Logo by Fiery - Contribute your own Logo!

END OF AN ERA, FRACTALFORUMS.COM IS CONTINUED ON FRACTALFORUMS.ORG

it was a great time but no longer maintainable by c.Kleinhuis contact him for any data retrieval,
thanks and see you perhaps in 10 years again

this forum will stay online for reference
News: Visit the official fractalforums.com Youtube Channel
 
*
Welcome, Guest. Please login or register. August 10, 2022, 04:58:00 PM


Login with username, password and session length


The All New FractalForums is now in Public Beta Testing! Visit FractalForums.org and check it out!


Pages: [1]   Go Down
  Print  
Share this topic on DiggShare this topic on FacebookShare this topic on GoogleShare this topic on RedditShare this topic on StumbleUponShare this topic on Twitter
Author Topic: Christmas Balls  (Read 837 times)
Description: Matchmaker Mandelbrot image and Julia sets
0 Members and 1 Guest are viewing this topic.
Pauldelbrot
Fractal Senior
******
Posts: 2592



pderbyshire2
« on: March 16, 2009, 10:40:28 AM »

This post contains many images, most of them 480x360 Julia images that link to larger sizes. Detailed statistics for a particular image can be had by clicking it; there are too many to list all the stats in this post. The slowest was Magnetic Force at nearly 10 hours. Christmas Balls itself only took half an hour; besides Magnetic Force, Vergence was slower. Those two Julias required aggressive anti-aliasing and the minimum iterations in Magnetic Force is 718.


The Matchmaker Mandelbrot b-plane with a = 0. This is the un-souped-up fractal that Mandelbrot himself studied at one time.

One layer is a greyscale gradient smoothed-iteration layer. Four more layers color particular subsets of the buds by filtering; for instance the red layer filters out the green and blue components of the underlying greyscale layer. All of the filtering layers color disjoint subsets of the buds, and together they color them all save for the "inside-out" one, radius 1 centered on infinity.

The grey area corresponds to zero being a stable fixed point and the only attractor. The red areas have a pair of equal-period attracting cycles and this period is odd; blue areas have a pair of equal-period attracting cycles and this period is even. Yellow areas have a single attractor whose period is not divisible by four. Green areas have a single attractor whose period is divisible by four.

The black area (no attractors at all) is perforated with swarms of minibrots. Minibrot cardioids come in every color except grey.

There are patterns in what colors occur where that are explained below and illustrated by numerous Julia sets.

The grey zone corresponds to the b-values of f(z,b) = \frac{z}{b(z^2 + 1)} for which zero attracts. It is easy to prove mathematically that zero is a fixed point and that it attracts for |b| > 1; 0 goes to 0/b = 0 so the first part is obvious, and the rest follows with the use of some elementary calculus: the derivative of f with respect to z is \frac{1 - z^2}{b(z^2 + 1)^2}, which at zero is 1/b, indicating stability for |b| > 1.

The colors of the other areas reflect which of four possible types of dynamics occur when zero is not an attractor. Black points have no attractors.

The critical points are at 1 and -1 (easily inferred from the derivative's numerator) and f(-z,b) = \frac{-z}{b(z^2 + 1)} = f(z,-b). Since negating z negates f(z,b), the critical points have symmetric orbits, exchanged by a 180 degree rotation. It follows that if one goes to zero both of them do, so in the grey zone zero is the only attractor.

It also follows that if there are two attractors, they are rotated images of one another, whereas if there is only one attractor, every point in it has its negative in it as well, and if the attractor isn't just the origin, the cycle is even in length.

Below is a Julia set from the grey zone; with only a single attractor and that a fixed point, the Julia sets out here are dusts.

Final Frontier

The red buds have a cycle of odd length, and therefore have two attractors.

Since negating b negates f(z,b), when you negate b, you negate the odd iterates but not the even iterates: f(z,-b) = -f(z,b), f(f(z,-b),-b) = -f(f(z,-b),b) = -f(-f(z,b),b) = -(-f(f(z,b),b) = f(f(z,b),b) and then it repeats.

If we have an odd length attracting cycle a1, a2, a3, ... then -a1, -a2, -a3, ... is the other attractor. If we then negate b, the iteration starting at a1 now goes a1, -a2, a3, ... and eventually hits the last element of the attractor un-negated, following which -a1, not a1. So a1 is now a member of an even-length cycle that consists of all the points of both of the old attractors.

Furthermore, this cycle is attracting: the derivative of f, given above, is an even function whose magnitude isn't altered by negating z or b, so even when it is applied over the whole cycle using the chain rule the same magnitudes get multiplied and only the sign may change. If those magnitudes were diminishing before, they still are. Or just note that the orbits of 1 and -1 are shuffled with each other but otherwise unchanged, so they still converge on the same set of points. It's just that those points form a single attractor instead of two.

