The Laguerre polynomials L_{n}(x) are defined through

L_{n}(x) = e^{x} · (\frac{d}{dx})^{n} (x^{n}e^{−x})

Theorem:

L_{n}(x) satisfies the recurrence relation

L_{0}(x) = 1

L_{n+1}(x) = (2n + 1 − x) · L_{n}(x) − n^{2} · L_{n−1}(x)

for n ≥ 0 and the differential equation x·Ln^{′′}(x)+(1−x)·Ln^{′}(x)+n·L_{n}(x) = 0 for all n ≥ 0.

Proof:

Let F := ∑_{n=0}^{∞} \frac{L_{n}(x)}{n!} · z^{n} be the exponential generating function of the sequence of polynomials. It is the diagonal series of the power series

G := ∑_{m,n=0}^{∞} \frac{1}{m!} · e^{x} · (\frac{d}{dx})^{m} (x^{n}e^{−x}) · y^{m} · z^{n}

Because the Taylor series development theorem holds in formal power series rings (see [1], section 2.16), we can simplify

G = e^{x} · ∑_{n=0}^{∞} ( ∑_{m=0}^{∞} \frac{1}{m!} · (\frac{d}{dx})^{m} (x^{n}e^{−x}) · y^{m}) · z^{n}
= e^{x} · ∑_{n=0}^{∞}(x + y)^{n}e^{−(x+y)} · z^{n}
= \frac{e^{−y}}{1 − (x + y)z}

We take over the terminology from the “diag rational” paper; here R = Q[x] and M = Q[[x]] (or, if you like it better, M = H(C), the algebra of functions holomorphic in the entire complex plane). G ∈ M[[y, z]] is not rational; nevertheless we can proceed similarly to the “diag series” paper. F (z^{2}) is the coefficient of t^{0} in

G(zt, \frac{z}{t}) = \frac{e^{−zt}}{1 − z^{2} − \frac{xz}{t}} ∈ M[[zt, \frac{z}{t} , z]] = M⟨⟨z, t⟩⟩

The denominator’s only zero is t = \frac{xz}{1−z^{2}} . We can write

e^{−zt} = e^{− \frac{xz^{2}}{1−z^{2}}} + (zt − \frac{xz^{2}}{1 − z^{2}} ) · P (z, t)

with P (z, t) ∈ Q[[zt, \frac{xz^{2}}{1−z^{2}} ]] ⊂ Q[[zt, x, z]] = M[[zt, z]] ⊂ M⟨⟨z, t⟩⟩. This yields – all computations being done in M⟨⟨z, t⟩⟩ –

G(zt, \frac{z}{t}) = \frac{e^{− \frac{xz^{2}}{1−z^{2}}}}{1 − z^{2} − \frac{xz}{t}} + \frac{zt}{1 − z^{2}} · P (z, t)
= \frac{1}{1 − z^{2}} · e^{− \frac{xz^{2}}{1−z^{2}}} · ∑_{j=0}^{∞} (\frac{x}{1 − z^{2}} \frac{z}{t} )^{j} + \frac{zt}{1 − z^{2}} · P (z, t)

Here, the coefficient of t^{0} is

F (z^{2}) = \frac{1}{1 − z^{2}} · e^{− \frac{xz^{2}}{1−z^{2}}}

hence

F (z) = \frac{1}{1 − z} · e^{− \frac{xz}{1−z}}

It follows that (1 − z)^{2} · \frac{d}{dz}F − (1 − x − z) · F = 0. This is equivalent to the claimed recurrence.

Starting from the closed form for F, we compute a linear relation for the partial derivatives of F. Write ∂_{x} = \frac{d}{dx} and Δ_{z} = z \frac{d}{dz}. One computes

F = 1 · F
(1 − z) · ∂_{x}F = −z · F
(1 − z)^{2} · ∂_{x}^{2}F = z^{2} · F
(1 − z)^{2} · Δ_{z}F = ((1 − x)z − z^{2}) · F
(1 − z)^{3} · ∂_{x}Δ_{z}F = (−z + xz^{2} + z^{3}) · F

Solve a homogeneous 4 × 5 system of linear equations over Q(x) to get

(1 − z)^{3} · ((1 − x) · ∂_{x}F + x · ∂_{x}^{2}F + Δ_{z}F ) = 0

Divide by the first factor to get

(1 − x) · ∂_{x}F + x · ∂_{x}^{2}F + Δ_{z}F = 0

This is equivalent to the claimed equation x·Ln^{′′}(x)+(1−x)·Ln^{′}(x)+n·L_{n}(x) = 0.

References

1 Bruno Haible: D-finite power series in several variables. Diploma thesis, University of Karlsruhe, June 1989. Sections 2.15 and 2.22.