The composition is palindromic, and hence
$$
G_\alpha(-t)=F(-\alpha t)F(-(1-2\alpha)t)F(-\alpha t)
=G_\alpha(t)^{-1}.
$$
Thus $G_\alpha$ is symmetric. Part c cancels the cubic term in its odd <formal logarithm>; symmetry forbids a fourth-degree term, so the next possible defect has degree five. Therefore
$$
\boxed{G_\alpha(t)=e^{t(A+B)}+O(t^5),}
$$
which makes the composition a fourth-order method.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 5 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-341/section-a/5/d/solution