Rodrigues’ formula for the Legendre polynomials

notes
calculus
Author

Kisung You

Published

August 10, 2022

Introduction

Legendre polynomials \(P_n (x)\) are solutions of the following Legendre’s differential equation

\[ (1-x^2) y'' - 2x y' + n(n+1) y = 0 \tag{1}\]

for some \(n \in \mathbb{N}\cup \{0\}\). An explicit, compact expression for the polynomials is provided by Rodrigues’ formula

\[ P_n (x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n. \tag{2}\]

This means that when \(P_n (x)\) is plugged in the position of \(y\) for Equation 1, it must satisfy the equality to 0. In this post, we show (a bit tedious) derivations to attain Equation 2.

Approach

I will proceed in two steps. Let \(f_n (x) = (x^2 - 1)^n\) then we first show that the \(n\)-th derivative of \(f_n (x)\) is a solution of Legendre equation. Then, we find a proper scaling factor of \(1/ 2^n n!\) to recover \(P_n (x)\) in line with a common constraint that \(P_n (x) = 1\) for all \(n\) when \(x=1\). For notational simplicity, we denote \(g^{(n)}\) for the \(n\)-th derivative of a function \(g(x)\), i.e,

\[ g^{(n)} = \frac{d^n}{dx^n} g(x). \]

Before proceeding, we need (generalized) Leibniz’s rule. Suppose we have \(n\)-times differentiable functions \(f(x)\) and \(g(x)\), then

\[ \frac{d^n}{dx^n} f(x) g(x) = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} f^{(n-k)} (x) g^{(k)} (x) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} f^{(n-k)} (x) g^{(k)} (x) \tag{3}\]

where the choice of \(f\) and \(g\) can help in reducing the number of terms when there exists a polynomial term. For example, when \(g(x) = x^2\), \(g^{(k)} = 0\) for all \(k \geq 3\).

Step 1. \(f_n^{(n)} (x)\) is one solution.

Our goal here is to show that \(f_n^{(n)}(x)\) is a solution for Equation 1. As a first step, let’s take derivative on \(f_n (x)\),

\[ \begin{align*} \frac{d}{dx} f_n (x) &= 2 n (x^2 - 1)^{n-1} x \\ &= 2nx (x^2 - 1)^{n-1}. \end{align*} \]

By multiplying \((x^2 - 1)\) both side of the above, we have

\[ (x^2 - 1) \frac{d}{dx} f_n (x) = 2nx (x^2 - 1)^n. \]

Now, differentiate both sides \((n+1)\) times, which leads to

\[ \begin{align*} \frac{d^{n+1}}{dx^{n+1}} \left[ \frac{d}{dx} f_n (x) \right] (x^2 -1) &= \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} \left( \frac{d}{dx} f_n (x) \right)^{(n+1-k)} (x^2-1)^{(k)} \\ &= \begin{pmatrix} n+1 \\ 0 \end{pmatrix} f_n^{(n+2)} (x) (x^2-1) + \begin{pmatrix} n+1 \\ 1 \end{pmatrix} 2x f_n^{(n+1)} (x) + \begin{pmatrix} n+1 \\ 2 \end{pmatrix} f_n ^{(n)} (x) \cdot 2 \\ &= (x^2-1) f_n^{(n+2)} (x) + 2(n+1)x f_n^{(n+1)} (x) + n(n+1) f_n^{(n)} (x) \end{align*} \]

for the left-hand side. We also have that

\[ \begin{align*} \frac{d^{n+1}}{dx^{n+1}} f_n(x) 2nx &= \begin{pmatrix} n+1 \\ 0 \end{pmatrix} f_n^{(n+1)} (x) 2nx + \begin{pmatrix} n+1 \\ 1 \end{pmatrix} f_n^{(n)} (x) 2n \\ &= 2nx f_n^{(n+1)} (x) + 2n(n+1)f_n^{(n)}(x). \end{align*} \]

Therefore, we have the following arrangement,

\[ \begin{equation*} \begin{gathered} (x^2 - 1) f_n^{(n+2)} (x) + 2x(n+1)f_n^{(n+1)} (x) + n(n+1)f_n^{(n)} (x) = 2nx f_n^{(n+1)} (x) + 2n(n+1) f_n^{(n)} (x) \\ (x^2-1) f_n^{(n+2)} (x) + 2x f_n^{(n+1)}(x) - n(n+1)f_n^{(n)} (x) = 0 \\ (1- x^2) f_n^{(n+2)} (x) - 2x f_n^{(n+1)}(x) + n(n+1)f_n^{(n)} (x) = 0 \end{gathered} \end{equation*} \]

where the last line is in the form of Equation 1 so that we have \(f_n^{(n)} (x)\) as a solution.

Step 2. Find a scaling factor.

Even though \(f_n^{(n)} (x)\) as a solution, we have a requirement for the standard Legendre polynomial that \(P_n (x)=1\) for \(x=1\). Let us take a closer look at \(f_n^{(n)}(x)\) when evaluated at \(x=1\).

\[ \begin{align*} f_n^{(n)} (x) &= \frac{d^n}{dx^n} (x^2-1)^n \\ &= \frac{d^n}{dx^n} (x+1)^n (x-1)^n \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \left( (x+1)^n \right)^{(k)} \left( (x-1)^n \right) ^{(n-k)} \quad \textrm{by the Leibniz's rule} \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac{n!}{(n-k)!} (x+1)^{n-k} \frac{n!}{k!} (x-1)^k \qquad (*) \\ &= n! \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac{n!}{(n-k)! k!} (x+1)^{n-k} (x-1)^k \\ &= n! \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix}^2 (x+1)^{n-k} (x-1)^k, \end{align*} \]

Since we want to evaluate \(f_n^{(n)} (x)\) at \(x=1\), the last line of equations above tells us that all the terms but \(k=0\) become zero,

\[ \begin{equation*} f_n^{(n)} (x=1) = n! \begin{pmatrix} n \\ 0 \end{pmatrix}^2 2^{n-0} = n! 2^n \end{equation*} \]

which finally leads to define \(P_n (x)\) as

\[ P_n (x) = \frac{1}{n! 2^n}f_n^{(n)}(x) = \frac{1}{n! 2^n} \frac{d^n}{dx^n} (x^2-1)^n \]

to fulfill the condition of \(P_n (x) =1\) for \(x=1\).

Citation

BibTeX citation:
@online{you2022,
  author = {Kisung You},
  title = {Rodrigues’ Formula for the {Legendre} Polynomials},
  date = {2022-08-10},
  url = {https://kisungyou.com/posts/note002-rodrigues-formula},
  langid = {en}
}
For attribution, please cite this work as:
Kisung You. 2022. “Rodrigues’ Formula for the Legendre Polynomials.” August 10, 2022. https://kisungyou.com/posts/note002-rodrigues-formula.