Rodrigues' Formula for Laguerre equation

[appmathstudent]

TL;DR

Rodrigues' Formula for Laguerre equation

This is exercise 12.1.2 a from Arfken's Mathematical Methods for Physicists 7th edition :Starting from the Laguerre ODE,

$$xy''+(1-x)y'+\lambda y =0$$

obtain the Rodrigues formula for its polynomial solutions $$L_n (x)$$

According to Arfken (equation 12.9 ,chapter 12) the Rodrigues formula is :

$$ y_n(x) = \frac {1}{w(x)}(\frac{d}{dx})^n[w(x)p(x)^n]$$

I found that $$w(x) = e^{-x}$$ and then :

$$L_n(x) = e^x (\frac{d}{dx})^n[e^{-x}x^n]$$

But the answer is ,according to Arfken and everywhere else I look,is :

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

I can't figure out exactly how $$ \frac{1}{n!}$$ appeared.
I think it might be related to the fact that $$ L_n(x) =\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{k!} \quad $$

Any help will be appreciated , thank you

 

[PeroK]

You can have any constant factor you want and the result is still a solution. The factor of $\frac 1 {n!}$ may simply be the convention.

 

[julian]

The $\frac{1}{n!}$ is the normalisation constant, it ensures that:

\begin{align*}
\int_0^\infty e^{-x} L_n (x) L_n (x) dx = 1
\end{align*}

Explicitly, it ensures that:

\begin{align*}
\frac{1}{(n!)^2} \int_0^\infty e^{x} \frac{d^n}{dx^n} (x^n e^{-x}) \frac{d^n}{dx^n} (x^n e^{-x}) dx = 1
\end{align*}

This can be verified by using Leibnitz and some integration by parts:

\begin{align*}
& \frac{1}{(n!)^2} \int_0^\infty \left( \sum_{k=0}^n (-1)^{n-k} \frac{n!}{k! (n-k)!} x^{n-k} \right) \frac{d^n}{dx^n} (x^n e^{-x}) dx
\nonumber \\
& = \frac{1}{n!} \int_0^\infty x^n e^{-x} dx
\nonumber \\
& = \frac{1}{n!} n! = 1 .
\end{align*}

I'll leave you to fill in the details.

 

[appmathstudent]

Thank you very much! I think I got it now