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appmathstudent

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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

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

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