Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 78 / 2 / b / Solution

For repeated small returns, ${\xi }_{n+1}=O(z_n^{\delta })$, so $\rho +{\xi }_n$ is close to the nonzero constant $\rho$. Retain its leading value in the second component. Combining $c\cos \vartheta +d\sin \vartheta$ into a single cosine, and absorbing the section size and fixed phases, gives the Shilnikov return mapChanging the sign of the logarithmic angle changes only the phase after using cosine's evenness. A generic reinjection has $A>0$; $\Phi$ can be represented modulo $2\pi$ by a positive value. The original PDF has exponent $\delta$, which the converted TeX mistakenly replaced by $2$.

This is an asymptotic scalar approximation, not an assertion that the full map has an exact one-dimensional invariant quotient. To see why it captures the thin geometry, differentiate the two components in part (a). Their determinant is of orderwhere $C$ involves $\Omega$ and the determinant of the global linear map. For $\delta >1/2$ this tends to zero, so the return is strongly area-contracting even when its longitudinal derivative is large. Its global quadratic corrections are $O(z^{2\delta })=o(z)$ in this range. For a small fixed point, also $\xi =O(z^{\delta })$, so ignoring $\xi$ contributes that same higher order. Multipassage iterates require all passage coordinates to stay in the small domain; the leading map describes their thin folded return geometry rather than controlling arbitrary points outside it.