Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 41A / b / iii / Solution

If $BC=CB$, the splitting commutator vanishes, but the backward-Euler resolvents still contributeThus commutativity alone does not change the order: the local error remains $O(k^2)$ and the global method remains first order, barring an additional special cancellation.