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I made an interesting discovery today (29 January 2005). I realized that it is possible to look at the Mandelbrot Set in a different way. Usually it is graphed by iterating the formula
and coloring each complexvalued pixel c based on how long it takes to escape a circle of radius 2. The resulting image looks like this:
z[0] = 0 
z[1] = z[0]^{2} + c = 0^{2} + (−1) = −1 
z[2] = z[1]^{2} + c = (−1)^{2} + (−1) = 0 
z[3] = z[2]^{2} + c = 0^{2} + (−1) = −1 
z[4] = z[3]^{2} + c = (−1)^{2} + (−1) = 0 
... 
My own homebrew Mandelbrot Set imager now allows me to type in an arbitrary algebraic expression f(z) in terms of the complex variable z, and generates a fractal based on iterating Newton's Method to find a solution for f(z)=0.
It occurred to me that it would be interesting to graph solutions for various orbital periods in the Mandelbrot formula. For example, if I wanted to find places where the Mandelbrot formula repeated every iteration (i.e. orbital period = 1), I would solve for z^{2} + z = z, or z^{2} = 0. Of course, this is a trivial case, since it is obvious that the only solution is z=0.
As one might expect, a trivial formula yields a trivial image, in this case not even a fractal... just a set of concentric circles:

For the orbital period of 2, we get our first nontrivial image. This time it is a fractal. The solutions are z = +1 and z = −1, which show up as the larger horizontal eyes. This is interesting, but there is no resemblance of the Mandelbrot Set (yet).

Now at orbital period 3, something interesting happens: we are starting to see a definite pear shape.
Also, because of the Fundamental Theorem of Algebra, there are 2^{p−1} complex solutions, where p is the orbital period, because each higher orbital period corresponds with a polynomial based on squaring the previous period's polynomial. The reason it is 2^{p−1} and not 2^{p} is because when the orbital period is 1, there is only 1 solution, and 1 = 2^{0}.



