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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Subsection 4.11.1 Laguerre’s Polynomial

Python Code:.

import numpy as np
from scipy.special import genlaguerre
from scipy.special import laguerre
import matplotlib.pyplot as plt
x = np.arange(-1.0, 5.0, 0.01)
fig, ax = plt.subplots()
ax.set_ylim(-5.0, 5.0)
ax.set_title(r’Laguerre polynomials $L_n$’)
for n in np.arange(0, 5):
ax.plot(x, laguerre(n)(x), label=rf’$L_{n}(x)$’)
plt.legend(loc=’best’)
plt.xlabel("x")
plt.ylabel(r’$L_n(x)$’)
plt.legend(loc=’lower right’)
plt.show()

Subsubsection 4.11.1.1 Generating Function for \(L_{n}(x)\)

The generating function for Laguerre’s polynomial is defined as
\begin{equation*} f(x,t)=\frac{e^{-xt/(1-t)}}{1-t}=\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{n!}t^{n}\quad |t| \lt 1. \end{equation*}

Proof.

We have
\begin{equation*} e^{-xt/(1-t)}=1-\frac{xt}{1-t}+\frac{1}{2!}\left(\frac{xt}{1-t}\right)^{2}+\cdots \end{equation*}
\begin{equation} =\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}x^{r}t^{r}}{(1-t)^{r}r!}\tag{4.11.9} \end{equation}
or,
\begin{equation*} \frac{e^{-xt/(1-t)}}{1-t}=\sum\limits_{r=0}^{\infty}(-1)^{r}\frac{t^{r}x^{r}}{r!(1-t)^{r+1}} \end{equation*}
\begin{equation*} =\sum\limits_{r=0}^{\infty}(-1)^{r}\frac{x^{r}t^{r}}{r!}(1-t)^{-(r+1)} \end{equation*}
\begin{equation*} =\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!}x^{r}t^{r}\left[1+(r+1)t+\frac{(r+1)(r+2)}{2!}t^{2}\right. \end{equation*}
\begin{equation*} \left.+\cdots+\frac{(r+1)(r+2)\cdots(r+s)}{s!}t^{s}+\cdots\right] \end{equation*}
\begin{equation*} =\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}x^{r}t^{r}}{r!}\sum\limits_{s=0}^{\infty}\frac{(r+1)(r+2)\cdots(r+s)}{s!}t^{s} \end{equation*}
\begin{equation} =\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}x^{r}t^{r}}{r!}\sum\limits_{s=0}^{\infty}\frac{(r+s)!}{r!s!}t^{s}\tag{4.11.10} \end{equation}
on putting \(r+s=n\text{,}\) the coefficient of \(t^{n}\) for fixed value of \(r\) is \(\frac{(-1)^{r}n!}{(r!)^{2}(n-r)!}x^{r}\text{,}\) as \(n-r=s\) and \(s \geq 0\text{.}\) Thus \(n-r \geq 0\) or, \(r \leq n\text{.}\) Hence the coefficient of \(t^{n}\) on RHS of equation (4.11.10) is
\begin{equation*} \sum\limits_{r=0}^{n}\frac{(-1)^{r}n!}{(r!)^{2}(n-r)!}x^{r} =\frac{L_{n}(x)}{n!} \end{equation*}
Therefore,
\begin{equation*} \frac{e^{-xt/(1-t)}}{1-t}=\sum\limits_{r=0}^{n}\frac{L_{n}(x)}{n!}t^{n} \end{equation*}
is the generating function of \(L_{n}(x)\text{.}\)

Subsubsection 4.11.1.2 Rodrigue’s Differential Formula for \(L_{n}(x)\)

\begin{equation*} L_{n}(x)=e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x}). \end{equation*}

Proof.

