# Tetration Forum

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From my considerations on n-periodic points I came to the question, what would the Kneser-method produce for the orbit over such periodic points, with fractional steps of some small order.

I think, the Kneser-method for base $\small b=\sqrt 2$ gives pretty smooth orbit/curves for iteration with fractional heights, so I used that base.

I can find n-periodic points of any order, at least (arbitrarily near) approximates (using my method discussed in the previous thread).
Assume for instance the set of 6-periodic points (call it for instance "6-cycle") having the following approximate values:

Code:
 p1=6.574747674571700333009302782482639405458 + 0.9684029672628275112988471898707035489652*I  p2=9.218591772304179221754557226368239753533 + 3.215625927832110139905357739034367499059*I p3=10.75598408660258595556412864882513957578 + 21.91051323491973949320807501979124161907*I p4=10.70582570926180333437881525870964054377 + 40.18332039944084859802348678029895039952*I p5=8.546967145179356295028740245483622451155 + 39.96464873483437244009418575109150854376*I p6=5.464835241819467227108868806348013192730 + 18.55139938049736513812617532327500721009*I

such that $\small{p_{k+1}=b^{p_k}}$ then, in a plot generated with Excel, we see the six points, and connected by some cubic-spline-interpolation, which is of course only by accident near to the closed curve generated by the fractional iterations using the Kneser- (or any other interpolation-) method.
See this picture first. I "normalized" the y-(imaginary-) axis by $\small \Im{(z)} / \tau$ where $\small \tau= \pi / \ln b$ *(for better comparability when this picture are made with another base)* :
[attachment=1413]
Note: the "key" for the cycles, mentioned in the legend of the plot, gives the periodic points in the inverse order (p1,p6,p5,p4,p3,p2) because the software uses logarithmizing instead of exponentiation)

The Kneser-iterations produce as well a smooth curve do far, but much different from the cubic spline; the light-blue segment even makes a big journey out-of-bounds of the plot here.

[attachment=1414]
The next segments of the orbit begin to evolve to "overcycle": the pink segment continues the light blue segment and even generates some spiral around the origin. The gold and the brown segments even don't continue the attaching segments smoothely.

[attachment=1415]

Here is a bigger view: it may give a bit better imagination, but is still far from exposing the full chaos:

[attachment=1417]

Problem: the source in Fatou.gp is difficult to edit. I'd like to be able to compute in higher precision (while not sending Pari/GP in the nirvana of neverending loops). Also it seems as if the orbits of -say- tet(z,1..2) is different or sometimes different from b^tet(z,0..1).

Questions:

- can someone check this curves in general by recomputing data with the Kneser-method?

- can someone check whether the Kneser-calculation can be made consistent at the switching points, where the segments don't follow smoothely?

- the chaotic ranges of the orbits are surely caused by the branch-cuts of the logarithm or by some other well known property alike. On the other hand, we've *infinitely many* n-periodic points, somehow randomly shuttered in the shown segment. For the Schröder-function this should be catastrophic: these are points, which cannot be iterated towards the fixpoints. What does this mean for the "moving towards the fixpoint"-part in the mechanism?
Hey, Gottfried could you explain more what you mean? Or, what you're intending to compute mathematically? This seems interesting, and I'm rather curious.

I get that you're using $n$ cycles,

$
\{z_i\}_{i=0}^{n-1}\,\,\text{s.t}\,\,\exp(z_i) = z_{i+1}\,\,\text{where}\,\,z_n = z_0\\
$

I get that you're describing this using the inverse Schroder function,

$
z_i = \Psi(L^i \xi)\,\,\text{for}\,\,\xi \in \mathbb{C}\\
$

And I assume you're trying to construct Kneser's solution; which is usually done about a fixed point, but somehow you mean to do it about a cycle? I'm genuinely curious as to how one would do that. Are we essentially just performing Kneser's method on,

$
$

Whereupon, we know this solution will still be Kneser (as far as I understand Kneser this will happen; correct me if I'm wrong, but can't one show this by uniqueness?).

