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S Oct 7, 2017 at 21:41 history suggested Adam CC BY-SA 3.0
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S Oct 7, 2017 at 21:41
Dec 29, 2011 at 10:45 vote accept Per Alexandersson
Sep 5, 2011 at 11:26 comment added Per Alexandersson Oh, yes it is required that the functions are contractive. However, I suspect that under certain milder assumptions, the attractor is still unique. (Numerical experiments suggests this, for example, the software Apophysis among other happily produces Julia sets from the maps above, even though there is no proof that I know of that says that it works). A set of functions that gives the Mandelbrot set suffices to me, even thought there is no proof that this really is the case.
Sep 4, 2011 at 23:25 history edited Pablo Shmerkin CC BY-SA 3.0
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Sep 4, 2011 at 23:12 history answered Pablo Shmerkin CC BY-SA 3.0