 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/18h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

x_{k} \geq 1

x_{k + 1} = \frac{1}{2} (x_{k} + \frac{a}{x _{k}}) \geq a \geq 1.

Thus all iterates remain in

[1, \infty)

Consider the strictly convex function

f (x) = \frac{x ^{3}}{3} - a x

[1, \infty)

. Since

f^{'} (x) = x^{2} - a

and

f^{''} (x) = 2 x

, its Newton minimization step is

x - \frac{f ^{'} ( x )}{f ^{''} ( x )} = \frac{1}{2} (x + \frac{a}{x}) .

Its unique minimizer is

x^{*} = a

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