I'm just curious if you could link me to threads, or explain what you mean to do here. I apologize if I'm being obtuse. I'm a horrible computer programmer. As much as I've lurked Sheldon's threads about fatou.gp; I wish it was just written in math, lol.

Do you mind if you explained the mathematics of what you're trying to do?

Sincerely, James
(04/29/2021, 03:02 AM)JmsNxn Wrote: [ -> ](...)
I get that you're describing this using the inverse Schroder function,

$
z_i = \Psi(L^i \xi)\,\,\text{for}\,\,\xi \in \mathbb{C}\\
$

And I assume you're trying to construct Kneser's solution; which is usually done about a fixed point, but somehow you mean to do it about a cycle? I'm genuinely curious as to how one would do that. Are we essentially just performing Kneser's method on,

$
$

Whereupon, we know this solution will still be Kneser (as far as I understand Kneser this will happen; correct me if I'm wrong, but can't one show this by uniqueness?).
(...)
Hi James,

I used the Kneser-solution implemented in Sheldon's "fatou.gp" program, starting at one of the 6 periodic points, say $z_0$ , and calculating fractional iterates from that point with iteration-height $h=0..1$ towards $z_1$ then $h=1..2$   towards $z_2$ and so on.  (If I've it correct at the moment, I had crosschecked the results with the simple  "polynomial method", which uses $h$-fractional powers simply of the truncated Carlemanmatrix for $b^x$  - which gives often good approximations for the Sheldon's Kneser-solution)

I did not see "Periodic points" discussed in our forum so far, and for the forward iteration $z_{k+1}=b^{z_k}$ it is for most bases not obvious to find periodic points at all, because of the divergent character of the iteration, say with $b=e=\exp(1)$ for which it is known that there are no attracting fixpoints (1-periodic points) and that n-periodic points of any order are repelling.
But if you use the backwards-iteration iteratively, applying the $\log()$, then that points (1-periodic or n-periodic) are attracting. With iterative application of the $log()$ for your fixpoint-iteration you only can find the primary fixpoint, and using the $log()$ at branches you find the secondary fixpoints - which is of course known here in the forum from the very beginning.
I extended that iteration by using the branches of the logarithm varying (but cyclic) in the sequence of iterations, so with a "key" of $K=[0,0,1]$ I do fixpoint iteration two times with $log()$ at the zero-branch, and the third time at the branch with $1*(2*Pi*I)$. For the example in the initial posting I used the "Key" $K=[1,2,2,1,0,0]$ for six consecutive iterations over the branched logarithm until periodic points are approximated well enough. Note, that the forward iteration $b^z$ "eats" the branching and we see actually the 6-periodic cycle without any obvious "key". (Note moreover, that Devaney has introduced the notion of a "string associated with" the iteration of exp(). and from a short discussion with W. Bergweiler (see references in my treatize) I've learned, that that "string" is exactly what I had introduced and named as "key".)
The easiest way to proceed, once one has the fractional interpolation between, say, $z_1$ and $z_2$ (in my example above $p_1$ and $p_2$ - using the letter $p$ for (p)eriodic points), is to simply exponentiate the Kneser-interpolated line one time, two times,... 5 times to get the other fractionally interpolated segments by the functional equation. However, we shall notice a blowing up of the curves of partial trajectories - much different to the spirals which occur if we use the 1-periodic (fix-) points and apply the functional equation beginning at some segment of the trajectory between neighboured $z_h$ and $z_{h+1}$. While this is surely the "natural" way to complete the full n-cycle I was curious, how Sheldon's implementation would perform, if I simply demand the full 6-cycle iteration by increasing the $h-$ iteration parameter from $0$ to $6$ (in steps of, say 1/10): not bad, really (!), but of course it cannot give a smooth interpolation when change of branch-index occurs while evaluating the trajectory when a change-of-branch occurs "on-the-road".
This inherent non-smooth change of branch-index over the logarithmizing can surely not give a meaningfull smooth curve for the fractionally iterated partial trajectory (=I think=), and it seems that singularities/jumps must occur, when one reduces the stepwidths of the iteration to small-enough extension - but I don't have really a deeper analysis of this yet...

