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A question asked me to find the fixed points of $F(x)=x^2-1.1$, then use the fact that these points were also solutions of $F^2(x)=x$ to find the cycle of the prime period 2 for F.

How do I go about this second part?

My work so far: To find the fixed points, I set $F(x)=x$ and used the quadratic formula to get $\frac{1+-\sqrt{5.4}}{2}$, then called the positive result $p_0$ and the negative one $p_1$. By graphical analysis, I saw that the positive point repelled on the right to infinity and on the left to $p_1$, but once I got close to $p_1$, I got an orbit. So now I know that the orbit is around the negative point.

My first thought was to use some number just a bit bigger than $p_1$, like, say $-1.5$, and then take iterations, but that wouldn't really be using the fact that "these points were also solutions of $F^2(x)=x$", would it?

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  • $\begingroup$ I think what they're getting at for the second part is that the solutions of $F^{2}(x)$ includes a genuine 2-cycle, but it also includes the fixed points. So getting those first lets you distinguish them from the cycles. $\endgroup$ Sep 18, 2014 at 4:27

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  1. compute fixed points of $F^2(x)$. It is a set X2 of points x :$$X2= \{x: x=F^2(x)\}$$ This set include fixed points of F and fixed points of F2 ( cycle of the prime period 2)
  2. compute set X1 of fixed points x of F : $$ X1= \{x: x=F(x)\}$$

  3. remove point of set X1 from set X2 then you wil have set of cycles with prime period 2

Generaly : roots of $F^p(x)=x$ are cycles for period p and its divisors

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