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Iterating at eta minor - JmsNxn - 07/22/2022
(07/21/2022, 05:04 PM)bo198214 Wrote: This time I wanted to know how the fixed points behave if the base moves along Shell-Thron boundary.
And what a wonderful dance is that! Daniel, when you talk of beauty ...
The indifferent fixed point moves on the red curve. But here one sees that one can not really assign, which of the two repelling fixed points is the other primary fixed point.
Ya you can really see the branching problem at \(\eta\). It's funny to think that it's actually more regular the closer it gets to \(e^{-e}\), it basically looks like \(\exp\)'s fixed point distribution.
I still get fuzzy headed thinking about Kneser for the Shell-thron values. But \(b = e^{-e}\) would have two "primary" repelling fixed points \(a^+,a^-\), such that:
\[
\lim_{\Im(z) \to \pm \infty} \text{tet}_{b}(z) = a^{\pm}\\
\]
While maintaining real valued right?
I wonder if the Beta method could help in this avenue at all.
For the boundary of shell-thron, the beta tetration can have arbitrary period \(2 \pi i / \lambda\) for \(\Re\lambda > 0\). And if you shrink \(\lambda \to 0\), it looked like it was converging, for \(\eta\) it looked to converge to the standard abel iteration. I wonder if \(b = e^{-e}\) would converge to something similar, or if it would converge to something more "Kneser" like, like what Tommy and I observed with \(b = e\) when you limit \(\lambda \to 0\).
Since \(a^\pm\) is an attracting fixed point of the function \(-\log(z)/e\), the orbits in the lower half/upper half planes, sufficiently far out, should converge to them right? Considering the principal branch only (similar to the orbits \(\log(z)\) converging to the Kneser fixed point pair in the upper/lower half planes). I don't think the beta method would find the abel iteration, I think it might converge to Kneser, but only at \(b = e^{-e}\), not sure about anywhere else on the boundary, the beta method got pretty chaotic there.And out side of Shell-thron I could never find a solid reasoning that the iteration converged as \(\lambda \to 0\)--but it did look like it was getting close.
I'll see if i can run some more experiments on \(b = e^{-e}\) for \(\lambda\) small, and see if it is converging to these fixed points in the upper lower half planes.
EDIT:
So yes, this is happening.
To over explain, let \(\beta_\lambda(s)\) be the beta function base \(b = e^{-e}\). Then:
\[
\beta_\lambda(s+1) = b^{\beta_{\lambda}(s)}/(1+e^{-\lambda s})\\
\]
is a holomorphic function on \(\mathbb{C}/\{z=j + 2\pi i k / \lambda\}\) for \(k \in \mathbb{Z}\) and \(j \ge 1\).
Then the tetration is:
\[
\text{tet}_\lambda(s+c) = \lim_{n\to\infty} \log^{\circ n}_b \beta_\lambda(s+n)\\
\]
This object is holomorphic on \(\mathbb{C}\) almost everywhere, and has period \(2 \pi i/\lambda\).
It does appear that for \(\lambda \approx 0.01\) that \(\text{tet}_\lambda(s) \to a^\pm\) when you take a limit \(|s| \to \infty\) with \(\pi / 2 < |\arg(s)| < \pi\). If you were to limit \(\lambda \to 0\) it would lose it's period, and would tend as \(\Im(s) \to \pm \infty\). So maybe this would be an avenue of constructing Kneser. It would be a cool way of checking to see if we can do this, because it gives a bit of heuristic that a similar process should work for \(b = e\). That's more difficult though, because the beta tetrations for arbitrary period are nowhere holomorphic (just smooth on the real line). We might just get lucky for \(b = e^{-e}\)...
EDIT2:
Holy hell this is making a lot of sense.
