You'll find things similar to a Julia set in an MSet (and vice-versa), due to the fact that the two can be considered 2D slices through a 4D set (where one 2D "axis" contains the Z0 value, and the other is C).
Due to several things, you are unlikely to find things involving squares or triangles.  First of all, the whole thing is "self-similar" (or "magnification independent") meaning that it look similar at any magnification - if you suddenly started getting Koch snowflakes, for example, it wouldn't be self-similar.  Secondly, complex math results in doing things like rotating and "smushing" angularly, which causes curves to appear.

No, the MSet that everybody knows and loves is constructed using complex math (which is 2D).  Complex math is a subset of quaternion math (as well as split-complex math, for what it's worth), but due to weird group theory issues, each "larger" math space has 2^n components (so there can be 2D, 4D, 8D, 16D, though each higher one loose useful properties - 4D quaternions are non-commutative (A x B is not necessarily equal to B x A), 8D octonions are non-associative (A x (B x C) is not necessarily equal to (A x B) x C), and 16D sedenions looses the ability to do division (there are sedenions A and B, neither of them zero, but A x B = 0 = B x A)).

One can formulate an arbitrary 3D set with specific operations for things like multiplication and addition, but they lack the "elegance" that complex, quaternions, octonions and sedenions.

Just a few thoughts about trying to make the "Mandelbulb" 3D object:

If you imagine such a beast, and do a slice through it, along two axis you should probably get the standard 2D MSet.  So if you're 3D coordinate system is (X,Y,Z) (with X being the "core" - up and down in the sketch you linked), this implies that:
(X,Y,0) ^ 2 ~= (X + i * Y) ^ 2
(X,0,Z) ^ 2 ~= (X + i * Z) ^ 2

i.e., the Y and Z axis are similar to the i component in complex numbers, and that when one is zero, you've got complex multiplication with the other two.

Now if you have a cross section where X = 0, you'll probably end up with something that looks like a rotated (by 45 degrees) version of the z = z ^ 5 + C.  This gives us:
(0,Y,Z) ^ 2 ~= (sqrt(2),sqrt(2)) * (Y + i * Z) ^ 5 ~= (sqrt(2),sqrt(2)) * (Z + i * Y) ^ 5.

If needed, you can negate any or all of the sqrt(2) parts, since rotation by 45 = 135 = 225 = 315
Expand all of those, and hopefully you'll get a series of consistent equations that allows you to solve the general (X,Y,Z)^2.  I'm not positive that such a thing is possible, or that you'll get useful results for the general case (i.e., we've only fixed the shape along 3 perpendicular slices, and the rest could bulge out or smear smooth).

Ingvar,

What applicaiton did you use to create those zooms? My application can't zoom as far as yours. My images start to fall apart due to floating point errors at page 11. (Plot width of about 4.0E-14. By 1.0E-14, the images are very badly degraded.)

The app also seems to do a pretty good job of creating color tables that show detail at a wide variety of magnifications.


Duncan C

I have found a bit of variety in the M-set
I have figured out where to go to find a julia set(sort of) made out of another julia set
It's still rendering, but I have a picture of what I mean.
animation coming soon

FractalMonster,

Cool image. What are the "specs"? (Plot type, coordinates, magnification)

Duncan C

fractalwizz,

Sorry, my mistake. I was indeed talking to you. Thanks for the coordinate info. That is a really deep zoom!