It seems to me (and I think it is somehow obvious), that the smoothness of the fractional iteration, for instance of the Kneser-method must break down at that exemplars in the n-cycle, where the key $K$ of branchindexes contains variation, say $[...,0,1,...]$ or $[...,2,5,...]$ or so, because the Kneser-solution for the interval between two points $z_1$ and $z_2$ should be dependent on the equality of branch-indexes.
Btw. with easier keys $K$ than that of the example in my opening post, say $K=[0,0,1]$ instead, I found the Sheldon's Kneser-solution remarkable nice, while of course the inconsistency could not really disappear.

----------------------------------

See further explanation/discussion in my initial treatize attached to this post (well, it deals prominently with the surprising observation of sheer existence of n-periodic points at all... but explains the method how to find such sets of points more detailed).

"[attachment=1468]" (periodic points initial notes)
Hi James,

a second read of your question makes me think that my previous reply has not been well to the focus of your question.

Perhaps this is a better one.
The existence of n-periodic points, when looking at the forward iteration $z \to \exp(z)$, gives rise to the assumption, the fractional interpolation could as well be implemented as it can be done by any two points on one trajectory.

But the backward iteration $z \to \log(z) + k \omega$ where $k$ gives the branch-index, between two points $z_j = \log(z_{j+1}) + K[j+1] \omega$ and $z_{j-1} = \log(z_{j}) + K[j] \omega$ where $K[j]<>K[j+1]$ cannot smoothly be interpolated: it were needed that also the jump between the two branchindexes $K[j]$ and $K[j+1]$ could be smoothed(*). But that would require the re-definition of the complex logarithm, so I think this can never be done.   So this has also consequences for the definition of restrictions of the Schroeder function and of its inverse: I think, the idea of the Schroeder-function as idealized/normed infinite iteration has no place in its mathematical derivation/representation for varying branchindexes of the log resp the varying entries of its associated Devaney-string. The best what we could do, in my opinion, with the Schroeder-function were to engineer it to work for other 1-periodic points/fixed points. But I've no idea how at all a variable branchindex could possibly be thought.

I've experimented with this on base of the most simple non-trivial "key" $K=[0,0,1]$ and used the fatou.gp and my "polynomial method" concurrently (without any visible difference) focusing the problem of the non-smooth step in the branch of the log between $z_1$ and $z_3$.   See the following picture (draft only so far)
[attachment=1469]
and perhaps more expressive
[attachment=1470]

We see here the ambiguity of continuously continuing the trajectory from $z_2$ to and beyond $z_1$ towards $z_0$ (more precisely to $z_0'$ where an adaption of the branchindex was not done)  respectively from $z_2$ to and beyond $z_0$ towards $z_1$.

--------------------------------------------

Perhaps we could try to make something out of it, if we connect the horizontal strip in the complex plane to make up a torus of perimeter $2 \pi$ and see, whether this allows meaningful curves, but a short exercise on such a thing seemed to me to lead simply to nothing and/or chaos...

---------------------------------------------

(*) in older times I've played around with the idea, whether it is possible to introduce a meaning to *fractional* branchindexes of the log, but with no avail. However don't have the link to the tetration-forum post available (searchable keyword something like "fixpointline" or so with some answer of Henryk, I'll insert it here when I've found it)
See: https://math.eretrandre.org/tetrationfor...hp?tid=422 Ufo:Fixpoint-line(?)
I'll take time to read over your answer more in depth later; but as I understand it, and as I'm boiling it down; it's a question of how the Kneser function chooses it's logarithms.

