= Solution
For a nearby orbit, linearized separation satisfies $\delta_{i+1}=f'(x_i)\delta_i$. Repeated use of the <chain rule> gives $\delta_n=\delta_0\prod_{i=0}^{n-1}f'(x_i)$ to first order. Taking the logarithm of its magnitude and dividing by $n$ gives the exponential growth rate
$$
\boxed{\lambda=\lim_{n\to\infty}\frac1n\log\left|\frac{\delta_n}{\delta_0}\right|=\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\log|f'(x_i)|.}
$$
The 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 $M=(f^p)'(x_0)=\prod_{i=0}^{p-1}f'(x_i)$, the remainder of fewer than $p$ terms has vanishing contribution to the long-time average and
$$
\boxed{\lambda=\frac1p\log|M|.}
$$
Linear 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.
Back to article page
