 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/30a/a/ii/solution

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

For every

N

f (ξ) = \sum_{n = 0}^{N} f_{n} (ξ - x_{0})^{n} + r_{N} (ξ), r_{N} (ξ) = o ((ξ - x_{0})^{N}) .

Integrating the finite sum gives

\int_{x_{0}}^{x} f (ξ) d ξ = \sum_{n = 0}^{N} \frac{f _{n}}{n + 1} (x - x_{0})^{n + 1} + \int_{x_{0}}^{x} r_{N} (ξ) d ξ .

For every

ε > 0

, sufficiently close to

x_{0}

one has

∣ r_{N} (ξ) ∣ \leq ε ∣ ξ - x_{0} ∣^{N}

, so the last integral is

o (∣ x - x_{0} ∣^{N + 1})

. This proves the integrated expansion.

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