Solution
= Solution
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.