Hausdorff dimension of the Mandelbulb

[Prokofiev]

          Area of surface of bulb is the product of the rotation of a 2d Mandelbrot around a 2d Mandelbrot (both with Hausdorff of 2).

The mandelbulb is not the product of 2 Mandelbrot sets.

Wouldn't this make the Hausdorff 4?

Certainly not, you can't just add dimensions. And, it can't be more than the space in which it lies.

Consider a 2d Mandelbrot spun in a circle (I know I've seen a few ):
    Mandelbrot Hausdorff of 2.
   Circle is Hausdorff of 1.
   Is the Hausdorff of the spun Mandelbrot 3?

Yes. Absolutely  . If you are interested in the subject I would advise you to read Falconer's "Fractal Geometry" (There is an entire chapter about the product of fractal sets)

[fractower]

M Benesi was not entirely off base when she/he described the product of two Mandelbrot sets as a 4D object. One just needs to use the outer product rather than some kind of rotation (There may be an outer product rotation as well, but I will ignore that for now since it hurts my head). The outer product of two column vectors is a 2D object known as a matrix. The outer produce of two matrices is a 4D tensor. We are not use to thinking of 2D objects such as the M set as a matrix, but it can be expressed as the sum of outer products of sins and cosins (Fourier transform). Granted an infiinite number would be required, but we should be use to infinite sums by now.

Try graphing sin(Ax)*cos(By) to get an idea of what an outer product of 2 continuous functions looks like.

[Calcyman]

M Benesi was not entirely off base when she/he described the product of two Mandelbrot sets as a 4D object.

The Cartesian product of two orthogonal Mandelbrot sets is four-dimensional. But the Mandelbulb, or any other 3D fractal, is certainly not the M-set²; such a set would lie in four-dimensional Euclidean space.

We are not used to thinking of 2D objects such as the M set as a matrix

That's because the Mandelbrot Set is not a matrix, and it is physically impossible to express the Mandelbrot Set as a matrix -- there are only Beth One matrices, and Beth Two sets over the complex plane. Placing them in one-to-one correspondence would conflict with Cantor.

[M Benesi]

          Area of surface of bulb is the product of the rotation of a 2d Mandelbrot around a 2d Mandelbrot (both with Hausdorff of 2).

The mandelbulb is not the product of 2 Mandelbrot sets. 

  Here is some code  (should make it clear, since this shows how the bulb works ... like the regular 2d set):

Code:

  sx,sy,sz are starting x,y, and z components; nx, ny, and nz are calculated components before pixel component addition;
r1=sqrt(sqr(sy)+sqr(sz));
r3=r1^-n;   //take it to the negative power to avoid division by zero error!

victor=complex(sx,r1);   // complex (x,y) is ChaosPro's command to create a complex number out of 2 real components x and y
bravo=complex(sy,sz);  //  the second part (1st,2nd) is the imaginary component

victor=victor^n;
bravo=bravo^n;

nx=real (victor);
ny=imag (victor) * real (bravo) *r3;  //  you have to divide out the magnitude (r3=r1^n) because it has already been applied
nz=imag (victor) * imag (bravo) *r3;  //  twice, once computing victor, once computing bravo

  then add in your pixel values.. 

  So you can see we have something that resembles two 2d sets influencing one another, almost like it's a product of 2 2d sets.... 

Wouldn't this make the Hausdorff 4?

Certainly not, you can't just add dimensions. And, it can't be more than the space in which it lies.

  So it's maximum hausdorff is 3, and it has to be 3.. although it has a higher apparent complexity than a spun brot...