The Legendre polynomials P_{n}(x) are defined through

P_{n}(x) = \frac{1}{2^{n}n!} · (\frac{d}{dx})^{n} (x^{2} − 1)^{n}

(For a motivation of the 2^{n} in the denominator, look at P_{n}(x) modulo an odd prime p, and observe that P_{n}(x) ≡ P_{p−1−n}(x) mod p for 0 ≤ n ≤ p − 1. This wouldn’t hold if the 2^{n} factor in the denominator weren’t present.)

Theorem:

P_{n}(x) satisfies the recurrence relation

P_{0}(x) = 1

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

for n ≥ 0 and the differential equation (1 − x^{2}) · Pn^{′′}(x) − 2x · Pn^{′}(x) + (n^{2} + n) · P_{n}(x) = 0 for all n ≥ 0.

Proof:

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

G := ∑_{m,n=0}^{∞} \frac{1}{2^{n}m!} · (\frac{d}{dx})^{m} (x^{2} − 1)^{n} · 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 = ∑_{n=0}^{∞} \frac{1}{2^{n}} · ( ∑_{m=0}^{∞} \frac{1}{m!} · (\frac{d}{dx})^{m} (x^{2} − 1)^{n} · y^{m}) · z^{n}
= ∑_{n=0}^{∞} \frac{1}{2^{n}} · ((x + y)^{2} − 1)^{n} · z^{n}
= \frac{1}{1 − \frac{1}{2} ((x + y)^{2} − 1) 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 rational; hence F is algebraic over R[z]. Let’s proceed exactly as in the “diag series” paper. F (z^{2}) is the coefficient of t^{0} in

G(zt, \frac{z}{t}) = \frac{2t}{2t − ((x + zt)^{2} − 1) z} = \frac{2t}{−z^{3} · t^{2} + 2(1 − xz^{2}) · t + (z − x^{2}z)}

The splitting field of the denominator is L = Q(x)(z)(α) where

α_{1/2} = \frac{1 − xz^{2} ± \sqrt{1 − 2xz^{2} + z^{4}}}{z^{3}}

α = α_{1} = \frac{2}{z^{3}} − \frac{2x}{z} + \frac{1 − x^{2}}{2} z + ⋯ ∈ Q(x)[[z]][\frac{1}{z}]

α_{2} = \frac{x^{2} − 1}{2}z + ⋯ ∈ Q(x)[[z]][\frac{1}{z}]

The partial fraction decomposition of G(zt, \frac{z}{t}) reads

G(zt, \frac{z}{t}) = −\frac{2}{z^{3}} · \frac{1}{α_{1} − α_{2}} · (\frac{α_{1}}{t − α_{1}} − \frac{α_{2}}{t − α_{2}} )

It follows that

F (z^{2}) = −\frac{2}{z^{3}} · \frac{1}{α_{1} − α_{2}} · (\frac{α_{1}}{0 − α_{1}} − 0) = \frac{1}{\sqrt{1 − 2xz^{2} + z^{4}}}

hence

F (z) = \frac{1}{\sqrt{1 − 2xz + z^{2}}}

It follows that (1 − 2xz + z^{2}) · \frac{d}{dz}F + (z − x) · 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 − 2xz + z^{2}) · ∂_{x}F = z · F

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

(1 − 2xz + z^{2}) · Δ_{z}F = (xz − z^{2}) · F

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

(1 − 2xz + z^{2})^{2} · Δ_{z}^{2}F = (xz + (x^{2} − 2)z^{2} − xz^{3} + z^{4}) · F

Solve a homogeneous 5 × 6 system of linear equations over Q(x) to get

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

Divide by the first factor to get

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

This is equivalent to the claimed equation (1−x^{2})·Pn^{′′}(x)−2x·Pn^{′}(x)+(n^{2} + n) · P_{n}(x) = 0.

References

1 Bruno Haible: D-finite power series in several variables. Diploma thesis, University of Karlsruhe, June 1989. Sections 2.14, 2.15 and 2.22.