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{\bf Question}

Show that $f:\ {\bf R} \longrightarrow {\bf R}:\ x+x^3$ has a
repelling fixed point at $x=0$ (although $f'(0)=1$).  What about
$x \mapsto x-x^3$?  What is the behaviour of $x \mapsto -x+x^3,\ x
\mapsto -x-x^3$ near the origin?
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{\bf Answer}

$\underline {f(x)=x+x^3}$ $\Rightarrow f(x)>x$ or $<x$ according
as $x>0$ or $<0$.  Thus $f^n(x)$ tends monotonically away from 0,
and in fact $f^n(x) \rightarrow \pm \infty$ (else $f^n(x)
\rightarrow l$ for some finite $l$, which then has $f(l)=l$: not
the case for $l \ne 0$).

$\underline {f(x)=x-x^3}$: attracting fixed pt. at 0, since $0<x<1
\Rightarrow 0<f(x)<x<1$ ($-1<x<0 \Rightarrow -1<x<f(x)<0$) so
$f(x) \rightarrow limit\ m$ which has $f(m)=m$ so $m=0$.

$\underline {f(x)=-x+x^3}$: \begin{eqnarray*} {\rm here}\ 0<x<1 &
\Rightarrow & -1<-x<f(x)<0\\ {\rm and}\ -1<x<0 & \Rightarrow &
0<f(x)<-x<1 \end{eqnarray*}

so $|x|<1 \Rightarrow |f(x)|<|x|$ so $|f^n(x)| \rightarrow l$ and
$|f(l)|=|l|$ so $l=0$: attracting.

Likewise $\underline {f(x)=-x-x^3}$: origin is repelling. (Last 2
cases with oscillation.)
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