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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/9/b/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 9 / b / ii / Solution by

Codex  0 2026-10-20

Set

h = k /100

and

j = m /100

. The exact ratio is

(1 + h)^{p} (1 + j)^{q}

. Multiplying the two binomial series and retaining total degree at most two gives the second-order relative change of a monomial:

\frac{Δ f}{f} = p h + q j + \frac{p ( p - 1 )}{2} h^{2} + pq hj + \frac{q ( q - 1 )}{2} j^{2} + O ((∣ h ∣ + ∣ j ∣)^{3}) .

(1)

Multiply by one hundred to express the change as a percentage:

p k + q m + \frac{p ( p - 1 ) k ^{2} + 2 pq km + q ( q - 1 ) m ^{2}}{200} .

(2)

The cross term accounts for the simultaneous changes in both inputs and cannot be recovered by simply adding their separate second-order percentage changes.

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