(Question) Continuous one-to-one functions and fractal connectivity

  • 11 Replies
  • 202 Views

0 Members and 1 Guest are viewing this topic.

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« on: October 02, 2018, 01:08:08 AM »
Can anyone with a more formal math education than myself please answer the following:

Given an entirely connected escape time iterated fractal (such as the Mandelbrot for z^2+c) if z is manipulated using a continuous one-to-one none divergent function on each iteration prior to feeding into the connected formula (in this case before squaring z) is the resulting fractal also guaranteed to be connected ?
Also if the extra manipulation is partially divergent is there then no guarantee that the result is connected ?

Intuitively I'd answer yes it will be connected and yes there's no guarantee if some divergence exists in the extra manipulation but I lack the formal background to come up with a definitive proof so I can't be 100% certain ;)
Maybe formal knowledge of topology can provide the answer......
The meaning and purpose of life is to give life purpose and meaning.

Offline gerrit

  • *
  • 3f
  • ******
  • Posts: 1536
« Reply #1 on: October 02, 2018, 01:49:29 AM »
No, a counterexample is \( z \leftarrow z^2-z,\ z \leftarrow z^2 +c \).

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« Reply #2 on: October 02, 2018, 02:02:50 AM »
No, a counterexample is \( z \leftarrow z^2-z,\ z \leftarrow z^2 +c \).

OK so if the added first function is divergent then we possibly have a fail, what if it's non-divergent everywhere ?

Offline gerrit

  • *
  • 3f
  • ******
  • Posts: 1536
« Reply #3 on: October 02, 2018, 02:17:41 AM »
OK so if the added first function is divergent then we possibly have a fail, what if it's non-divergent everywhere ?
What do you mean by "divergent"?

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« Reply #4 on: October 02, 2018, 09:19:03 AM »
By "divergent" I mean if applied as an iterated formula on its own as f(z) then it would have some areas "outside" i.e. with the infinity attractor, rather than something either scale invariant or convergent, and non-drifting (e.g. x+1 would drift to the infinity attractor).
Please note that maybe that f(z) should involve the usual "c" but I'm not sure ;)

Offline hobold

  • *
  • Fractal Phenom
  • ****
  • Posts: 57
« Reply #5 on: October 02, 2018, 11:55:18 AM »
Given an entirely connected escape time iterated fractal (such as the Mandelbrot for z^2+c) if z is manipulated using a continuous one-to-one none divergent function on each iteration prior to feeding into the connected formula (in this case before squaring z) is the resulting fractal also guaranteed to be connected ?
In short: no, there is no such guarantee.

This is due to the fact that we iterate infinitely often, at least conceptually, and the transition from finite to infinite can break smoothness and continuity, and thus connectivity can be lost.

Let me try a simple illustrative example: a one dimensional "smooth step" function f(x) on the interval [0 .. 1]. The necessary constraints are:
f(0) := 0
f(1) := 1
f'(0) := 0
f'(1) := 0
and we don't want it to wildly zigzag inbetween. You could pick the cubic polynomial determined by the above four constraints, or you could construct something smoother.

Iterating this function, i.e. f(f(x)) is still a smooth step. This is true for any finite number n of iterations. But the slope of the iterated function at x = 0.5 will keep rising as n increases. After transcending n to infinity, the iterated function's slope is infinite. It is now a hard step, a discontinuity. It will tear apart the two ends that it used to smoothly connect in the finite case.


(Things are really a bit more complicated than this example, because I would also need to prove that any smooth function connecting both ends must have slope > 1 at some point in its interval of definition. And then I would have to prove that this suffices even if that "steep point" is not at 0.5, and so on and so forth.)

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« Reply #6 on: October 02, 2018, 05:22:20 PM »
I knew this would be more complicated than I thought at first ;)

What if the extra function is a. Convergent (like a Newton) or b. Scale invariant like a simple rotation such as z^p/magnitude(z)^(p-1) ?

