... I especially wonder whether the Mandelbrot set would be in any way "visible" in this dimension plot.
Every Julia set with nonempty interior has Hausdorff dimension 2. These are indexed by attracting Mandelbrot cycles, so the interior of the Mandelbrot set has value 2 in this map. I think every repelling Mandelbrot cycle indexes a Julia set with 1 <= Hausdorff dimension < 2, and disconnected Julia sets (indexed by the Mandelbrot complement) have Hausdorff dimension < 1. -- Scott
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun- bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Monday, August 01, 2011 1:46 PM To: Dan Asimov; math-fun Subject: Re: [math-fun] Hausdorff dimension of Julia sets
That's an excellent question, and I haven't seen it posed before. First of all, one must know that a given Julia set in fact has a well-defined dimension (the boundary of the Mandelbrot set doesn't), but I think this is well-established. I don't know what technology there is for computing dimensions of Julia sets, and of course I especially wonder whether the Mandelbrot set would be in any way "visible" in this dimension plot.
On Mon, Aug 1, 2011 at 4:26 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Has anyone seen a graph {(c, H(c)) in R^3 | c in C=R^2} of the Hausdorff dimension H(c) of the Julia set of f(z) = z^2 + c, c in C ?
--Dan
Sometimes the brain has a mind of its own.