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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 17B / a / Solution

By the Leibniz rule,

\frac{d ^{n}}{d x ^{n}} (x^{n + 1/2} e^{- x}) = e^{- x} \sum_{j = 0}^{n} (j n) (- 1)^{n - j} \frac{d ^{j}}{d x ^{j}} x^{n + 1/2} .

After multiplication by

(- 1)^{n} x^{- 1/2} e^{x}

, the term

j = 0

is

x^{n}

, and every other term has lower degree. Thus

p_{n}

is a monic polynomial of degree

n

. Direct calculation gives

p_{0} (x) = 1, p_{1} (x) = x - \frac{3}{2}, p_{2} (x) = x^{2} - 5 x + \frac{15}{4} .

These are the monic normalization of the generalized Laguerre polynomials with parameter

1/2

.

Solved by gpt-5.6-sol high.

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