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Need help with Julia sets

  1. Dec 20, 2009 #1
    I need someone to help explain to me in simple terms how Julia sets work. I understand how the equations governing the Mandelbrot set work, but am finding Julia sets to be a little more complex and difficult to understand.
     
  2. jcsd
  3. Dec 20, 2009 #2

    HallsofIvy

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    Actually, finding Julia sets requires exactly the same computation as the Mandlebrot set.

    Both involve the iteration [itex]z_{n+1}= z_n^2+ c[/itex]. With the Julia sets you are given a fixed "c" (and we refer to the Julia sets as "Jc") and the Julia set consists of all [itex]z_0[/itex] such that that iteration converges. The Mandlebrot set consists of all c such that the iteration with [itex]z_0= 0[/itex] converges. (Some texts say "the iteration with [itex]z_0= c[/itex]". Of course, if you start with [itex]z= 0[/itex] you immediately get [itex]z_1= c[/itex] so the convergence is the same either way.)

    If I wanted to draw the Julia set, Jc, I would set up a double loop to step through every possible z0= x+iy and check each to see if the sequence converges. If I wanted to draw the Mandlebrot set, I would set up a double loop to step through every possible c= x+ iy and check to see if the sequence starting with z0= 0 converges.

    The Mandlebrot set, by the way, "indexes" the Julia sets. If c is a complex number well within the Mandlebrot set, then Jc will be a single "blob" with boundary less complicated the deeper inside it is (If c= 0, J0 is simply a disk). If c is near the boundary of the Mandlebrot set, Jc will be a single connected set with a fractal boundary. If c is just outside the boundary of the Mandlebrot set, Jc is a number of disjoint piecess. If c is far outside the boundary of the Mandlebrot set, Jc is a "dust".

    Also, while Mandlebrot worked for IBM, Gaston Julia lived around the beginning of the 19 th century and did all computations by hand!
     
    Last edited by a moderator: Dec 21, 2009
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