David Lowenstein previously wrote:

http://www.hypercomplex.org/quats.htm http://www.mathcs.sjsu.edu/faculty/rucker/cubic_mandel.htm

and some other links in messages before that. Thanks Dave, for letting me know about Rudy Rucker's very informative site. I learned quite a lot from reading it, and am grateful for your sharing it. Specifically concerning the form of the cubic mandelbrots, I do have a quibble though. It is this: He points out that a change of basis (which doesn't do anything visually fundamental to the fractal) can let you eliminate one of the coefficients in the process transformation z->z^3 + A*z^2 + B*z + C, much as you canonically normalize the leading coefficient to one since all that does is not-visually-interesting magnification. However, he chooses to eliminate the quadratic coefficient. I think that is the wrong one to eliminate from the point of view of visual fractal exploration. Of course if all you're going to do is look at cubic Mandelbrots only, it doesn't matter which of {A,B,C} you zero out, and A is the one eliminated first when solving the zeroes of a cubic polynomial equation, so maybe he chose to zero out A just out of habit. OTOH, the classic Mandelbrot has a nonzero quadratic term in its process transformation z->z^2 + C. You could not therefore make a generalized formula mix between the quadratically-zeroed form and the classic Mandelbrot which would vary all the way to the classic Mandelbrot. That to me is a drawback; I like to be able to see the canonical form of a higher order formula subsume a lower order one so that visually you can generate the lower order one as a special case. Then you can proceed slowly from one type to the other by gradually changing coefficients. In order to enable this view, the first order term B is the term that ought to be zeroed out and consider z -> z^3 + A*z^2 + C to be an intermediate form of the cubic Mandelbrot. Why is that IMO, a preferable form? Well, now you can normalize on the first two coefficients instead of just the first one to get: cubic Mandelbrot transform: z -> cos(A)*z^3 + sin(A)*z^2 + C which lets uses just 2 coefficients to show all visually different sets, is normalized to elide the affine copies, and reduces to the classic Mandelbrot in the special case of coefficient A = pi/2+0i. That's obviously just my opinion based on a desire for facile exploration of the fractal space, and not an assertion that there is otherwise any reason choose which coefficient to zero out. Sincerely, Hiram Berry