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Question on Rodrigues' equation in Legendre polynomials.

  1. Jan 27, 2010 #1
    I have problem understand in one step of deriving the Legendre polymonial formula. We start with:

    [tex]P_n (x)=\frac{1}{2^n } \sum ^M_{m=0} (-1)^m \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]

    Where M=n/2 for n=even and M=(n-1)/2 for n=odd.


    For 0<=m<=M

    [tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]


    For M<m<=n

    [tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = 0[/tex]



    [tex]P_n (x)=\frac{1}{2^n n!} \sum ^M_{m=0} (-1)^m \frac{n!)}{m!(n-m)}\frac{d^n}{dx^n}x^2n-2m [/tex](1)


    [tex]\Rightarrow P_n (x)=\frac{1}{2^n n!}\frac{d^n}{dx^n} \sum ^n_{m=0} (-1)^m \frac{n!)}{m!(n-m)}(x^2)^{n-m} [/tex](2)


    Notice the [tex]\sum^M_{m=0}[/tex] change to [tex]\sum^n_{m=0}[/tex] from (1) to (2). Can anyone explain this to me?
     
  2. jcsd
  3. Jan 28, 2010 #2

    tiny-tim

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    Hi yungman! :smile:

    (in LaTeX, ^ or _ has to be followed by {} unless there's only one character :wink:)

    It' s because of the line before (1) …

    m=M+1n {…blah…} dn/dxn x2n-2m = 0. :smile:
     
  4. Jan 29, 2010 #3
    Thanks.
     
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