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{\bf Question}

Show that $C(x)=\ds\frac{1}{4}(x+2)$ defines a topological
conjugacy between $f(x)=4x(1-x)$ and $g(x)=2-x^2$. Deduce that g
has chaotic orbits.
\medskip

{\bf Answer}

$\left.\begin{array} {rcl} f \circ C(x) & = & 4 \cdot
\ds\frac{1}{4}(x+2)\left(1-\ds\frac{1}{4}(x+2)\right)=
\ds\frac{1}{4}(x+2)(2-x)\\C \circ g(x) & = &
\ds\frac{1}{4}92-x^2+2)=\ds\frac{1}{4}(4-x^2)=\ds\frac{1}{4}(2+x)(2-x)
\end{array}\right\}$

so $f \circ C=C \circ g$.

Since $f$ has orbits which are bounded and not asymptotically
periodic the same holds for $g$.  For positive Lyapunov exponents
we use the fact that $C'(x)=\ds\frac{1}{4}$ so from
$g(x)=C^{-1}(f(C(x)))$ we get $g'(x)=4f'(C(x))=f'(C(x))$, so the
Lyapunov exponent of the g-orbit of $x$ is the same as the
Lyapunov exponent of the f-orbit of $C(x)$.

[We could not expect this for a general conjugacy $C$ if e.g.
$C'(x) \rightarrow 0$ somewhere.]
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