Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 109 / 3 / a

Solution

Let

D (a)

denote the stated constant term and suppose every

a_{i} > 0

. The rational identity

\sum_{k = 1}^{n} \prod_{j \neq = k} (1 - \frac{X _{k}}{X _{j}})^{- 1} = 1

can be verified after clearing denominators, or by Lagrange interpolation. Multiplying it by the Dyson product and taking constant terms gives the recursion

D (a_{1}, \dots, a_{n}) = \sum_{k = 1}^{n} D (a_{1}, \dots, a_{k} - 1, \dots, a_{n}) .

M (a) = \frac{( a _{1} + \dots + a _{n} )!}{a _{1} ! \dots a _{n} !}

obeys the same recursion by the multinomial form of Pascal's identity.

a_{k} = 0

, taking the constant term in

X_{k}

forces the zero term from every factor involving

X_{k}

and reduces the expression to the

(n - 1)

-variable Dyson product with

a_{k}

omitted. The same boundary reduction holds for

M (a)

. Finally

D (0, \dots, 0) = M (0, \dots, 0) = 1

. Induction on

n

and on

a_{1} + \dots + a_{n}

D (a) = M (a) .

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