s31415


« Reply #15 on: November 29, 2009, 05:49:59 PM » 

Another alternative method of getting a true 3D Mandelbrot I've been considering is a 3D version of trigonometry using an idea that I've never actually seen described or used but I think is sound (I know I'm not rigorous enough for a proof): Consider a rightangled tetrahedron (which is an object that can by defined simply by (x,0,0),(0,y,0) and (0,0,z) it's fairly obvious that any two such objects are similar in the same way as triangles if the ratios of the *areas* of the sides stay unchanged. Considering this I thought that maybe we could define a specific rightangled tetrahedron by any two known ratios of areas  for example if we define the areas of the sides as xy, xz and yz for the small sides and xyz as the base (=="hypoteneuse") then if we know xy/xyz and either xz/xyz or yz/xyz then we have defined a set of selfsimilar tetrahedrons. Is this correct ? If so it probably gives an alternative method of defining rotation in 3D and squaring, multplying etc. once we have a version of Pythagoras for tetrahedrons relatng the areas. Edit: OK I know  make that the ratios of the square roots of the areas Hi David, The areas of the faces of your tetrahedron are A_1 = xy/2, A_2 = yz/2, A_3 = zx/2 and something a bit more difficult to compute for the last face. But the tetrahedron is completely determined by x,y,z, and provided you don't allow negative value, knowing the areas allows to recover x,y and z. For instance x = sqrt(2A_1 A_3/A_2). So it's really equivalent to consider the sides of the tetrahedron or the areas of its faces. I'm not sure how you want to use this... For the mathematically inclned, we can think of the trick for squaring vectors as a "covering" of the 2sphere S^2 by another 2sphere, ie a map from S^2 to S^2 such that the preimage of each point contains two points (or n points for the power n Mandelbulb). We saw that all the solutions found so far (involving doubling the angles in spherical coordinates) produced disconinuous maps. There is a reason for this. In the mathematical litterature, a covering is always a continuous applications, and we may wonder if there exists coverings of the sphere by the sphere. The answer is no. Connected coverings over a space X are classified by the subgroups of the fundamental group of this space ( http://en.wikipedia.org/wiki/Fundamental_group ), see theorem 1.38 here: http://www.math.cornell.edu/~hatcher/AT/ATch1.pdfThe fundamental group of the sphere is trivial, so it admits no nontrivial connected coverings. Edit: There can be branched coverings of the sphere by the sphere, see for instance http://www.jstor.org/stable/2046685?origin=crossrefBut I couldn't find any analytic formula. I'll think a bit more about it. Best, Sam


« Last Edit: November 29, 2009, 06:15:47 PM by s31415 »

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David Makin


« Reply #16 on: November 29, 2009, 07:00:48 PM » 

Another alternative method of getting a true 3D Mandelbrot I've been considering is a 3D version of trigonometry using an idea that I've never actually seen described or used but I think is sound (I know I'm not rigorous enough for a proof): Consider a rightangled tetrahedron (which is an object that can by defined simply by (x,0,0),(0,y,0) and (0,0,z) it's fairly obvious that any two such objects are similar in the same way as triangles if the ratios of the *areas* of the sides stay unchanged. Considering this I thought that maybe we could define a specific rightangled tetrahedron by any two known ratios of areas  for example if we define the areas of the sides as xy, xz and yz for the small sides and xyz as the base (=="hypoteneuse") then if we know xy/xyz and either xz/xyz or yz/xyz then we have defined a set of selfsimilar tetrahedrons. Is this correct ? If so it probably gives an alternative method of defining rotation in 3D and squaring, multplying etc. once we have a version of Pythagoras for tetrahedrons relatng the areas. Edit: OK I know  make that the ratios of the square roots of the areas Hi David, The areas of the faces of your tetrahedron are A_1 = xy/2, A_2 = yz/2, A_3 = zx/2 and something a bit more difficult to compute for the last face. But the tetrahedron is completely determined by x,y,z, and provided you don't allow negative value, knowing the areas allows to recover x,y and z. For instance x = sqrt(2A_1 A_3/A_2). So it's really equivalent to consider the sides of the tetrahedron or the areas of its faces. I'm not sure how you want to use this... For the mathematically inclned, we can think of the trick for squaring vectors as a "covering" of the 2sphere S^2 by another 2sphere, ie a map from S^2 to S^2 such that the preimage of each point contains two points (or n points for the power n Mandelbulb). We saw that all the solutions found so far (involving doubling the angles in spherical coordinates) produced disconinuous maps. There is a reason for this. In the mathematical litterature, a covering is always a continuous applications, and we may wonder if there exists coverings of the sphere by the sphere. The answer is no. Connected coverings over a space X are classified by the subgroups of the fundamental group of this space ( http://en.wikipedia.org/wiki/Fundamental_group ), see theorem 1.38 here: http://www.math.cornell.edu/~hatcher/AT/ATch1.pdfThe fundamental group of the sphere is trivial, so it admits no nontrivial connected coverings. Edit: There can be branched coverings of the sphere by the sphere, see for instance http://www.jstor.org/stable/2046685?origin=crossrefBut I couldn't find any analytic formula. I'll think a bit more about it. Best, Sam With respect to the ratios of the roots of the areas of the sides of the tetrahedron I was basically thinking that in the same way that a complex number is magnitude*(cos(theta)+i*sin(theta) then getting an equivalent to pythagoras for the areas of the sides of a tetrahedron would give us a "3D triginometry" such that our 3 valued number is magnitude*(side1/base + i*side2/base + j*side3/base), with side 1 being sqrt(x*y/2), side2 sqrt(y*z/2), side3 sqrt(x*z/2), as you say the area of the base is not so simple (I haven't tried to derive it, I was hoping someone would know the best formula for it for a rightangled tetrahedron in tems of x,y,z). Exactly how to handle signs correctly in "3D trig" would be interesting



