Logo by mrob - 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 us on facebook
 
*
Welcome, Guest. Please login or register. October 02, 2018, 09:00:51 AM


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: Can it be done at all  (Read 1203 times)
0 Members and 1 Guest are viewing this topic.
Tglad
Fractal Molossus
**
Posts: 703


WWW
« on: December 04, 2009, 09:13:14 AM »

If we accept that a 3d equivalent of the mandelbrot would want to be a conformal mapping that maps the 3d space onto itself, then we run into this: http://en.wikipedia.org/wiki/Liouville's_theorem_(conformal_mappings)

It states that in 3d you can only generate conformal mappings out of scale, rotation, translation, reflection and sphere inversion which means replacing the magnitude of all vectors with r*r/magnitude for some r.
So a conformal mapping might be: invert, translate by (1,0,0), invert, scale by 2, rotate around y axis 90 degrees.

So that's the set we have, no doubling of angles, or squaring the magnitude.
My tests show that any combination of these produce completely smooth boundaries.. which would suggest there is no _direct_ 3d analogy of a mandelbrot.

But there are a couple of loose ends:
1. non-smooth transformations seem to allow more variation
2. Maybe if there are no helpful conformal mappings then 3d mandelbrot would be the single mapping that is closest to conformal.
Logged
BradC
Safarist
******
Posts: 85



« Reply #1 on: December 04, 2009, 02:30:16 PM »

Maybe something related to this: http://en.wikipedia.org/wiki/Quasiconformal_mapping, but I don't know anything about it.
Logged
kram1032
Fractal Senior
******
Posts: 1863


« Reply #2 on: December 04, 2009, 07:52:32 PM »

Maybe I didn't understand some things but doubling a rotation to me is pretty much equivalent to rotation (actually it's equal. Only the amount of rotation changes, depending on where the point is found) and sqaring the radius is replacing the magnitude... (though I'm not sure if that could be written as "r*r/magnitude for some r")
Logged
Tglad
Fractal Molossus
**
Posts: 703


WWW
« Reply #3 on: December 05, 2009, 06:40:52 AM »

>> doubling a rotation to me is pretty much equivalent to rotation (actually it's equal. Only the amount of rotation changes, depending on where the point is found)
doubling a rotation stretches out any shape, just rotating the whole coordinates doesn't stretch anything.
This is what is so special about the 2d mandelbrot, the Z^2 doesn't stretch any shape (that is small enough) no matter where it is in the complex plane.

BradC, quasi-conformal maps sound like they might be way to go. It seems to mean that stretching is allowed but it is finite / bounded.
Logged
kram1032
Fractal Senior
******
Posts: 1863


« Reply #4 on: December 05, 2009, 02:35:12 PM »

ah, I see...
That makes sense smiley
Logged
Tglad
Fractal Molossus
**
Posts: 703


WWW
« Reply #5 on: December 15, 2009, 12:08:18 AM »

So the set of operations listed above seems unlikely to be able to produce interesting fractals as it doesn't have two attractors, everything seems to converge on one point (a single branch point I guess).
The other option is to find the *most* conformal 3d fractal, such as http://www.fractalforums.com/3d-fractal-generation/a-new-3d-mandelbrot-like-fractal/?action=post;quote=9827;num_replies=9;sesc=d3a594eba608ec813a9dad70a2e9fdf2
which is probably about '2 quasi conformal' meaning small spheres map to ellipsoids with eccentricity (long radius / short radius) no greater than 2.
Does anyone know of any 3d mandelbrot-like fractal with conformality < 2 ? (1 is fully conformal like 2d mandelbrot, mandelbulb is infinity-quasi conformal).
Even so, the above link shows that it needs to be more than just nearly conformal, the mapping probably needs to be smooth (infinitely differentiable)... and I'm not sure that it is even possible to have a continuous 3d mapping with conformality < 2.

So what other options? Perhaps in 3d one shouldn't look for it being conformal in euclidean space, but conformal in some other space.
Logged
kram1032
Fractal Senior
******
Posts: 1863


« Reply #6 on: December 15, 2009, 12:18:28 AM »

@ infinitely differentiable...

x² can be derived twice
and each iteration doubles the differentiability...

so at iteration infinite, just as the standard mandelbrot, the whole set should become infintely differentiable.
Or not differentiable at all as the graph has infinite k along the border of the set.
But that then should be the same for the standard mandelbrot.
« Last Edit: December 15, 2009, 02:48:39 PM by kram1032 » Logged
lycium
Fractal Supremo
*****
Posts: 1158



WWW
« Reply #7 on: December 15, 2009, 01:10:43 AM »

@ infinitely differentiable...
there, you quoted it!

x² can be derived twice
and each iteration doubles the derivability...
aaaaaaaaaaargh  lips are sealed
Logged

kram1032
Fractal Senior
******
Posts: 1863


« Reply #8 on: December 15, 2009, 02:49:19 PM »

don't see what you mean  praying
Logged
Pages: [1]   Go Down
  Print  
 
Jump to:  


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 0.148 seconds with 26 queries. (Pretty URLs adds 0.01s, 2q)