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I've seen a julia set zoom but it is not nearly as interesting as a mandelbrot zoom. I also have not seen corresponding julia sets for zooms in the mandelbrot deeper than the original image. I'm wondering if it is possible to locate some form of a mini mandelbrot inside the julia set.

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not directly related to the question, but did you know that the converse is true, ie that there appears copies of some julia set in some parts of the mandelbrot set ? –  Glougloubarbaki Aug 31 at 18:17

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up vote 7 down vote accepted

This answer has two parts: a partial negative and a partial positive.

Why Julia sets are typically much simpler than the Mandelbrot set

Julia sets of rational functions can be computed using an inverse iteration technique that shows them to be something close to self-similar. This helps explain the extreme regularity displayed when zooming into most Julia sets. For example, here we zoom in to the Julia set for $f(z)=z^2-1$ increasing the magnification by a factor of the Golden ratio with each step.

enter image description here

Unlike this, the Mandelbrot set gets more complicated with each step in magnification, rather than staying at the same basic level of complexity.

Julia sets that contain part of a Mandelbrot set

Nonetheless, these inverse images are not strict self-similarities (just not so far off for many functions) so more complicated behavior is possible. In fact, the Julia sets of some rational functions do indeed contain some Mandelbrot type behavior. Here's the Julia set of

$$f(z) = \frac{z^3-z}{1+4 z-z^2}$$

enter image description here

Note the appearance of something that looks like the period two bulb of the Mandelbrot set. I believe that there is, in fact, a genuine correspondence here but that the true nature is not yet fully understood. It appears to be related to the fact that the period two bulb is missing from the parameter space for a related family of functions:

$$f_a(z) = \frac{z^3-z}{1+az-z^2}$$

The study of this family is ongoing by Jane Hawkins and her students at UNC Chapel Hill and you can find more in her papers, particularly the ones that have the term "rational maps" in the title.

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(warning, heuristic comment). from the article you cited, the Hausdorff dimension of $f_4$ seem to be strictly less than two. convincing numerical simulations seem to indicate that the boundary of the mandelbrot set has positive lebesgue measure (shishikura proved it has hausdorff dimension 2). so if we believe this numerical experiment, there can be no quasiconfomal copies of the boundary of the mandelbrot set in your julia set (or in any julia set of null area). of course that's a very crude obstruction, and it doesn't rule out all homeomorphic copies, but –  Glougloubarbaki Aug 31 at 15:32
    
experience proves that in complex dynamics all known homeomorphic copies are in fact quasiconformal. so for my money, I'd say no ! –  Glougloubarbaki Aug 31 at 15:33

The Mandelbrot set has nonempty interior, so if a Julia set contained a Mandelbrot set, then necessarily that Julia set would have nonempty interior. As soon as a Julia set has nonempty interior, however, it must be all of the Riemann sphere.

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I've seen some of your other comments on complex dynamics which are typically spot-on. In this particular case, your argument doesn't imply that a Julia set cannot contain part of the boundary of Mandelbrot set. You might be interested in my answer in this regard. –  Mark McClure May 30 '13 at 18:34
    
@MarkMcClure: Very interesting, thank you! –  froggie May 31 '13 at 0:45

Are there mini-mandelbrots inside the julia set?

The question is a bit misleading because of the term "mini-mandelbrot".

The characteristic buddha-shape of the Mandelbrot set, is an artifact of the iteration which produces it, aka: iteration of $f(z)=z^2-c$, with $z_0=0$ and $c\in\mathbb{C}$.

Julia sets are produced using a different iteration, with $f(z)=z^2-c$, with FIXED $c$ and $z\in\mathbb{C}$.

As such, the EXACT buddha shape is not replicated in Julia sets, rather, a deeper connection shows up: The Julia set for a given $c$, replicates the local behavior of the Mandelbrot map around this $c$.

If the $c$ picked belongs to a mini-mandelbrot in the Mandelbrot map, then the corresponding Julia iteration will produce a connected Julia set, which will have at its center an artifact corresponding to the mini-mandelbrot where the value $c$ was chosen from.

The central artifact in the Julia set doesn't replicate the buddha shape, rather, it is a solid island, usually having a Blancmange boundary.

An example is given below:

Pick a $c=-0.745723970937-0.109963562329i$ (center of mini-mandelbrot in Sea Horse valley)

enter image description here

and the corresponding Julia set for this set, will have a corresponding Blanmange island at its center which is replicated elsewhere: enter image description here

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the mathematical content of that answer is null –  Glougloubarbaki Aug 31 at 15:18
    
@Glougloubarbaki: WhoOps! My apologies, but the mini-mandelbrots on Julia sets actually correspond to these small islands, as per standard theory. Perhaps you could improve the answer, by explaining that the rotational behavior of the two maps in the two processes are not the same, hence the difference in shape. Good luck. –  ioannis galidakis Aug 31 at 15:31
    
the second image is a bit misleading. the black components are topological disks (they are the bounded components of the complementary of the julia set). thus their topology has absolutely nothing to do with the mandelbrot set. so the julia set does not "replicate" anything from the mandelbrot set. however, something is true in the opposite direction : at certain parameters c (called Misiurewicz parameters) the mandelbrot set contains near c small copies of the julia set of $z \mapsto z^2+c$ –  Glougloubarbaki Aug 31 at 19:19
    
@Glougloubarbaki: I of course agree with what you are saying, with one small qualification: My term "replication" concerns the behavior of the Mandelbrot set ONLY in a small neighborhood of c. Obviously there is no exact replication, since the images are not homeomorphic (Julia sets have a point symmetry, while no neighborhood around such c points on the Mandelbrot set has, except at the x-axis). That's what Mandelbrot MAP means: I can look at c and predict (roughly) the shape of the corresponding Julia set. In any case, my apologies for any misunderstandings. –  ioannis galidakis Aug 31 at 20:15
    
any neighborhood of any point in the mandelbrot set will meet an infinity of copies of the mandelbrot set. so I'm not sure I understand what you mean. –  Glougloubarbaki Aug 31 at 21:21

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