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kram1032


« Reply #17 on: November 29, 2009, 09:13:16 PM » 

the last area would be x² or y² or z² if I did everything correct (which actually shows your pythagoras in some kind of way: x²=y²=z² ) and the volume is abs(x*y*z)/6



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Tglad
Fractal Bachius
Posts: 590


« Reply #18 on: November 30, 2009, 12:52:56 AM » 

>>In the mathematical litterature, a covering is always a continuous applications, and we may wonder if there exists coverings of the sphere by the sphere. The answer is no. I haven't read the link, but as far as I can see the answer is yes. The simple map that doubles the longitude on a sphere is continuous (C(0) continuity). Any two points an arbitrarily small distance apart will end up no more than twice that distance apart. Perhaps you mean it is discontinuous in some other way.



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s31415


« Reply #19 on: November 30, 2009, 01:03:06 AM » 

Yes, the matter is a little bit more complicated than what I explained. The map you're refering to is continuous, indeed, but it has two branch points, that is two points which have a single preimage, at the north and the south pole. Such a generalized notion of covering is called a "branched covering". The theorem cited above is valid only for genuine coverings which do not have any branch point.
If we allow branched covering, then the one you mention is the only one of degree 2 (up to continuous deformations such as moving the branch points around). It is not very interesting from the point of view of the Mandelbulb, because it acts trivially on the coordinate parallel to the axis of rotation. For higher powers, associated with branched covering of degree larger than 2, there might be more interesting possibilities, I haven't looked into it yet.



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David Makin


« Reply #20 on: November 30, 2009, 01:35:49 AM » 

I should add that I'm suggesting considering it specifically in terms of the areas because in 3D we're essentially dealing with the surface of a sphere rather than the circumference of a circle



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David Makin


« Reply #21 on: November 30, 2009, 02:16:32 AM » 

the last area would be x² or y² or z² if I did everything correct (which actually shows your pythagoras in some kind of way: x²=y²=z² ) and the volume is abs(x*y*z)/6 Methinks you made a mistake somewhere with the area of the base



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Tglad
Fractal Bachius
Posts: 590


« Reply #22 on: November 30, 2009, 04:14:00 AM » 

>> The map you're refering to is continuous, indeed, but it has two branch points, So does the 2d mandelbrot, where i=0... so branch points aren't necessarily a bad thing. >> the one you mention is the only one of degree 2 (up to continuous deformations such as moving the branch points around). It is not very interesting from the point of view of the Mandelbulb True, but most mandelbulb attempts don't just doublecover the sphere, they quadruple cover the sphere. Afterall, in order to get any patch on the sphere to double in width and in length, you need to quadruplecover. That's why the standard mandelbulb doubles the longitude and also doubles the latitude.
Doubling the latitude is kind of problematic since it isn't a strict 2>1 mapping... but it isn't the cause of the whipped cream effect... I'm not sure what causes that... but its might be because doubling the longitude actually leaves a whole semicircle untouched... but that's a guess.



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s31415


« Reply #23 on: November 30, 2009, 04:19:13 AM » 

>> So does the 2d mandelbrot, where i=0... so branch points aren't necessarily a bad thing.
What do you mean by i = 0 ? Recall that we are forgetting the radial direction. In the 2d case, we have a covering of degree two of a circle by a circle, without any branch point.
>>True, but most mandelbulb attempts don't just doublecover the sphere, they quadruple cover the sphere.
That's right. So investigating coverings of higher degree might be interesting.



