 Solution

ID: past-exam-of-the-analysis-course-of-the-university-of-oxford/2009/b/4/i/solution

(e_{n})

has

⟨ e_{n}, e_{m} ⟩ = δ_{nm}

and dense linear span; equivalently, the only vector orthogonal to every

e_{n}

is zero. The Parseval identity says

∥ x ∥^{2} = \sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2}

, with expansion

x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}

in the Hilbert norm.

Put

G (t, z) = (1 - 2 z t + z^{2})^{- 1/2}

and

D = 1 - 2 z t + z^{2}

G_{t} = z D^{- 3/2}, G_{tt} = 3 z^{2} D^{- 5/2}, G_{z} = (t - z) D^{- 3/2}, G_{zz} = - D^{- 3/2} + 3 (t - z)^{2} D^{- 5/2} .

Substitution yields

(1 - t^{2}) G_{tt} - 2 t G_{t} + z^{2} G_{zz} + 2 z G_{z} = 0

. Extracting the coefficient of

z^{n}

(1 - t^{2}) P_{n}^{''} - 2 t P_{n}^{'} + n (n + 1) P_{n} = 0,

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