 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/7d/a/solution

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

For any

n

with

Re z > - n

F (z) = \frac{F ( z + n )}{z ( z + 1 ) \dots ( z + n - 1 )} .

The definitions agree on overlaps by the recurrence, giving a meromorphic continuation to

C

. Its only singularities are simple poles at

0, - 1, - 2, \dots

. Since

F (1) = 1

, repeated use of the recurrence gives

res_{z = - n} F (z) = \frac{( - 1 ) ^{n}}{n !} .

These residues are nonzero, so all the listed singularities really are simple poles.

 New to topics? Read the docs here!