 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-1/a1/a/v/solution

Project the expansion onto one normalized basis function. Orthogonality removes every other coefficient:

\int_{0}^{a} Ψ_{m} (x)^{*} f (x) d x = n = 1 \sum \infty a_{n} \int_{0}^{a} Ψ_{m} (x)^{*} Ψ_{n} (x) d x = n = 1 \sum \infty a_{n} δ_{mn} = a_{m} .

Consequently,

a_{n} = ⟨ Ψ_{n} ∣ f ⟩ = \frac{2}{a} \int_{0}^{a} f (x) sin \frac{nπ x}{a} d x .

This is a Fourier sine series in normalized form. For a square-integrable

f

, completeness gives convergence in the norm of the function space; pointwise convergence requires the usual additional regularity. If

f

∣ a_{n} ∣^{2}

and

\sum_{n} ∣ a_{n} ∣^{2} = 1

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