 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/4/d/solution

F [e^{- t^{2}}] = π e^{- ω^{2} /4}, F^{- 1} [2 e^{- ω^{2} /4}] = \frac{2}{π} e^{- t^{2}} \equiv q (t) .

F [q^{''}] = - ω^{2} q

1 + ϵ ω^{2}

corresponds to

q - ϵ q^{''}

. Since

q^{''} = (2/ π) (4 t^{2} - 2) e^{- t^{2}}

s (t) = \frac{2}{π} (1 + 2 ϵ - 4 ϵ t^{2}) e^{- t^{2}} .

This is real and even, as required by the given spectrum. Its time integral is two for every

ϵ

, consistently with

s (0) = 2

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