\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Consider the tent map $t:\ I \longrightarrow I$ and its
itineraries.  Find that end-points of the subinterval of $I$
consisting of all those points whose itinerary begins LRR, and
likewise for LRRLRR.  Find a point $x_n$ whose itinerary begins
LRRLRR $\cdots$ LRR ($n$ times).  Hence find a point of period 3
for $T$, and verify directly from $T$ that its period is 3.  Give
a point of period 3 for the logistic map $G(x)=4x(1-x)$.
\medskip

{\bf Answer}

For the tent map $T$ the interval LRR is
$\left[\ds\frac{1}{4},\ds\frac{3}{8}\right]$ i.e. the third of 8
subintervals.  Hence the interval LRRLRR is the third of 8
subintervals, i.e.
$\left[\ds\frac{18}{64},\ds\frac{19}{64}\right]$.

Continuing we see that the point with itinerary LRRLRRLRR $\cdots$
is:

$$\ds\frac{2}{8}+\ds\frac{2}{8^2}+\ds\frac{2}{8^3}+ \cdots =
2\ds\frac{\frac{1}{8}}{1-\frac{1}{8}}=2\ \cdot
\ds\frac{1}{7}=\ds\frac{2}{7}$$

$\left(=1-\left[\ds\frac{5}{8}+\ds\frac{5}{64}+\cdots\right]=1-5
\cdot \ds\frac{1}{7}=1-\ds\frac{5}{7}=\ds\frac{2}{7}.
\surd\right)$

\underline {Check}: $T\left(\ds\frac{2}{7}\right)=\ds\frac{4}{7};\
T\left(\ds\frac{4}{7}\right)=2\left(\ds\frac{3}{7}\right)=\ds\frac{6}{7};\
T\left(\ds\frac{6}{7}\right)=2\left(\ds\frac{1}{7}\right)=\ds\frac{2}{7}$.

Since we have a conjugacy

\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(8,3)
\put(0,0){$I$} \put(2,0){$I$} \put(0,2){$I$} \put(2,2){$I$}
\put(0.2,0.1){\vector(1,0){1.8}} \put(0.2,2.1){\vector(1,0){1.8}}
\put(0.1,1.9){\vector(0,-1){1.6}}
\put(2.1,1.9){\vector(0,-1){1.6}} \put(1,0.2){$G$}
\put(1,2.2){$T$} \put(0.2,1){$C$} \put(2.2,1){$C$}
\end{picture}
\end{center}


\bigskip



with $C(x)=\ds\frac{1}{2}(1-\cos \pi x)$ a point of period 3 for
$G$ is $C\left(\ds\frac{2}{7}\right)=\ds\frac{1}{2}\left(1-\cos
\ds\frac{2\pi}{7}\right)$
\end{document}