Re: [math-fun] The Real 3D Mandelbrot Set
Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity. Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is: z(0) = c0 z(i) = z(i-1)^2 + c1 The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1. Unfortunately, having a 2-dimensional retina makes it difficult for me to visualize such a set. Being three dimensional can be very frustrating. Tom
I cannot understand the basis for the assumption that there is a "real" 3D Mandelbrot set.
--Dan
Sometimes the brain has a mind of its own.
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On 8/9/11, Tom Karzes <karzes@sonic.net> wrote:
Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity.
Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is:
z(0) = c0 z(i) = z(i-1)^2 + c1
The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1.
Unfortunately, having a 2-dimensional retina makes it difficult for me to visualize such a set. Being three dimensional can be very frustrating.
Tom
You wouldn't want to be (spatially) 4-D. It messes up the planetary dynamics something rotten, or so I understand. WFL
This is ancient history - 3D/4D Julibrots have been known of and rendered since the early days of Fractint. Here's an animation of many ways of viewing the 4D Julibrot: http://www.youtube.com/watch?v=gr-ul7sZDwc Also when Daniel White and others at http://www.fractalforums.com/ say "the true 3D Mandelbrot" (or 4D+) they actually mean a (Mandelbrot-style) fractal using the standard iteration of z^2+c that is fractal in all directions on the surface i.e. it has no "whipped cream" - clearly so far there is no 3D+ number form for which z^2+c fulfils this - in fact such a thing may not be possible in 3D+ - at least it seems unlikely with respect to "sensible" number forms ;) Daniel White and Paul Nylander's Mandelbulb (z^p+c) does however come close at higher powers - say p>=6 or so: http://makinmagic.deviantart.com/gallery/?offset=72#/d2f2kkz On 9 Aug 2011, at 00:13, Tom Karzes wrote:
Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity.
Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is:
z(0) = c0 z(i) = z(i-1)^2 + c1
The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1.
Unfortunately, having a 2-dimensional retina makes it difficult for me to visualize such a set. Being three dimensional can be very frustrating.
Tom
I cannot understand the basis for the assumption that there is a "real" 3D Mandelbrot set.
--Dan
Sometimes the brain has a mind of its own.
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* Tom Karzes <karzes@sonic.net> [Aug 11. 2011 07:51]:
Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity.
Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is:
z(0) = c0 z(i) = z(i-1)^2 + c1
The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1.
[...]
Neat! Back in 1993 (or so) I used quaternions, only to become slightly diappointed because all cut planes containing the real axis produce the usual image 8-) The other cuts looked "random" to me, and that is what I observe with most "new" attempts. It might be more interesting to visualize the Julia sets. Maybe some live in a proper 3D subspace and no projection or cut would be required. Another thing is to look at finite fields. Of course, nothing escapes anymore, so one may rather look at the decomposition into orbits of different size. Take GF(q^2) to get 2D right from the start, q being a prime, or maybe a prime power (dunno whether GF(2^(2*k)) could possibly give something nice). My attempts of this back then, again, produced rather "random" images. However I did know too little about the (well known) theorems of elementary field theory.
In GF[2^K], f(X) = X^2 is a linear map, and g(X) = X^2+C is a linear map with a constant offset. This produces a relatively simple iteration theory. Rich ---- Quoting Joerg Arndt <arndt@jjj.de>:
* Tom Karzes <karzes@sonic.net> [Aug 11. 2011 07:51]:
Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity.
Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is:
z(0) = c0 z(i) = z(i-1)^2 + c1
The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1.
[...]
Neat!
Back in 1993 (or so) I used quaternions, only to become slightly diappointed because all cut planes containing the real axis produce the usual image 8-)
The other cuts looked "random" to me, and that is what I observe with most "new" attempts.
It might be more interesting to visualize the Julia sets. Maybe some live in a proper 3D subspace and no projection or cut would be required.
Another thing is to look at finite fields. Of course, nothing escapes anymore, so one may rather look at the decomposition into orbits of different size. Take GF(q^2) to get 2D right from the start, q being a prime, or maybe a prime power (dunno whether GF(2^(2*k)) could possibly give something nice). My attempts of this back then, again, produced rather "random" images. However I did know too little about the (well known) theorems of elementary field theory.
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participants (5)
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David Makin -
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Joerg Arndt -
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Tom Karzes