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}
}