Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 2 / b

Solution

Suppose for contradiction that some

q \in P_{n}

satisfies

∥ f - q ∥_{\infty} < a_{*} := min_{i} a_{i} .

t_{i}

p (t_{i}) - q (t_{i}) = [f (t_{i}) - q (t_{i})] - (- 1)^{i} a_{i} .

Because

∣ f (t_{i}) - q (t_{i}) ∣ < a_{i}

, the values

p (t_{i}) - q (t_{i})

have strictly alternating signs. The intermediate value theorem therefore gives at least one root of

p - q

in each of the

n + 1

intervals

(t_{i}, t_{i + 1})

. A nonzero polynomial of degree at most

n

cannot have

n + 1

distinct roots. If

p - q

were identically zero, its error at

t_{i}

would be

a_{i} \geq a_{*}

, also a contradiction. Hence

E_{n} (f) \geq 1 \leq i \leq n + 2 min a_{i} .

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