If we fix a logarithm, and we take a cycle of $z_k$; it is not necessarily true that,

$
\log(z_j) = z_{j-1}\\
$

But, there exists a logarithm where this is true; it's some trailing multiple of $2\pi i$ away. However, the Kneser function knows to do this; such that the map $\log$ will necessarily choose the right branch (or rather, the right sequence of branches). I presume by your Key; you mean a sequence of numbers $(k_0,...,k_{n-1})$ such that,

$
\log_{k_j}(z_j) = z_{j-1}\\
$

And,

$
\log_{k_j}(z) = \log(z) + 2\pi i k_j\,\,\text{for the principal branch of }\log\\
$

So I'm seeing this as a goal to describe what sequence of logarithms work. Correct me if I'm way off here; I'm just trying to understand.

Do cycles stay as cycles indefinitely in Kneser's iteration (I believe this can't happen as $\Re(z) \to -\infty$ it causes $\text{tet}_K(z) \to L,L*$; and this limit should be satisfied clearly.) So Kneser's "keys" if you will will eventually degrade and no longer work; but as you do the forward iteration we get the usual weird cluster of repelling cycles. And if we pull back from the forward iteration; we have to keep track of which logarithm, when.

Are you asking if we can use different keys to construct a varied Tetration? Or at least a different method of computation?

I'll read your treatise tomorrow; need to go to bed tonight. Thank you for your explanation though; I'll read it more carefully tomorrow.

Regards, James
Hi James -

yes, you've got correctly what I meant with the logarithmizing and the "key".

But then you formulate
(04/30/2021, 05:32 AM)JmsNxn Wrote: [ -> ]However, the Kneser function knows to do this; such that the map $\log$ will necessarily choose the right branch (or rather, the right sequence of branches).

Hmm, I don't know how I should understand this...

(04/30/2021, 05:32 AM)JmsNxn Wrote: [ -> ]So I'm seeing this as a goal to describe what sequence of logarithms work. Correct me if I'm way off here; I'm just trying to understand.
Here is perhaps one source of misunderstanding. No, I just give a "key", say $K=[0,0,1]$.
Then initialize, say $z_1=1+î$ and repeat $z_{1+3k+3}= \log(\log(\log(z_{1+3k})))+1*C$ (where $C=2 *\pi *î$) until convergence.
I find, by this iteration the three 3-periodic-points $p_1 = z_{1+3n}$,$p_2 = z_{2+3n}$,$p_3 = z_{3+3n}$
This is then the material I look at.

I can then use the Sheldon's Kneser-implementation (or my simpler "polynomial method" for approximation) to compute, say, 99 points on a curve between $p_1$ and $p_2$ meaning 101 points with iterationheight differences of 1/100 between each. This is a discrete orbit of the idealized continuous fractional iteration between $p_1$ and $p_2$, let's call the vector of data "line12".

Naturally, by the functional equation I should be able to calculate the according fractional iterates between $p_2$ and $p_3$ just by using the functional equation by doing $line23=\exp(line12)$ and then should continue $line31=\exp(line23)$

This vectors line12,line23,line32 should then equal the data computed by the tetration-function say $\tet(p_1,h)$ for the iterationheight $h=0..3$ in steps of 1/100. I expected, that the Kneser-implementation/"polynomial method" would simply produce that data for the fractional heights.

My thoughts are no further innovative or so. No new method there. I simply observe, that the curves come out funny, if not chaotic, and I try to locate the source of the problem.  I can't however deny, that my suspicion grows, that the problem is a principal one - but don't know, and possibly simply some restrictions in our function-definition might be sufficient.

One idea is to problematize the behave of the (Kneser-) $\tet(p1,h)$ function for *negative* h. So $\tet(p1,-1)$ should equal $\tet(p1,+2)$ and $\tet(p1,-0.01)$ should equal $\tet(p1,+2.99)$ but this doesn't happen, because for the fractional negative heights, the fractional logarithm "does not know" ;-) which & when it should adapt branch index...

