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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$\begingroup$ 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 ? $\endgroup$– AlbertAug 31, 2014 at 18:17
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3 Answers
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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.
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
$$f(z) = \frac{z^3-z}{1+4 z-z^2}$$
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.
answered May 30, 2013 at 18:30
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$\begingroup$ (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 $\endgroup$– AlbertAug 31, 2014 at 15:32
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$\begingroup$ experience proves that in complex dynamics all known homeomorphic copies are in fact quasiconformal. so for my money, I'd say no ! $\endgroup$– AlbertAug 31, 2014 at 15:33
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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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1$\begingroup$ 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. $\endgroup$ May 30, 2013 at 18:34
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Due to the Julia set being much simpler than the Mandelbrot (it never actually repeats smaller copies of itself), there’s no Mandelbrots in there.
