Solution

Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 53 / 2 / a / Solution

Period-doubling cascade

= Super-stable periodic orbit
{synonym}

A <periodic orbit> of a differentiable <interval map> is superstable when its <periodic-orbit multiplier> vanishes. A cycle passing through a <critical point> has zero multiplier because the <derivative> product includes $f'(c)=0$. For the nonconstant quadratic family $f_\mu(x)=1-\mu x^2$, a superstable $2^n$ cycle satisfies $f_\mu^{2^n}(0)=0$ and does not have a smaller critical period.

// Target: dynamical-systems.bigb

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 Incoming links: Superstable periodic orbit