# Homework Help: Help with Julia set

1. Dec 21, 2011

### chinye11

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. Dec 21, 2011

### jackmell

Re: Help with Complex quadratic forms

So you have

$$y_{n+1}=\lambda y_n(1-y_n)$$

and you want to convert that to:

$$z_{n+1}=z_n^2+c$$

$$y_n=az_n+b$$

or:

$$a z_{n+1}+b=\lambda (a z_n+b)\left(1-(a z_n+b)\right)$$

Now, can you figure what a and b have to be so that the $z_n$ term drops out and you're left with
$$z_{n+1}=z_n^2+c$$

Tell you what though, I don't think this is happenin' for me because I arrive at an expression of $z\to z^2-2.6$ 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
3. Dec 23, 2011

### chinye11

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.

4. Dec 23, 2011

### jackmell

I thought the linear transformation above would work for the general quadratic case. Doesn't it? Have you tried converting:

$$h z_{n+1}+k=a(h z_n+k)^2+b(h z_n+k)+c$$

Oh yeah, wanna' post your code? Ok if you don't want to though.

Last edited: Dec 23, 2011