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Author Topic: infinite zoom of the mandelbrot?  (Read 3265 times)
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kram1032
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« Reply #15 on: May 26, 2010, 11:01:44 PM »

Good that you made that clear smiley

If a hexagon is thought of as the shape of the densest squashed packing of circles, then the 3d equivalent of a hexagon would be the octahedron, since that is what you get by using the optimal 'face centred cubic' packing and squashing it... incidentally octahedrons also have 6 corners.
http://www.wolframalpha.com/input/?i=octahedron this is not quite the shape of a sphere packing.
http://www.wolframalpha.com/input/?i=Rhombic+dodecahedron this one is for the cubic packing.
http://en.wikipedia.org/wiki/Trapezo-rhombic_dodecahedron and this is for hexagonal packing.

As the hexagonal packing is called... hexagonal, I'd say, a Trapezo-rhombic Dodecahedron is about as close to a 3D-hexagon as you could get smiley

Now I wonder, what kinds of kaleidoscopic fractals could be done with those two shapes smiley

EDIT: Ok, I'm now 99% sure that I even found the 4D-equivalent which has some nice and interesting properties smiley
http://en.wikipedia.org/wiki/Kissing_number_problem <- the Kissing Number for the 4D-Case is 24. That means, a central sphere can be surrounded by up to 23 other spheres.
http://www.answers.com/topic/24-cell <- the 4D-equivalent of the Rhombic Dodecahedron is the 24-cell. It has 24 vertices and http://www.answers.com/topic/icositetrachoronic-tetracomb it tiles the euclidean R^4 hyperspace, which is somewhat of a requirement, I guess...
It's pretty obvious that this is the 4D-variant. Now, while that's easy to assume by what we see, it must be ridiculously hard to proof it for sure...
« Last Edit: May 26, 2010, 11:23:33 PM by kram1032 » Logged
Tglad
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« Reply #16 on: May 27, 2010, 07:31:04 AM »

I stand corrected Kram. Interesting 3d shapes.

ikmitch and bohold, I don't think this is true. I don't think the mandelbrot gets 'denser' as you zoom in, I've seen images zoomed to e^100 which don't look much different than e^10.
I have a little theory, inspired by your 'the midgets become more isolated' point, and I'm talking just about boolean images here...

Towards the z=-2 the minibrots get closer and closer to replicating the main mandelbrot, i.e. the thin 'wires' that connect them (or emminate from them) get proportionally thinner and thinner. As a result, you can approximate the main mandelbrot to an arbitrary degree of accuracy.
Therefore, the main mandelbrot (and any image within it that you choose to render) can equivalently be seen as just a minibrot at your required accuracy... and so the conclusion is-
 Any image of the mandelbrot set is also a pixel accurate infinitely deep zoom; you can zoom out forever without reaching the main mandelbrot.
« Last Edit: May 27, 2010, 07:36:25 AM by Tglad » Logged
hobold
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« Reply #17 on: May 28, 2010, 06:32:43 PM »

It depends towards what kind of point you are zooming. If you zoom into some sort of antenna tip, then the shapes will continue to thin out as described. There is a result that Mandelbrot images converge to Julia images as you zoom in, because the minibrots' relative size tends towards zero diameter. In the end, all minibrots have vanished and only the embedded julias remain, at which point you have a julia set image.

However, if you zoom into the point where the main cardioid just touches the largest disk, you are obviously centered on a border point (moving an infinitesimal step along +i or -i takes you outside). An infinitely deep zoom in here will end up with an "all black" image.

I think the latter case is true whenever you zoom into a border point that touches a cardioid or disk. The former case is true for points that do not touch any solid components. I am not sure if that leaves room for a third case.
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kram1032
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« Reply #18 on: May 28, 2010, 11:15:51 PM »

I just testet around a little bit:
%3D%28z[n-1]%29%C2%B2-2]http://www.wolframalpha.com/input/?i=z[n]%3D%28z[n-1]%29%C2%B2-2
It seems like for some cases you could find a generic set smiley
Just go lim n->infinity for a continuous function which probably is way simpler than for a series of integers.

I wonder how to solve systems like that... for -2 it seems pretty simple. if you use a generic c, no solution is found. Neither so for -1 smiley

However I'm not actually sure if those solutions are really correct:
-3 should be diverging, right?
%3D%28z[n-1]%29%C2%B2-3]http://www.wolframalpha.com/input/?i=z[n]%3D%28z[n-1]%29%C2%B2-3
« Last Edit: May 29, 2010, 10:04:22 AM by kram1032 » Logged
pseudogenius
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« Reply #19 on: May 29, 2010, 01:54:56 AM »

Great idea, if we can solve the recurrence relation for a certain point and points infinitely close to it, then we can take n->infinity and get the real infinite zoom of it!  smiley

Here's how we might start solving for certain points
http://en.wikipedia.org/wiki/Recurrence_relation

P.S. kram, for -3 wolfram is using z(0)=1 instead of 0, so that is why its not diverging
« Last Edit: May 29, 2010, 02:03:56 AM by pseudogenius » Logged
kram1032
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Posts: 1863


« Reply #20 on: May 29, 2010, 10:03:47 AM »

ah, yeah, didn't notice that one smiley

The ideal solution would be z_{n+1}={z_{n}}^2+c_1 and z_0=c_2

with c_1 and c_2 being aribitary variables.
That would include the whole Mandelbrot and Juliaset.

Btw, taking n->infinity wont be too simple either: -2 featured a cosine. Infinity of cosine is defined as [-1,1] which isn't really helpful^^
However, from that equation it's easy to find the period of that point, I guess smiley
(Of course, -2 is rather trivial, here)
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pseudogenius
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« Reply #21 on: May 30, 2010, 09:58:15 PM »

I don't think there is any hope in finding a solution in general, Here's what wolfram says:

"A quadratic map is a quadratic recurrence equation of the form
x_{n+1}={a_2}x_n^2+{a_1}x_n+{a_0}
While some quadratic maps are solvable in closed form (for example, the three solvable cases of the logistic map), most are not."

We still might be able to take n->infinity, but it'll have to be done without using regular limits.

Anybody have any thoughts on how to do this?
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kram1032
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Posts: 1863


« Reply #22 on: May 30, 2010, 10:29:40 PM »

I guess, if it was solvavle, someone would have done it years ago... Grin with closed eyes
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