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

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- Solution iii

Solution

If

BC = CB

, the splitting commutator vanishes, but the backward-Euler resolvents still contribute

\frac{k ^{2}}{2} (B^{2} + C^{2}) + O (k^{3}) .

Thus 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.

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