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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/5/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5 / b / Solution by

Codex  0 2026-10-18

Expand both sides through second order as

ϵ \to 0

, holding

X, Y

fixed. The product is

e^{ϵ X} e^{ϵ Y} = I + ϵ (X + Y) + ϵ^{2} (\frac{X ^{2}}{2} + X Y + \frac{Y ^{2}}{2}) + O (ϵ^{3}) .

Put

Z = ϵ (X + Y) + ϵ^{2} [X, Y] /2

. Its exponential is

e^{Z} = I + ϵ (X + Y) + \frac{ϵ ^{2}}{2} {[X, Y] + (X + Y)^{2}} + O (ϵ^{3}) = I + ϵ (X + Y) + ϵ^{2} (\frac{X ^{2}}{2} + X Y + \frac{Y ^{2}}{2}) + O (ϵ^{3}) .

Thus

e^{ϵ X} e^{ϵ Y} = e^{ϵ X + ϵ Y + ϵ^{2} [X, Y] /2} + O (ϵ^{3}) .

This is the quadratic part of the Baker--Campbell--Hausdorff formula. The remainder estimate is in any fixed finite-dimensional matrix norm, and is an asymptotic statement rather than an exact formula for arbitrary finite

ϵ

.

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