Iterating at eta minor

[JmsNxn]

To be honest this is a bit too much to digest for me, and unfortunately in the next time I will not have time to further participate at the forum. Anyways, the stuff sounds really intriguing.

I apologize, I like to throw up everything I can think of. Just so I can see it written out, so that maybe when I look at it again something new will come to me. I am not the most concise writer, lol.

Particularly about the period dial. So you can change the regular iteration at fixed point 2 (base \(\sqrt{2}\)) into the regular solution at fixed point 4, just by turning the period dial from \(2\pi i/\log(\log(2))\) to \(2\pi i/\log(\log(4))\) ?! Very interesting!

And then turning it to \(\infty\) you would get the base-continued crescent/Paulsen iteration (i.e. something that is not real on the real line but nearly (supersmall imaginary part))?

Unfortunately no, to both of those examples. But there is a period dial. But this is largely because you chose \(\sqrt{2}\). If you take \(b =\sqrt{2}\) for example, then you can create a tetration with arbitrary period \(2 \pi i /\lambda\) so long as \(\Re(\lambda) > - \log \log(2) = 0.36651292058166432701243915823266946945426344783711\). So \(\log\log(4)\) nor \(0\) satisfies this. The period \(2 \pi i/\log\log2 \) is PRECISELY the boundary of the beta method. That's the biggest you can make the period. You can only make it smaller from here.

The exact formula is if \(e^{\mu p} = p\) (at the primary fixed point in the Shell thron--\(b=e^{\mu}\) in S-T ) then we can make a tetration with period \(2 \pi i/\lambda\) so long as \(\Re(\lambda) > -\log|\mu p|\). This means on the boundary of Shell-Thron we can make \(\lambda \to 0\), which is why it's great.

When you take \(b = e\) for instance, you can make \(\Re(\lambda) > 0\), and although the tetration only converges as an asymptotic series (hard to explain how this is derived), when you let \(\lambda \to 0\) THAT is converging to what looks like the Crescent iteration. So the crescent iteration seems to only be discoverable OUTSIDE of the Shell thron region. Inside, a natural boundary forms at regular iteration's period. This was a conjecture by Sheldon, and I spent months proving it, it didn't prove to be too hard, and largely just reduced into the convergence of a series.

So essentially, the "dial" has a max setting, and within shell thron, the "max setting" is the regular iteration; on the boundary of shell thron, it is a period of \(i\infty\), and out side of Shell-Thron, it is again \(i\infty\). Outside of shell thron, when the dial hits the max setting, then it is the crescent iteration (? again it's a conjecture, but it's fairly well supported. Especially when you use more clever numerical procedures, like Tommy's Gaussian). Hope that makes sense...

Then I have a question about this assertion:

Iterating about periodic points is pretty standard for the repelling case, (a bit more complicated with neutral). The thing is, that the local iterations can never contain the periodic points.

What exactly do you mean by iterating about the periodic points?

Just want to make some remarks about the "Kneser iteration" - for me it is not interchangeable with the crescent iteration, but is a very special case. It means to finding a real analytic iteration if there is no real fixed point by using a conjugated fixed point pair. In this special case one can make use of the Riemann mapping theorem. If however we are not on the real line anymore, but just have two arbitrary fixed points with a fundamental region, Kneser's construction does not work anymore (or at least Paulsen and me don't see a way to generalize Kneser's construction) in that case Riemann's mapping theorem is not sufficient anymore, but one needs the *measurable* Riemann mapping theorem)

Sorry, I was just reaffirming what I had said before. We can take \(f(p_0) = p_1\) and \(f(p_1) = p_1\), then we can iterate \(f^{\circ 2}(p_0) = p_0\) using the regular iteration. That's all I meant. The thing is, this is an iterate of \(f^{\circ 2}\) and doesn't produce \(f\) in any form or shape.

And yes, okay. I'll adjust my vernacular. The crescent iteration is the general idea, which I presume can be generalized to other functions etc etc... Kneser is just for the case \(b > \eta\). And it's specific to a real valued criterion. Gotcha!

the branch cut at \(\eta\)  happens exactly along \((\eta,\infty)\) 

Dunno what you mean "happens" - isn't the branch cut something you set instead of something that happens?

Oh sorry, I meant I would choose the branch cut a long here. We'd probably have the same problem as with the crescent iteration, where \(\sqrt{2}\) has a non trivial imaginary part along the real line, and it's positive approaching from the top, and negative approaching from the bottom. I just meant that we could choose it to happen along \((\eta,\infty)\). But also, it's important to remember that \(x^{1/x}\) has a second order branch cut at \(e\), this means there are "two branches"--the highest order non zero taylor coefficient is 2. This is sort of the language you'll see in Riemann surface stuff. Similar to how \(x^2\) has "two branches" at zero \(\sqrt{x}, -\sqrt{x}\). I think we're going to see a similar problem at about \(\eta\), where there are actually two tetration branches here. Sure we can move the branch cut around and around and everything should be fine, but i'm not so sure. Maybe we can have real valued on \(\eta,\infty\), then it's not real valued on \(1,\eta\). Or we can have real valued on \((1,\eta)\) then it's not real valued on \((\eta,\infty)\). Which would imply a very different scenario than "just choosing where the branch cut is". It'd be comparable to adding \(2 \pi i\) to \(\log\).

I also wonder why you only consider \(\eta_-\) to be the Suez canal, should it not behave like all the other parabolic fixed points which don't explode (in terms of creating more fixed points when perturbed). Did you try the beta method on non-real parabolic fixed points?

