Laguerre polynomial

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Laguerre derivative square identity ( $\int_{0}^{\infty} t (L_{n}^{'} - L_{n}) (L_{m}^{'} - L_{m}) e^{- t} = (n + 1) \int_{0}^{\infty} L_{n} L_{m} e^{- t}$ )

For polynomial solutions of the Laguerre differential equation, put

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

. The equation gives

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

. Integrating

\int t D_{n} (L_{m}^{'} - L_{m}) e^{- t}

by parts gives the displayed identity; the factor

t

kills the endpoint at zero and exponential decay kills the endpoint at infinity. Its left side is symmetric in

n, m

, so subtracting the swapped identities proves orthogonality for distinct degrees. This provides a direct energy proof of orthogonality for the Laguerre polynomials without assuming their norm formula.

Laguerre orthonormality ( $\int_{0}^{\infty} L_{n} (x) L_{m} (x) e^{- x} d x = δ_{nm}$ )

 0  0

The ordinary Laguerre polynomials, normalized by

L_{n} (0) = 1

, are orthonormal for the weighted inner product displayed above. Their coefficient formula is

L_{n} (x) = \sum_{k = 0}^{n} (- 1)^{k} (k n) x^{k} / k!

. To prove the relation, use Rodrigues' formula

e^{- x} L_{n} = (n!)^{- 1} (x^{n} e^{- x})^{(n)}

and integrate by parts

n

times against

L_{m}

. For

n > m

its

n

th derivative vanishes. At zero every derivative of

x^{n} e^{- x}

of order less than

n

vanishes, and at infinity the exponential kills every polynomial boundary term. For

n = m

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

, so the integral is

(n!)^{- 1} \int_{0}^{\infty} x^{n} e^{- x} d x = 1

. Symmetry covers

m > n

. Thus these are orthonormal without an extra normalization factor.

Generating function for Laguerre polynomials ( $\sum_{n = 0}^{\infty} L_{n} (x) h^{n} = (1 - h)^{- 1} e^{- x h / (1 - h)}$ )

 0  0

The generating function holds for

∣ h ∣ < 1

and fixed

x

. From Rodrigues' formula,

L_{n} (x) = \sum_{j = 0}^{n} (j n) (- x)^{j} / j!

. Sum first over

n \geq j

, using

\sum_{n \geq j} (j n) h^{n} = h^{j} / (1 - h)^{j + 1}

, and then over

j

to obtain the exponential function. The absolute-value double series is finite for

∣ h ∣ < 1

, justifying the rearrangement. Alternatively, the weighted inner product and the Laplace transform of a Laguerre polynomial extract the coefficient

h^{n}

Laplace transform of a Laguerre polynomial ( $L [L_{n}] (p) = (p - 1)^{n} / p^{n + 1}$ )

 0  0

For the ordinary Laguerre polynomial

L_{n} (t) = \sum_{j = 0}^{n} (j n) (- t)^{j} / j!

, apply the Laplace transform to each term, using

L [t^{j}] (p) = j! / p^{j + 1}

for

Re p > 0

. The binomial theorem gives

L [L_{n}] (p) = \frac{1}{p} \sum_{j = 0}^{n} (j n) (- p^{- 1})^{j} = \frac{1}{p} (1 - p^{- 1})^{n} .

(1)

This proves the formula for every nonnegative integer

n

. In particular

L_{2} (t) = 1 - 2 t + t^{2} /2

has transform

(p - 1)^{2} / p^{3}

Deriving the Laguerre equation from Rodrigues' formula ( $x L_{n}^{''} + (1 - x) L_{n}^{'} + n L_{n} = 0$ )

 0  0

Set

v = x^{n} e^{- x}

. Direct differentiation gives

x v^{'} = (n - x) v

. Apply the Leibniz rule to differentiate

n + 1

times:

x v^{(n + 2)} + (n + 1) v^{(n + 1)} = (n - x) v^{(n + 1)} - (n + 1) v^{(n)} .

(1)

Hence

x v^{(n + 2)} + (x + 1) v^{(n + 1)} + (n + 1) v^{(n)} = 0

. By Rodrigues' formula,

v^{(n)} = n! e^{- x} L_{n}

, so the next two derivatives are

n! e^{- x} (L_{n}^{'} - L_{n})

and

n! e^{- x} (L_{n}^{''} - 2 L_{n}^{'} + L_{n})

. Substitution gives the displayed Laguerre differential operator on polynomials eigenvalue equation. This derivation also covers

n = 0

Normalization of the ordinary Laguerre polynomials ( $\int_{0}^{\infty} e^{- x} L_{n} (x)^{2} d x = 1$ )

 0  0

The Laguerre polynomial normalized by

L_{n} (0) = 1

has coefficients

L_{n} (x) = \sum_{j = 0}^{n} (- 1)^{j} (j n) x^{j} / j!

. By the Leibniz rule, this is equivalent to Rodrigues' formula,

e^{- x} L_{n} (x) = D^{n} (e^{- x} x^{n}) / n!

. For any polynomial

q

of degree below

n

n

applications of integration by parts give

\int_{0}^{\infty} e^{- x} L_{n} q d x = 0

. All endpoint terms vanish: derivatives of

e^{- x} x^{n}

of order below

n

vanish at zero and decay at infinity. Since

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

, the same calculation gives

\int_{0}^{\infty} e^{- x} L_{n}^{2} d x = \frac{( - 1 ) ^{n}}{n !} \int_{0}^{\infty} e^{- x} x^{n} L_{n}^{(n)} d x = \frac{Γ ( n + 1 )}{n !} = 1.

(1)

Thus these orthogonal polynomials are already an orthonormal set for the weighted inner product with weight

e^{- x}

Generalized Laguerre polynomial

 0  0

The generalized Laguerre polynomial

L_{n}^{(α)}

is orthogonal on

[0, \infty)

for the weight

x^{α} e^{- x}

when

α > - 1

Normalization of generalized Laguerre polynomials ( $L_{n}^{(m)} (0) = (n n + m), ∥ L_{n}^{(m)} ∥_{x^{m} e^{- x}}^{2} = \frac{( n + m )!}{n !}$ )

 0  0

For integers

m, n \geq 0

, the standard Rodrigues' formula

L_{n}^{(m)} = e^{x} x^{- m} D^{n} (x^{n + m} e^{- x}) / n!

has the endpoint value and squared norm displayed above. The lowest-power term after

n

differentiations is

(n + m)! x^{m} / m!

, giving the endpoint value after division by

n! x^{m}

. For the norm, integrate by parts

n

times against the same polynomial; its leading coefficient is

(- 1)^{n} / n!

, so the final derivative is

(- 1)^{n}

. All boundary terms vanish and the remaining Gamma function integral is

(n + m)! / n!

. Normalizing the value at zero to one divides the polynomial by

(n n + m)

and its squared norm by the square of that factor. Only when

m = 0

n = 0

do these normalizations already coincide.

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