Iterating at eta minor
#1
(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.
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Messages In This Thread
Iterating at eta minor - by JmsNxn - 07/22/2022, 01:17 AM
RE: Iterating at eta minor - by bo198214 - 07/24/2022, 12:43 PM
RE: Iterating at eta minor - by bo198214 - 07/25/2022, 04:01 PM
RE: Iterating at eta minor - by bo198214 - 07/25/2022, 04:27 PM
RE: Iterating at eta minor - by JmsNxn - 07/25/2022, 08:19 PM
RE: Iterating at eta minor - by bo198214 - 07/26/2022, 07:19 AM
RE: Iterating at eta minor - by JmsNxn - 07/25/2022, 09:06 PM
RE: Iterating at eta minor - by JmsNxn - 07/26/2022, 02:42 AM
RE: Iterating at eta minor - by JmsNxn - 07/28/2022, 12:21 AM
RE: Iterating at eta minor - by JmsNxn - 07/28/2022, 03:06 AM
RE: Iterating at eta minor - by JmsNxn - 07/29/2022, 05:18 AM
RE: Iterating at eta minor - by bo198214 - 07/31/2022, 08:24 PM
RE: Iterating at eta minor - by JmsNxn - 08/01/2022, 10:41 PM
RE: Iterating at eta minor - by JmsNxn - 08/02/2022, 02:03 AM
RE: Iterating at eta minor - by JmsNxn - 08/03/2022, 06:43 AM
RE: Iterating at eta minor - by JmsNxn - 08/05/2022, 02:01 AM
Iterating at eta minor - by bo198214 - 07/24/2022, 12:19 PM

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