Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What change of variables would trtansform the logistic equation into the Mandelbrot equation $z_{n+1}=z_n^2+c$?

share|improve this question
1  
AND: since we have different answers here, the next step is to check your answer! –  GEdgar Jan 30 '13 at 14:22
    
Since the OP is still present on the site, they might wish to answer my comment on @distantTransformer's answer below. –  Did Sep 11 '13 at 16:45
add comment

3 Answers 3

up vote 0 down vote accepted

$$ z= i \cdot s \cdot (x - 0.5),\quad c = 0.5 \cdot \sqrt{s} $$

share|improve this answer
    
For the record, I do not understand this answer. If either the answerer or the author of the question could explain. –  Did Feb 2 '13 at 10:48
add comment

Well, you can do the next trick:

$$z_n=f(x_n)$$

$$f(sx_n(1-x_n))=f(x_{n+1})=z_{n+1}=z_n^2+c = f(x_n)^2+c$$

So we want our transformation to fulfill:

$$f(sx_n-sx_n^2)-(f(x_n))^2= c$$

I guess you can check for $$f(x)=Ax+B$$, and then solve to find $A,B$.

$$A(sx-sx^2)+B-(A^2x^2+2ABx+B^2) = c$$ Because this should be valid for all $x$, we should have: $$-As-A^2=0$$ $$-2BA+As=0$$ $$B-B^2=c$$ $A\neq 0$, thus: $A=-s$ and $B=s/2$, and we should also have $s/2-s^2/4=c$

share|improve this answer
add comment

Using the change of variable $z_n=s\cdot(\frac12-x_n)$ yields $z_{n+1}=z_n^2+c$ with $c=\frac12s-\frac14s^2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.