---
title: "Rodrigues' formula for Legendre polynomial"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:rodrigues-formula-legendre-polynomial"
source: https://www.bctengineer.com/q/semester_2/mathematics/rodrigues-formula-legendre-polynomial
site: BCT Engineering
---

# Rodrigues' formula for Legendre polynomial

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2067 Mangsir, 5 marks
- Video: [Rodrigue's formula explained simple and easy way](https://www.youtube.com/watch?v=sJXcrLFx81s) — EASY MATHS EASY TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/rodrigues-formula-legendre-polynomial
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

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

## Short answer

Let v=(x²−1)ⁿ. Differentiating (x²−1)v′-2nx v=0 n+1 times shows v^{(n)} satisfies Legendre's equation. At x=1, v^{(n)}(1)=2ⁿ n!, so Pₙ(x)=(1/(2ⁿ n!))dⁿ/dxⁿ(x²−1)ⁿ.

## Answer

**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:

```text
(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:

```text
(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 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The proof of Rodrigues' formula consists of:
   A. Show w = dⁿ/dxⁿ (x^2−1)ⁿ satisfies Legendre's equation, then fix the constant
   B. Expanding (x^2−1)ⁿ by binomial theorem only
   C. Differentiating Pₙ n times
   D. Using orthogonality of Pₙ
   Answer: A. w is a polynomial solution of Legendre's equation, so w = c Pₙ. (Key point: Strategy)

2. For v = (x^2 − 1)ⁿ, which relation is differentiated (n + 1) times by Leibniz?
   A. (x^2 − 1)v' + 2nx v = 0
   B. v' = 2nx v
   C. (x^2 − 1)v'' = 2n v
   D. (x^2 − 1)v' − 2nx v = 0
   Answer: D. Since v' = 2nx(x^2−1)^(n−1), multiplying by (x^2−1) gives that relation. (Key point: Starting relation)

3. Why is w a constant multiple of Pₙ?
   A. Because w has degree 2n
   B. Because w(0) = 0
   C. It is a polynomial solution of Legendre's equation, and the only one is a multiple of Pₙ
   D. Because Legendre's equation has one solution
   Answer: C. The only polynomial solution of Legendre's equation of order n is cPₙ. (Key point: Why w = cPₙ)

4. Using Pₙ(1) = 1, the constant c equals:
   A. n!
   B. 2ⁿ n!
   C. 2ⁿ
   D. 1/(2ⁿ n!)
   Answer: B. At x = 1 only one term survives: w(1) = n!·2ⁿ. (Key point: The constant)

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