Subsection2.7.3A chaotic quadratic
Let \(f(x)=x^2-2\text{.}\) We now show that \(f\) is semi-conjugate to the doubling map \(d\) under the semi-conjugacy \(\varphi(x)=2\cos(2\pi x)\text{.}\) As a result, \(\varphi\) maps all the orbit types that \(d\) has to an orbit of \(f\) with similar properties. Thus, \(f\) is chaotic.
Claim2.7.8
The map \(f(x)=x^2-2\) is semi-conjugate to the doubling map \(d(x)=2x\bmod1\) under the semi-conjugacy \(\varphi(x)=2\cos(2\pi x)\text{.}\)
We must simply show that \(f\circ\varphi=\varphi\circ d\text{,}\) so let's compute. First,
\begin{equation*}
f(\varphi(x)) = 2(2\cos(2\pi x))^2 - 2 = 4\cos^2(2\pi x)-2.
\end{equation*}
Well, that was easy. The next part is a little trickier - we just need to apply a couple of trig identities and use the fact that we can drop the mods inside the squared trig functions due to the symmetries of those functions. \begin{align*} \varphi(d(x)) & = 2\cos(2\pi(2x \bmod 1)) \\ & = 2(\cos^2(\pi (2 x \bmod 1))-\sin^2(\pi (2 x \bmod 1))) \\ & = 2(\cos^2(2\pi x)-\sin^2(2\pi x)) \\ & = 2(\cos^2(2\pi x)-(1-\cos^2(2\pi x))) \\ & = 2(2\cos^2(2\pi x)-1) \\ & = 4\cos^2(2\pi x)-2 \\ \end{align*}
Again, the key fact about semi-conjugacy is that \(\varphi\) maps orbits of \(d\) to orbits of \(f\text{.}\) Thus, since \(d\) has a dense orbit \(f\) too has a dense orbit. Here's a concrete example illustrating this idea.
Example2.7.9
Find a point of period 11 for the chaotic quadratic \(f(x)=x^2-2\text{.}\)
SolutionFirst, it's easy to find an orbit of period 11 for the doubling map. One example is
\begin{equation*}
0_{\dot 2}00000000001 = \sum_{k=1}^{\infty}\frac{1}{2^{11k}} = \frac{1}{2047}.
\end{equation*}
The point behind conjugacy is that \(\varphi(1/2047) = 2\cos(2\pi/2047)\) will be a point of period 11 for \(f\text{.}\) The reader is advised to check this numerically!