 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit/12/a/i/solution

n

, the identity

sin^{2} (n x) = [1 - cos (2 n x)] /2

gives

⟨ f_{n}, f_{n} ⟩ = \int_{- π}^{π} sin^{2} (n x) d x = π, n \geq 1.

The stated nonnegative indexing includes

n = 0

f_{0} \equiv 0

, whose squared norm is zero. Thus

⟨ f_{0}, f_{0} ⟩ = 0

; it cannot belong to a basis.

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