 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/4/b/solution

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts gives

f^{'} (k) = \int_{R} f^{'} (x) e^{- ik x} d x = ik f (k) .

f^{(m)} = (ik)^{m} f

. On the other hand, differentiation under the integral gives

\frac{d ^{m} f}{d k ^{m}} = \int_{R} (- i x)^{m} f (x) e^{- ik x} d x,

and therefore

x^{m} f = i^{m} f^{(m)}

Write

p_{n} (x) = \sum_{j = 0}^{n} a_{j} x^{j}

, with

a_{n} \neq = 0

. Applying the second rule and part (a) gives

F [p_{n} (x) g (x)] (k) = 2 π \sum_{j = 0}^{n} a_{j} i^{j} g^{(j)} (k) = q_{n} (k) e^{- k^{2} /2},

where

q_{n} (k) = 2 π j = 0 \sum n a_{j} i^{j} r_{j} (k) .

Only the

j = n

term can contribute degree

n

. Its leading coefficient is

2 π a_{n} (- i)^{n} \neq = 0

, so

q_{n}

has exactly the same degree

n

, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of

p_{n}

are real.

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