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{\bf Question}

What are the options for the Lyapunov exponents of bounded orbits
of $f_a(x)=ax(1-x)$ whene (i)a=2.5,\ (b)a=3.1?

Show that if $a>2+\sqrt 5$ then every orbit has positive Lyapunov
exponent (if it exists).
\medskip

{\bf Answer}

All orbits of $f$ which remain bounded are attracted to the fixed
point $x_*=1-\ds\frac{1}{2.5}=0.6$, except for 0 which remains at
0, and 1 with $f(1)=0$. $f'(0.6)=2.5(1-1.2)=-0.5$ and $f'(0)-2.5$
so the only options for Lyapunov exponents are \underline {$\ln
(2.5)$} for (0,1) or \underline {$\ln (0.5)$}$<0$ for all other
points of [0,1].

For $g$ there are repelling fixed points at 0 and
$x_*=1-\ds\frac{1}{3.1}=0.68$ approx, with Lyapunov exponents

\underline {$\ln|g'(0)|=\ln(3.1)>0$} and \underline
{$\ln|g'(0.68)|=\ln(1.1)>0$}, and an attracting 2-cycle $\{p,q\}$
with $p,\ q$ roots of $x^2-1.32x+0.43$.

Then
$(g^2)'(p)=g'(q)g'(p)=(3.1)^2(1-2q)(1-2p)=(3.1)^2(1-2(p+q)+4pq)=(3.1)^2(1-2.54+1.72)=0.77$,
so the Lyapunov exponent for $p$ and hence for all points which
lie in (0,1) but do not land on $x_*$ is \underline
{$\ds\frac{1}{2} \ln(0.77)=\ln(0.88)$ approx,\ $<0$}.

$\begin{array}{l}
\rm{For\ } h(x)=ax(1-x) \rm{\ with\ } a>4 \textrm{ the graph looks like:}
\end{array}
\begin{array}{c}
\epsfig{file=314-2-2.eps, width=20mm}
\end{array}$

so the Lyapunov exponent of any bounded orbit will be $>0$
\underline{if} the slope of the graph is everywhere $>1$ or $<-1$.

Now the graph cuts the top of the square where
$x=\ds\frac{1}{2}\left(1 \pm \sqrt{1-\frac{4}{a}}\right)$, where
the slope is $h'(x)=\mp \sqrt{a^2-4a}$. This gives $|h'(x)|>1$
when $a^2-4a>1:\ a>2+\sqrt 5$.
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