From: rwinther@my-dejanews.com
Subject: Re: Associated Laguerre polynomials: Rodrigues formulae
Date: Sat, 20 Mar 1999 00:39:26 GMT
Newsgroups: sci.math.num-analysis,sci.math

In article <36F28592.63F4FCCD@hermes.cam.ac.uk>,
  Yen Lee Loh <yll21@hermes.cam.ac.uk> wrote:
> Hello!
> There seem to be two Rodrigues formulae for the associated Laguerre
> polynomials L(n,k,x), both equally valid, one involving more
> differentiations than the other
> (*have omitted factors of (-1)^n or n!)
>
>     L(n,k,x) = e^x D^(n+k)[x^n e^(-x)]
>     L(n,k,x) = e^x x^(-k) D^n [x^(n+k) e^(-x)]
>
> There is also the definition in terms of the ordinary Laguerre
> polynomials
>
>     L(n,k,x) = D^k L(n+k,x)
>              = D^k [e^x D^(n+k) [x^(n+k) e^(-x)] ]
>
> Would somebody like to comment on this? ... e.g., how do you convert
> between the different formulae?  The situation seems analogous to the
> one with associated Legendre functions re. earlier post ...
>
> Thank you
> Yen Lee.
>
Formally, D^(n+k)[x^n e^(-x)] is given by

  Sum_{j=0}^{n+k} (n+k)!/[j!(n+k-j)!] D^(j)[x^n]*D^(n+k-j)[e^(-x)]

but because D^(j)[x^n] = 0 for j > n, the upper limit on the sum may be
replaced by n. Evaluating the effects of the D's in the sum, we get

  Sum_{j=0}^n (n+k)!/[j!(n+k-j)!] (n!/(n-j)!) x^(n-j)*(-1)^(n+k-j)*e^(-x)

Similarly,  x^(-k) D^n [x^(n+k) e^(-x)] is given by

  x^(-k)*Sum_{j=0}^n  n!/[j!(n-j)!] D^(j)[x^(n+k)]*D^(n-j)[e^(-x)]

which evaluates to

  Sum_{j=0}^n  n!/[j!(n-j)!] [(n+k)!/(n+k-j)!] x^(n-j)*(-1)^(n-j)*e^(-x)

Hmmm... looks like they differ by a factor of (-1)^k


Ron Winther

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