 Solution

For a nearby orbit, linearized separation satisfies ${\delta }_{i+1}=f^{\prime }(x_i){\delta }_i$. Repeated use of the chain rule gives ${\delta }_n={\delta }_0{\prod }_{i=0}^{n-1}{f}^{\prime }(x_i)$ to first order. Taking the logarithm of its magnitude and dividing by $n$ gives the exponential growth rateThe infinitesimal-perturbation limit is taken before interpreting long-time growth; finite separations eventually leave the linear regime. The limit is assumed to exist, with a limsup used when needed.

On a $p$-cycle, blocks of $p$ successive derivatives repeat. If $M=(f^p{)}^{\prime }(x_0)={\prod }_{i=0}^{p-1}{f}^{\prime }(x_i)$, the remainder of fewer than $p$ terms has vanishing contribution to the long-time average andLinear attraction means $|M|<1$, giving $\lambda <0$. A superstable cycle has $M=0$, meaning at least one derivative in the cycle vanishes. Its infinitesimal return multiplier is zero and the corresponding logarithmic exponent is $-\infty$. This concerns the derivative cocycle; actual finite perturbations can converge by higher-order rather than literally vanishing in one step.