Rodrigues’ formula of the Legendre Polynomials

The Rodrigues’ formula for the Legendre polynomials is a mathematical expression that provides a method to calculate any Legendre polynomial using differentiation.

It is given by

$P_l(x)=\frac{1}{2^ll!}\frac{d^l}{dx^l}(x^2-1)^l\;\;\;\;\;\;l=0,1,2,\cdots\;\;\;\;\;\;348$

with the first few Legendre polynomials being $P_0(x)=1,P_1(x)=x,P_2(x)=\frac{1}{2}(3x^2-1)$ .

To derive eq348, let $y=(x^2-1)^l$ . The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}-l2x(x^2-1)^l(x^2-1)^{-1}=0$ or equivalently,

$(x^2-1)\frac{d}{dx}(x^2-1)^{l}=l2x(x^2-1)^{l}\;\;\;\;\;\;\;\;349$

Using Leibniz’s theorem to differentiate eq349 by $l+1$ times with respect to $x$ gives

$(1-x^2)\frac{d^2}{dx^2}\frac{d^l}{dx^l}(x^2-1)^{l}-2x\frac{d}{dx}\frac{d^l}{dx^l}(x^2-1)^{l}+l(l+1)\frac{d^l}{dx^l}(x^2-1)^l=0\;\;\;\;\;\;\;\;350$

Comparing eq350 with eq332 reveals that $y=\frac{d^l}{dx^l}(x^2-1)^l$ is a solution to the Legendre differential equation. However, this solution does not produce the conventional Legendre polynomials of $P_0(x)=1,P_1(x)=x,P_2(x)=\frac{1}{2}(3x^2-1)$ and so on.

Question

Prove by induction that $\frac{d^k}{dx^k}(x+1)^l=\frac{l!}{(l-k)!}(x+1)^{l-k}$ for $0\leq k\leq l$ , and hence, $\frac{d^{l-k}}{dx^{l-k}}(x-1)^l=\frac{l!}{(k)!}(x-1)^{k}$ .

Answer

For $k-0$ , we have $\frac{d^0}{dx^0}(x+1)^l=(x+1)^{l}$ . Assuming that $\frac{d^k}{dx^k}(x+1)^l=\frac{l!}{(l-k)!}(x+1)^{l-k}$ is true, then,

$\frac{d^{k+1}}{dx^{k+1}}(x+1)^l=\frac{l!(l-k)}{(l-k)!}(x+1)^{l-k-1}=\frac{l!}{\[l-(k+1)\]!}(x+1)^{l-(k+1)}$

So, $\frac{d^k}{dx^k}(x+1)^l=\frac{l!}{(l-k)!}(x+1)^{l-k}$ holds for all $0\leq k\leq l$ . Similarly, $\frac{d^k}{dx^k}\[x+(-1)\]^l=\frac{l!}{(l-k)!}\[x+(-1)\]^{l-k}$ . If we let $k=l-k$ , then $\frac{d^{l-k}}{dx^{l-k}}\[x+(-1)\]^l=\frac{l!}{k!}\[x+(-1)\]^{k}$ .

To derive the factor $\frac{1}{2^ll!}$ in eq348, we again use Leibniz’s theorem to find the $l$ -th derivative of $y$ with respect to $x$ :

$\begin{align}\frac{d^l}{dx^l}(x^2-1)^l &=\frac{d^l}{dx^l}(x+1)^{l}(x-1)^l\\&=\sum_{k=0}^l\begin{pmatrix}l\\k\end{pmatrix}\frac{l!}{(l-k)!}(x+1)^{l-k}\frac{l!}{k!}(x-1)^k\\&=l!\sum_{k=0}^l\begin{pmatrix}l\\k\end{pmatrix}^2(x+1)^{l-k}(x-1)^k\end{align}$

If we evaluate $\frac{d^l}{dx^l}(x^2-1)^l$ at $x=1$ , only the $k=0$ term in $l!\sum_{k=0}^l\begin{pmatrix}l\\k\end{pmatrix}^2(x+1)^{l-k}(x-1)^k$ survives because the binomial theorem defines $0^0=1$ , resulting in

$y(1)=\biggr\[\frac{d^l}{dx^l}(x^2-1)^l\biggr\]_{x=1}=2^ll!$

Since the conventional Legendre polynomials are defined by the condition $P_l(1)=1$ , the solution to the Legendre differential equation in terms of $y$ must be $\frac{1}{2^ll!}y$ , which is equivalent to eq348.

Rodrigues’ formula of the Legendre Polynomials

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