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Author Topic:  Julia and Mandelbrot sets w or w/o Lyapunov sequences  (Read 6197 times)

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Offline marcm200

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Julia and Mandelbrot sets w or w/o Lyapunov sequences
« on: March 19, 2019, 03:20:46 PM »
I tried to combine the Lyapunov idea of using a sequence, e.g. AB to set the value of a disturbance parameter with Julia sets. But I wanted to be as close to the original Julia set / Mandelbrot definition, so the Julia set formula zn+1=zn2+c now can take two different values c1 and c2, which one depends on the sequence position. After each iteration the sequence position gets incremented (the sequence is infinitely long and repeats itself, just like the Lyapunov algorithm).

I started some initial trials with the simplest sequence AB. I'm sure I'm not the first one to do that (claude will know, right?), but I'm quite excited after the first results (I hope my program is correct, at least, it produces nice images).

I used as bailout for escape the 2-disc, although I do not know whether this would have any meaning here, but I checked also the disc around the origin with radius 100 and the images didn't change, so at least in the resolution and maxiteration I use, it's stable (but, numerical evidence might be misleading).

I like the variety of outcomes: subtle changes from the one-c-value base Julia sets, images appearing out of nowhere, or very strong differences between the base sets and the combined one. And I only checked a couple dozen randomly generated c value pairs.

File description: The large part in the images is the c1/c2 Julia/Lyapunov set, the two smaller images at the right picture border show the single c-value Julia sets, the filename contains the coordinates of the c1/c2 values used.

Has anyone experience with that Julia / Lyapunov crossover?

EDIT 2019 Dec 6: I changed the topic's title as there are sets with and without Lyapunov variations.

Linkback: https://fractalforums.org/image-threads/25/julia-and-mandelbrot-sets-w-or-wo-lyapunov-sequences/2696/
« Last Edit: December 06, 2019, 10:26:00 PM by marcm200, Reason: Title change to reflect type of Julia sets »

Offline marcm200

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #1 on: March 20, 2019, 03:43:50 PM »
I did a sweep over the plane generating a bunch of images with the sequence AB. Here are some more examples.

The first one is quite common: Smaller copies of the dominant structure lying swirling around getting smaller and smaller.

The second one seems to be a disconnected set of objects, but I have to zoom in to see whether there is (I assume so) a bridge between the blocks.

The last two are among the highest repetive structures I've seen so far. They consist of blocks that look like segments of an arthropode arranged in swirls of different lengths.

I would now like to color these images. If someone has suggestions, that would help me a lot (I have to code it into my program)..

Offline marcm200

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #2 on: March 20, 2019, 03:48:25 PM »
And as was done for the Mandelbrot set I tried constructing a 4D Lyapunovbrot set and visualize it. I used the real or imaginary part of c2 is a color index and the other three coordinates as spatial positions. Judged were the pairs c1, c2 as for the Mandelbrot set, whether the origin is bounded or not.

The images look kind of odd. One like a saddle, one like planes intersecting each other. And the color did not help at all. I think it's too much information. So the next attempt will have one of the 4 coordinates fixed (maybe combining a complex number c1 and a real number c2 for a start) and then rather try different values of the fixed parameter.

Offline marcm200

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #3 on: March 22, 2019, 10:24:37 AM »
I checked some AABB sequence for the Julia/Lyapunov crossover. The shapes often seem to be comprised of two or more underlying basic shapes (parameters are stated in the image). And there are quite a number of those pointy ones at the upper left where I really like to zoom into.

The one in the lower left does not seem to have many structure on the boundary. Is this still a fractal? Or can the usage of alternating between values distroy that property? Has anyone read literature for that topic?

Offline pauldelbrot

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #4 on: March 22, 2019, 03:32:07 PM »
It has bumps that have smaller bumps and still smaller bumps, so yes, it's fractal. :)

Offline marcm200

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #5 on: March 25, 2019, 10:08:03 AM »
There is a feature with connectedness from classic Julia sets that seems different when calculating sequence based sets.

Often there are three large objects that appear to be connected within themselves. From what I understand, Julia sets are either totally connected (one big object) or totally disconnected (dust). If the screen does not lie, then we here have a mixture.

The picture shows such a 3-object Julia set with values cA=-0.793275+1.43797i, cB=1.98607+2.39872i and sequence ABA. Nothing special at first sight, but what I find very intriguing is the magnitude of cB, it's well above 2. And if I recall right, every Julia set of that type is disconnected (correct me, if I'm wrong). So the dust property can somehow be reversed to connectedness. I'm currently trying to find more examples of that behaviour to see which values can be "rescued" from falling into dust.
(The color is just some experiment with orbit distance).


Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #6 on: March 25, 2019, 02:10:21 PM »
Found a citation:
Quote
Connectedness and Stability of Julia Sets of the Composition of Polynomials of the Form z2+cn
Rainer Brück
Journal of the London Mathematical Society 61(02):462 - 470 · April 2000
https://dx.doi.org/10.1112/S0024610799008509
I found the fulltext on sci-hub
« Last Edit: March 25, 2019, 02:55:36 PM by claude »

Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #7 on: March 25, 2019, 03:39:13 PM »
Julia sets are either totally connected (one big object) or totally disconnected (dust).

True for quadratic polynomials, but not in general AFAIK.

