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.

ii

Solution

Suppose

\sum_{j = 0}^{N} c_{j} ϕ_{j} (x) = 0

throughout the punctured neighbourhood, and let

k

be the least index with

c_{k} \neq = 0

. Part (i) implies

ϕ_{j} = o (ϕ_{k})

for every

j > k

. Dividing by

ϕ_{k}

and taking

x \to x_{0}

therefore gives

0 = c_{k} + \sum_{j = k + 1}^{N} c_{j} \frac{ϕ _{j}}{ϕ _{k}} ⟶ c_{k},

a contradiction. Hence every

c_{j}

vanishes and

ϕ_{0}, \dots, ϕ_{N}

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