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Subsection 2.9.2 A 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 orbits of \(d\) has to orbits of \(f\) with similar properties. Thus, \(f\) is chaotic.

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.

Example 2.9.7

Find a point of period 11 for the chaotic quadratic \(f(x)=x^2-2\text{.}\)

Solution

First, 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!

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