Sketch the graph of $F_4^n(x)$ on the unit interval, where $F_4(x) = 4x(1-x)$. Conclude that $F_4$ has at least $2^n$ periodic points of period $n$.

Question

Sketching the graph I found that $F_4$ has exactly $2^{n-1}$ point wich prime period $n$. My $F_4^n$ graph look like a $|sin(x)|$ graph with $2^n$ intersections on $Id$, but I removed the $2^{n-1}$ to remove points that have period less then n. What is wrong?

Answer 1

Given the following functions: $\varphi(x)=-4x+2$, $\varphi^{-1}(x)=\frac{x-2}{-4}$ and $g(x)=x^2-2$ we have: $$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$ or $$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x) \tag{1}$$

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Further $g(x)$ can be written as $$g(x)=(\gamma^{-1} \circ T_2\circ \gamma)(x)$$ where $T_2(x)=2x^2-1$ (a Chebyshev polynomials), $\gamma(x)=\frac{x}{2}$ and $\gamma^{-1}(x)=2x$. Then $$F_4(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_2\circ \gamma \circ \varphi)(x)$$ or $$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_2^{\circ n}\circ \gamma \circ \varphi)(x) \tag{2}$$

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Chebyshev polynomials satisfy $$T_n(T_m(x))=T_{nm}(x) \Rightarrow T_2^{\circ n} (x)=T_{2^n}(x)$$ or $$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_{2^n}\circ \gamma \circ \varphi)(x) \tag{3}$$

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Periodic points of period $n$ are fixed points of $\circ n$ composition. Fixed points of $T_{2^n}(x)$ are fixed for $F_4^{\circ n}(x)$ "through" $\gamma(x)$ and $\varphi(x)$, both linear functions, i.e.

Proposition 1. $T_{2^n}(x)=x \iff F_4^{\circ n}(y)=y$

where $y=(\varphi^{-1} \circ \gamma^{-1} )(x)$

We should also observe that if $y \in [0,1]$ then $$x=(\gamma \circ \varphi) (y) = -2y+1 \in [-1,1] \tag{4}$$

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For Chebyshev polynomials we have $T_m(\cos{\alpha})=\cos{(m\alpha)}$. From $(4)$ we can assume that $\exists \alpha$ s.t. $x=(\gamma \circ \varphi) (y)=\cos{\alpha}$. As a result, for proposition 1 we will be looking at $$T_{2^n}(\cos{\alpha})=\cos{(2^n \alpha)}=\cos{\alpha}$$ which means for $k\in\mathbb{Z}$ $$2^n \alpha = \alpha +2k\pi \Rightarrow \alpha =\frac{2k\pi}{2^n-1}$$ or $$2^n \alpha = -\alpha +2k\pi \Rightarrow \alpha =\frac{2k\pi}{2^n+1}$$ and we found 2 sequences $$x_{1,k} = T_{2^n}(x_{1,k})=T_2^{\circ n} (x_{1,k}) \text{, where } x_{1,k}=\cos{\left(\frac{2k\pi}{2^n-1}\right)}$$ $$x_{2,k} = T_{2^n}(x_{2,k})=T_2^{\circ n} (x_{2,k}) \text{, where } x_{2,k}=\cos{\left(\frac{2k\pi}{2^n+1}\right)}$$

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Applying proposition 1 $$y=(\varphi^{-1} \circ \gamma^{-1} )(x)=\frac{1-x}{2}$$ and considering that $\frac{1-\cos{\alpha}}{2}=\sin^2{\frac{\alpha}{2}}$, we have $$y_{1,k} =\sin^2{\left(\frac{k\pi}{2^n-1}\right)}$$ $$y_{2,k} =\sin^2{\left(\frac{k\pi}{2^n+1}\right)}$$

Some of these values will repeat, for example $$y_{1,2^n-k}=y_{1,k-1} \text{, so we can limit } k=\overline{1,2^{n-1}}$$ $$y_{2,2^n-k}=y_{2,k+1} \text{, so we can limit } k=\overline{0,2^{n-1}-1}$$ which makes $2^n$ altogether.

