 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/31d/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For fixed

n

r_{n} = f - \sum_{j = 0}^{n} a_{j} ϕ_{j}, r_{n + 1} = f - \sum_{j = 0}^{n + 1} a_{j} ϕ_{j} .

r_{n} = o (ϕ_{n})

and

r_{n + 1} = o (ϕ_{n + 1})

, while subtraction gives

r_{n} = a_{n + 1} ϕ_{n + 1} + r_{n + 1} .

Because

a_{n + 1} \neq = 0

\frac{r _{n}}{ϕ _{n + 1}} ⟶ a_{n + 1} .

For

x

sufficiently close to

x_{0}

this denominator is therefore nonzero, and

\frac{ϕ _{n + 1}}{ϕ _{n}} = \frac{r _{n} / ϕ _{n}}{r _{n} / ϕ _{n + 1}} ⟶ \frac{0}{a _{n + 1}} = 0.

Thus $ϕ_{n + 1} = o (ϕ_{n})$ for every $n$ , so $(ϕ_{n})$ is an asymptotic sequence.

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