Timeline for On the relevance of the property $\exp^{\circ a}(\exp^{\circ b}(z))=\exp^{\circ a+b}(z)$ for the *fractional* iteration ("tetration")
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 12, 2022 at 22:36 | comment | added | Richard Diagram | Last comment, I apologize to the math overflow staff. Referring to my original answer: If a superlogarithm $\text{slog}(z)$ is analytic on $\mathbb{R}$; then it is analytic at $x_0 \in \mathbb{R}$, which means $\text{slog}(z+x_0)$ is a holomorphic function for $N = \{|z| < \delta\}$. The orbits $O = \bigcup_n \exp^{\circ n}(N)$ is dense in $\mathbb{C}$--and additionally has infinite measure. Thereby, $\text{slog}$ is holomorphic on $O$ (which is almost everywhere in $\mathbb{C}$ under a Lebesgue area measure), by the rule $\text{slog}(\exp(z)) = \text{slog}(z) + 1$. | |
| Nov 12, 2022 at 22:24 | comment | added | Richard Diagram | So my advice, as long winded as it is, to test if a point $z=z_0$ is a periodic point, look at: $\text{slog}_K(z+z_0)$, and see if $z$ has poor Taylor data, has obviously too large values. If the taylor series looks like $O(c^k)$ for some $0 < c$, we're not at a periodic point. If the taylor series looks like $O(c^k k!^{\delta})$, for some $\delta>0$--then this is a periodic point. This works for ANY superlogarithm; so I could test this with the Kneser method, or Andy's slog. And again, the key is to look at taylor data, because the point values may be deceptive. Weird singularities and all | |
| Nov 12, 2022 at 22:23 | comment | added | Richard Diagram | If I wanted to detect a periodic point, I'm really not sure. I have run my own code to test something similar with the "beta method" (from the tetration forum)--singularities happen at periodic points with it too. The best way, by far, is to use Kneser's Slog. The function $\text{slog}_K(z)$ is holomorphic almost everywhere in $\mathbb{C}$. Where it isn't, is precisely when $z=z_0$ is a periodic point. This gets tricky, as as a global function, the slog isn't technically holomorphic at every other points but periodic points. But it is locally holomorphic... | |
| Nov 12, 2022 at 22:20 | comment | added | Richard Diagram | You should definitely have infinite periodic points near $L$. I'm pretty sure, about every fixed point of $\exp$ you have uncountably infinite periodic points arbitrarily close to every fixed point. This seems like a contradiction, but it's not, $\exp^{\circ s_0}(z)$ for a fixed $s_0$ is holomorphic almost everywhere in a neighborhood about $z = L$--where almost everywhere is intended on an $\mathbb{R}^2 = \mathbb{C}$ Lebesgue measure. Everywhere it isn't holomorphic near $L$ is some kind of periodic point. As to detecting these... | |
| Nov 11, 2022 at 9:32 | comment | added | Gottfried Helms | RD - nice to see you back! Recently I've put the problem this way: given a point $z_0$, maybe arbitrarily near the fixpoint $L$ or maybe far away. How do you detect that it is a periodic point? For instance: if you have a series to describe the point and its assumed-to-be continuous trajectory how could that coordinate and a coordinate of an (arbitrarily) near non-periodic point determine such different evaluation of the series? (I've seen, that periodic points occur in any epsilon-neighbourhood of $L$ if the period-length $n$ is taken arbitrarily long) | |
| Nov 11, 2022 at 2:10 | comment | added | Richard Diagram | ALSO! The function $\exp^{\circ s}(z)$ is not holomorphic for a neighborhood $N$ of which $L^{\pm} \in N$. So you definitely find many periodic points near $L$, and that's perfectly fine according to Kneser. | |
| Nov 11, 2022 at 2:06 | comment | added | Richard Diagram | If $\text{slog}(z) + s = -k < -1$ then $\text{slog}(z) + s + n = n-k \ge -1$, which equates to $\exp^{\circ n}(\exp^{\circ s}(z))$. I'm wondering if you could rephrase your question, or pm me a more developed question. I am a little confused, plus it's been a year and a third of a year, lol. | |
| Nov 11, 2022 at 2:03 | comment | added | Richard Diagram | @GottfriedHelms It's been a year and four months, Gottfried! And, honestly, Gottfried, I do not have a clue. I will say that the function $\exp^{\circ s}(z)$ has some very very bizarre oddities. But to be simple, it's helpful to remember that $\exp^{\circ s}(z) = \text{tet}(\text{slog}(z) + s)$. And the true singularities appear when $\text{slog}(z) + s = \infty$ or they equal a negative integer less than $1$. Periodic points instantly cause a singularity, but there are still other singularities you have to account for--which ironically, can be removed by taking $\exp(z)$ a bunch of times... | |
| Sep 27, 2022 at 13:54 | comment | added | Gottfried Helms | ... we can add a Newton-iteration with quadratic convergence to come near to the true value as much as we want. So how do we make sure, our point $z$ that we want iterate fractionally is not nearer to a periodic point than to the primary (or selected) fixpoint? | |
| Sep 27, 2022 at 13:52 | comment | added | Gottfried Helms | Hmm, if it is a question of a neighbourhood around a fixpoint, and only outside of this neighbourhood the fractional iteration (be it Schröder, Kneser, Walker or Kousnezov...) is affected because of the existence of the periodic point in the near, I think I can claim that this neighbourhood is arbitrarily small, because one can identify periodic points in arbitrary small neighbourhood of the fixpoint. Such points can be found by iteration, say $z_1=1+I$ and iterate that 2-step-sequence ($z_1=\log(z_1)+2 \pi î $ -> $z_1=\log^{\circ 1000}(z_1)$ ->...). After this iteration approximates well, ... | |
| Sep 27, 2022 at 13:38 | comment | added | Gottfried Helms | Hi, 4 monthes later I give it new read, and I think now I've got it correct. Thanks for your nice elaborate answer - your texts are always good to read :-) | |
| Sep 27, 2022 at 13:05 | vote | accept | Gottfried Helms | ||
| May 5, 2021 at 4:24 | comment | added | Richard Diagram | The argument is designed for any tetration, not just Kneser's, any tetration. What you are describing in your limit process can't work in a neighborhood of the points. It can't produce a holomorphic function. It can produce tetration on a line, but not a domain; at least not a domain containing all 3 periodic points. That's what I'm saying here. The function $\exp^{\circ z}(\xi)$ can't be holomorphic at all three points; in no manner. So the discrepancy you are seeing, in (2.2) and (2.1) can't happen. That's the argument I'm presenting. sorry, been a long night, I'll check back in 2 days, lol. | |
| May 5, 2021 at 4:15 | comment | added | Gottfried Helms | Thanks for the workout! I'll need some time to digest this, though. Just to clear one aspect: it might have been unfortunate to introduce the term "Kneser-interpolation" at all, so we might shift to discuss pros and contras of this method. But I could have used even linear interpolation for the initial (in the graphic sea-green line $l_{12}$) and the problem of non-periodicity of the trajectory and thus of discrepance in (2.1) and (2.2) would occur in the same way. But again: I really appreciate to find any coherent/coherence-restoring point of view! | |
| May 5, 2021 at 3:07 | history | edited | Richard Diagram | CC BY-SA 4.0 |
added 10 characters in body
|
| May 5, 2021 at 3:01 | history | edited | Richard Diagram | CC BY-SA 4.0 |
Added a summary of what iterations about periodic points look like.
|
| May 5, 2021 at 1:26 | history | answered | Richard Diagram | CC BY-SA 4.0 |