 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/p/p3/d/solution

By part (c), the best polynomial of degree at most

2

is odd, hence has the form

p_{2} (x) = c x

. The orthogonal projection condition with

q = x

gives

c = \frac{( x ^{2011} , x )}{( x , x )} = \frac{\int _{- 1}^{1} x ^{2014} d x}{\int _{- 1}^{1} x ^{4} d x} = \frac{2/2015}{2/5} = \frac{1}{403} .

The residual is odd, so its weighted inner products with the even polynomials

1

and

x^{2}

vanish automatically. Thus all three normal equations hold and

p_{2} (x) = \frac{x}{403} .

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