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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-2/10g/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 10G / i / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows

⟨ g, f ⟩ = \overline{⟨ f, g ⟩}

. Thus it is a Hermitian form.

f (z) = \sum_{j = 0}^{N} a_{j} z^{j}

, Fourier orthogonality gives

⟨ f, f ⟩ = \sum_{j, k} a_{j} \overline{a_{k}} \frac{1}{2 π} \int_{0}^{2 π} e^{i (j - k) θ} d θ = \sum_{j = 0}^{N} ∣ a_{j} ∣^{2} .

(1)

This is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

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