1. ## Lagurre Polynomials

Hello all,

$\displaystyle Q_{\ell}$ ($\displaystyle \ell \in \mathbb{N}_{0}$) are the Lagurre polynomials. I would like to derive the following expression:

$\displaystyle Q_{\ell}(x)=\ell! \sum_{k=0}^{\ell}\frac{(-1)^{k}}{(k!)^{2}(l-k)!}x^{k}$

So far I have shown that the coefficients in any power series solution $\displaystyle u(x)=\sum_{k=0}^{\infty}c_{k}x^{k}$ of $\displaystyle x\frac{d^{2}u}{du^{2}}+(1-x)\frac{du}{dx}+\ell u=0$ will satisfy that:

$\displaystyle c_{k+1}=\frac{k-l}{(k+1)^{2}}c_{k}$

Furthermore I have shown that $\displaystyle c_{\ell+1}=c_{\ell+2}=\ldots=0$

Now, using indcution, I would like to show that:

$\displaystyle c_{k}=\frac{\ell!(-1)^{k}}{(k!)^{2}(\ell-k)!} \hspace{1cm}(*)$

I have that $\displaystyle c_{0}=1$ for $\displaystyle k=0$. Thus the basis for the induction is established. Now I assume that $\displaystyle (*)$ is true for a fixed $\displaystyle k$ and want to derive the case for $\displaystyle k+1$. I have done the following:

$\displaystyle c_{k+1}=\frac{\ell!(-1)^{k+1}}{((k+1)!)^{2}(\ell-(k+1))!} =\frac{\ell!(-1)^{k}(-1)^{1}}{(k+1)^{2}(k!)^{2}((\ell-k)-1)!}$

I am stuck here. Suggestions would be greatly appreciated.

Thanks.

2. Originally Posted by surjective
using induction, I would like to show that:

$\displaystyle c_{k}=\frac{\ell!(-1)^{k}}{(k!)^{2}(\ell-k)!} \hspace{1cm}(*)$

I have that $\displaystyle c_{0}=1$ for $\displaystyle k=0$. Thus the basis for the induction is established. Now I assume that $\displaystyle (*)$ is true for a fixed $\displaystyle k$ and want to derive the case for $\displaystyle k+1$. I have done the following:

$\displaystyle c_{k+1}=\frac{\ell!(-1)^{k+1}}{((k+1)!)^{2}(\ell-(k+1))!} =\frac{\ell!(-1)^{k}(-1)^{1}}{(k+1)^{2}(k!)^{2}((\ell-k)-1)!}$

I am stuck here. Suggestions would be greatly appreciated.
You're almost there! Notice that $\displaystyle ((l-k)-1)! = \frac{(l-k)!}{l-k}$. So your expression for $\displaystyle c_{k+1}$ is equal to $\displaystyle \frac{\l!(-1)^{k}(-1)(l-k)}{(k+1)^{2}(k!)^{2}(l-k)!} = \frac{k-l}{(k+1)^2}c_k$ , as required.

Note 1. Instead of starting with the desired formula for $\displaystyle c_{k+1}$ and then showing that it is equal to $\displaystyle \frac{k-l}{(k-1)^2}c_k$, it is more usual when writing out a proof by induction to reverse that process. In other words, start with the formula (*) for $\displaystyle c_k$ and then show that $\displaystyle \frac{k-l}{(k-1)^2}c_k$ is equal to the desired formula for $\displaystyle c_{k+1}$.

Note 2. Laguerre's name has an e after the u.