Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 30A / a

Solution

The statement

f (x) \sim \sum_{n = 0}^{\infty} f_{n} (x - x_{0})^{n}

x \to x_{0}

means that for every

N \geq 0

f (x) - \sum_{n = 0}^{N} f_{n} (x - x_{0})^{n} = o ((x - x_{0})^{N}) .

Write the remainders after order

N

r_{N} = o ((x - x_{0})^{N})

for

f

and

s_{N} = o ((x - x_{0})^{N})

for

g

. Then

f (x) + g (x) - \sum_{n = 0}^{N} (f_{n} + g_{n}) (x - x_{0})^{n} = r_{N} (x) + s_{N} (x) = o ((x - x_{0})^{N}) .

Hence $h = f + g$ has the termwise sum of the two asymptotic expansions.

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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