Results 1 to 2 of 2

Thread: Lagurre Polynomials

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    148

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by surjective View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: Apr 7th 2011, 12:38 PM
  2. Polynomials
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 27th 2010, 12:15 PM
  3. Replies: 7
    Last Post: Jan 8th 2010, 03:13 AM
  4. Polynomials Help
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Mar 30th 2009, 10:29 AM
  5. Replies: 5
    Last Post: Nov 29th 2005, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum