📂Functions

Laguerre Polynomials' Rodrigues' Formula

Formulas

The explicit formula for the Laguerre polynomials is as follows.

$$ L_{n}(x) = \frac{1}{n!}e^{x}\frac{ d ^{n}}{ dx^{n} }(x^{n}e^{-x}) \tag{1} $$

Description

The formula above is referred to as the Rodrigues’ formula for Laguerre polynomials. Originally, the term Rodrigues’ formula denoted the explicit form of the Legendre polynomials, but it later became a general term for formulas expressing the explicit form of special functions represented by polynomials. Writing down the first few polynomials yields the following.

$$ \begin{align*} L_{0}(x) &= 1 \\ L_{1}(x) &= -x+1 \\ L_{2}(x) &= \frac{1}{2}\left( x^{2}-4x+2 \right) \\ L_{3}(x) &=\frac{1}{6}\left( -x^{3}+9x^{2}-18x+6 \right) \\ & \vdots \end{align*} $$

Proof¹

Strategy: We need to show that $(1)$ satisfies the Laguerre differential equation.

For convenience, let’s introduce the differential operator $D=\frac{d}{dx}$, which is $Df=\frac{ d f}{ d x }=f^{\prime}$, and we will use both expressions appropriately in the proof.

Let the Rodrigues’ formula be $f(x)=\frac{e^{x}}{n!}\frac{ d ^{n}}{ d x^{n} }(x^{n}e^{-x})$, and let $v=x^{n}e^{-x}$. First, we will show that $xv^{\prime}=(n-x)v$ occurs. $$ \begin{align*} && v^{\prime}&=nx^{n-1}e^{-x}-x^{n}e^{-x} \\ \implies && xv^{\prime}&=nx^{n}e^{-x}-xx^{n}e^{-x} \\ && &=(n-x)v \end{align*} $$

Now differentiate both sides $n+1$ times. First, differentiating the left side, we obtain the following by the Leibniz rule.

$$ \begin{align*} D^{n+1}(xv^{\prime}) &= \sum \limits _{k=0}^{n} \frac{(n+1)!}{(n+1-k)!k!}(D^{k}x)(D^{n+1-k}v^{\prime}) \\ &= \sum \limits _{k=0}^{1} \frac{(n+1)!}{(n+1-k)!k!}(D^{k}x)(D^{n+1-k}v^{\prime}) \end{align*} $$

The second equality holds because when $k \ge 2$, $D^{k}x=0$, dissolving the sigma notation yields the result below.

$$ \begin{align*} D^{n+1}(xv^{\prime}) &= xD^{n+1}v^{\prime}+(n+1)D^{n}v^{\prime} \\ &= x(D^{n}v)^{\prime \prime} +(n+1)(D^{n}v)^{\prime} \end{align*} $$

Similarly, the right side yields the following.

$$ \begin{align*} D^{n+1}\left[ (n-x)v \right] &= \sum \limits _{k=0}^{n}\frac{(n+1)!}{(n+1-k)!k!}\left[D^{k}(n-x)\right] (D^{n+1-k}v) \\ &=\sum \limits _{k=0}^{1}\frac{(n+1)!}{(n+1-k)!k!}\left[D^{k}(n-x)\right] (D^{n+1-k}v) \\ &=(n-x)D^{n+1}v+(n+1)(-1)D^{n}v \\ &= (n-x)(D^{n}v)^{\prime}-(n+1)D^{n}v \end{align*} $$

Therefore, we obtain the following.

$$ \begin{align*} && D^{n+1}(xv^{\prime}) &= D^{n+1}[(n-x)v] \\ \implies && x(D^{n}v)^{\prime \prime} +(n+1)(D^{n}v)^{\prime} &= (n-x)(D^{n}v)^{\prime}-(n+1)D^{n}v \end{align*} $$

Organizing this gives the result below.

$$ x(D^{n}v)^{\prime \prime} +(1+x)(D^{n}v)^{\prime}+(n+1)D^{n}v=0 $$

However, by assumption, the following holds.

$$ D^{n}v=\frac{ d ^{n}}{ dx^{n} }(x^{n}e^{-x})=n!e^{-x}f(x) $$

Substituting this into the equation above, we obtain the result below.

$$ x[n!e^{-x}f(x)]^{\prime \prime}+(1+x)[n!e^{-x}f(x)]^{\prime}+(n+1)[n!e^{-x}f(x)]=0 $$

Fully expanding the differentiation yields the following. The process is simple; hence, it is omitted.

$$ n!e^{-x} \left[ xf^{\prime \prime}(x)+(1-x)f^{\prime}(x)+nf(x) \right]=0 $$

Since $n!e^{-x} \ne 0$, the expression inside the brackets must be $0$.

$$ xf^{\prime \prime}(x)+(1-x)f^{\prime}(x)+nf(x)=0 $$

This is the Laguerre differential equation, thus, $f(x)$ is a solution to the Laguerre differential equation, the Laguerre polynomial.

$$ f(x)=\frac{e^{x}}{n!}\frac{ d ^{n}}{ d x^{n} }(x^{n}e^{-x})=L_{n}(x) $$

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