Test is coming up very soon, and I just couldn't get hold of this one suggested problem!!
Consider the dynamical defined by iterating the function $f(x)=1-x^2$, that is, for a given initial point $x_0$, we consider sequences <$x_n$> obtained by iterating,
$x_{n+1} = f(x_n), n \in N$
The question asks the following
a) show that if $x_0 \in (0,1)$, then $x_n \in (0,1)$ for all $n \in N$
b) Find the fixed points $x = f(x)$. call $p_0$ the one in $(0,1)$.
c) calculate $f^{(2)}=f(f(x))$. find all period 2 points which are not already fixed points of $f(x)$. call them $p1,p2$ with $p_1 < p_2$
d) show that $f^{(2)}(x) < x$ for $0 < x < p_0$ and $f^{(2)}x > x$ for $p_0 <x <1$
e) assume $x_0 \in (0,p_0)$ show that the subsequence has $x_{2k}$ $\in (0,p_0)$ for all $k$, and is strictly monotone decreasing, $x_{2k+2} < x_{2k}$
All I have so far is,
fixed points of $f(x)$ is $1-x^2 =x$ => $x^2+x-1=0$ => $x=\frac{-1 +-\sqrt5}{2}$
and since $f(x) = 1-x^2$, $f^2(x) = f[f(x)]$ = $2x^2-x^4$ and since we need to solve $f^2(x) = x$ we have $x^4-2x^2+x=0$ where we can factor it out as $x(x-1)(x^2+x-1)$ to get points of period 2 of $f$ be $0,1$ and $(-1+-\sqrt5)/2$
but the problem i'm facing is that those last two values are the fixed points of $f$ aren't they? (as question C) asked, i need to find 2 points that are not already fixed points...)
Let's go from the beginning though, how do i show a)?
sorry in advance, I am very lost in this question, and don't really have anything more..
a) By induction:
$$\begin{align} x_n\in(0,1)&\implies 0<x_n<1 \\ &\implies 0<x_n^2<1 \\ &\implies 1-1<1-x_n^2<1-0 \\ &\implies 0<x_{n+1}<1 \\ &\implies x_{n+1}\in(0,1) \end{align}$$
b) Your analysis is correct, but only one of the two solutions of your equation is in the interval $(0,1)$. So the only fixed point solving $f(x)=x$ is $$p_0=\frac{\sqrt 5-1}{2}$$ which is the golden ratio.
c) Your analysis is correct, but only three solutions to $f(f(x))=x$ are in the interval $[0,1]$, and they are $0,1,\frac{\sqrt 5-1}{2}$. That last value is $p_0$, so we get $$p_1=0, \quad p_2=1$$
Note that $p_1$ and $p_2$ are not actually in $(0,1)$. That does not seem to affect the later parts of your problem.
We can see those fixed points graphically, by plotting the intersections of $y=f(x)$ and $y=f(f(x))$ with the line $y=x$.
I believe that answers your stated questions. I hope you can continue and do parts d) and e).