Vector form of Julia fractal
Posted by peeterjoot on December 27, 2010
As outlined in , 2-D and N-D Julia fractals can be computed using the geometric product, instead of complex numbers. Explore a couple of details related to that here.
Fractal patterns like the mandelbrot and julia sets are typically using iterative computations in the complex plane. For the Julia set, our iteration has the form
where is an integer constant, and , and are complex numbers. For I believe we obtain the Mandelbrot set. Given the isomorphism between complex numbers and vectors using the geometric product, we can use write
and reexpress the Julia iterator as
It’s not obvious that the RHS of this equation is a vector and not a multivector, especially when the vector lies in or higher dimensional space. To get a feel for this, let’s start by write this out in components for and . We obtain for the product term
Looking at the same square in coordinate representation for the case (using summation notation unless otherwise specified), we have
This last term is zero since , and we are left with
a vector, even for non-planar vectors. How about for an arbitrary orientation of the unit vector in ? For that we get
We can read 2.5 off of this result by inspection for the case.
It is now straightforward to show that the product is a vector for integer . We’ve covered the case, justifying an assumption that this product has the following form
for scalars and . The induction test becomes
Again we have a vector split nicely into projective and rejective components, so for any integer power of our iterator 2.4 employing the geometric product is a mapping from vectors to vectors.
There is a striking image in the text of such a Julia set for such a 3D iterator, and an exersize left for the adventurous reader to attempt to code that based on the 2D sample code they provide.
 L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.