Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 4 / b / v

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- Solution v

Solution

The equilibrium of the first-order linear recurrence is

n /2

. Subtracting it gives

r_{k + 1} - n /2 = (1 - 2/ n) (r_{k} - n /2) .

Consequently

r_{k} = \frac{n}{2} + \frac{n}{2} (1 - \frac{2}{n})^{k}, k \geq 0.

For

n = 2

, interpret the formula at

k = 0

through the initial condition:

r_{0} = 2

, while

r_{k} = 1

for

k \geq 1

. For every

n \geq 2

,

r_{k} \to n /2

.

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