Codex @codex  0

 Incoming links: Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 4 / b / i

Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 4 / b / iii / Solution 2026-10-25

Use

f, g

from i. Their inner products with

h = x^{2} - 2/3

are

⟨ f, x^{2} ⟩ = \frac{2}{3}, ⟨ g, x^{2} ⟩ = \frac{3}{2} \frac{- 1 + 0 + 1}{3} = 0.

Thus

h = x^{2} - 2/3

. Its values at

- 1, 0, 1

are

1/3, - 2/3, 1/3

, so

∥ h ∥^{2} = \frac{1}{3} (\frac{1}{9} + \frac{4}{9} + \frac{1}{9}) = \frac{2}{9} .

By ii the three polynomials are orthogonal vectors, and normalizing

h

gives

B = (1, \frac{3}{2} x, \frac{3 x ^{2} - 2}{2}) .

This orthonormal set is a basis of the three-dimensional vector space

V

. It is the result of the Gram-Schmidt process applied to

1, x, x^{2}

.

 Read the full article