Series solution of Legendre's equation

Legendre's equation: (1 − x²)y″ − 2xy′ + n(n + 1)y = 0. x = 0 is an ordinary point, so a power-series solution y = Σ aₖxᵏ exists.

Substitution

y′ = Σ k aₖ x^(k−1), y″ = Σ k(k−1) aₖ x^(k−2).

Σ k(k−1)aₖ x^(k−2) − Σ k(k−1)aₖ xᵏ − 2Σ k aₖ xᵏ + n(n+1)Σ aₖ xᵏ = 0

Shift the first sum (k → k + 2) and collect the coefficient of xᵏ:

(k + 2)(k + 1)aₖ₊₂ − [k(k − 1) + 2k − n(n + 1)]aₖ = 0.

Since k(k − 1) + 2k − n(n + 1) = k(k + 1) − n(n + 1) = (k − n)(k + n + 1),

aₖ₊₂ = − (n − k)(n + k + 1) / ((k + 1)(k + 2)) · aₖ

Coefficients

a₀ and a₁ are arbitrary.

a₂ = − n(n+1)/2! · a₀
a₃ = − (n−1)(n+2)/3! · a₁
a₄ = (n−2)n(n+1)(n+3)/4! · a₀
a₅ = (n−3)(n−1)(n+2)(n+4)/5! · a₁

Solution

y = a₀y₁(x) + a₁y₂(x), where

y₁ = 1 − n(n+1)/2! x² + (n−2)n(n+1)(n+3)/4! x⁴ − …
y₂ = x − (n−1)(n+2)/3! x³ + (n−3)(n−1)(n+2)(n+4)/5! x⁵ − …

Both series converge for |x| < 1.

When n is a non-negative integer, one of the series terminates (y₁ for even n, y₂ for odd n), giving a polynomial, the Legendre polynomial Pₙ(x) after suitable choice of the constant. For example n = 2 gives y₁ = 1 − 3x², proportional to P₂ = (3x² − 1)/2.