Brand New Method for True 3D Fractals.

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marcm200

• Fractal Frankfurter
• Posts: 532

Re: Brand New Method for True 3D Fractals.

« Reply #15 on: October 08, 2019, 10:59:59 AM »
Well...the way that's worded I can't really work with it.
? ? ?
The classical Mandelbrot set partitions the complex plane into c values with connected and disconnected sets. So I think a 3D Mset should partition the space also into two parts (not necessarily by the property of (dis)connectivity). I was curious what property your object separates.

As far as the truly claim goes - I'm very confident that I got it right. It's a bold claim but I made this a goal of mine like 13 years ago and put more than enough thought into it lol. But if I'm wrong that's fine too it happens. Either way I'm having fun
Agreed. I was wrong many times in the past (and lately) - and learnt a lot from it.
13 years - that's what I call persistance.

I tried WolframAlphas Jacobian tool to generate the derivative matrix for the substituted form: (I used one letter variables: x := zr, y := zri, z := zr, d := cr, e := cri, f := ci).
If I understood everything correctly in wikipedia, the critical points for a multivariate function are where the Jacobian is undefined or zero (correct my if I'm wrong).

If I plug in x=0, y=0, z=0 I get a matrix with a remaining e-term in the lower right field. Would that mean, the origin is a critical point only in case e (cri) is zero? And if so, would that also mean, that, dependent on the c-vector, the critical point changes like in the 2D z²+c*z+c case?

TylerSmith

• Fractal Fanatic
• Posts: 24

Re: Brand New Method for True 3D Fractals.

« Reply #16 on: October 08, 2019, 11:34:28 AM »
Look at the code again. Especially the first one I posted where it was split up. It's a function that runs once and outputs two results and uses one of the two results along with additional data to get the second and third results by running the same function again.

The critical points of the first time it's run are the same as those of the Mandelbrot set. The critical points of the second time it's run are the same as the mandelbrot set.

The substituted formula is literally substituting the imaginary result of the first mandelbrot set into the real component of the second mandelbrot set.

Trying to understand what that does once the iterations start, I get a bit lost myself.

But it's basically two sets.

Set one. We take the imaginary component from set one and plug it in to the real component from set two.

That's why I called it zri -> z real imaginary because it's an imaginary value pretending to be real.

So the behavior of both sets is the same as the regular mandelbrot set.

When it comes to how the two interact once they start iterating, I think they basically spiral and pass values back and forth and that spiral follows the mandelbrot shape...if that makes any sense.

I feel like my own explanation isn't quite on the money there but my brain hurts lol. I just programmed a volumetric renderer from scratch so that I could post the last few images I posted.

? ? ?
The classical Mandelbrot set partitions the complex plane into c values with connected and disconnected sets. So I think a 3D Mset should partition the space also into two parts (not necessarily by the property of (dis)connectivity). I was curious what property your object separates.
Agreed. I was wrong many times in the past (and lately) - and learnt a lot from it.
13 years - that's what I call persistance.

I tried WolframAlphas Jacobian tool to generate the derivative matrix for the substituted form: (I used one letter variables: x := zr, y := zri, z := zr, d := cr, e := cri, f := ci).
If I understood everything correctly in wikipedia, the critical points for a multivariate function are where the Jacobian is undefined or zero (correct my if I'm wrong).

If I plug in x=0, y=0, z=0 I get a matrix with a remaining e-term in the lower right field. Would that mean, the origin is a critical point only in case e (cri) is zero? And if so, would that also mean, that, dependent on the c-vector, the critical point changes like in the 2D z²+c*z+c case?

marcm200

• Fractal Frankfurter
• Posts: 532

Re: Brand New Method for True 3D Fractals.

« Reply #17 on: October 08, 2019, 12:50:02 PM »
I get what you're doing: Input (x,y,z): first iteration is a complex multiplication with (x+i*y)²+(cx,cy). Second iteration is taking the imaginary of the first output as new real and the original z as imaginary and performing a complex multiplication again.

So the input is a 3D vector, (_old), the output is a 3D vector (_new), so the critical points are 3D vectors themselves. But IMO, looking at the individual iteration's (2D) critical points (both times 0,0}), will not provide one with the critical point(s) of the performed 3D function. I think {0,0,0} is not in general a critical point for every choice of cx,cy,cz.

Could you provide the c-parameters for the Julia sets you posted earlier (as a positive control for my implementation)? Thanks.

Have you made a check list of what properties of the classical 2D Mandelbrot set your transformation also possesses and/or a comparison to other tricomplex squaring formulas?

• 3f
• Posts: 1897

Re: Brand New Method for True 3D Fractals.

