# Show that the familiar logistic map $x_{n+1} = sx_n(1 - x_n)$, can be recoded into the form $x_{n+1} = x_n^2 + c$.

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

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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

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

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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

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 $s/2-s^2/4=c$

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

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