= Logistic inverse branches contract in an angular coordinate
{title2=$\operatorname{Lip}(h^{-1}\circ g_{L,R}\circ h)\leq\mu^{-1/2}<1$}
For the real <logistic map> $f(x)=\mu x(1-x)$ with $\mu>4$, let $g_L,g_R$ be its two <inverse branches> from $[0,1]$ to the disjoint subintervals on which $0\leq f(x)\leq1$. With the <homeomorphism> $h(u)=\sin^2(\pi u/2)$, these become
$$
G_L(u)=\frac1\pi\arcsin\left(\frac2{\sqrt\mu}\sin\frac{\pi u}{2}\right),\qquad G_R(u)=1-G_L(u).
$$
Direct <differentiation> gives
$$
|G_{L,R}'(u)|=\frac{\cos(\pi u/2)}{\sqrt\mu\sqrt{1-(4/\mu)\sin^2(\pi u/2)}}\leq\mu^{-1/2},
$$
since $1-(4/\mu)\sin^2v\geq\cos^2v$. Thus length-$n$ <itinerary cylinders> have angular diameter at most $\mu^{-n/2}$, and ordinary diameter at most $(\pi/2)\mu^{-n/2}$ because $|h'|\leq\pi/2$. The <nested interval theorem> then gives exactly one point for each infinite itinerary. This proves the coding by the one-sided <full shift> even when the ordinary-coordinate <inverse branches> are not contractions.
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