 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/41e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

u_{m}^{n} = r^{n} e^{im θ}

, put

q = 4 μ sin^{2} \frac{θ}{2}, 0 \leq q \leq 4 μ .

The recurrence becomes

r^{n + 1} = (1 - \frac{3}{2} q) r^{n} + \frac{1}{2} q r^{n - 1},

p (r) = r^{2} - (1 - \frac{3}{2} q) r - \frac{1}{2} q .

Its roots are real. Moreover,

p (1) = q, p (- 1) = 2 (1 - q), \frac{r _{1} + r _{2}}{2} = \frac{1}{2} - \frac{3}{4} q .

The root-location criterion in the question therefore puts both roots in

[- 1, 1]

exactly when

0 \leq q \leq 1

4 μ \leq 1

, so

the scheme is stable if and only if μ \leq \frac{1}{4} .

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