### Julia and Mandelbrot sets w or w/o Lyapunov sequences

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• 3f
• Posts: 1746

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #45 on: January 13, 2020, 04:55:29 PM »
With that formula, if any coordinate is ever zero then x is zero one iteration later, and if x ever attains zero it is stuck there and the further evolution of y and z follows the rules: y <- -z2, z <- -(y5 + z4 + 0.5), which can be turned into an expression determining next z from both current and previous z.

Eyeballing this it looks like if x is zero and both y and z have small magnitudes (well below 1) the point is trapped. So, all points with one coordinate zero and the other two of sufficiently small magnitude would be trapped, which fits the three intersecting round-cornered squares in your image. The round-cornered squares are likely the shapes whose boundaries are solutions to implicit equations of degree 4 or 5 in the two initially-zonzero coordinates that determine whether z, with its -0.5 every iteration, ever manages to get outside of the unit circle, whereupon the point should escape. Since these shapes are "simple" it's likely the case that either z gets out in the first couple of iterations or else it never does (given one of the coordinates started out zero).

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #46 on: January 15, 2020, 04:59:32 PM »
@pauldelbrot: Thanks for the analysis. I computed the set with interval arithmetics, but couldn't detect interior so far.

Here's another image that shows a (half-)plane - but this time, the IA version detected interior (purple = immediate basin of a fix point, looks a bit like an anvil).

The image contains the positive parts of the coordinate axis, introduced after computing the set: x (blue), y (green), z (red).

So there is a sharp stop of not-escaping points of an xz-parallel plane around y=0 (I have not checked that the plane lies exactly at y=0). I hope to find something similar for the previous image as well.

{x,y,z}_new := { 2xz-2yz , y+3x²y-y4 , y²-z² } + { -0.5 , 0 , 0 }

• 3f
• Posts: 1746

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #47 on: January 15, 2020, 06:37:36 PM »
With that last one, if y ever becomes zero it's stuck there so I'd bet good money the planar feature is in fact exactly on the plane y = 0.

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #48 on: January 15, 2020, 06:57:26 PM »
With that last one, if y ever becomes zero it's stuck there so I'd bet good money the planar feature is in fact exactly on the plane y = 0.
Now that you point it out, it's obvious to me. I should maybe look at what I'm copy&paste-ing into my symbolic interval arithmetics parser to get the bounding box formulas. Thanks!

#### marcm200

• Fractal Frankfurter
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#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #49 on: January 17, 2020, 11:32:28 AM »
"Submarine"

Some strings (muscle fibers?) are accumulating in the center of this Julia set - and then there is this unexpected arc. Quite an interesting shape combination, having two very distinct features so close together. The side view looks like a submarine

I think the colors (trafassel's direct RGB summation) here roughly reflect the orbits: A reddish point moves predominantly around larger x-coordinates and in the y/z plane, the greenish ones with larger y and smaller x,z.

Color method: Rgb.red of orbit starting point (=visible pixel) = sum over all orbit points: |x coordinate of orbit point| divided by squared length of point, and at the end taking the fractional part times 255 gives the direct rgb.red value; green and blue use y and z coordinates.

{x,y,z}_new := { x²-y²-z² , 4xyz , y+4x3y-4y3x } + { 0.75 , 0 , -1 }

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #50 on: January 22, 2020, 08:58:13 PM »
"Highway traffic junction"

Very filigree triplex Julia set in the 4-cube (500 max it, non-escaped points, uncolored).

Interesting is how vastly different the object appears from different observer positions.

{x,y,z}_new := { x4-y4-z² , x²-y² , x²-y²-z² } + { 0 , -1 , 0 }

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #51 on: January 25, 2020, 07:10:52 PM »
I am currently looking for polynomial Julia sets having two (moderately long) cycles, where the periodic points of one cycle lie completely in the convex hull of the other.

The image below is the closest I've come so far.

z^7+A*z+c
with A=(12441960-31177622*i) * 2^-25
and c=(-49152+11141120*i) * 2^.25

All but two (red arrows) periodic points of the yellow cycle lie in the purple-enclosed region (iteration of the purple periodic points, so not directly the convex hull, but close enough for my purposes).

A small parameter variation of A and c (+-0.001%) did not change the position of the periodic points toward my goal.

I have to check the images I've computed from last year, but from the top of my head, I don't recall having seen a set with that property before.

Does anyone know parameters for such a set?

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #52 on: January 27, 2020, 05:06:53 PM »
First Julia set where one cycle (period-16, green) lies completely in the convex hull of another (period-6, yellow).

z^3+A*z+c
with c=(-3366912+36853760*i) * 2^-25
A=(-2747128-11828042*i) * 2^-25

The image was found by a point-sampling brute-force approach.

It seems that not only the periodic points are within the convex hull but also the complete immediate basins, but I haven't robustly verified that.

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #53 on: January 28, 2020, 10:29:37 AM »
"One in two"

An almost-square (yellow, period 4) enclosed here by two cycles of length 4 and 8.

z^4+A*z+c
c=(-6438912+5703680*i) * 2^-25
A=(835009+33871213*i) * 2^-25

This inscribed-feature seems quite common and easy to find. It would be interesting to see the subset of the corresponding parameter space {A,c}, where this appears, but that's currently out of my scope.

#### marcm200

• Fractal Frankfurter
• Posts: 620

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #54 on: March 24, 2020, 10:10:10 AM »
"Steps to cycle"

A cubic Julia set with just a fix point (red). The non-immediate basin Fatou components are colored according to how many iterations it takes them to reach the fix point's immediate basin.

It looks like a reaction cascade in particle physics: A yellow S-quark (upper right) is exchanging a blue gluon with the quark's fat supersymmetric cyan partner.

z^3+A*z+c

c=(-9191424-23142400*i) * 2^-25
A=(-26803078-16292030*i) * 2^-25

I was wondering: It seems that the size of the Fatou component and the distance to the fix point influence how many steps it takes (something like, in general, if smaller, it takes longer; if nearer it takes longer). Does anyone have articles about that?

• 3f
• Posts: 1746

#### Re: Julia and Mandelbrot sets w or w/o Lyapunov sequences

« Reply #55 on: March 24, 2020, 05:53:37 PM »
No, but I can perhaps offer some insight. The immediate basin will have a finite set of preimages (for quadratic Julia sets, only two, and in general governed by the degree of the polynomial or rational mapping), and one of those will be the immediate basin itself. The other preimages will likewise each have finitely many preimages. So they are iterationally organized as an nary tree, save that the root (the immediate basin) has only n - 1 branches from it.

So right away we see that there will be a small, finite number of images that are only a few hops from the immediate basin. After plotting those, all the other images, winding down into spirals and heading out to distant limit points and such, will have to be larger numbers of hops away. Anywhere there is an infinite regress of smaller and smaller basin-images the trend along that regress must of necessity be toward greater and greater hop-distance from the immediate basin.

Another thing: regions where the map is contracting are mostly in immediate basins around attractor member points. So it tends to be expanding at non-immediate basins, and wherever there's a spiral of these the map is going to be climbing up out of the spiral along them.

Also, if the map went down into a spiral of such in a regular fashion, it would never reach the immediate basin and what you'd have instead is a wandering domain. Those are proven to only occur in transcendental functions, not rational (or by extension polynomial) maps.

So as a matter of statistical fact you'll see precisely what you observed, though there can be limited and finite exceptions, such as one non-immediate basin being bigger than the immediate one, or a spiral where though the trend during iteration is toward bigger non-immediate basins it goes "two steps forward, one step back" repeatedly in that regard.

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