 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-8/1/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

∣ f (n) ∣ = O (1/∣ n ∣)

f^{'} (n) = in f (n) .

\sum_{n \neq = 0} n^{2} ∣ f (n) ∣^{2} = \sum_{n \neq = 0} ∣ f^{'} (n) ∣^{2} \leq ∥ f^{'} ∥_{2}^{2} .

n \neq = 0 \sum ∣ f (n) ∣ \leq n \neq = 0 \sum n^{2} ∣ f (n) ∣^{2}^{1/2} n \neq = 0 \sum \frac{1}{n ^{2}}^{1/2} \leq \frac{π}{3} ∥ f^{'} ∥_{2} .

Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has

f^{'} \in L^{2}

, all hypotheses of part (iii) hold and its Fourier series converges uniformly.

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