 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-3/5a/solution

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

\tilde{g}_{σ}' (k) = - σ^{2} k \tilde{g}_{σ} (k)

and

\tilde{g}_{σ} (0) = 1

, so

\tilde{g}_{σ} = e^{- σ^{2} k^{2} /2}

f * g = \tilde{f} \tilde{g}

; multiplying the two Gaussian transforms and inverting yields

g_{σ_{1}^{2} + σ_{2}^{2}}

 New to topics? Read the docs here!