« Reply #18 on: October 08, 2019, 06:32:12 PM »
I get what you're doing: Input (x,y,z): first iteration is a complex multiplication with (x+i*y)²+(cx,cy). Second iteration is taking the imaginary of the first output as new real and the original z as imaginary and performing a complex multiplication again.
So:
$$c = voxel\\ (x,y,z)=(0,0,0)\\ x+iy \leftarrow (x+iy)^2 + c_x + ic_y\\ y+iz \leftarrow (y+iz)^2 + c_y + ic_z$$
y is updated twice, unlike (x,z). Why not add
$$z+ix \leftarrow (z+ix)^2 + c_z + ic_x$$
so we'll have more symmetry in the axis.

TylerSmith

• Fractal Fanatic
• Posts: 24

Re: Brand New Method for True 3D Fractals.

« Reply #19 on: October 08, 2019, 08:24:43 PM »
You're probably right.
For the C values I don't remember which ones I used. They were all the same somewhere between -0.3 and -0.4.

I get what you're doing: Input (x,y,z): first iteration is a complex multiplication with (x+i*y)²+(cx,cy). Second iteration is taking the imaginary of the first output as new real and the original z as imaginary and performing a complex multiplication again.

So the input is a 3D vector, (_old), the output is a 3D vector (_new), so the critical points are 3D vectors themselves. But IMO, looking at the individual iteration's (2D) critical points (both times 0,0}), will not provide one with the critical point(s) of the performed 3D function. I think {0,0,0} is not in general a critical point for every choice of cx,cy,cz.

Could you provide the c-parameters for the Julia sets you posted earlier (as a positive control for my implementation)? Thanks.

Have you made a check list of what properties of the classical 2D Mandelbrot set your transformation also possesses and/or a comparison to other tricomplex squaring formulas?

TylerSmith

• Fractal Fanatic
• Posts: 24

Re: Brand New Method for True 3D Fractals.

« Reply #20 on: October 08, 2019, 08:27:36 PM »
I wasn't going for symmetry. Not sure I understand the math the way you're showing it. I'm more used to coding it than writing it out that way.

So:
$$c = voxel\\ (x,y,z)=(0,0,0)\\ x+iy \leftarrow (x+iy)^2 + c_x + ic_y\\ y+iz \leftarrow (y+iz)^2 + c_y + ic_z$$
y is updated twice, unlike (x,z). Why not add
$$z+ix \leftarrow (z+ix)^2 + c_z + ic_x$$
so we'll have more symmetry in the axis.

TylerSmith

• Fractal Fanatic
• Posts: 24

Re: Brand New Method for True 3D Fractals.

« Reply #21 on: October 08, 2019, 08:32:26 PM »
I guess I was thinking of that along the lines of how the mandelbrot set isn't symmetrical. In my mind that's like making a 2D set and thinking hey, why not make y absolute so that both sides are the same and losing half the set. I may be wrong though. Let me give it a try.

So:
$$c = voxel\\ (x,y,z)=(0,0,0)\\ x+iy \leftarrow (x+iy)^2 + c_x + ic_y\\ y+iz \leftarrow (y+iz)^2 + c_y + ic_z$$
y is updated twice, unlike (x,z). Why not add
$$z+ix \leftarrow (z+ix)^2 + c_z + ic_x$$
so we'll have more symmetry in the axis.

TylerSmith

• Fractal Fanatic
• Posts: 24

Re: Brand New Method for True 3D Fractals.

« Reply #22 on: October 08, 2019, 08:42:23 PM »
Okay, I plugged it in and got this trippy looking thing.

So:
$$c = voxel\\ (x,y,z)=(0,0,0)\\ x+iy \leftarrow (x+iy)^2 + c_x + ic_y\\ y+iz \leftarrow (y+iz)^2 + c_y + ic_z$$
y is updated twice, unlike (x,z). Why not add
$$z+ix \leftarrow (z+ix)^2 + c_z + ic_x$$
so we'll have more symmetry in the axis.

• 3f
• Posts: 1897

Re: Brand New Method for True 3D Fractals.

« Reply #23 on: October 08, 2019, 10:12:20 PM »
I tried WolframAlphas Jacobian tool to generate the derivative matrix for the substituted form:
I got this Jacobian
Code: [Select]
[ 2*x, -4*y^2,  4*y*z][ 2*y,  4*x*y, -4*x*z][   0,    2*z,    2*y]Crit. point is where all eigenvalues are zero (not just Jacobian I think). (0,0,0) here if I'm right.

marcm200

• Fractal Frankfurter
• Posts: 532

Re: Brand New Method for True 3D Fractals.

« Reply #24 on: October 08, 2019, 11:03:33 PM »
How did you get rid of the cri (or e) term? If the substituted form below is correct, there is a term 4*zr_old*zri_old*cri (or 4*x*y*e in one-letter form) - shouldn't that have e in the partial derivatives for dx and dy? Or did I substitute the one-letter version in a wrong way?
Code: [Select]
zri_new = ( 2 * zr_old * zri_old + cri ) * ( 2 * zr_old * zri_old + cri ) - zi_old * zi_old + cri;
Below are some Julia sets of my implementation, point-sampled. It is not an exhaustive display of shapes, just some objects I like, especially the level of detail and the surface structure.

