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Sep 27, 2022 at 13:05 vote accept Gottfried Helms
May 5, 2021 at 4:20 history edited Gottfried Helms CC BY-SA 4.0
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May 5, 2021 at 1:26 answer added Richard Diagram timeline score: 2
May 4, 2021 at 7:47 history edited Gottfried Helms CC BY-SA 4.0
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May 4, 2021 at 7:24 history edited Gottfried Helms CC BY-SA 4.0
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May 3, 2021 at 11:43 comment added Gottfried Helms (...) I thought about understanding those as "branches", but did not come to a conclusion so far.
May 3, 2021 at 11:43 comment added Gottfried Helms @GeraldEdgar - the keyword "branchpoint" might be suitable. If we look at one of the 3-periodic points, say $p_1$ then the fractional iterates $\exp^{\circ h}(p_1)$ for, say , $h=0 \ldots 0.1$ form a different curve from that for $h=3 \ldots 3.1$ while the "starting point" with $h=0$ and $h=3$ is in both cases $p_1$. We see expansion in the distances and in the form of the initial lines. If we draw many such partial curves , i.e. $h=6 \ldots 6.1$, $h=9 \ldots 9.1$ then we get various directions from the point $p_1$ away. (...)
May 3, 2021 at 11:33 comment added Gottfried Helms @GeraldEdgar - Well, the examples are calculated using some Kneser software-implementation in our tetration-forum. Because it is computationally expensive, and -at least- for me yet intractable I actually use a much simplified method, but which seems to approximate the Kneser-solution to sufficiently many digits. Because it simply uses truncated Carlemanmatrices and their fractional powers I call it "polynomial method" But the selection of the interpolation method is actually unimportant: as the problem appears to me it occurs inherently in any interpolation method for the tetration.
May 3, 2021 at 11:27 comment added Gottfried Helms @AlexeyUstinov - I use this as notation for the fractional iteration of the $\exp()$: the $a$'th iterate of the $\exp()$-function beginning at $z$ (where $z=\exp^{\circ 0}(z)$) . The use of the circle is to allude to the small circle in the notation $f \circ f$ and is in use in a fairly wide community.
May 3, 2021 at 11:26 comment added Gerald Edgar So: if $z_1$ is a $3$-periodic point, then (by your reasoning) so is $z_{1,1} :=\exp^{\circ 1/2} z_1$. But probably $\exp^{\circ 3} z_{1,1}$ is not $z_{1,1}$ itself, but lies on a different branch of $\exp^{\circ 1/2}$; making $z_{1,1}$ actually period $6$ in the Riemann surface.
May 3, 2021 at 11:20 comment added Gerald Edgar It seems Kneser worked only on $\exp^{\circ 1/2} z$. It is good on the real line. But was criticized because it is not single-valued in the complex plane. From your picture, I conjecture that every $n$-periodic point is a branch point for $\exp^{\circ a}$ (most $a$) making this a complicated thing in $\mathbb C$.
May 3, 2021 at 11:13 comment added Alexey Ustinov What is $\exp^{\circ a}(z)$?
May 3, 2021 at 10:44 comment added Gottfried Helms @GeraldEdgar: regarding your second comment: I don't know more, at the moment, than that this seems to be problematic for the set $\Bbb P_n$ of n-periodic points ($n \gt 1$ of course). Don't have an idea at the moment, whether, and in case of yes: how, there could follow something for the full set $\Bbb C$
May 3, 2021 at 10:16 history edited Gottfried Helms CC BY-SA 4.0
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May 3, 2021 at 10:09 history edited Gottfried Helms CC BY-SA 4.0
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May 3, 2021 at 10:04 comment added Gottfried Helms [concerning your first comment] Dear Prof. Edgar - I'll put a picture in an answerbox, because this comment box is not sufficient. It illustrates the three periodic points $p_1=0.908668534315 + 0.676249885471 î$,$p_2= 1.93500786337 + 1.55280058174 î$, $p_3=0.124597586009 + 6.92297735843 î]$, also showing the Kneser-interpolation between $[p_1,p_2]$ and the subsequent partial trajectories taken by functional equations.
May 3, 2021 at 10:03 comment added Gerald Edgar ...Then we conclude that there is no family of functions satisfying $(1)$ with $\exp^{\circ 1} z = \exp z$, $\exp^{\circ 0} z = z$, defined for all $z \in \mathbb C$, continuous in the real variable $a$. For the known methods of tetration, what is the domain in $\mathbb C$?
May 3, 2021 at 9:27 comment added Gerald Edgar Perhaps show us existence of a $3$-periodic point. (of course complex, not real) That fact would not involve fractional iteration.
May 3, 2021 at 8:31 history asked Gottfried Helms CC BY-SA 4.0