<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Show that the 2-cycle that exists for $g_a(x)=ax(1-x)$ when $a&gt;3$
changes from attracting to repelling at $a=1+\sqrt 6$.
\medskip

{\bf Answer}

If the period-2 orbit is $\{p,q\}$ then
${g_a^2}'(p)={g_a}'(p){g_a}'(q)=a^2 (1-2p)(1-2q)=a^2
(1-2(p+q)+4pq)$. Now $p,\ q$ (together with $0,\
1-\ds\frac{1}{a}$: fixed points of $g_a$) are the roots of
$g_a^2(x)=x$, i.e. $a(ax(1-x))(1-ax(1-x))=x$

i.e. $x(a^2(1-x)(1-ax+ax^2)-1)=0$,

i.e. $x=0$ or $a^3x^3-2a^3x^2+a^2(1+a)x+(1-a^2)=0.$

Sum of roots: $2=p+q+(1-\ds\frac{1}{a})$, so
$p+q=\ds\frac{1}{a}+1$.

Product of roots:
$\ds\frac{-1-a^2}{a^3}=pq\left(1-\ds\frac{1}{a}\right)$,\ so
$pq=\ds\frac{1}{a}\left(\ds\frac{1}{a}+1\right)$.

So from above we see \begin{eqnarray*} {g_a^2}'(p) &amp; = &amp;
a^2\left(1-2\left(\ds\frac{1}{a}+1\right)+\ds\frac{4}{a}\left(\ds\frac{1}{a}+1\right)\right)\\
&amp; = &amp; -a^2-2a+4+4a\\ &amp; = &amp; -a^2+2a+4. \end{eqnarray*}

Period-2 orbit $\{p,q\}$ becomes repelling when ${g_a^2}'(p)=-1$,
i.e. $-a^2+2a+4=-1$: $a=1 \pm \sqrt 6$ (only the + sign is
relevant).
\end{document}
</pre></body></html>