope you can visualize it. It rotates the petals a small amount if you only affect the original multiplier.
Try writing:
\[
e^{z}1 2z =  z + \sum_{k=2}^\infty \frac{z^k}{k!}\\
\]
And observe how things get permuted in comparison to:
\[
e^{z} 1\\
\]
It is very much the difference between \(z \mapsto z\) and \(z \mapsto z\). The negative z's send to positive, and the positive to negatives, rather than the clean behaviour of \(z \mapsto z\). This is the "swirling" I'm referring to. It's literally just the action of the unit of the circle \(1\).
I don't have much time lately, but I'll try to keep this conversation going
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09/10/2022, 07:19 AM
(This post was last modified: 09/10/2022, 07:40 AM by Daniel.)
Bo, this is very helpful. It substantiates my understanding of the dynamics of "indifferent" or irrationally neutral fixed points. Since my general Taylor's series of iterated functions is based on the summation of geometrical series, it is only natural to use the modulus. As far as Siegal in tetration, Robert Devaney claimed that the exponential map couldn't have a Siegal disk because it is exponentiation is entire. I believe that irrationally neutral dynamical systems are where the "small divisor problem" is considered.
Daniel
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09/10/2022, 02:14 PM
(This post was last modified: 09/10/2022, 02:18 PM by bo198214.)
@James
I think this totally does not work. One can only talk about petals for multiplier 1  and Milnor only does that  or show me where he does differently. For other unit roots it does not work, i.e. you can not have continuous iterates or petals there.
Just consider the following, let the multiplier be a qth root of unity, then by Milnor we know that \({\rm valit}[f^{\circ q}] = qm \) for some integer m>0.
Say for this case just m=1. We have q attractive petals of \(f^{\circ q}\), so if we start with \(z_0\) in one attracting petal then the sequence of iterates of \(z_0\) \(f^{\circ 0}(z_0)\dots f^{\circ q1}(z_0)\) orbits the fixed point 1 time and \(f^{\circ q}(z_0)\) arrives in the same petal as we started.
If we would now have continuous iterates \(f^{\circ t}(z_0)\) for t=0...q it would be a curve \(\gamma(t)\) connecting all the previously mentioned integer iterates. Particularly it would cross all the attractive and repelling petals of \(f^{\circ q}\).
For the next round t=q...2q the image of the curve would just be \(\gamma(q..2q)=f^{\circ q}(\gamma(0..q))\) and so on for further rounds.
So with higher t it would bulge out at the repelling petals and sucked in at the attractive petals.
So this curve gets more wild the greater t is and the closer parts of the curve come to the fixed point.
In no way could \(\lim_{t\to\infty}f^{\circ t}(z_0)=0\)  but wouldnt that be required if you want some regular iteration at 0?
In the following picture I show the iteration of \(f(z)=z+z^2\) on \(z_0=0.25\) where I connect the first two iterates with a *hyptothetical* curve (but any curve would suffer from the effects I described above) and continue this curve by applying f. One can clearly see that the petals are those of \(f^{\circ 2}(z)=z  2z^3 + z^4\), i.e. 2 attractive petals on the real axis and two repelling petals along the imaginary axis.
@Daniel
The irrational indifferent case is not so much my concern in the moment because it can  from the formal powerseries standpoint  be treated like the hyperbolic case.
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(06/09/2021, 10:10 AM)Leo.W Wrote: (06/09/2021, 12:40 AM)JmsNxn Wrote: ...
If I'm interpreting you right; do you mind sourcing me this result? This is absolutely breathtaking.
...
Finally we compare (4)&(7), we complete the proof, showing that \( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity
Tahdah!
ref/Almost all of these results were concluded by Ramanujan, what I did is just adding more details since his paper was like, filled with "Q.E.D."
A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \)
Leo
"\( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity"
this is exactly what I am talking about with semigroup addition homomo , the conjecture
https://math.eretrandre.org/tetrationfor...p?tid=1640
https://math.eretrandre.org/tetrationfor...p?tid=1639
the escape equation and the displacement equation.
https://math.eretrandre.org/tetrationfor...p?tid=1641
In a way it was all about iteration velocity, since that imo gives the semigroup homomo.
So I will start to look at things from the julia equation perspective !!
Notice the julia equation is usually only considered for parabolic fixpoints but we can go around that.
Also the julia equation is not unique so we are not done yet.
I had the idea of connecting these ideas with andrews slog.
( and its borel transform )
Lots of work to do and mysteries to solve but im getting on a road. I have some inspiration now !

" A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \) "
Im not sure what you mean by that ??
What is that suppose to mean ?
Is it interesting ?
Are those iteration accelerations or so ??
Im confused.

Im thinking about generalizing Julia equation.
Julia equation is about iteration velocity and a somewhat linear displacement idea.
I was thinking about the quadratic analogue...
regards
tommy1729
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(09/18/2022, 10:37 PM)tommy1729 Wrote: (06/09/2021, 10:10 AM)Leo.W Wrote: (06/09/2021, 12:40 AM)JmsNxn Wrote: ...
If I'm interpreting you right; do you mind sourcing me this result? This is absolutely breathtaking.
...
Finally we compare (4)&(7), we complete the proof, showing that \( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity
Tahdah!
ref/Almost all of these results were concluded by Ramanujan, what I did is just adding more details since his paper was like, filled with "Q.E.D."
A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \)
Leo
"\( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity"
this is exactly what I am talking about with semigroup addition homomo , the conjecture
https://math.eretrandre.org/tetrationfor...p?tid=1640
https://math.eretrandre.org/tetrationfor...p?tid=1639
the escape equation and the displacement equation.
https://math.eretrandre.org/tetrationfor...p?tid=1641
In a way it was all about iteration velocity, since that imo gives the semigroup homomo.
So I will start to look at things from the julia equation perspective !!
Notice the julia equation is usually only considered for parabolic fixpoints but we can go around that.
Also the julia equation is not unique so we are not done yet.
I had the idea of connecting these ideas with andrews slog.
( and its borel transform )
Lots of work to do and mysteries to solve but im getting on a road. I have some inspiration now !

" A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \) "
Im not sure what you mean by that ??
What is that suppose to mean ?
Is it interesting ?
Are those iteration accelerations or so ??
Im confused.

Im thinking about generalizing Julia equation.
Julia equation is about iteration velocity and a somewhat linear displacement idea.
I was thinking about the quadratic analogue...
regards
tommy1729
In other words studying
f( exp(x) ) = exp(x) f(x).
or maybe
f( (1 + x/n)^n ) = (1 + x/n)^n f(x)
for large n or limit n.
**
f( exp(0) ) = exp(0) f(0).
giving f(1) = f(0).
hmmm ...
f( (1+0)^n ) = f(1) = (1 + 0)^n f(0).
getting there too fast ??
I will think more b4 i post more musings.
regards
tommy1729
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inspired by the julia equation and the andrew slog and velocity iteration , this makes me wonder ...
should we modify the andrew slog equation and solve instead for
1/f(x) = 1/(f(exp(x))  1)
by using the taylor for 1/(x  1) and avoiding divisions by 0 ?
just an idea.
regards
tommy1729
