 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-66/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

S = A + B

and

E (t) = F (t) - e^{tS}

. Since

F (0) = I

E (0) = 0

. Differentiate the ordered exponential products, using that each matrix commutes with its own exponential:

F^{'} (t) - SF (t) = \frac{1}{2} {[e^{tB}, A] e^{t A} + [e^{t A}, B] e^{tB}} =: D (t) .

Here the commutator convention is

[X, Y] = X Y - Y X

; this fixes both signs. The error satisfies

E^{'} = SE + D

E (t) = \frac{1}{2} \int_{0}^{t} e^{(t - x) S} {[e^{x B}, A] e^{x A} + [e^{x A}, B] e^{x B}} d x .

A, B

was not needed for the identity itself; it will be used to bound their exponentials in part (c).

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