Form of Laguerre Equation/Laguerre Polynomials
It will be the last of the special polynomials we are going to introduce in this chapter. The Laguerre Equation is a type of second-order linear ODE with the form of
\begin{align}
xy^{\prime\prime} + (1-x)y’ + \nu y &= 0 \tag{1}
\end{align}
\( \nu \) is the eigenvalue that is almost always an integer. It clearly has a regular singularity at \( x = 0 \), and now we set out to find the corresponding Frobenius solution there. First, recall the indicial equation (8) in this tutorial, which is
\begin{align}
m^2 + (1-1)m + 0 &= 0 \\
m^2 &= 0 \tag{2}
\end{align}
So there is only one repeated exponent \( m = 0 \) and thus one possible Frobenius solution (which reduces to the usual series solution in this case). Substituting (4)-(6) in the aforementioned tutorial into (1), we have
\begin{align}
x \sum_{n=0}^{\infty} n(n-1)a_nx^{n-2} + (1-x)\sum_{n=0}^{\infty} na_nx^{n-1} + \nu\sum_{n=0}^{\infty} a_nx^n &= 0 \\
\begin{aligned}\sum_{n=0}^{\infty} n(n-1)a_nx^{n-1} + \sum_{n=0}^{\infty} na_nx^{n-1} \\
-\sum_{n=0}^{\infty} na_nx^{n} + \nu\sum_{n=0}^{\infty} a_nx^n\end{aligned} &= 0 \\
\begin{aligned}\sum_{n=0}^{\infty} (n+1)na_{n+1}x^{n} + \sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n} \\ -\sum_{n=0}^{\infty} na_nx^{n} + \nu\sum_{n=0}^{\infty} a_nx^n\end{aligned} &= 0 \tag{3}
\end{align}
So the recurrence relation is
\begin{align}
(n+1)na_{n+1} + (n+1)a_{n+1} -na_n + \nu a_n &= 0 \\
(n+1)^2a_{n+1} -(n-\nu) a_n &= 0 \\
a_{n+1} &= \frac{n-\nu}{(n+1)^2} a_n \tag{4}
\end{align}
Under the condition that \( \nu \) is an integer, the series breaks off when \( n = \nu \) and becomes
\begin{align}
y &= 1 + \frac{0-n}{(1)^2}x + \frac{0-n}{(1)^2}\frac{1-n}{(2)^2}x^2 + \cdots + \frac{(-1)^{n}(n)(n-1)\cdots(1)}{(n!)^2}x^n \\
&= \sum_{m=0}^{n} \frac{(-1)^mn!}{(n-m)!(m!)^2}x^m \tag{5}
\end{align}
The above expression is a polynomial, denoted as the Laguerre polynomial of order \(n\), \( L_n(x) \). The normalization factor is just that \( L_n(0) = 1 \), equivalent to (5) where \(a_0 = 0\). The first few Laguerre polynomials are
\begin{align}
\begin{aligned}
L_0(x) &= 1 & L_1(x) &= -x + 1 \\
L_2(x) &= \frac{1}{2!}(x^2 -4x +2) & L_3(x) &= \frac{1}{3!}(-x^3 + 9x^2 -18x + 6)
\end{aligned} \tag{6}
\end{align}
Laguerre Equation as a Sturm-Liouville Problem
The Laguerre Equation is similar to other previous equations which can be put in the Sturm-Liouville Form, but there are some differences. The desired form is derived by multiplying by the integrating factor \( e^{-x} \):
\begin{align}
xe^{-x}y^{\prime\prime} + (1-x)e^{-x}y’ + \nu e^{-x} y &= 0 \\
(xe^{-x}y)’ + \nu e^{-x} y &= 0 \tag{7}
\end{align}
Here \( p(x) = xe^{-x} \) and this strongly hints that the natural interval for Hermicity is \( [0, \infty) \), the positive \(x\)-axis. All the results about Sturm-Liouville operators then apply to the finite Laguerre polynomials. Since most of the properties resemble the ones derived for other special polynomials, we will just list them out and the readers are asked to verify them.
Orthogonality
With the weighting \( \rho = e^{-x} \), the orthogonality statement is
\begin{align}
\int_0^{\infty} e^{-x} L_{n}(x)L_{k}(x) &= 0 & \text{as \(n \neq k\)} \tag{8}
\end{align}
Rodrigues’ Formula
The corresponding Rodrigues’ formula is
\begin{align}
L_n(x) &= \frac{e^x}{n!}\frac{d^n}{dx^n} (x^n e^{-x}) \tag{9}
\end{align}
This can be shown by directly expanding (9) using Leibniz Product Rule and comparing it to (5).
