I Rodrigues' Formula for Laguerre equation

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The discussion revolves around deriving the Rodrigues formula for the Laguerre polynomial solutions from the Laguerre ordinary differential equation. The user identifies the weight function as \( w(x) = e^{-x} \) and attempts to express \( L_n(x) \) but struggles with the normalization factor \( \frac{1}{n!} \). This factor is explained as a normalization constant that ensures the orthogonality condition of the Laguerre polynomials over the interval \([0, \infty)\). The normalization is verified through integration and the use of Leibniz's rule. Ultimately, the user confirms their understanding of the derivation and the role of the normalization constant.
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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
 
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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.
 
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.
 
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Likes PeroK and appmathstudent
Thank you very much! I think I got it now
 
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