\(b = e^{-e}\) has a second order neutral fixed point at \(1/e\). The function:
\[
b^{z + 1/e} - 1/e = -z + O(z^2)\\
\]
So since \((-1)^2 = 1\) there exists exactly 4 petals about this fixed point, two attracting... and two repelling... So if you focus on the attracting case, you get an abel expansion. Using the repelling fixed point pair will produce a real valued (because the real line is a boundary of petals) Kneser-type iteration. This is very fascinating. I think I should dig deeper into the beta method and \(e^{-e}\)...
Also, Bo, Catullus proposed that this base should have its own symbol. As it's directly opposite of \(\eta\), is it fine if we call \(e^{-e} = \eta_{-}\) or something similar... I get tired of writing \(e^{-e}\). Someone also proposed an upside down \(\eta\), but I'm not sure how to typeset that efficiently, and it sounds like a headache. But would be awesome.
RE: The Different Fixed Points of Exponentials - bo198214 - 07/22/2022
To be honest I have also no better idea than to call it \(\eta_-\)
I made another animation showing the fundamental region and its image.
Though its not "officially" proven I think that, as long we have a fundamental region (a region bounded by a curve \(\ell\) connecting the two fixed points and its image \(f(\ell)\)) and the function \(f\) is (holomorphic and) injective on that region then there exists the perturbed Fatou coordinates (aka Abel function) \(\Phi\) that is injective on the region and it is the only Abel function up to an additive constant. Alternatively one can also use the uniqueness criterion of Paulsen for the superfunction (inverse of the Abel function) that it tends to fixed point 1 when imaginary part goes to \(+\infty\) and to fixed point 2 if the imaginary part goes to \(-\infty\).
And I guess this not only valid for two repelling fixed points, but for any fixed point pair.
In the following animation the black line is the connection from the indifferent fixed point to the primary fixed point, call it \(\ell\). The blue line is \(b^\ell\) and the green line is \(b^{b^\ell}\). I.e. we have the fundamental region between the black and blue line and its image between the blue and green line. So we can visually check for injectivity.
So \(b^z\) seems to be injective on our chosen fundamental region up to the indifferent fixed point at \(\eta_-\), just a bit further (which is the last part of the animation) it is not anymore (if we stick to the repelling fixed point corresponding to index 1 of the LambertW function. If we change however to fixed point with index -1, then it is just the conjugated case, and everything works out fine)
I didnt find any constellation of injectivity on the fundamental region if the indices of the fixed points differ more than 1, particularly not for the two repelling fixed point close to (the indifferent fixed point) \(\eta_-\).
RE: The Different Fixed Points of Exponentials - bo198214 - 07/23/2022
Though the two repelling fixed points at \(\eta_-\) are really seductive to do the Kneser/Perturbed fatou coordinates on, they seem not even to have a fundamental region. This animation shows the fundamental region of the fixed points with index -1 and 1 (close to \(b=\eta_-\)).
RE: The Different Fixed Points of Exponentials - bo198214 - 07/23/2022
I tested also combinations like index 1 and 2. They have a fundamental region however the image is not a region (\(b^z\) not injective).
These are the fundamental regions, they look odd already.
but with the mapping of the fundamental region you can not see much anymore, only that it is not injective:
So the conjecture here would be that only the fixed point combinations (0,1) and (0,-1) have fundamental regions with \(b^z\) injective on it.
More precisely (0,1) only in the case \(b\) is in the upper half plane and (0,-1) only in the case of the lower half plane.
RE: The Different Fixed Points of Exponentials - JmsNxn - 07/24/2022
Hmmm, that's very interesting. So since this fundamental domain isn't injective, the idea of the riemann mapping would fail, and the Kneser construction would be nulled right? At least, that's how I interpret this.
Very fascinating. So you'd want to take the attracting petal of \(b = e^{-e}\) which includes \(0\) and just construct the Abel function about an indifferent fixed point and call it that then? So that there wouldn't even be a Kneser construction.
