The three substeps have total length $2\alpha+(1-2\alpha)=1$. Their leading cubic defects add, so cancellation requires
$$
2\alpha^3+(1-2\alpha)^3=0.
$$
Taking the real cube root and solving gives
$$
\boxed{\alpha=\frac1{2-2^{1/3}}.}
$$
This is the coefficient in the <higher-order composition of a symmetric splitting>; the middle substep $1-2\alpha$ is negative.

Solution

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

 Solution

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