True 3D mandelbrot fractal (search for the holy grail continues)

[David Makin]

It should be remembered that Pythagoras (although used for distance calculations) actually relates areas rather than distances

[jehovajah]

I think the problem lies in the fact that i is a unary operator defined on the plane, and j and k have followed suit. To define a 3d operator is perhaps as straight forward as

i on a = ia and iia = -a and iiia =-ia and iiiia =a

i on (a,b)= (-b,a) ii(a,b) = (-a,-b) iii(a,b) = (b,-a)  iiii(a,b) = (a,b)


i on (a,b,c) = (-c,a,b,)  ii(a,b,c) = (-b,-c,a)  iii(a,b,c) = (-a,-b,-c,)  iiii(a,b,c) = (c,-a,-b,) iiiii(a,b,c) = (b,c,-a) iiiiii(a,b,c) = (a,b,c,)

We then use z in the same notation for triplex but we need a triplex transform for z²


In we use a matrix transform  0  -1 and a definition (a,b)(c,d) = (ac - bd, bc +ad)
                                                1   0


can we generalise this to punch out the above definitions for i on the triplex?

[kram1032]

@David Makin
that then would mean, that the plane-version would act on cubes....

that way, there indeed would be a ....

[David Makin]

@David Makin
that then would mean, that the plane-version would act on cubes....

that way, there indeed would be a <Quoted Image Removed>....

I was thinking of maybe something in terms of x^2*y, y^2*z and z^2*x and/or x*y^2, y*z^2, z*x^2

[kram1032]

I guess, trying that geometrically first would be simpler than directly doing the math...

[David Makin]

I guess, trying that geometrically first would be simpler than directly doing the math...

If I remember correctly Pythagoras' formula was based on a geometric proof.

[kram1032]

Yup, it was.

That's why I guess, the 3D variant might need the same first, if it's valid at all...

though, by now there are thousands of proofs both geometrical and mathematical, afaik

[jehovajah]

For I designed a transform based on the transform in the following way:

(a,b,c) vx (d,e,f)  

(ad - be - cf, ae + bd, af +cd, bf + ce) using as a manipulation /construction space.

This i reduce to

(A,B,C,D) =(ad - be - cf, ae + bd, af +cd, bf + ce)

Using pairs from the construction bracket in the transform i obtain 6 building blocks

1 (AA - BB, 2AB)

2 (AA - CC, 2AC)

3 (-CC - BB, 2BC)

4(AA + DD, 2AD)

5 (DD - BB, 2BD)

6 (DD - CC, 2CD)

THE unary OPERATORS i and j are used to inform the manipulations so that i² = j² = -1 and (ij)² =+1.


Now my intention was to rotate the planes xy, xz, yz by this construction and i assumed that was what was happening until i rechecked the construction principles. The yz plane is not the same as the other two planes with the unary operators i and j operating on the axes. Under the transform the yz plane is sheared to the xij plane whatever that is. It may be a vortex surface.

so the first constructed transform is mistaken in two counts. The manipulations were faulty and i will show the correct manipulations; but the design was mistaken as it was not tranforming to a map of geometrical space.


The expansions are as follows for the right handed form

AA = (ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef

BB = (ae)² + 2abde + (bd)²

CC = (af)² + 2acdf + (cd)²

2AB = 2(a)²de + 2ab(d)² - 2ab(e)² - 2(b)²de - 2acef - 2bcdf

2AC = 2(a)²df + 2ac(d)² - 2ac(f)² - 2(c)²df - 2abef - 2bcde

2BC = 2(a)²ef + 2bc(d)² + 2abdf + 2acde

Now the first posted construction was based on combining blocks 1,2,3

supposedly giving

*{AA - BB + AA - CC, 2AB + BB - CC,  2AC + 2BC}

=> {AA - BB/2 - CC/2, AB + BB/2 - CC/2, AC + BC }.                                                    


 [ in fact it should be {AA - BB/2 - CC/2, AB - BB/2 - CC/2, AC + BC } due to an error in the original formulation of block 3]

So clearly (when i expand it) my original manipulations were wrongly copied from page to page to screen.