Normalization Factor
The normalization relation for Laguerre polynomials is
\begin{align}
\int_0^{\infty} e^{-x} L_{n}(x)L_n(x) &= 1 \tag{10}
\end{align}
Generating Function
The generating function for Laguerre polynomials is
\begin{align}
G(x,t) = \frac{e^{-xt/(1-t)}}{1-t} = \sum_{n=0}^{\infty} L_n(x) t^n \tag{11}
\end{align}
(see exercise below)
Recurrence Relations
Some notable recurrence relations are (as listed in the textbook Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence)
\begin{align}
\begin{aligned}
(n+1) L_{n+1}(x) &= (2n+1-x) L_n(x) -nL_{n-1}(x) \\
L_{n-1}(x) &= L_{n-1}'(x) -L_{n}'(x) \\
xL_{n}'(x) &= nL_{n}(x) -nL_{n-1}(x)
\end{aligned} \tag{12}
\end{align}
(also see exercise below)
Exercise
By differentiating the generating function (11) with respect to \(x\) and \(t\), deduce the form of recurrence relations as in (12). Then, by manipulating (12), show that such \( L_n \) will satisfy the original Laguerre equation (1), and so in turn establish the validity of (11).
Answer
Differentiating (11) against \(t\) yields
\begin{align}
\frac{\partial}{\partial t}(\frac{-xt}{1-t})\frac{e^{-xt/(1-t)}}{1-t} + \frac{e^{-xt/(1-t)}}{(1-t)^2} &= \sum_{n=0}^{\infty} nL_n(x) t^{n-1} \\
-\frac{xe^{-xt/(1-t)}}{(1-t)^3} + \frac{e^{-xt/(1-t)}}{(1-t)^2} &= \sum_{n=0}^{\infty} nL_n(x) t^{n-1} \\
\frac{(1-x-t)e^{-xt/(1-t)}}{(1-t)^3} &= \sum_{n=0}^{\infty} nL_n(x) t^{n-1} \\
\end{align}
Using (11) itself and rearranging leads to
\begin{align}
\frac{1-x-t}{(1-t)^2} \sum_{n=0}^{\infty} L_n(x) t^n &= \sum_{n=0}^{\infty} nL_n(x) t^{n-1} \\
(1-x-t) \sum_{n=0}^{\infty} L_n(x) t^n &= (1-t)^2 \sum_{n=0}^{\infty} nL_n(x) t^{n-1} \\
(1-x) \sum_{n=0}^{\infty} L_n(x) t^n -\sum_{n=0}^{\infty} L_n(x) t^{n+1} &= \sum_{n=0}^{\infty} nL_n(x) t^{n-1} -2\sum_{n=0}^{\infty} nL_n(x) t^{n}+\sum_{n=0}^{\infty} nL_n(x) t^{n+1} \\
(1-x) \sum_{n=0}^{\infty} L_n(x) t^n -\sum_{n=0}^{\infty} L_{n-1}(x) t^{n} &= \sum_{n=0}^{\infty} (n+1)L_{n+1}(x) t^{n} -2\sum_{n=0}^{\infty} nL_n(x) t^{n}+\sum_{n=0}^{\infty} (n-1)L_{n-1}(x) t^{n} \\
\end{align}
where we have adjusted the indices accordingly. Equating the coefficients then gives the first relation:
\begin{align}
(1-x) L_n(x) -L_{n-1}(x) &= (n+1)L_{n+1}(x) -2nL_n(x) + (n-1)L_{n-1}(x) \\
(n+1) L_{n+1}(x) &= (2n+1-x) L_n(x) -nL_{n-1}(x)
\end{align}
The second relation is similarly obtained by differentiating (11) against \(x\):
\begin{align}
-\frac{te^{-xt/(1-t)}}{(1-t)^2} &= \sum_{n=0}^{\infty} L_n'(x) t^n \\
-t\sum_{n=0}^{\infty} L_n(x) t^n &= (1-t)\sum_{n=0}^{\infty} L_n'(x) t^n \\
-\sum_{n=0}^{\infty} L_n(x) t^{n+1} &= \sum_{n=0}^{\infty} L_n'(x) t^n -\sum_{n=0}^{\infty} L_n'(x) t^{n+1}\\
-\sum_{n=0}^{\infty} L_{n-1}(x) t^{n} &= \sum_{n=0}^{\infty} L_n'(x) t^n -\sum_{n=0}^{\infty} L_{n-1}'(x) t^{n}
\end{align}
So
\begin{align}
L_{n-1}(x) &= L_{n-1}'(x) -L_{n}'(x)
\end{align}
Differentiating the first relation with respect to \(x\) and using the second relation, we arrive at the third relation:
\begin{align}
(n+1) L_{n+1}'(x) &= (2n+1-x) L_n'(x) -L_n(x) -nL_{n-1}'(x) \\
(n+1) (L_{n}'(x)-L_{n}(x)) &= (n+1-x) L_n'(x) -L_n(x) -nL_{n-1}(x) \\
xL_n'(x) &= nL_n(x) -nL_{n-1}(x) \\
\end{align}
Differentiating the last relation and applying the second relation, we recover the Laguerre equation:
\begin{align}
xL_{n}^{\prime\prime}(x) + L_n'(x) &= nL_{n}'(x) -nL_{n-1}'(x) \\
xL_{n}^{\prime\prime}(x) + L_n'(x) &= nL_{n}'(x) -n(L_{n-1}(x) + L_{n}'(x)) \\
xL_{n}^{\prime\prime}(x) + L_n'(x) &= xL_{n}'(x) -nL_{n}(x) \\
xL_{n}^{\prime\prime}(x) + (1-x) L_n'(x) + nL_{n}(x) &= 0
\end{align}
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