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Tglad
Fractal Bachius
Posts: 590


« Reply #24 on: November 30, 2009, 05:07:16 AM » 

>> In the 2d case, we have a covering of degree two of a circle by a circle, without any branch point. The circle at i=0,r=1 doesn't move under the squaring operation (doubling the angle) The poles (and in fact all the way down the greenwich meridean) don't move when you double the longitude.
I'm just saying that the poles on a sphere aren't really different to this point on a circle... are they? So such points aren't necessarily a bad thing.



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s31415


« Reply #25 on: November 30, 2009, 06:14:45 AM » 

A branch point is a point whose preimage has a single element instead of 2, it has nothing to do with the point being left fixed by the map. In the case of the circle, you can check that each point has exactly two preimages, these are the two square roots of the corresponding complex number when you embedd the circle into the complex plane. I don't know if branch points are a bad thing or not, whatever this means, I'm just saying there are none in the case of the double covering of a circle by a circle, and there are two of them in the case of a double covering of a sphere by a sphere. But there is actually a much, much better way to think about all this. Consider first the usual Mandelbrot set. We can see the complex plane as a sphere after adding a point at infinity http://en.wikipedia.org/wiki/Riemann_sphereNow the map z > z^2 is a realization of the double covering we spoke before. The two branch points are 0 and the point at infinity. But this is not just a random double covering, this covering is conformal (it preserves the angles). http://en.wikipedia.org/wiki/Conformal_mapThe higher power mandelbrot sets correspond similarly to coverings of higher degree. Now here is what I could consider as a mathematically meaningful generalization of the Mandelbrot set. Just as the complex plane can be turned into a 2 sphere by adding a point at infinity, we can "compactify" the three dimensinonal Euclidian space R^3 into a 3sphere by adding a point at infinity. Now suppose that we have a covering of the 3sphere by the 3sphere which  is conformal  has a branch point at infinity. (To ensures that the point at infinity stays there and that we get a map from R^3 to R^3.) I've no idea if such a thing exists... While there are a lot of conformal maps in 2 dimensions, they are much more rare in 3 dimensions. I think that the property of being conformal is crucial if you want to avoid the "whipped cream" effect. The map currently used in the Mandelbulb is not conformal.



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Tglad
Fractal Bachius
Posts: 590


« Reply #26 on: November 30, 2009, 07:48:21 AM » 

Very interesting stuff. "A map of the extended complex plane (which is conformally equivalent to a sphere) onto itself is conformal if and only if it is a Möbius transformation" Are you saying that the Z^2 operation is a Mobius transformation? (visual here http://en.wikipedia.org/wiki/File:Mob3delipopp200.png) There are some things going click in my head here... the mobius visual looks a lot like electromagnetism, and it has been quoted that mandelbrots have been observed in electromagnetic fields (hence a 3d mandelbrot likely exists in some form). Also the mobius looks a lot like a torus, which is a shape that can go to double cover with no branch points (I think). >> Now suppose that we have a covering of the 3sphere by the 3sphere which >>  is conformal >>  has a branch point at infinity. Isn't that the quaternion multiply? Hmmm... I'm definitely out of my depth here.


« Last Edit: November 30, 2009, 07:52:22 AM by Tglad, Reason: answered my own question »

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s31415


« Reply #27 on: November 30, 2009, 08:06:27 AM » 

The Moebius transformations are the global conformal transformations of the Riemann sphere. But there are also "local" transformations which map open sets conformally onto open sets, but which might have poles at some isolated points. in particular, any holomorphic function of z (such as z^2), is locally conformally invariant.
You can test these claims very easily: take a square gird, transform it by any holomorphic function and check that the lines (which are now curves) still intersect orthogonally.


« Last Edit: November 30, 2009, 08:09:40 AM by s31415 »

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kram1032


« Reply #28 on: November 30, 2009, 03:48:14 PM » 

the last area would be x² or y² or z² if I did everything correct (which actually shows your pythagoras in some kind of way: x²=y²=z² ) and the volume is abs(x*y*z)/6 Methinks you made a mistake somewhere with the area of the base Oh yes, I did... I kinda used the dotproduct instead of the crossproduct :S It's That makes more sense



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David Makin


« Reply #29 on: November 30, 2009, 05:51:07 PM » 

Oh yes, I did... I kinda used the dotproduct instead of the crossproduct :S It's <Quoted Image Removed> That makes more sense I guess that's correct because that's the result I got too after looking on Wiki



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