 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3/12f/iii/b/solution

p = t - λ r

using part (a). At every node,

f (y_{j}) - p (y_{j}) = λ (- 1)^{j}, ∥ f - p ∥_{L^{\infty} (Y)} = ∣ λ ∣.

λ = 0

, the error vanishes. Otherwise any degree-at-most-

n

q

with strictly smaller error would make

q - p

have alternating nonzero signs at the increasingly ordered nodes (reversed globally if

λ < 0

). It would then have at least

n + 1

distinct roots by the intermediate value theorem, impossible. Hence

p = t - λ r minimizes ∥ f - q ∥_{L^{\infty} (Y)} over de g q \leq n .

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