 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-3/15a/b/solution

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

⟨ x δ^{'}, f ⟩ = ⟨ δ^{'}, x f ⟩ = - (x f)^{'}_{x = 0} = - f (0) = ⟨ - δ, f ⟩ .

Hence $x δ^{'} (x) = - δ (x)$ as distributions.

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