I have already the conjecture that the fixed point of base \(\eta\) is the only indifferent fixed point where the regular iteration is not analytic. Some computations are quite supporting that the regular iteration powerseries at parabolic fixed points have convergence radius (while it is quite a difficult proof that the regular iteration powerseries of \(e^x-1\) and hence of \(\eta^x\) has zero convergence radius). I am not completely sure, but does this mean that all the Fatou-Coordinates in the petals are just one function? I also guess this has to do with the \(\eta\)-fixed point being the only "exploding" one.
And maybe this in turn would lead to a proof that the Shell-Thron boundary is permeable except at \(\eta\).

Oh, I see it as the "suez canal" because it is real valued! That is the only reason why I think it's important. It's the only other real valued parabolic point. I did try the beta method on a few other parabolic points, and it works very similarly. It's still very chaotic; and the weak Julia set can become more troublesome, but it does work, again, a.e. under an area measure in \(\mathbb{C}\).

Hmmm so if I'm understanding your question. Let's take \(e^{z}-1\), the left half plane is the attracting petal, and the right half plane is the repelling petal. So there are two different abel functions to the left or the right (eta or cheta iteration, basically). When the fixed point is parabolic but of order \(n\), so that the lyapunov multiplier is \(e^{2 \pi i k/n}\) where \((k,n) = 1\) (k and n are coprime), then, I'll double check milnor, but there should be 2n petals, n attracting, n repelling, and if I'm remembering correct, all the repelling petals abel functions are different versions of each other--and similarly for all the attracting. I forget how milnor describes this, I apologize but parabolic isn't my strong suit, but I do believe they are essentially the same function. Because:

\[
\alpha(f^{\circ n}(z)) = \alpha(z) + 1\\
\]

Has only two solutions, the repelling version and the attracting version (eta or cheta). But I'll need to double check this. Milnor can be a slow digest, and I can't remember all the details.

EDIT:

I forgot the fundamental lemma-- the union of all the petals about zero is a neighborhood (it contains a disk about zero). So when you talk about the repelling/attracting petals--you are talking about two different partitions of a disk about zero. Wherein, there are n branches about zero for the repelling case, and n branches about zero for the attracting case. No different than the n branches of \(x^{1/n}\). Or as \(\eta\) behaves; except you will see an x^2n type behaviour, as opposed to an x^2 kind of behaviour you see at \(\eta\).

Do we btw have a strict proof that any iteration of a non-trivial function can never be holomorphic at two fixed points (except for integer iterates)? I think I saw you reasoning about that topic, but I was not sure how strict it was ... this would be surely worth a dedicated thread.

I follow this from Milnor primarily. And I will try to dig up a solid proof. I know he has a few small remarks explaining why you can't have a coordinate change (as he calls it) about two fixed points. It's phrased as to normality though. Where in, a function can only have an Abel (I believe we call it Fatou coordinate, in this forum, and even in Milnor) coordinate change (or a Schroder coordinate change) about a fixed point. It's easier to explain with Schroder, but more difficult for Abel. WLOG we can assume that one \(p\) is attracting, and \(p'\) is in the Fatou set or in the Julia set. We can be assured of this by considering \(f^{-1}\) instead of \(f\). By which, A schroder coordinate change can only be valid on the Fatou set; by which \(p\) and \(p'\) are in disconnected domains, separated by a Julia set. Therefore there exists no Schroder coordinate change (and hence no holomorphic iteration about both).

For an Abel coordinate change it's more technical. But the way Milnor describes how an abel function about a parabolic point is constructed is complete, in the sense that any Abel function will be of this form. It's possible to have it holomorphic in a half plane, so that the codomain is \(\mathbb{C}_{\Re(z) <> 0}\) depending on if we are in an attracting or repelling petal. Any other abel function is equivalent up to a riemann mapping (? this might be a bit more nuanced). So you can draw out that \(\pm\infty\) is the fixed point; and the boundary of the Half plane, is precisely the boundary of the petal.

Note this only applies to parabolic fixed poionts, for abel functions like in Kneser's case, we are doing something more involved; but still we can map it to an abel function that fits these criteria, where now it's not a petal about the fixed point, but the fatou set about \(L^{\pm}\) (fatou set of \(\log\)).

I do have a proof somewhere, I'll find it, but it is a feature of fractional iterations. It can be tricky though to observe it, especially because Milnor is a much more topological book; so it's sort of hidden in the sense that you can partition fixed points surrounded by Julia sets.

To construct the regular iteration (which is kneser's iteration) about \(b = \eta_- - 0.5\), we can think ...

Öhm, I don't know how the Kneser iteration (in what sense ever) can be the same as the regular iteration .... please explain!

Oh, I seem to have misremembered here. Sorry, I meant that if you use Kouznetsov's interpretation of the regular iteration, then it is Kneser's iteration at \(b= e\). At least this is how Kouznetsov presents it in his text book. So I made the false equivalency "Regular iteration = Kouznetsov's regular iteration = Kneser's iteration" (which would follow from Paulsen's uniqueness claim). So I think I just used the wrong meaning of regular iteration, I apologize. I should've said it's just Kouznetsov's regular iteration--not Kneser's iteration. And Kouznetsov's regular iteration will agree with kneser on \(b > \eta\), according to Paulsen, which would be where my confusion came from. I apologize, misspoke here.