[math-fun] Hausdorff dimension of Julia sets
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.
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.
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... 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.
Scott is talking about *filled* Julia sets, while I took Dan to be asking about *unfilled* ones. It's probably worth explaining the difference. When you investigate the dynamics of (a certain large class of) functions of the complex plane under repeated application, you find that different starting points can have (again, skating around some pathologies) three distinct behaviors. Their orbits can converge to a finite limit cycle, or they can diverge to infinity, or they can dance around chaotically forever. If you call the point at infinity a special sort of limit cycle (by projectively completing, or compactifying) the complex plane, adding the point at infinity, then the second behavior is re-envisioned as a special case of the first: convergence to the point at infinity. This leaves two behaviors: convergence to a limit cycle, or the chaotic dance. Each limit cycle (including the point at infinity) has a basin of attraction; any starting point in a cycle's basin will converge to that cycle. These basins are all two-dimensional open sets. The chaotic regime is confined to their boundary. Whose boundary? It doesn't matter! No matter which basin you pick, you get the same boundary; the basins all touch each other everywhere along their borders. Unless there are only two limit cycles, this unintuitive behavior demands a fractal solution. The classic, "unfilled" Julia set is the set of points with chaotic behavior; its complement, the union of the attractive basins of all the limit cycles, is the Fatou set; the former is the boundary of the latter. This kind of Julia set always has dimension less than 2; the Fatou set takes up most of the complex plane, only leaving room for a lacy measure-zero fractal Julia set along its border. One of the basins can be viewed as special: the one that feeds the point at infinity. The complement of this basin is the "filled" Julia set of the sort Scott was visualizing; its dimension is 2 when there are other basins, and less than 2 (a Cantor dust, in fact) when the point at infinity is the only finite limit cycle. The question of what the map of dimensions would look like is interesting in both cases. On Mon, Aug 1, 2011 at 5:50 PM, Huddleston, Scott < scott.huddleston@intel.com> wrote:
... 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.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The *Julia set* is not the same as the *boundedness set* for a polynomial map. The Julia set is the set of points such that every neighborhood eventually spreads out to cover a dense subset of C. It does not include points that are attracted to a periodic cycle. Hausdorff dimension of the Julia set is not continuous in general: in particular, it fails to be continuous at any parameter where there is a rationally indifferent periodic point, that is, a periodic point where the derivative of the return map is a root of unity. These are points where mini-Mandelbrot sets are attached. For a sequence of points converging to such a point, the Julia set of the limit is always contained in the limit of Julia sets, but the limit of Julia sets can be much bigger. You can see this by playing around with a Julia set visualizer, (if you use a mac with recent operating system, there are good programs such as fractalworks available from the App Store). However, Hausdorff dimension is continuous as long as the critical point tends toward an attracting periodic cycle, infinity allowed. So, I think you'd see the Mandelbrot set as basically the discontinuity set for the graph of Hausdorff dimension, since rationally indifferent periodic cycles are presumed to be dense in the Mandelbrot set. Bill Thurston On Aug 1, 2011, at 5:50 PM, Huddleston, Scott wrote:
... 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.
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participants (4)
-
Allan Wechsler -
Bill Thurston -
Dan Asimov -
Huddleston, Scott