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

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

Use the fixed exponents

p, q

in the original PDF. Assume positive

a, b

so arbitrary real powers and a relative change are defined. Set

h = k /100

. With

b

fixed, the exact new-to-old ratio is

(1 + h)^{p}

. The binomial series, or the one-variable case of the second-order relative change of a monomial, gives

100 \frac{Δ f}{f} = 100 [p h + \frac{p ( p - 1 )}{2} h^{2} + O (h^{3})] .

Therefore the approximate percentage change through second order is

p k + \frac{p ( p - 1 )}{200} k^{2} .

For negative

k

this same signed expression describes a decrease.

ii

Solution

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}) .

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} .

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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