Using calculus, I want to determine all the fixed points of the function $f^3$ where $f$ is given by:$$ f:[0,1]\rightarrow[0,1];\;f(x)=4x(1-x) $$and such that those fixed points are not fixed points of the functions $f^2$ and $f$. If we can not determine them explicitly, are there any argument to justify that such points exist? Thanks in advance.
$\endgroup$3 Answers
$\begingroup$The fixed points of a function $F$ are simply the solutions of $F(x)=x$ or the roots of $F(x)-x$. The function $f(x)=4x(1-x)$, for example, are $x=0$ and $x=3/4$ since $$4x(1-x)-x = x\left(4(1-x)-1\right) = x(3-4x).$$ Geometrically, these are the points of intersection between the graphs of $y=f(x)$ and $y=x$, as shown here:
The simplest way to demonstrate the existence of fixed points of $f^3$ that are not fixed points of $f$ is to simply sketch the graphs of $y=x$, $y=f(x)$, and $y=f(f(f(x)))$ together.
Note that, in addition to the two fixed points of $f$, we've picked up $6$ new points of intersection between $y=f^3(x)$ and $y=x$. These are actually two separate orbits of period 3. As Michael correctly points out in his answer, these can be characterized algebraically as the solutions of $f^3(x)=x$ or $f^3(x)-x=0$. You should understand that dealing with this type of equation is often tedious. Using Mathematica to factor the equation, I found that $$f^3(x)-x = -x (4 x-3) \left(64 x^3-112 x^2+56 x-7\right) \left(64 x^3-96 x^2+36 x-3\right)$$ Thus, those two orbits are the roots of the irreducible cubic polynomials seen in the factorization.
Finally, as $2$ and $3$ are relatively prime, the only common fixed points of $f^2$ and $f^3$ are the fixed points of $f$ itself.
One more comment: If you really want a by hand computation of the fixed points of $f^3$, there is an elegant way to obtain them using a dynamical conjugacy between $f$ and the squaring function $z^2$, but I'm guessing that's a bit more involved than you're looking for.
$\endgroup$ $\begingroup$You have a functional form for $f$, so write down (work out) $f^2$ and $f^3$. Set
$f^3(x) = x$
and solve for $x$. If you also solve $f(x) = x$ and $f^2(x) = x$ you will find which fixed points of $f^3$ belong to $f$ and $f^2$ as well, and you can discard these.
$\endgroup$ 1 $\begingroup$Note that the solutions for $f^n(x)=x$ are equivalently the solutions to the equation $\sin^2(t)=\sin^2(2^nt)$, where $x=\sin^2(t)$. Therefore such fixed points exist in $2^n$-touples for each value of $n$ and the fixed points are the set
$$ \{\sin^2\left(\frac{k\pi}{2^n+1}\right), \sin^2\left(\frac{(k-1)\pi}{2^n-1}\right)\} $$ for $k\in \mathbb{N}, k\leq 2^{n-1}$
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