The pictures in my recent post take a simpler "key" than that in my initial post ($K=[0,0,1]$) to display things more focused. I can also post the actual data set for the pictures if this would make my considerations more transparent.

update
After some more working with the data as described, I believe I have a proof, that a continuous curve, connecting all the three periodic points $p1,p2,p3,p1,...$ in the sense of representing the trajectory of continuous iteration, cannot exist:
If there is such a curve, then each point on one partial curve connecting a pair of points (for instance $p1,p2$), must as well be 3-periodic. But this would mean the number of 3-periodic points would be uncountably infinite (and for this single example of a 3-period only). But it has been proved elsewhere, that the whole number of n-periodic points for each n is only countably infinite.
/end update

Kind regards-
Gottfried
(04/30/2021, 09:51 PM)Gottfried Wrote: [ -> ]Hi James -

yes, you've got correctly what I meant with the logarithmizing and the "key".

But then you formulate
(04/30/2021, 05:32 AM)JmsNxn Wrote: [ -> ]However, the Kneser function knows to do this; such that the map $\log$ will necessarily choose the right branch (or rather, the right sequence of branches).

Hmm, I don't know how I should understand this...

......

update
After some more working with the data as described, I believe I have a proof, that a continuous curve, connecting all the three periodic points $p1,p2,p3,p1,...$ in the sense of representing the trajectory of continuous iteration, cannot exist:
If there is such a curve, then each point on one partial curve connecting a pair of points (for instance $p1,p2$), must as well be 3-periodic. But this would mean the number of 3-periodic points would be uncountably infinite (and for this single example of a 3-period only). But it has been proved elsewhere, that the whole number of n-periodic points for each n is only countably infinite.
/end update

Kind regards-
Gottfried

Hey, Gottfried!

Your update is much more what I was driving at. If we take a key, and talk about producing an iteration;

$
\log_k(z_k) = z_{k-1}\\
$

And drawing a spline between them--then it must disagree with Kneser at some point, because Kneser eventually tends towards the value $L$ as we iterate it. Which is, it no longer satisfies the periodic structure. So, as you are proving this by saying it produces uncountable periodic points; I'm saying something slightly different.

That, for an appropriate sequence of logarithms (that Kneser's solution some how knows):

$
\log^{\circ k} \text{tet}_K(z) \to L,L^*\,\,\text{as}\,\,k\to\infty\\
$

So if you were to attempt to do the inverse iteration to construct this, where its centered around the behaviour,

$
\log_k(z_k) = z_{k-1}\\
$

And we create splines between them; this inherently shouldn't work, right? Because eventually there will be a value $s_j$ such that,

$
\text{tet}_K(s_j) = z_j\\
$

But,

$
\text{tet}_K(s_j-1) = \log \text{tet}_K(s_j) \neq z_{j-1}\\
$

And this would mean, as your iteration is stable about these periodic points; there should be some sort of discrepancy when you create the spline between the periodic points. As this spline will be stable under the orbits of $\log_k$ (with the appropriate key); but as we can see; Kneser's solution is NOT stable under any key about any periodic points.

At least, I think that makes sense. Sorry; I'm not the best with Kneser's solution, or the existing code. As I lack much education in computer programming; it's difficult for me to intuit the fatou.gp program.

I apologize if I'm speaking nonsense, lol.

Regards, James.
Hi Jamens -

The shown spline in the first picture was only to give an idea what my problem was, which I could not exactly pinpoint then.   It has nothing do to with any thinkable method for interpolation, or, in other words, for a tetration solution - just a sketch to show that I think, the curve of the trajectory of the continuous tetration periodically through the set of the three 3-periodic points would have some problem to be periodic itself.

Now I see, that my problem is much more simple to denote:
let's define the three points $p_1$, $p_2$, $p_3$ which are 3-periodic under $\exp()$.

