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Oct 7, 2017 at 21:54 comment added Mr Pie Look at Numberphile concerning the Mandelbrot Set. It's a YouTube Channel.
Mar 6, 2015 at 20:28 answer added Sascha timeline score: 1
Dec 29, 2011 at 10:45 vote accept Per Alexandersson
Sep 4, 2011 at 23:20 history edited j.c. CC BY-SA 3.0
edit title, tags, capitalize abbreviations
Sep 4, 2011 at 23:12 answer added Pablo Shmerkin timeline score: 20
Sep 4, 2011 at 12:45 history edited Per Alexandersson CC BY-SA 3.0
changed some words
Sep 4, 2011 at 7:31 history edited Per Alexandersson CC BY-SA 3.0
better definition
Sep 4, 2011 at 7:27 comment added Per Alexandersson Yes, something like that. I mean, the Julia set is the fixed set for a certain Hutchinson operator, en.wikipedia.org/wiki/Hutchinson_operator with the two functions given above.
Sep 3, 2011 at 21:26 comment added Ryan Budney What do you mean by "create"? If $J_c$ is the Julia set corresponding to $c\in \mathbb C$, let $X_{c,n}$ be the set you create by taking the $n$-th iterate of your 2-valued function $z\longmapsto \sqrt{z−c}$, with the initial iterate being $z_0=0$. Do you want "create" to mean that $\cap_{n=1}^\infty \overline{\cup_{k=n}^\infty X_{c,k}} = J_c$ or something like that?
Sep 3, 2011 at 19:13 history asked Per Alexandersson CC BY-SA 3.0