Answer 2

Incomplete and with typos ... but contains a few hints and references.

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Given the following functions: $$\varphi(x)=-4x+2,\varphi^{-1}(x)=\frac{x-2}{-4}, g(x)=x^2-2$$ we have: $$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$ or $$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x)$$ Further to this, function $\gamma(x)=x+\frac{1}{x}$ has the property that $$(g \circ \gamma)(x)=x^2+\frac{1}{x^2}=\gamma(x^2)\Rightarrow (\gamma^{-1} \circ g \circ \gamma)(x)=x^2\Rightarrow \\g(x)=(\gamma \circ t \circ \gamma^{-1})(x), t(x)=x^2$$ As a result $$F_4(x)=(\varphi^{-1} \circ \gamma \circ t \circ \gamma^{-1} \circ \varphi)(x)$$ or $$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma \circ t^{\circ n} \circ \gamma^{-1} \circ \varphi)(x) \tag{1}$$ Effectively, by introducing $\gamma$ we have to exclude $x=0$ perioding point, which is not of period $n$ though.

Periodic points of $t(x)=x^{2}$ are periodic for $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$. Complex roots of $x^{2^n-1}=1$ are periodic points for $t(x)$ (and $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$) and, except $x=1$, all the other roots have period $n$ (there is an entire chapter in this book, starting with page 6, dedicated to this function). To see all these working, let's take $x=\sin^2{\frac{\alpha}{2}} \in (0,1)$ then (I will omit the details as those are easy to check) $$\varphi(x)=2\cos{\alpha}$$ $$\gamma^{-1}\left(2\cos{\alpha}\right)=e^{i\alpha}$$ $$t^{\circ n}\left(e^{i\alpha}\right)=e^{i2^{n}\alpha}$$ $$\gamma\left(e^{i2^{n}\alpha}\right)=2\cos{\left(2^{n}\alpha\right)}$$ $$\varphi^{-1}\left(2\cos{\left(2^{n}\alpha\right)}\right)=\sin^2{\left(2^{n-1}\alpha\right)}$$ or $$F_4^{\circ n}\left(\sin^2{\frac{\alpha}{2}}\right)=\sin^2{\left(2^{n-1}\alpha\right)} \tag{2}$$ Complex roots of $z^{2^n-1}=1$ are of the form $$z_k=\exp{\left(\frac{2i\pi k}{2^n-1}\right)},k=\overline{0..2^n-1}$$ thus taking $$\alpha=\frac{2\pi k}{2^n-1}$$ we have $$F_4^{\circ n}\left(\sin^2{\frac{\pi k}{2^n-1}}\right)=\sin^2{\left(2^{n-1}\frac{2\pi k}{2^n-1}\right)}=\sin^2{\left(\frac{(2^{n}-1+1)\pi k}{2^n-1}\right)}=\\ \sin^2{\left(k\pi + \frac{\pi k}{2^n-1}\right)}=\sin^2{\frac{\pi k}{2^n-1}},k=\overline{0..2^n-1} \tag{3}$$ Note: we could simply apply (from $(2)$) $\sin^2{\frac{\alpha}{2}}=\sin^2{\left(2^{n-1}\alpha\right)} \Rightarrow \frac{\alpha}{2}+k\pi=2^{n-1}\alpha$ and arrive to the same conclusion, but without conjugacy functions reaching this point would require having a good intuition.

However $$F_4^{\circ n}\left(\sin^2{\frac{\pi (2^n-k)}{2^n-1}}\right)=\sin^2{\left(\frac{\pi (2^n-k)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (2^n-1-k+1)}{2^n-1}\right)}=\\ \sin^2{\left(\pi - \frac{\pi (k-1)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}=F_4^{\circ n}\left(\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}\right) \tag{4}$$ this last statement reduces the number of periodic points with period $n$ to $2^{n-1}$.

Note: $(4)$ can also simply be obtained from $\frac{2\pi k}{2^n-1}=\frac{2\pi m}{2^n-1} + p\pi$, where $0\leq k,m \leq 2^n-1$ and concluding that $p=1$ and $k-m=2^n-1$.