So the buds 180 degrees opposite red buds have a single attracting even-length cycle; for a given b-value in one of these buds this attractor contains the same points as the pair of odd-length cycles you get with the point -b in the opposite red bud. These buds I've called "splitter" buds, because negating b splits the attractor into two attractors. They are colored yellow in the image of the Mandelbrot set at the top of this post. Here are some illustrative Julia sets, one from the edge of the largest yellow bud and one with that one's b negated (image flipped horizontally):

Golden
Space Frontier

The left image is all one attracting basin, colored with the same golden gradient throughout, unlike the right image, which has two distinct basins (red and dark-purplish).

Whenever there's only one attractor, it's an even-length cycle invariant under negation, but it isn't always a "splitter". In particular, a "splitter"'s cycle length has to be congruent to 2 modulo 4, since its length is double an odd number. So, what happens if a cycle whose length is divisible by four is subjected to negation of alternating points? The result ultimately depends on how points and their negatives are distributed in an even-length cycle.

In fact, if fk(a) = -a, then since negating the input negates the output, fk(-a) = a. If a cycle contains a and -a, then, they are halfway around the cycle from one another. Therefore if the cycle length is not divisible by 4, so k is odd, negating alternate iterates causes fk(a) = a and therefore produces two separate odd cycles (so all such attractors are splitters, the converse of a result obtained above). If, however, the cycle length is divisible by four and so k is even, fk(a) is still -a, and the cycle remains unchanged in length and set of member points. However, if the cycle originally went a1, a2, a3, ... -a1, -a2, -a3, ... it now goes a1, -a2, a3, ..., -a1, a2, -a3, ...; it has been rearranged. I call these attractors "shufflers", because negating b shuffles the points in the attractor but doesn't change the period or membership. "Shuffler" buds therefore lie opposite other "shuffler" buds under a 180 degree rotation of the Mandelbrot set. I have colored these buds green in the Mandelbrot image at the top of this post.

The following Julia set images demonstrate a "shuffler" at work: the top images show the Julia set from a tiny green bud close to the attachment point of the giant red bud to the grey outer zone, with smoothed iterations coloring, and its opposite number; the bottom images show the same with phase coloring instead. Though they look identical under smoothed iterations coloring, the phase coloring clearly shows how the order of visitation of components of the immediate attracting basin is shuffled.

Vergence

Finally, it is possible to have two disjoint even-length attracting cycles. Unlike the odd case, negating every other point won't make the cycles join into a single double-length cycle. Instead, the two attractors will exchange an alternating subset of their members. These I call "mixer" buds, because the points of two attractors get mixed by negating b. Mixer buds are colored blue in the Mandelbrot image at the top of this post, and they are always opposite other mixer buds.

The following two Julia sets show the basins of two opposite mixer buds (second image flipped horizontally):

Interlock I
Interlock II

Dark blue is one basin and light blue is the other in both cases. Note how some Fatou components belong to different basins after b is negated.

Bifurcations of these attractors have various predictable patterns. Where a smaller bud buds off a larger bud the associated attracting cycle has its period multiplied by a factor as each point in the attractor becomes a group of points. For example, if we had a 2-cycle x, y and it trifurcates we get a 6-cycle x1, y1, x2, y2, x3, y3, with x1, x2, and x3 near x and y1, y2, and y3 near y.

Where there are two symmetric attractors, both must undergo identical bifurcations. Pairs of odd cycles bifurcate into either pairs of odd cycles or pairs of even cycles, so one expects alternate blue and red buds attached to red buds, which is exactly what we get. Similarly, pairs of even cycles only bifurcate into pairs of even cycles, so blue buds have only blue buds attached. The behavior of yellow buds can be predicted from the b-negation rules and the behaviors of red and blue buds: they have yellow and blue buds where the red ones have red and blue buds.

Green buds, interestingly, have the same behavior again, blue buds at even multipliers and still green at odd ones. This is harder to explain (and prove) but consider any even-length self-symmetric cycle at all, q, r, s, ... -q, -r, -s, ... bifurcating with any multiplier. The first part of the larger cycle will be q1, r1, s1, ..., (-q)1, (-r)1, (-s)1, ... with (-q)1 not necessarily the same thing as -(q1). But it WILL be -(qx) for some x.

In fact, the cluster of new attracting points about members of the old cycle must slowly make one revolution as one completes a circuit around the old cycle, so that after (-z)1 (say) comes q2. So half way around (at -q) it has rotated half-way, and (-q)1 should be close to -(q1) (since the two near-180-degree rotations should about cancel). If the multiplier is 2, it has to be -(q1) as -(q2) is nowhere nearby, so the cycle breaks into two: after (-z)1 comes q1 again rather than q2, so the x2s form a separate cycle. If the multiplier is higher it cannot be -(q1) as the number of separate cycles would be more than the number of available critical points. It must therefore be -(q2) or -(qN) with N the multiplier, i.e. there is a shift of either 1 or -1 modulo N on reaching the "negative half" of the cycle. Which means there is a shift of 2 or -2 modulo N on each complete cycle. Which means that if 2 and N are not relatively prime (that is, if N is even) the cycle returns to its starting point after exhausting only half the points, and again the attractor has split in two.

