Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 3 / b

There is a normalization conflict in the source: the printed Rodrigues' formula gives the standard Generalized Laguerre polynomial

P_{n} = L_{n}^{(m)}

with

P_{n} (0) = (n n + m)

, which is generally not one. The norm requested here corresponds to that Rodrigues normalization. A solution normalized to one at zero is

P_{n} = P_{n} / (n n + m)

w = x^{m} e^{- x}

and

p = x^{m + 1} e^{- x}

. The equation becomes

(p P_{n}^{'})^{'} + n w P_{n} = 0

. Multiply the equations for

P_{n}, P_{k}

by the opposite polynomial, subtract and integrate. The boundary term

p (P_{n}^{'} P_{k} - P_{k}^{'} P_{n})

vanishes at zero and infinity, so

(n - k) \int_{0}^{\infty} w P_{n} P_{k} d x = 0.

This proves orthogonality for

k \neq = n

under either normalization.

For the squared norm, use the printed Rodrigues' formula and integrate by parts

n

\int_{0}^{\infty} w P_{n}^{2} d x = \frac{1}{n !} \int_{0}^{\infty} P_{n} D^{n} (x^{n + m} e^{- x}) d x = \frac{( - 1 ) ^{n}}{n !} \int_{0}^{\infty} P_{n}^{(n)} x^{n + m} e^{- x} d x .

All boundary terms vanish: the required derivatives have a positive power of

x

(- 1)^{n} / n!

, so

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

\int_{0}^{\infty} x^{m} e^{- x} P_{n} (x)^{2} d x = \frac{( n + m )!}{n !} .

For the source's alternative unit-at-zero normalization the correct answer instead is

\int_{0}^{\infty} x^{m} e^{- x} P_{n} (x)^{2} d x = \frac{( n + m )!}{n ! ( n n + m ) ^{2}} .

Both cannot equal the printed norm when

m, n > 0

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