Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 2 / a

For the nodes

c_{1} = α

and

c_{2} = 1

ℓ_{1} (t) = \frac{1 - t}{1 - α}, ℓ_{2} (t) = \frac{t - α}{1 - α} .

Direct integration gives

a_{ij} = \int_{0}^{c_{i}} ℓ_{j} (t) d t

and therefore

A = \frac{α ( 2 - α )}{2 ( 1 - α )} \frac{1}{2 ( 1 - α )} - \frac{α ^{2}}{2 ( 1 - α )} \frac{1 - 2 α}{2 ( 1 - α )} .

b_{j} = \int_{0}^{1} ℓ_{j} (t) d t, b^{T} = (\frac{1}{2 ( 1 - α )}, \frac{1 - 2 α}{2 ( 1 - α )}) .

This exposes a sign error in the printed tableau: its lower-right entry is shown as

(- 1 + 2 α) / [2 (1 - α)]

. With

1 - 2 α

in that position, the tableau is exactly the claimed collocation method. Taken literally, the printed weights satisfy

b^{T} 1 = α / (1 - α)

, so the method is not even consistent unless

α = 1/2

and cannot be a collocation method for general

α

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