 Solution

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

x^{n + 1}

r_{j}

c_{j} = \frac{1}{\prod _{k \neq = j} ( y _{j} - y _{k} )} .

With the increasing enumeration, its sign is

(- 1)^{n + 2 - j}

. Therefore all terms

(- 1)^{j} c_{j}

have the same sign

(- 1)^{n + 2}

, and

R := \sum_{j} (- 1)^{j} c_{j} \neq = 0

. The leading coefficient of

t

T := \sum_{j} f (y_{j}) c_{j}

. Canceling the coefficient of

x^{n + 1}

gives the unique value

λ = \frac{T}{R} = \frac{\sum _{j} f ( y _{j} ) / \prod _{k \neq = j} ( y _{j} - y _{k} )}{\sum _{j} ( - 1 ) ^{j} / \prod _{k \neq = j} ( y _{j} - y _{k} )} .

Both interpolants have degree at most

n + 1

, so this single cancellation is equivalent to

de g (t - λ r) \leq n

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