Superstable periodic orbit

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 $f^{\prime }(c)=0$. 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