 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2009/a/2/d/solution

The best approximation in the

L^{2}

span {ϕ_{0}, ϕ_{1}}

. Therefore

p_{1} = \frac{( x ^{k} , ϕ _{0} )}{∥ ϕ _{0} ∥ _{2}^{2}} ϕ_{0} + \frac{( x ^{k} , ϕ _{1} )}{∥ ϕ _{1} ∥ _{2}^{2}} ϕ_{1} .

Using

\int_{0}^{1} x^{k} d x = 1/ (k + 1)

(x^{k}, ϕ_{1}) = \frac{1}{k + 2} - \frac{1}{2 ( k + 1 )} = \frac{k}{2 ( k + 1 ) ( k + 2 )} .

p_{1} (x) = \frac{2 ( 1 - k ) + 6 k x}{( k + 1 ) ( k + 2 )} .

The error is orthogonal to every linear polynomial, so the Pythagorean identity proves optimality and uniqueness. At

k = 0

, this gives

p_{1} = 1

; at

k = 1

, it gives

p_{1} = x

. Both are exact because the target itself then belongs to

P_{1}

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