We have the generating function for \(L_{n}(x)\) is
\begin{equation*} \frac{e^{-xt/(1-t)}}{1-t}=\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{n!}t^{n} \end{equation*}
Differentiating \(n\) times w. r. to ’t’, we get -
\begin{equation*} \frac{d^{n}}{dt^{n}}\left(\frac{e^{-xt/(1-t)}}{1-t}\right)= L_{n}(x)+L_{n+1}(x)t+\cdots \end{equation*}
or,
\begin{equation*} e^{x}\frac{d^{n}}{dt^{n}}\left[(1-t)^{-1}e^{-x/(1-t)}\right] \end{equation*}
\begin{equation} =L_{n}(x)+L_{n+1}(x)t+\cdots \tag{4.11.11} \end{equation}
But,
\begin{equation*} \frac{d}{dt}\left[(1-t)^{-1}e^{-x/(1-t)}\right]=\frac{1-x-t}{(1-t)^{3}}e^{-x/(1-t)} \end{equation*}
or,
\begin{equation*} \lim\limits_{t \to 0}\frac{d}{dt}\left[(1-t)^{-1}e^{-x/(1-t)}\right]=(1-x)e^{-x}=\frac{d}{dx}(xe^{-x}) \end{equation*}
similarly,
\begin{equation*} \lim\limits_{t \to 0}\frac{d^{2}}{dt^{2}}\left[(1-t)^{-1}e^{-x/(1-t)}\right]=\frac{d^{2}}{dx^{2}}(x^{2}e^{-x}) \end{equation*}
In general,
\begin{equation*} \lim\limits_{t \to 0}\frac{d^{n}}{dt^{n}}\left[(1-t)^{-1}e^{-x/(1-t)}\right]=\frac{d^{n}}{dx^{n}}(x^{n}e^{-x}) \end{equation*}
Hence from equation (4.11.11), we have
\begin{equation*} e^{x}\lim\limits_{t \to 0}\frac{d^{n}}{dt^{n}}\left[(1-t)^{-1}e^{-x/(1-t)}\right]=e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})=L_{n}(x). \end{equation*}
Thus,
\begin{equation*} L_{n}(x)= e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x}) \end{equation*}
This is the Rodrigue’s formula for \(L_{n}(x)\text{.}\)

Subsubsection 4.11.1.3 Recurrence Relation for \(L_{n}(x)\)

  1. \begin{equation*} (1+2n-x)L_{n}=n^{2}L_{n-1}+L_{n+1} \end{equation*}

    Proof.

    We have the generating function
    \begin{equation} \frac{e^{-xt/(1-t)}}{1-t}=\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{n!}t^{n} \qquad |t| \lt 1.\tag{4.11.12} \end{equation}
    Differentiating equation (4.11.12) w. r. to ’t’, we get -
    \begin{equation*} \frac{1-x-t}{(1-t)^{3}}e^{-xt/(1-t)}=\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n-1)!}t^{n-1} \end{equation*}
    or,
    \begin{equation*} (1-x-t)\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n)!}t^{n}=(1-t)^{2}\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n-1)!}t^{n-1} \end{equation*}
    or,
    \begin{equation*} (1-x)\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n)!}t^{n}-\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n)!}t^{n+1} \end{equation*}
    \begin{equation*} =\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n-1)!}t^{n-1}-2\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n-1)!}t^{n}+\sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{(n-1)!}t^{n+1} \end{equation*}
    Equating the coefficient of \(t^{n}\) on both sides, we get -
    \begin{equation*} (1-x)\frac{L_{n}}{(n)!}-\frac{L_{n-1}}{(n-1)!}=\frac{L_{n+1}}{(n)!}-2\frac{L_{n}}{(n-1)!}+\frac{L_{n-1}}{(n-2)!} \end{equation*}
    or,
    \begin{equation*} (1-x) L_{n}-nL_{n-1}=L_{n+1}-2nL_{n}+n(n-1)L_{n-1} \end{equation*}
    [on multiplying by \(n!\)] Or,
    \begin{equation*} \therefore (1+2n-x)L_{n}=n^{2}L_{n-1}+L_{n+1} \end{equation*}
  2. \begin{equation*} L'_{n}=n[L'_{n-1}-L_{n+1} \end{equation*}

    Proof.

    Differentiating above equation (4.11.12) w. r. to ’x’, we get -
    \begin{equation*} -(1-t)^{-1}\frac{1}{(1-t)}e^{-xt/(1-t)}=\sum\limits_{n=0}^{\infty}\frac{L'_{n}}{(n)!}t^{n} \end{equation*}
    or,
    \begin{equation*} -t\sum\limits_{n=0}^{\infty}\frac{L_{n}}{(n)!}t^{n}=(1-t)\sum\limits_{n=0}^{\infty}\frac{L'_{n}}{(n)!}t^{n} \end{equation*}
    or,
    \begin{equation*} -\sum\limits_{n=0}^{\infty}\frac{L_{n}}{(n)!}t^{n+1}=\sum\limits_{n=0}^{\infty}\frac{L'_{n}}{(n)!}t^{n}-\sum\limits_{n=0}^{\infty}\frac{L'_{n}}{(n)!}t^{n+1} \end{equation*}
    Equating the coefficient of \(t^{n}\) on both sides, we get -
    \begin{equation*} -\frac{L_{n+1}}{(n-1)!}=\frac{L'_{n}}{n!}-\frac{L'_{n-1}}{(n-1)!} \end{equation*}
    \begin{equation*} L'_{n}=nL'_{n-1}-nL_{n-1} \end{equation*}
    or,
    \begin{equation*} \therefore L'_{n}=n[L'_{n-1}-L_{n-1}] \end{equation*}
  3. \begin{equation*} xL'_{n}=n[L_{n}-n^{2}L_{n-1} \end{equation*}

    Proof.