I imagine you'd be able to derive the decay to the fixed points in the upper-lower planes, so it'd almost be like a pseudo Kneser? You never actually use a Kneser Riemann Mapping, you just do the Abel iteration and:
\[
\begin{align*}
\alpha^{-1}(0) &= 1\\
\alpha^{-1}(z) \,\,&\text{is holomorphic on}\, \mathbb{C}/(-\infty,-2]\\
\alpha^{-1} &: (-2,\infty) \to (-\infty, 1/e)\\
\alpha^{-1}(z) &\to 1/e \,\text{as } \Re(z) \to \infty\\
\alpha^{-1}(z) &\to a^{\pm}\,\text{as}\, \Im(z) \to \pm \infty\\
e^{-e\alpha^{-1}(z)} &= \alpha^{-1}(z+1)\\
\end{align*}
\]
Where \(a^\pm\) are the fixed points of \(e^{-ez}\) with nonzero imaginary part, with the least imaginary part.
I mean, this construction should be entirely possible, and it looks an awful like Kneser, but if the riemann mapping approach is nonsensical cause there's no fundamental domain, it sounds more like a "pseudo"-Kneser.
I guess the main question would be how to ensure holomorphy at precisely this point in \(b\). Would we get another branching problem like at \(\eta\)? That doesn't really make sense though, unless the branch cut formed by \(\eta\) were more disastrous than we think, which maybe spawns at \(\eta\) but continues to \(\eta_-\). Perhaps it goes all the way to \(0\).
Paulsen never explains how to take the branch cut in his construction, but I wouldn't be surprised if it resulted in something like:
\[
b \in [0,\eta]\\
\]
This would be really odd though, as holomorphy, about say, \(\sqrt{2}\) shouldn't be a problem for Kneser... right? But maybe there would be a branching problem in \(b\) about these points that we haven't encountered yet.
I also haven't familiarized myself enough with Kneser, so I may be saying stupid things. But no fundamental domain means no Riemann mapping... right?
What if, Kneser is perfectly continuable at the point \(b = \eta_-\), but there exists no fundamental domain? In the sense that, the standard Abel iteration is Kneser at \(\eta_-\), this would also follow for \(\eta\) as a limit, but there's trouble doing that, because it's a branching point in not only the iterated exponential, but also in \(b = y^{1/y}\). \(\eta_-\) has all the benefits of being a real valued, neutral fixed point, without the trouble of being a critical point.
I mean, what if for complex tetration \(\eta_-\) gave us a view of how standard petal/Ecalle iteration about an attracting basin, relates to the Kneser method about two fixed points. In a similar manner that \(\eta\) does. But in this case \(\eta \uparrow \uparrow z \to \infty\) as we let \(\Re(z) \to - \infty\) (the iterated log largely diverges). With \(\eta_-\) we have nearby attracting fixed points for \(\log\) and if we iterate we aproach them. And not only that.
\[
\log_{\eta_-}^{\circ n}(z) = a^{\pm}\,\,\text{depending on if}\,\,\pm\Im(z) > 0\\
\]
But not only that, there exists a real valued abel iteration about an indifferent fixed point. THERE'S NO NEED TO DO THE RIEMANN MAPPING, HENCE NO FUNDAMENTAL DOMAIN!
But nonetheless \(b\) is holomorphic at \(\eta_-\)! Because, the standard Abel iteration about the attracting petal that includes \(0\) (I mean this as there are 4 petals about \(1/e\) for the \(\eta_-^z\), and therefore 4 Abel functions on each petal, which can and cannot be continued together--there's one petal that \(0\) is in, use that Abel iteration (Think of this like if we choose the eta iteration or the cheta iteration, but now we have 4 choices--choose the one with 0)).
Think of it like this. Take the standard Abel iteration of \(b = \eta_-\) and let's force it to be real as \(b\) moves on the real line. What if a way of talking about Kneser is a perturbation of the abel solution at \(b = \eta_-\). Very much how you spoke of perturbed parabolic points (forget the exact words). But we do it at \(b = \eta_-\) instead. And then \(b = \eta\) is a branching problem.