Found another paper:
Quote
Quartic Julia sets including any two copies of quadratic Julia sets
Koh Katagata
Discrete & Continuous Dynamical Systems - A, 2016, 36 (4) : 2103-2112. doi: 10.3934/dcds.2016.36.2103

If the Julia set of a quartic polynomial with certain conditions is neither connected nor totally disconnected, there exists a homeomorphism between the set of all components of the filled-in Julia set and some subset of the corresponding symbol space. The question is to determine the quartic polynomials exhibiting such a dynamics and describe the topology of the connected components of their filled-in Julia sets. In this paper, we answer the question, namely we show that for any two quadratic Julia sets, there exists a quartic polynomial whose Julia set includes copies of the two quadratic Julia sets.

Keywords: polynomial-like maps, Julia sets, quasiconformal/quasiregular maps.

The sequence AB defines a quartic polynomial, maybe there's a way to transform c_A c_B in a way to get a desired pair of Julia sets, investigating...

Another paper by the same author which looks interesting:
Quote
Entire functions whose Julia sets include any finitely many copies of quadratic Julia sets
Koh Katagata
2017 Nonlinearity 30 2360

We prove that for any finite collection of quadratic Julia sets, a polynomial and a transcendental entire function exist whose Julia sets include copies of the given quadratic Julia sets. In order to prove the result, we construct quasiregular maps with required dynamics and employ the quasiconformal surgery to obtain the desired functions.
« Last Edit: March 25, 2019, 03:55:13 PM by claude, Reason: more »

Offline marcm200

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #8 on: March 25, 2019, 05:00:45 PM »
Thanks, claude - again.

The sequence AB defines a quartic polynomial, maybe there's a way to transform c_A c_B in a way to get a desired pair of Julia sets, investigating...

I'm not sure I understand. I can perform the sequence AB by hand and get a power 4 formula for zn+2=zn4+(2*cA)*zn²+cA²+cB. But wouldn't then the orbit only contain "half" of the points visited (skipping every other)? And that can also be done for the normal case without Lyapunov sequence (or just the sequence A), but those are not quartic in nature.
What am I missing here?


Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #9 on: March 25, 2019, 07:23:55 PM »
If one half of the orbit escapes to infinity, the other half must do also, so it doesn't matter so much.  Julia set is boundary of filled-in Julia set for this case, I think, so the usual escape-time algorithm applies.  The Julia set and fillled-in Julia set are invariant under the quartic, but not necessarily under each quadratic.  This also means sequence AB may be different from sequence BA.

The process I had in mind was choosing two parameters c_1 c_2 for their quadratic Julia sets, then transforming them some how to c_A c_B such that sequence AB induces a quartic whose Julia set contains both Julia sets.  Kind of like a high level design tool - "give me seahorse spirals mixed with elephant dust, please!"

Offline pauldelbrot

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #10 on: March 25, 2019, 07:32:02 PM »
If one half of the orbit escapes to infinity, the other half must do also, so it doesn't matter so much.  Julia set is boundary of filled-in Julia set for this case, I think, so the usual escape-time algorithm applies.  The Julia set and fillled-in Julia set are invariant under the quartic, but not necessarily under each quadratic.  This also means sequence AB may be different from sequence BA.

The process I had in mind was choosing two parameters c_1 c_2 for their quadratic Julia sets, then transforming them some how to c_A c_B such that sequence AB induces a quartic whose Julia set contains both Julia sets.  Kind of like a high level design tool - "give me seahorse spirals mixed with elephant dust, please!"

You mean, a tool like this?

http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15594

Mind you, that one is a rational function rather than a low-degree polynomial.

Offline ThunderboltPagoda

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #11 on: March 25, 2019, 07:33:53 PM »
Often there are three large objects that appear to be connected within themselves. From what I understand, Julia sets are either totally connected (one big object) or totally disconnected (dust). If the screen does not lie, then we here have a mixture.

The picture shows such a 3-object Julia set [...]

I'm not sure if I understand correctly what you mean here. The image in question has small objects around the three large objects. And perhaps they have small objects around them in turn, and so on, so there is dust?

Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #12 on: March 25, 2019, 08:54:05 PM »
You mean, a tool like this?
Yep!

Trying to reverse engineer the paper from the abstract (haven't managed to find the full text gratis), so far I got as far as there being 2 fixed points z_1,z_2 of the quartic, near z_1 it behaves locally like z^2+c_1, and near z_2 it behaves locally like z^2+c_2, but with affine conjugacies a_i z + b_i my attempted algebra collapses everything to 0=0 which means i probably missed some important step...

Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #13 on: March 25, 2019, 08:55:37 PM »
I'm not sure if I understand correctly what you mean here. The image in question has small objects around the three large objects. And perhaps they have small objects around them in turn, and so on, so there is dust?
those objects have non-empty interior, so they are not dust.  "dust" means totally disconnected, no point has a neighbour

Online claude

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Re: Julia and Mandelbrot sets using Lyapunov sequences
« Reply #14 on: March 26, 2019, 02:51:33 AM »
https://math.stackexchange.com/questions/3162591/a-quartic-whose-julia-set-includes-two-quadratic-julia-sets I tried some algebra, but not got something automatic yet (needs numerical root finding in 6 complex variables, plus a way to choose the best solution (the one with lambda_k closest to each other).  These quartics are not in general reachable by any AB sequences, though.