Advanced Chemistry

September 23, 2024

Laguerre polynomials

Laguerre polynomials $v(x)$ are a sequence of polynomials that are solutions to the Laguerre differential equation:

$x\frac{d^2v}{dx^2}+(1-x)\frac{dv}{dx}+nv=0\;\;\;\;\;\;\;\;420$

where $n$ is a constant.

When $x=0$ , eq420 simplifies to $\frac{dv}{dx}+nv=0$ . The solution to this first-order differential equation is $v(x)=Ae^{-nx}$ , which can be expressed as the Taylor series $v(x)=A\sum_{n=0}^{\infty}\frac{(-nx)^n}{n!}$ . This implies that eq420 has a power series solution around $x=0$ . To determine the exact form of the power series solution to eq420, let $v(x)=\sum_{j=0}^{\infty}a_jx^j$ .

Substituting $v$ , $v'$ and $v''$ in eq420 yields

$x\sum_{j=0}^{\infty}j(j-1)a_jx^{j-2}+(1-x)\sum_{j=0}^{\infty}ja_jx^{j-1}+n\sum_{j=0}^{\infty}a_jx^{j}=0\;\;\;\;\;\;\;\;421$

$\sum_{j=0}^{\infty}j^2a_jx^{j-1}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

Setting $j=j+1$ in the first sum,

$\sum_{j=-1}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=\sum_{j=0}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

$\sum_{j=0}^{\infty}\biggr\[(j+1)^2a_{j+1}+(n-j)a_j\biggr\]x^{j}=0\;\;\;\;\;\;\;\;422$

Eq422 is only true if all coefficients of $x$ in is 0 (see this article for explanation). So, $(j+1)^2a_{j+1}+(n-j)a_j=0$ , or equivalently,

$a_{j+1}=\frac{j-n}{(j+1)^2}a_j\;\;\;\;\;\;\;\;423$

Eq423 is a recurrence relation. If we know the value of $a_0$ , we can use the relation to find $a_1,a_2,...$ .

$\boldsymbol{\mathit{j}}$	Recurrence relation
$0$	$a_1=-na_0$
$1$	$a_2=\frac{1-n}{(1+1)^2}a_1=\frac{n(n-1)}{1^22^2}a_0$
$2$	$a_3=\frac{2-n}{(2+1)^2}a_2=-\frac{n(n-1)(n-2)}{1^22^23^2}a_0$
$\vdots$	$\vdots$

Comparing the recurrence relations, we have

$a_k=(-1)^k\frac{n(n-1)(n-2)\cdots (n-k+1)}{1^22^2\cdots k^2}a_0$

$a_k=(-1)^k\frac{n!}{(k!)^2(n-k)!}\;\;\;\;\;\;\;\;424$

where $a_0=1$ by convention (so that $L_n(0)=1$ ).

Letting $k=j$ in eq424 and substituting it in $v(x)=\sum_{j=0}^{n}a_jx^j$ yields the Laguerre polynomials:

$L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j\;\;\;\;\;\;\;\;425$

where we have replaced $v(x)$ with $L_n(x)$ .

The first few Laguerre polynomials are:

$L_0(x)=1$

$L_1(x)=-x+1$

$L_2(x)=\frac{1}{2}(x^2-4x+2)$

$L_3(x)=\frac{1}{6}(-x^3+9x^2-18x+6)$

Next article: Rodrigues’ formula for the Laguerre polynomials

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September 23, 2024

Rodrigues’ formula for the Laguerre polynomials

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

It is given by

$L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}(x^ne^{-x})\;\;\;\;\;\;\;\;428$

To prove eq428, we apply Leibniz’ theorem as follows:

$L_n(x)=\frac{e^x}{n!}\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{d^kx^n}{dx^k}\frac{d^{n-k}e^{-x}}{dx^{n-k}}$

Substituting $\frac{d^kx^n}{dx^k}=\frac{n!}{(n-k)!}x^{n-k}$ and $\frac{d^{n-k}e^{-x}}{dx^{n-k}}=(-1)^{n-k}e^{-x}$ in the above equation and rearranging yields

$L_n(x)=\sum_{k=0}^n(-1)^{n-k}\frac{n!}{k!(n-k)!(n-k)!}x^{n-k}$

Letting $j=n-k$ , we have $L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j$ , which is the expression for the Laguerre polynomials

Question

How do we change the variable in the summation $\sum_{k=0}^{n}$ by letting $j=n-k$ ?

Answer

When $k=0$ , $j=n$ , and when $k=n$ , $j=0$ . So, $\sum_{k=0}^{n}=\sum_{j=n}^{0}$ . Reversing the summation order, $\sum_{k=0}^{n}=\sum_{j=0}^{n}$ .

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