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{\bf Question}

Let $f(x)=\frac{1}{2}(3x-x^3)$.  Show that the 2-cycle $\left\{\pm
\sqrt{5} \right\}$ is repelling.  Solve the inequality
$|f(x)|&gt;|x|$ and also $|f(x)|&lt;|x|$ to show there are {\underline
no} periodic points apart from $\left\{\pm \sqrt{5} \right\}$ and
the fixed points $\{0, \pm 1\}$.

\medskip

{\bf Answer}

$f'(x)=\ds\frac{3}{2}(1-x^2)$, so $f'(\pm \sqrt
5)=\ds\frac{3}{2}(-4)=6$.

Hence $(f^2)'(\sqrt 5)=f(\sqrt 5)(-\sqrt 5)=36&gt;1$ so the 2-cycle
$\{\sqrt 5,-\sqrt 5\}$ is repelling.

From the graph of $f$ (or considering $f(x)&gt;x, f(x)&lt;-x$ etc.) we
see that $|f(x)|&gt;|x|$ when $0&lt;|x|&lt;1$ or $|x|&gt;\sqrt 5$.

\begin{center}
\epsfig{file=314-1-1.eps, width=50mm}
\end{center}

Graphical iteration shows $x \mapsto 1$ or $-1$ in the first case,
and $x \mapsto \infty$ in the second.  If $1&lt;|x|&lt;\sqrt5$ then
$|f(x)|&lt;|x|$, so either eventually $|f^m(x)|&lt;1$ (so $f^n(x)
\rightarrow 1$ or $-1$) or $|f^m(x)| \rightarrow limit\ l \geq
1$:\ in this case $|f(l)|=|l|$ so again $l \pm 1$. So no more
periodic orbits.


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