The problem I forsee with this approach is that the branch cuts will look like \(b \in (-\infty,0) \cup \(\eta,\infty)\). This might be weird but this was something I was very interested in, made a thread recently.
Either way, I guess, I'm asking if we could think of \(\eta_-\) as a complex valued \(\eta\). I think this might help a lot. It's like Kneser's \(\eta\).
RE: The Different Fixed Points of Exponentials - JmsNxn - 07/24/2022
I thought I'd add my drawings of tetration for base \(b = \eta_-\).
This is I believe the \( 2 \pi i\) periodic tetration, drawn about the singularity as it moves in the real line. Here's a zoom in on the very sinusoidal curves which spawn as you go further and further to the right (enclosing about \(1/e\)):
And if you increase the imaginary argument you get:
Now when you graph this in the complex plane, it gets a little weird. Essentially in the strip \(- \pi < \Im(z) < \pi\), this the important case, and serves as a "they all kinda work like this". Which you can see in this graph, it's all red, excusing a sea of green or purple. This function is still holomorphic here, it's just a branch cut you can visibly see. Then the goal of "beta can be turned into Kneser" is to stretch this strip into the entire complex plane.
Now the singularities you see in the top and bottom are native to the \(\beta\) method. We always have singularities at precisely \(\Im(z) = \pi i + j\) for all \(j \in \mathbb{Z}\). These are precisely not seen in this graph. But, this function is \(2 \pi i\) periodic. Therefore the top and bottom parts of this function have a direct discontinuity from top to bottom. This results in a branch cut directly along the line \(\Im(z) = \pi i\). These are the singularities inherited by \(\beta\).
Then there are iterated log singularities, which appear at the boundary of the green and red/purple and red sea.
ALSO NOTE THAT GREEEN IS \(a^-\) AND PURPLE IS \(a^+\). Stretch the red domain and we apprach the fixed points.
Thought I'd make a graph explaining what I mean by the singularity is somewhere, and the "false singularity which is really just a julia set" makes it look like the singularity is somewhere else. I'm going to use one of my stock photos of \(b = \sqrt{2}\) and the weak julia set as described in my paper.
The singularity is at the left most point white point, and the white area is how my code fails/describing what I called the weak julia set. This was the same thing as the Julia set but sort of modified in terms for \(\beta\) functions. This is where we hit a fractal boundary. So picture a fractal-like branch cut with a lot of large and small values, but no actual "non-holomorphy" except for on the branching fractal that appears.
This fractal extends from the green sea to the purple sea to the red sea, and that's where we see our "obvious singularities". It's really weird with \(\eta_-\).
Again the white area is as good an approximation as I could get. Sheldon even said that's not the actual fractal right? And he argued for perturbation theory and everything, and we concurred the actual fractal should be a lightning bolt kind of thing.Which as I showed, is actually measure zero, under an Lebesgue area measure.
In the above graph the singularities are precisely on the boundary of the sea of green and sea of purple. And there's precisely a branch parallel to the real line (remember these graphs are \(2 \pi i\) periodic), and this is a graph on a window \(|\Im(z)| < \pi\). So there's a boundary at \(\pi i\) of purple and green. And the singularities are here. But there exists iterated log errors which are branching errors. But they aren't inherent to \(\beta\). They are fractals produced by the \(\beta\) method which result in an explosion half the graph down... Like a lightning fractal.
RE: The Different Fixed Points of Exponentials - JmsNxn - 07/24/2022
To get an even better look.
I circled about where the singularities are, but they cause a lightning bolt fractal in about the sea.
Actually the sea of green (purple), means the function \(\beta\) is very large, and therefore iterated \(\log_{\eta_-}\) just takes us to the fixed point faster.