But now i realise i have not combined like with like and so have to construct the following transform from blocks 1and 2 which i fear will be even less interesting than my mistaken one

{AA -BB/2 - CC/2. AB, AC}

[jehovajah]

AA = (ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef

BB = (ae)² + 2abde + (bd)²

CC = (af)² + 2acdf + (cd)²

2AB = 2(a)²de + 2ab(d)² - 2ab(e)² - 2(b)²de - 2acef - 2bcdf

2AC = 2(a)²df + 2ac(d)² - 2ac(f)² - 2(c)²df - 2abef - 2bcde

2BC = 2(a)²ef + 2bc(d)² + 2abdf + 2acde


So the corrected vx is as follows

(a, b, c) vx (d, e, f)  = ((ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef -((ae)² + 2abde + (bd)²)/2 - ((af)² + 2acdf + (cd)²)/2,
                                       (a)²de + ab(d)² - ab(e)² - (b)²de - acef - bcdf + ((ae)² + 2abde + (bd)²)/2 - ((af)² + 2acdf + (cd)²)/2,

                                               (a)²df + ac(d)² - ac(f)² - (c)²df - abef - bcde + (a)²ef + bc(d)² + abdf + acde)

This gives  
   (x, y, z)² = (x, y, z) vx (x, y, z) = (x⁴ +  y⁴ + z⁴ - 4x²y² - 4x²z² + 2y²z²,

                                                     (2x³y - 2xy³ - 2xyz² + 2x²y² - 2x²z²,

                                                                (2x³z -  2xz³ - 2xy²z + 4x²yz   )

I would like to see if that makes alot of difference to the rendering please David if you have the time.

[jehovajah]

AA = (ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef

BB = (ae)² + 2abde + (bd)²

CC = (af)² + 2acdf + (cd)²

2AB = 2(a)²de + 2ab(d)² - 2ab(e)² - 2(b)²de - 2acef - 2bcdf

2AC = 2(a)²df + 2ac(d)² - 2ac(f)² - 2(c)²df - 2abef - 2bcde

2BC = 2(a)²ef + 2bc(d)² + 2abdf + 2acde

The left handed formulation requires the following

DD = (bf)² + 2bcef + (ce)²

2AD =  2abdf + 2acde - 2bc(e)² - 2bc(f)² - 2(b)²ef   - 2(c)²ef

2BD =2ac(e)² + 2(b)²df + 2abef + 2bcde

2CD = 2ab(f)² + 2(c)²de + 2bcdf + 2acef


Now from the 6 blocks i can construct 3 linked formulations abbreviated as followa

{AA - BB/2 - CC/2, AB, AC}                           RIGHT HANDED

{(AA + DD - BB - CC)/2, BC + AD}                 A MIRROR PLANE  at to the xy plane but here plotted in xy to have a look

{DD - BB/2 - CC/2, CD, BD}                          LEFT HANDED


THIS STRUCTURE i think might be interesting even if the mandelbrots are not. Particularly the plane as it may contain reflections of details in the brots not visible in the 3d brots.

Any way is any one interested as i am in seeing what this system looks like? I will expand it and find the vx for (x, y, z) if you are.

[jehovajah]

David you said that the cut down quaternion would lathe and i think paul shows such a rendering, so can you explain what is happening in these renderings? The first is a 3d representation of z=z² +c. The second and third if it shows are

z= z²- 2yzij +c which should be the same as

real(z)²-imag(z)²-imaj(z)² +2*real(z)*imag(z)*i+2*real(z)*imaj(z)*j+c

which is .

[David Makin]

David you said that the cut down quaternion would lathe and i think paul shows such a rendering, so can you explain what is happening in these renderings? The first is a 3d representation of z=z² +c. The second and third if it shows are

z= z²- 2yzij +c which should be the same as

zre²-zimagi²-zimagj² +2*zre*zimagi*i+2*zre*zimagj*j+c

which is <Quoted Image Removed>.

I don't see how "-2yzj" is the same as "+2xzj" ?

Did you remember to avoid variable corruption ?

I mean you can't do:

zri = zri^2+cri
zj = 2*real(zri)*zj + cj

You have to do (for example)

zj = 2*real(zri)*zj + cj
zri = zri^2 + cri

[Dranorter]

Hi all, I apologize for not keeping up with this (or Mandelbulb discussion), but I hope this thread is an OK place to post an interesting 3D mandelbrot set-like render I ran into.


http://www.hevanet.com/bradc/MiscMathStuff.html

It is probably too chaotic overall, but I thought looking at the bifurcation diagram was interesting.

[BradC]

Hey I recognize that picture

[kram1032]

hmmm....

looks much like a solidified Version of a rotated Buddhagram by Melinda Green...

Or the mix of a Mandelbrot and a logistic map (which are closely related anyway)