 Solution

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

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

Every

x_{0} \in [1, \infty)

works. Indeed,

x_{1} \geq a

, and whenever

x_{k} \geq a

0 \leq x_{k + 1} - a = \frac{( x _{k} - a ) ^{2}}{2 x _{k}} \leq x_{k} - a .

The iterates from

k = 1

onward therefore decrease to a limit, and the recurrence forces that limit to be

a

There is also an explicit error bound. Put

q = \frac{x _{0} - a}{x _{0} + a} < 1.

A direct calculation gives

\frac{x _{k + 1} - a}{x _{k + 1} + a} = (\frac{x _{k} - a}{x _{k} + a})^{2} .

Hence, for

k \geq 1

∣ x_{k} - a ∣ = \frac{2 a q ^{2^{k}}}{1 - q ^{2^{k}}} \leq \frac{2 a}{1 - q} q^{2^{k}} .

This double-exponential decay is consistent with the Newton bound in part (a). For the chosen objective, the supplied factorization also gives

f (x) - f (a) = \frac{1}{3} (x - a)^{2} (x + 2 a),

so convergence of the objective and convergence of the iterates are equivalent on

[1, \infty)

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