What's remarkable is the branch cut is parallel to \(\mathbb{R}\). And that there's no mixing in the green/purple sea. This means \(\Im(z) \to \pm \infty\) should behave just as we like when we grow the candy strip. It looks like a weird kind of Kneser.
ALSO, try to remember this is all the \(2 \pi i\) periodic solution. You can make \(Ti\) periodic solutions; let \(T\to\infty\) and it looks just like the Abel solution.
Iterating at eta minor - bo198214 - 07/24/2022
This is a continuation of this thread, and I moved the previous posts from there to here.
RE: Iterating at eta minor - bo198214 - 07/24/2022
Is I am not as good with reading complex function graphs I reduced the problem to the real line.
The function \(\eta_-^x\) has a parabolic fixed point at 1/e with multiplier -1 which is a/the second root of unity.
In general it is a standard consideration if you have a m-th root of unity as multiplier to just consider the m-th iteration of the function.
Because if we have an iteration \((f\circ f)^t\) then we also have an iteration \(f^t = (f\circ f)^{\frac{t}{2}}\).
So I did that and was graphing how the parabolic fixed point behaved under perturbation:
And we see this is quite different from \(b=\eta\).
But particularly I was wondering: where do those extra attracting fixed points - left and right for \(b=e^{-e}-0.005\) - come from?!
These are not fixed points of \(b^z\), because it does not have any attracting fixed points (b is outside the Shell-Thron region).
And then it became clear to me, these are periodic points!
So one can iterate at periodic points! Gottfried, sorry that I never attentively read your posts about the periodic points!
Seems they are quite important as another source of iterating the exponentials.
(beside regular iteration, and - how about we call Kneser/perturbed Fatou coordinates "double regular iteration"?! Because the iteration is so to say done at 2 fixed points - double regular iteration.)
Indeed these two extra fixed points of \(b^{b^x}\) are a 2-cycle of \(b^x\):
\(b^{0.527385987834363}= 0.228746810954529\)
RE: Iterating at eta minor - bo198214 - 07/25/2022
Because I was always wondering why is there no parabolic implosion/explosion at all the other parabolic fixed points (except b=eta)?!
The reason seems to be that we did not consider fixed cycles.
This becomes visible when we take \(f^{\circ N}\) as the function in question, suddenly it has much more fixed points, because a fixed N-cycle turns into a fixed point under \(f^{\circ N}\).
So I was looking at parabolic fixed points for bases on the Shell-Thron boundary and perturbed them perpendicular to the Shell-Thron boundary.
But not considering \(f(z)=b^z\), but considering \(f^{\circ N}\) for \(e^{2\pi i /N}\) being the multiplier of the fixed point.
And voilá we see the ex-/im-plosion there!
This is plosion at \(b_2=\eta_-\), i.e. \(b_2=\exp\left(e^{\phi i - e^{\phi i}}\right)\) with \(\phi=2\pi/2 = \pi\).
The fixed point \(z_2 = e^{e^{\pi i}}=1/e\) has multiplier \(-1=e^{\pi i}\)
The animation shows \(f^{\circ 2}(z) -z\) while \(b\) breaks through the STB at \(b_2\) from inside to outside (however with such a small amount that you can not see it in the animation, the middle fixed point breaks through the red curve from inside to outside).
I think you need to reload the page if you want to see the animation again.
The read curve are the fixed points with absolute value of the multiplier being 1.
Additionally to the two real fixed points (we saw in the previous image) we can see in this image also the two complex fixed points:
This is the plosion for \(\phi=2\pi / 3\) showing \(f^{\circ 3}(z)-z\):
And this is the plosion for \(\phi = 2\pi /4 = \pi/2 \) showing \(f^{\circ 4}(z)-z\):
The fixed point turns into N+1 fixed points.
So theoretically one could do an iteration at any of these fixed points/cycles maybe even at two of them (if they have a fundamental region).
This again opens the question whether one can continue a regular iteration through the STB.
But actually I am already again deviating from the thread's topic XD.