Rodrigues' formula for Legendre polynomial

Rodrigues' formula gives the Legendre polynomial in closed form. The proof shows that the right side satisfies Legendre's equation, and then fixes the constant by Pₙ(1) = 1.

Step 1: Legendre's equation

Let v = (x² − 1)ⁿ. Then v′ = 2nx(x² − 1)ⁿ⁻¹, so

(x² − 1)v′ − 2nx v = 0.

Differentiate (n + 1) times using Leibniz's theorem, writing v⁽ᵏ⁾ for the kth derivative:

(x²−1)v⁽ⁿ⁺²⁾ + (n+1)(2x)v⁽ⁿ⁺¹⁾ + [(n+1)n/2](2)v⁽ⁿ⁾
   − 2n[x v⁽ⁿ⁺¹⁾ + (n+1)v⁽ⁿ⁾] = 0

Put w = v⁽ⁿ⁾ = dⁿ/dxⁿ (x² − 1)ⁿ:

(x²−1)w″ + 2x w′ + n(n+1)w − 2n(n+1)w = 0
(1−x²)w″ − 2x w′ + n(n+1)w = 0

So w is a solution of Legendre's equation. It is a polynomial of degree n, and the only polynomial solution is a constant multiple of Pₙ. Hence w = c Pₙ(x).

Step 2: Find the constant c

Write (x² − 1)ⁿ = (x − 1)ⁿ(x + 1)ⁿ and differentiate n times at x = 1 by Leibniz's theorem. Every term still containing a factor (x − 1) vanishes at x = 1; only the term where all n derivatives hit (x − 1)ⁿ survives:

w(1) = n! · (x + 1)ⁿ|ₓ₌₁ = n! · 2ⁿ.

Since Pₙ(1) = 1, c = w(1) = 2ⁿ n!. Thus

dⁿ/dxⁿ (x² − 1)ⁿ = 2ⁿ n! Pₙ(x).

Hence Pₙ(x) = (1/(2ⁿ n!)) dⁿ/dxⁿ (x² − 1)ⁿ.

Check, n = 2: (1/8) d²/dx² (x⁴ − 2x² + 1) = (12x² − 4)/8 = (3x² − 1)/2 = P₂ ✓.