 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/27h/b/solution

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

Since

θ_{n}

is even,

θ_{n} (ξ) = 2 \int_{0}^{n} (1 - \frac{x}{n}) cos (x ξ) d x .

For

ξ \neq = 0

θ_{n} (ξ) = \frac{2 ( 1 - c o s ( n ξ ))}{n ξ ^{2}} = n (\frac{s i n ( n ξ /2 )}{n ξ /2})^{2} .

ξ = 0

the continuous limiting value is the area of the triangle,

θ_{n} (0) = \int_{- n}^{n} (1 - \frac{∣ x ∣}{n}) d x = n .

In particular,

θ_{n} \geq 0

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