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Laguerre polynomials $L_n(x)$ can be calculated using the Rodriguez formula

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

Show that $L_n(x)$ in the Rodriguez formula satisfy Laguerre's eguation

$x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+(1-x)\frac{\mathrm{d}y}{\mathrm{d}x}+ny=0$, following the steps below: Let $v=x^n e^{-x}$. First show that $x\frac{\mathrm{d}v}{\mathrm{d}x}=(n-x)v$. Then derive that (2):

$$x\frac{\mathrm{d}^{n+2}v}{\mathrm{d}x^{n+2}}+(1+x)\frac{\mathrm{d}^{n+1}v}{\mathrm{d}x^{n+1}}+(n+1)\frac{\mathrm{d}^nv}{\mathrm{d}x^n}=0$$

Next show that $y(x)=\frac{e^x}{n!}\frac{\mathrm{d}^nv}{\mathrm{d}x^v}$ is a solutin of Laguerre's differential eguation using (2)

I showed that $x\frac{\mathrm{d}v}{\mathrm{d}x}=(n-x)v$ but I can't derive (2). Any ideas?

Jean-Claude Arbaut
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user113747
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1 Answers1

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Starting with $x\frac{\mathrm{d}v}{\mathrm{d}x}=(n-x)v$ and taking the $(n+1)^{st}$ derivative left and right, you get $$x\frac{\mathrm{d}^{n+2}v}{\mathrm{d}x^{n+2}} + \binom{n+1}{1} \frac{\mathrm{d}^{n+1}v}{\mathrm{d}x^{n+1}} = (n-x)\frac{\mathrm{d}^{n+1}v}{\mathrm{d}x^{n+1}} - \binom{n+1}{1} \frac{\mathrm{d}^{n}v}{\mathrm{d}x^{n}}$$ or $$x\frac{\mathrm{d}^{n+2}v}{\mathrm{d}x^{n+2}} + (1+x)\frac{\mathrm{d}^{n+1}v}{\mathrm{d}x^{n+1}} + (n+1)\frac{\mathrm{d}^{n}v}{\mathrm{d}x^{n}} = 0$$ which is exactly what you are looking for.

Use has been made of Leibniz' rule:

$$\frac{\mathrm{d}^{n}(fg)}{\mathrm{d}x^{n}} = \sum_{k=0}^n \binom{n}{k}\frac{\mathrm{d}^{k}f}{\mathrm{d}x^{k}}\frac{\mathrm{d}^{n-k}g}{\mathrm{d}x^{n-k}}$$

Maestro13
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