 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/32d/a/solution

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

Integration by parts gives the recurrence

γ (x, y) = y^{x - 1} e^{- y} + (x - 1) γ (x - 1, y) .

Iterating

N

times yields

γ (x, y) = y^{x - 1} e^{- y} \sum_{n = 0}^{N - 1} (x - 1) (x - 2) \dots (x - n) y^{- n} + R_{N},

where the empty product for

n = 0

is one and

R_{N} = (x - 1) (x - 2) \dots (x - N) γ (x - N, y) .

For fixed

x

, the remainder has the order of the first omitted term as

y \to \infty

. Therefore

a_{0} (x) = 1, a_{n} (x) = j = 1 \prod n (x - j) (n \geq 1) .

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