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Homework Help: Help with Julia set

  1. Dec 21, 2011 #1
    Help with Complex quadratic forms

    1. The problem statement, all variables and given/known data
    The question is to plot the julia set of the logistic map f [x] = 4.4x(1-x). I know how to plot the graph of z^2 + c for any given c using mathematica and i'm pretty sure its possible to convert any logistic map to the form z^2 + c but i was wondering if someone could help me with this conversion, how to and the theory with it?

    2. Relevant equations
    Logistic Map: r x (1 - x)
    Desired Form: z^2 +c

    3. The attempt at a solution
    Translation 1
    c = [( r / 2) - ( ( r ^ 2 ) / 4) ] gives answer 2.89
    c = [ 1 - ( ( r -1)^ 2 / 2 ) ] gives answer -2.64

    So far i have found two different answers 2.89 and -2.64 using two different translation i have found on the internet and was wondering if someone could help me.
    Last edited: Dec 21, 2011
  2. jcsd
  3. Dec 21, 2011 #2
    Re: Help with Complex quadratic forms

    So you have

    [tex]y_{n+1}=\lambda y_n(1-y_n)[/tex]

    and you want to convert that to:


    How about the linear transformation:



    [tex]a z_{n+1}+b=\lambda (a z_n+b)\left(1-(a z_n+b)\right)[/tex]

    Now, can you figure what a and b have to be so that the [itex]z_n[/itex] term drops out and you're left with

    Tell you what though, I don't think this is happenin' for me because I arrive at an expression of [itex]z\to z^2-2.6[/itex] and that is way far to the left of the Mandelbrot set which means the Julia set is really, really small and my Mathematica code starts chocking at around -1.4. Probably I'm doing something wrong though.
    Last edited: Dec 21, 2011
  4. Dec 23, 2011 #3
    Re: Help with Complex quadratic forms

    no i had a miscalculation in one of my calculations the correct answer is I believe - 2.64 and my mathematica code is plotting it however I have to zoom in to see the Julia set, it corresponds to a piece of the thinned out part of the mandelbrot set seen on the left of it as you said.

    While I have a chance do you mind if i ask if you know a translation from the general quadratic form: ax^2 + bx + c

    to the one requested above zn+1=z2n+c

    thanks very much for the help.
  5. Dec 23, 2011 #4
    I thought the linear transformation above would work for the general quadratic case. Doesn't it? Have you tried converting:

    [tex]h z_{n+1}+k=a(h z_n+k)^2+b(h z_n+k)+c[/tex]

    Oh yeah, wanna' post your code? Ok if you don't want to though.
    Last edited: Dec 23, 2011
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