 Solution

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

P_{n} (- t) = (- 1)^{n} P_{n} (t)

f

is orthogonal to every even-degree Legendre polynomial. For

n \geq 1

, integrate the self-adjoint form of the Legendre differential equation from zero to one:

[(1 - t^{2}) P_{n}^{'} (t)]_{0}^{1} + n (n + 1) \int_{0}^{1} P_{n} (t) d t = 0.

Thus

\int_{0}^{1} P_{n} = P_{n}^{'} (0) / (n (n + 1))

. Differentiating the generating function in

t

and then setting

t = 0

gives

\sum_{n} P_{n}^{'} (0) z^{n} = z (1 + z^{2})^{- 3/2}, P_{2 m + 1}^{'} (0) = (- 1)^{m} \frac{( 2 m + 1 )!}{2 ^{2 m} ( m ! ) ^{2}} .

Combining these identities with the normalization of

p_{n}

yields

⟨ f, p_{2 m} ⟩ = 0, ⟨ f, p_{2 m + 1} ⟩ = (- 1)^{m} \frac{4 m + 3}{2} \frac{( 2 m )!}{2 ^{2 m} m ! ( m + 1 )!} .

∥ f ∥_{2}^{2} = 2

2 = \sum_{m \geq 0} \frac{4 m + 3}{2} \frac{(( 2 m )! ) ^{2}}{2 ^{4 m} ( m ! ( m + 1 )! ) ^{2}} .

Therefore the requested sum equals

4.

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