 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/32c/b/solution

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

Since

\frac{x}{x - 1} = \frac{1}{1 - x ^{- 1}} \sim \sum_{j = 0}^{\infty} x^{- j},

we have

f (x) \sim x^{- 1} + n = 1 \sum \infty x^{- n + 1} e^{- x} = n = 0 \sum \infty ϕ_{n} (x) .

Thus

a_{n} = 1

for every

n \geq 0

Explicitly,

\frac{f}{ϕ _{0}} = 1 + \frac{x ^{2} e ^{- x}}{x - 1} ⟶ 1.

For

n \geq 1

, the remainder after terms

0, \dots, n - 1

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

Its ratio to

ϕ_{n}

tends to one, verifying every coefficient formula.

Because

e^{- x} = o (x^{- N})

for every

N

, the exponentially small part is invisible to the power sequence

ψ_{n} = x^{- n}

. Hence

f (x) \sim x^{- 1}

with coefficient one at

ψ_{1}

and all other power coefficients zero.

 New to topics? Read the docs here!