Let $f: I \rightarrow I$ be a continuous one-dimensional map of an interval $I \subset \mathbb{R}$. Explain what is meant by saying that $f$ has a horseshoe.

A map $g$ on the interval $[a, b]$ is a tent map if

(i) $g(a)=a$ and $g(b)=a$;

(ii) for some $c$ with $a<c<b, g$ is linear and increasing on the interval $[a, c]$, linear and decreasing on the interval $[c, b]$, and continuous at $c$.

Consider the tent map defined on the interval $[0,1]$ by

$$f(x)= \begin{cases}\mu x & 0 \leqslant x \leqslant \frac{1}{2} \\ \mu(1-x) & \frac{1}{2} \leqslant x \leqslant 1\end{cases}$$

with $1<\mu \leqslant 2$. Find the corresponding expressions for $f^{2}(x)=f(f(x))$.

Find the non-zero fixed point $x_{0}$ and the points $x_{-1}<\frac{1}{2}<x_{-2}$ that satisfy

$$f^{2}\left(x_{-2}\right)=f\left(x_{-1}\right)=x_{0}=f\left(x_{0}\right) .$$

Sketch graphs of $f$ and $f^{2}$ showing the points corresponding to $x_{-2}, x_{-1}$ and $x_{0}$. Indicate the values of $f$ and $f^{2}$ at their maxima and minima and also the gradients of each piece of their graphs.

Identify a subinterval of $[0,1]$ on which $f^{2}$ is a tent map. Hence demonstrate that $f^{2}$ has a horseshoe if $\mu \geqslant 2^{1 / 2}$.

Explain briefly why $f^{4}$ has a horseshoe when $\mu \geqslant 2^{1 / 4}$.

Why are there periodic points of $f$ arbitrarily close to $x_{0}$ for $\mu \geqslant 2^{1 / 2}$, but no such points for $2^{1 / 4} \leqslant \mu<2^{1 / 2}$ ? Explain carefully any results or terms that you use.