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Math Help - Fixed points of the Mandelbrot set

  1. #1
    Super Member Deadstar's Avatar
    Oct 2007

    Fixed points of the Mandelbrot set

    Bit of a long read here sorry...

    I've just finished writing this huge project on complex dynamics and I have to give a presentation on it on Wednesday.

    I'm just going through my project and I realized I'm not quite sure if I have this one part right...

    It's about finding the fixed points of the Mandelbrot set (if there is any! It's not exactly a standard set!)

    Now the way you normally do this is to set f(z) = z but with the Mandelbrot set, z_0 = 0 so what we're really dealing with is a sequence like this...

    z_0 = 0,
    z_1 = c,
    z_2 = c^2 + c,

    So to find fixed points we set the above to be 0 (as 'z' = 0 from z_0 = 0).

    So we get, by solving the above equations to be equal to 0 we get...

    For period 1 (i.e fixed) points
    z_1 = c = 0

    For period 2 points
    z_2 = c^2 + c = 0
    => c = -1 or 0 (discard 0 as it is a fixed point) so we get...

    c=-1 which is the centre of the circular area to the left of the main cardioid.

    We can take this further by solving for period 3 points in the same way and we get (ignoring the fixed point)
    -0.1225611669 - 0.7448617670i,
    -0.1225611669 + 0.7448617670i

    Which are the centres of the discs at the top and bottom of the main cardioid and also the 'centre' (centre being equivalent to the (0,0) position on the main cardioid) of the mini Mandelbrot set way over to the left.

    If you plug these back into the z^2 + c you find that 3 iterations will lead you back to your value so it seems to be true...

    For reference see this picture.

    The numbers correspond to the periodic cycle of that area.

    So anyone got any ideas on this? Are these in fact the fixed/periodic points of the set or am I going about this completely the wrong way. I would hate to start talking about this then get shot down under lecturer questions at the end of my talk...

    Just another thought...
    Perhaps I should be setting f(z) = z^2 + c = \mathbf{c} and solving like that? Don't think so though as I'll end up with c's and z's so no way to solve for values.
    Last edited by Deadstar; March 22nd 2010 at 01:54 PM.
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