    Differentiating the recurrence relation (1) w. r. to ’x’, we get -
    \begin{equation} (1+2n-x)L'_{n}-L_{n}=n^{2}L'_{n-1}+L'_{n+1}\tag{4.11.13} \end{equation}
    Replacing \(n\) by \(n+1\) in relation (2), we get -
    \begin{equation*} L'_{n+1}=(n+1)[L'_{n}-L_{n}] \end{equation*}
    or,
    \begin{equation} L'_{n+1}=(n+1)L'_{n}-(n+1)L_{n}]\tag{4.11.14} \end{equation}
    Also, from relation (2), we have
    \begin{equation} L'_{n-1}=L_{n-1}+\frac{1}{n}L'_{n}\tag{4.11.15} \end{equation}
    Substituting equations (4.11.14) and (4.11.15) in equation (4.11.13), we get -
    \begin{equation*} (1+2n-x)L'_{n}-L_{n}=n^{2}[L_{n-1}+\frac{1}{n}L'_{n}] \end{equation*}
    \begin{equation*} \qquad +(n+1)L'_{n}-(n+1)L_{n} \end{equation*}
    or,
    \begin{equation*} (1+2n-x)L'_{n}-L_{n}=n^{2}L_{n-1}+nL'_{n}+nL'_{n}+L'_{n}-nL_{n}-L_{n} \end{equation*}
    or,
    \begin{equation*} (1+2n)L'_{n}-xL'_{n}=n^{2}L_{n-1}+(1+2n)L'_{n}-nL_{n} \end{equation*}
    or,
    \begin{equation*} -xL'_{n}=n^{2}L_{n-1}-nL_{n} \end{equation*}
    or,
    \begin{equation*} \therefore xL'_{n}=nL_{n}-n^{2}L_{n-1} \end{equation*}

Subsubsection 4.11.1.4 Orthogonal Property of Laguerre Polynomials

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\frac{L_{m}(x)}{m!}\frac{L_{n}(x)}{n!}\,dx = \delta_{m,n} =\begin{cases} 0, & \text{if}\quad m \neq n\\ 1, & \text{if}\quad m = n \end{cases} \end{equation*}

Proof.

We have
\begin{equation*} \sum\limits_{m=0}^{\infty}\frac{L_{m}(x)}{m!}s^{m}=\frac{1}{1-s}e^{-xs/(1-s)} \end{equation*}
and
\begin{equation*} \sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{n!}t^{n}=\frac{1}{1-t}e^{-xt/(1-t)} \end{equation*}
\begin{equation*} \therefore \sum\limits_{m,n=0}^{\infty}e^{-x}\frac{L_{m}(x)}{m!}s^{m}\frac{L_{n}(x)}{n!}t^{m}s^{m}t^{n} \end{equation*}
\begin{equation*} =e^{-x}\frac{1}{(1-s)(1-t)}e^{\frac{-xs}{1-s}}e^{\frac{-xt}{1-t}} \end{equation*}
thus,
\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\frac{L_{m}(x)}{m!}\frac{L_{n}(x)}{n!}\,dx = \text{coefficient of}\quad s^{m}t^{n}\quad \text{in} \end{equation*}
\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\frac{1}{(1-s)(1-t)}e^{\frac{-xs}{1-s}}e^{\frac{-xt}{1-t}}\,dx. \end{equation*}
Now,
\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\frac{1}{(1-s)(1-t)}e^{\frac{-xs}{1-s}}e^{\frac{-xt}{1-t}}\,dx \end{equation*}
\begin{equation*} =\frac{1}{(1-s)(1-t)}\int\limits_{0}^{\infty}e^{-x\left[1+\frac{t}{1-t}+\frac{s}{1-s}\right]}\,dx \end{equation*}
\begin{equation*} =\frac{1}{(1-s)(1-t)}\frac{1}{[1+\frac{s}{1-s}+\frac{t}{1-t}}\left[e^{-x}\{[1+\frac{s}{1-s}+\frac{t}{1-t}\}\right]_{0}^{\infty} \end{equation*}
\begin{equation*} =\frac{1}{(1-s)(1-t)}\frac{(1-s)(1-t)}{(1-s)(1-t)+s(1-t)+t(1-s)}[0-1] \end{equation*}
\begin{equation*} =\frac{1}{1-st}=(1-st)^{-1}=1+st+(st)^{2}+\cdots+(st)^{n}+\cdots \end{equation*}
in which coefficient of \(s^{m}t^{n}\) is \(0\) if \(m \neq n\) and \(1\) if \(m=n\text{.}\) Hence,
\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\frac{L_{m}(x)}{m!}\frac{L_{n}(x)}{n!}\,dx = \delta_{m,n} =\begin{cases} 0, & \text{if}\quad m \neq n\\ 1, & \text{if}\quad m = n \end{cases} \end{equation*}
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