In the first case I'm now thinking that if said convergence at any pixel is greater during iteration than the divergence for the following function (in normally divergent areas of the following function) then there's a chance it will be disconnected anyway i.e. normal divergent areas of the original not necessarily connected could become convergent, however in the scale invariant case....since the angle change is essentially a mod() function with no scale change in magnitude at all and a fixed rotation change for a given value over the iterations ? (unless the angle scaling is increased deliberately over iterations - might try that ;) )

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« Reply #7 on: October 02, 2018, 05:41:47 PM »
In short: no, there is no such guarantee.

This is due to the fact that we iterate infinitely often, at least conceptually, and the transition from finite to infinite can break smoothness and continuity, and thus connectivity can be lost.

(In addition to my last post)
What about practically rather than conceptually ? i.e. assuming finite iterations ?

Offline hobold

  • *
  • Fractal Phenom
  • ****
  • Posts: 57
« Reply #8 on: October 02, 2018, 11:14:14 PM »
Functions that neither contract nor expand anywhere are safe. But that limits you to rotations, translations, and mirroring. In the 1D case, these are all the functions with slope equalling 1 or -1 everywhere.

I am not sure what happens with functions that are not expanding anywhere, i.e. analogous to 1D with slope <= 1.0 everywhere. On first glance they might be okay, but I wonder if there is a downside. After all, such functions globally shrink whichever sub-volume of space you push through them.

Doing only a finite number of iterations in practice ... hmm. I cannot predict that. We could be lucky that all level sets look smooth and concentric, and only the boundary of the attractor looks fragmented. Or we could get the opposite: the level sets look fragmented after merely a few iterations, but under extreme magnification there could be thin filaments connecting those fragments.

Offline FractalDave

  • *
  • Uploader
  • *
  • Posts: 167
    • Makin Magic Fractals
« Reply #9 on: October 02, 2018, 11:34:39 PM »
Functions that neither contract nor expand anywhere are safe. But that limits you to rotations<snip>

OK - in fact I was (apparently) finding that in practice - for 3D (x,y,z) scaling the angle of y/z around x prior to treating (x,y,z) as a quaternion with zero for w seems to result with consistently connected fractals provided the quaternionic fractal formula used on its own results in a connected fractal. But any power scaling of the magnitude of y/z as say y+iz does seem to result in disconnected sections.

Offline gerrit

  • *
  • 3f
  • ******
  • Posts: 1536
« Reply #10 on: October 03, 2018, 02:00:28 AM »
Applying \( f(z) = ze^{i|z|} \) before each M-set iteration seems to produce a connected fractal, at least as far as I've been able to check visually.

Offline gerrit

  • *
  • 3f
  • ******
  • Posts: 1536
« Reply #11 on: October 03, 2018, 06:15:12 AM »
It does not seem true that any norm preserving function applied  to z before the usual M-set iteration results in a connected fractal.
I considered \( f(z) = z e^{ir/(|z|-q)} \), Attached uf141 has \( q = 1/4 -i/10 \) and uf142 \( q=1/4-i/4 \).
For both \( r = 1/2 \). The first is not connected, the second is.


xx
"Time Span"

Started by cricke49 on Fractal Image Gallery

0 Replies
120 Views
Last post August 02, 2018, 07:05:21 AM
by cricke49
xx
Birdie Style

Started by gannjondal on Fractal Image Gallery

1 Replies
210 Views
Last post May 08, 2018, 02:39:37 PM
by who8mypnuts
xx
Buddhabrot-style Burning Ship [65536x24576]

Started by programagor on Fractal Image Gallery

12 Replies
260 Views
Last post October 10, 2018, 02:12:39 AM
by 3DickUlus
xx
GIMP/GMIC/Continuous Droste Filter

Started by Kalter Rauch on ChaosPro

19 Replies
334 Views
Last post April 28, 2018, 03:35:16 AM
by Kalter Rauch
xx
Greetings... I have a question! :D

Started by vrana on Meet & Greet

6 Replies
388 Views
Last post December 25, 2018, 01:07:51 PM
by FractalDave