Then let's define the point $p_{1.1}=tet(p_1, 0.1)$ by (any method of) tetration with heights $h$ over the reals.
Then, one one hand, we'll have $p_1=tet(p_1,3)=tet(p_1,6)=tet(p_1,9)=...$ which is periodic
but on the other hand we'll have $p_{1.1}=tet(p_1,0.1)\ne tet(p_1,3.1)\ne tet(p_1,6.1)\ne tet(p_1,9.1)...$ which thus cannot be periodic.

This can also be shown, if we apply the functional equation: $p_{1.1} \ne \exp^{\circ 3}(p_{1.1})$ and which is already known elsewhere. In a paraphrasing formulation: "if $p_1$ is a 3-periodic point, then there are no other 3-periodic points in an epsilon-neighbourhood of $p_1$" (which would be required by the assumption that the trajectory of the fractional tetration were as well periodic). It seems, I've just rediscovered the concept of the Devaney's "hairs" where he says, that any point in any small epsilon-neighbourhood of an n-periodic point diverges to infinity when iterated under $\exp()$. This again means, there are no fractional iterations in the near of $p_1$ (or $p_2$ or $p_3$) which itself can be periodic.
In a single statement:

The continuous curve, produced by fractional tetration, with increasing iteration-height $h$  through n-periodic points cannot itself be n-periodic but diverges (chaotically) to infinity for all points except for the n-periodic points themselves (likely paraphrasing Devaney's "hairs" with the here made own observations).
This statement concerns all methods of tetration, be it Kneser, "regular" or whatever.

update: This means also, that the order of iterated exponentiation is no more irrelevant. We have, with a 3-periodic point $p_1$, that we get the destruction of the equality

$\exp^{\circ 3}(\exp^{\circ 0.1}(p_1)) \ne \exp^{\circ 0.1}(\exp^{\circ 3}(p_1))$ where the rhs is by 3-periodicity $=\exp^{\circ 0.1}(p_1)$

, which is what I'd never expected with the construction of the tetration. endupdate

For me, this is catastrophic for the concept of tetration, and maybe this explains also my difficulties to understand the problem at all, and to get a sufficient grip on it, when I got some instinct regarding possible problems in tetration after getting aware of n-periodic points at all in the middle of last year.

What do you think after this more precise focusing of the problem?

If this is not only a trivia (negating/correcting my instinct), I think I should better make a separate post about this...

Kind regards -

Gottfried
Oh yes, I am well aware of that. Never heard it called a Devaney hair before though, lol.

I think I was misunderstanding something. I was referring to the fact that iterated log's have attractive cycles at these points. So if you draw a spline in a certain manner; and expect the iterated logarithms to be stable on these splines; then it cannot be Kneser's Tetration. Because Kneser's tetration tends to L,L* as we iterate the logarithm; and doesn't cluster towards any periodic points. Which, I wasn't so sure, but it seems like this idea may force the iterated log to be normal on the spline (given the correct key). I think I may be missing something though... I guess it could still produce Kneser's solution; so long as it's not normal for whatever Kneser's key is in a neighborhood of that point. Where I'm referring to kneser's key as the key such that,

$
\log_{s_k} \text{tet}_K(s_k) = \text{tet}_K(s_k -1)\\
$

And after writing this... I think I see what my error was; I was assuming the algorithm used to draw the spline was in itself a pull back using logarithms. But that's not the case; you're just using the pull back to find the periodic points; and then constructing the spline using fatou.gp (or your polynomial method). That was a silly mistake of me. For some reason I thought you were somehow drawing the spline using the pull back; and that it looked like it would make the iterated log converge about that periodic point. But you're not doing that at all.

That was silly of me. Now it makes more sense, a lot more sense, what you're trying to do. Lol, I get it now. You can ignore my comments, I was misunderstanding something greatly! lolll

Regards, James
I've composed a question on math-overflow yesterday. Some comments for clarification, no answers so far.

See https://mathoverflow.net/questions/391772/

Gottfried
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