My favourite is the one on the lower right. It seems as if it is planar in nature (or very thin).
« Last Edit: October 08, 2019, 11:18:53 PM by marcm200 »

• 3f
• Posts: 1897

Re: Brand New Method for True 3D Fractals.

« Reply #25 on: October 09, 2019, 02:44:47 AM »
How did you get rid of the cri (or e) term?
I just took the product of the 2 Jacobians of the two transformations, neither depends on c.
Attached 2 close ups of 2D parameter space slices at cz=1/3 with original formula and one with added extra transform.

Why is this "true 3D" whereas all the other 3D stuff Mandelbulb/box etc. is not "true"?

mclarekin

• Fractal Frankfurter
• Posts: 628

Re: Brand New Method for True 3D Fractals.

« Reply #26 on: October 09, 2019, 06:03:55 AM »
interesting maths you guys ,  some of it a bit beyond me

[quoteWhy is this "true 3D" whereas all the other 3D stuff Mandelbulb/box etc. is not "true"?][/quote]

it is just the different way that  we can read english i.e this is a brand new  method for making  "other" true 3D fractals, opposed to a brand new method to make pseudo 3D fractals

marcm200

• Fractal Frankfurter
• Posts: 532

Re: Brand New Method for True 3D Fractals.

« Reply #27 on: October 09, 2019, 09:01:23 AM »
@gerrit: A math question I can't figure out: The derivative of f(g(x)) is f'(g(x))*g'(x) by chain rule. Does a similar rule exist for Jacobian matrices? Do I calculate J(f) and the substitute the variables with their g-transformed version, and then multiply this with J(g)? Or do I multiply the Jacobians themselves? Or does no such rule exist and one has to use functions in their explicit form?

@TylerSmith: How does the shape of your object change if you follow a different point than the origin? (I followed some non-critical paths in the 2D case. The result were some strange shapes forum link). Would be interesting to see how this one changes.

• 3f
• Posts: 1897

Re: Brand New Method for True 3D Fractals.

« Reply #28 on: October 09, 2019, 06:36:08 PM »
@gerrit: A math question I can't figure out: The derivative of f(g(x)) is f'(g(x))*g'(x) by chain rule. Does a similar rule exist for Jacobian matrices?
Yes, exactly the same formula replacing derivatives with matrices. I didn't do it right though (had f(x) instead of f(g(x))). I think this is correct(using matlab):
Code: [Select]
[              2*x,             -2*y,            0][ 4*y*(cy + 2*x*y), 4*x*(cy + 2*x*y),         -2*z][            4*y*z,            4*x*z, 2*cy + 4*x*y]Has det=0 at (x,y,z)=0 but one eigenvalue depends on cy and is non-zero generally. Full formulas for eigenvalues are
Code: [Select]
ev =                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        (2*cy)/3 + (2*x)/3 + (4*cy*x)/3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3) + (4*x*y)/3 + (8*x^2*y)/3 - ((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)/(32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3) (2*cy)/3 + (2*x)/3 + (4*cy*x)/3 - (3^(1/2)*((32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3) + ((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)/(32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3))*1i)/2 - (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3)/2 + (4*x*y)/3 + (8*x^2*y)/3 + ((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)/(2*(32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3)) (2*cy)/3 + (2*x)/3 + (4*cy*x)/3 + (3^(1/2)*((32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3) + ((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)/(32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3))*1i)/2 - (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3)/2 + (4*x*y)/3 + (8*x^2*y)/3 + ((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)/(2*(32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (((32*x^3*y^2)/3 + (4*cy*x)/3 - (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^2/9 + (8*cy*x^2)/3 + (8*cy^2*x)/3 + (8*cy*y^2)/3 + (8*x^2*y)/3 + (16*x*y^3)/3 + (16*x^3*y)/3 + (8*x*z^2)/3 + (32*cy*x^2*y)/3)^3 + (32*x^2*y^4 + 32*x^4*y^2 + 8*x^2*z^2 + 8*y^2*z^2 + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^2)^(1/2) + (2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)^3/27 - ((2*cy + 2*x + 4*cy*x + 4*x*y + 8*x^2*y)*(8*cy^2*x + 32*cy*x^2*y + 8*cy*x^2 + 4*cy*x + 8*cy*y^2 + 32*x^3*y^2 + 16*x^3*y + 8*x^2*y + 16*x*y^3 + 8*x*z^2))/6 + 8*cy^2*x^2 + 8*cy^2*y^2 + 32*cy*x*y^3 + 32*cy*x^3*y)^(1/3))

• 3f
• Posts: 1897

Re: Brand New Method for True 3D Fractals.

« Reply #29 on: October 09, 2019, 09:05:37 PM »
Handsome Julia set on z=0 slice.

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