Past exam of the analysis course of the University of Oxford / 2008 / B / 6

Solution

Write

L_{n} (t) = \sum_{k = 0}^{n} a_{k} t^{k}

(k + 1)^{2} a_{k + 1} + (n - k) a_{k} = 0.

Taking

a_{0} = 1

yields

L_{n} (t) = k = 0 \sum n (- 1)^{k} (k n) \frac{t ^{k}}{k !} .

The top coefficient is

(- 1)^{n} / n! \neq = 0

, so this is a real polynomial of degree exactly

n

Put

D_{n} = L_{n}^{'} - L_{n}

. Integration by parts, with vanishing endpoint terms because of the factor

t

\int_{0}^{\infty} t D_{n} D_{m} e^{- t} d t = - \int_{0}^{\infty} [(t D_{n} e^{- t})^{'} + t D_{n} e^{- t}] L_{m} d t = - \int_{0}^{\infty} [D_{n} + t D_{n}^{'}] L_{m} e^{- t} d t = (n + 1) \int_{0}^{\infty} L_{n} L_{m} e^{- t} d t .

The last step uses

D_{n} + t D_{n}^{'} = t L_{n}^{''} + (1 - t) L_{n}^{'} - L_{n} = - (n + 1) L_{n}

. The left side is symmetric in

m, n

. Interchanging them therefore gives

(n - m) \int L_{n} L_{m} e^{- t} = 0

, proving orthogonality for distinct indices. These integrals are exactly

⟨ φ_{n}, φ_{m} ⟩

, and each

φ_{n}

is square-integrable and nonzero. Thus they form an orthogonal sequence in

L^{2} (0, \infty)

Multiplying the equations by

z^{n}

and summing gives

t G_{tt} + (1 - t) G_{t} + z G_{z} = 0.

Conversely equality of power-series coefficients recovers every polynomial equation. For the proposed expression,

\frac{G _{t}}{G} = - \frac{z}{1 - z}, \frac{G _{tt}}{G} = \frac{z ^{2}}{( 1 - z ) ^{2}}, \frac{G _{z}}{G} = \frac{1}{1 - z} - \frac{t}{( 1 - z ) ^{2}} .

Substitution makes the PDE left side zero. Its coefficient of

z^{n}

has value one at

t = 0

, so it corresponds to the normalization

L_{n} (0) = 1

chosen above.

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