In particular, this predicts not only the same behavior for yellow buds noted above (the even-multiplier smaller buds are blue), but the observed behavior for green ones as well. (With these, all attractor periods in the whole bifurcation "family tree" will be divisible by four.)

It also predicts that the attractors for blue buds attached to yellow or green ones will be interleaved with each other, which is exactly the behavior demonstrated by some stunning Julia set images I've generated.

The above pair, Interlock I and Interlock II, are from blue buds attached to, respectively, the big red bud and the big yellow bud; in the yellow case, the 2-cycle has become a pair of interleaved 4-cycles, as indicated by alternate light and dark petals (light blue is one attracting basin, dark blue the other). In the red case, two separate attracting fixed points bifurcated into 4-cycles, which remained clustered.

What happens if the bifurcation is by a factor of three instead is illustrated below:

Lava Shore
Froth

The two fixed points become two three-cycles (left); the 2-cycle opposite becomes a single 6-cycle (right; all one color indicates single attracting basin).

Below is a Julia set from a "mixer" bud attached to a "shuffler" bud -- in fact, the same "shuffler" bud that produced Vergence, which is reproduced to its left below:

Vergence
Magnetic Force

Again, note that when an even cycle splits the basins become interleaved as tightly as possible. A 1-cycle split into 32 (Vergence), and then this split into two 32-cycles. Each point of the original 32-cycle split in two, but as per the model above, one wound up belonging to each of the new attractors, so instead of the 32 petals becoming 64 with one group of 32 and then a second group of 32, or even with two interleaved groups of 32, each petal became a chain of beads belonging alternately to the two attractors. The largest beads in each petal contain attractor points, and the attractors form two concentric rings of 32 points each.

The Julia sets of f(z,b), in general, have a number of symmetries, some deriving from the same reasoning as the above. For one thing, repelling periodic points are still repelling periodic points after negating b or z, so the Julia set itself, as the closure of this set, is unchanged under both symmetries. In particular, it is always 2-fold rotationally symmetric about the origin. f(1/z,b) = f(z,b) so it is also inversion-symmetric. The usual conjugacy symmetry applies, such that conjugating b mirrors the Julia set in the y-axis. (Which the twofold symmetry makes equivalent into mirroring in the x-axis; the mirrored images above were actually mirrored by negating the imaginary part of b). Negating the real part of b amounts to negating b and then conjugating, so it shuffles shufflers, mixes mixers, swaps splitters with pairs of odd cycles, and mirrors in the y-axis all at once. The M-set is therefore y-axis symmetric, and x-axis symmetric except for red and yellow buds exchanging, which reasoning is borne out by the image at the top of this post.

The biggest red bud corresponds to two symmetric attracting fixed points; in the corresponding big yellow bud one gets single 2-cycles. The biggest two green buds (top and bottom of set) have single 4-cycles. The black area (no attractors at all) is perforated with swarms of minibrots. Minibrot cardioids come in every color except grey. 180-degree rotation (and x-axis mirroring) again must swap red and yellow and fix green and blue. The Julia sets in these are foams surrounding "bubble" basins, as demonstrated with our final pair of Julia images:

Gold Nuggets
Dark Matter

As usual, each image link leads to a 1024x768 image, with a 2048x1536 image available by following one more link from there, and a 2048x1536 lossless PNG available upon request.

All images are, as usual, freely redistributable and usable subject to the Creative Commons Attribution license, version 3.0.
Logged

Pages: [1]   Go Down
  Print  
 
Jump to:  

Related Topics
Subject Started by Replies Views Last post
Pytagora's balls Mandelbulb3D Gallery DarkBeam 0 376 Last post March 24, 2011, 07:37:06 PM
by DarkBeam
Field of balls Mandelbulb3D Gallery miguelnpg 0 463 Last post May 15, 2011, 07:38:12 AM
by miguelnpg
Twit Twoo Balls Mandelbulb3D Gallery JoaGoo 0 228 Last post January 05, 2012, 04:10:43 PM
by JoaGoo
Curls Loops and Balls Images Showcase (Rate My Fractal) thom 0 506 Last post February 26, 2012, 05:54:44 AM
by thom
Valleys of the Balls Movies Showcase (Rate My Movie) trumanbrown 2 556 Last post April 29, 2016, 11:57:52 AM
by 1Bryan1

Powered by MySQL Powered by PHP Powered by SMF 1.1.21 | SMF © 2015, Simple Machines

Valid XHTML 1.0! Valid CSS! Dilber MC Theme by HarzeM
Page created in 0.231 seconds with 27 queries. (Pretty URLs adds 0.007s, 2q)