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Legendre polynomials formula

  1. Oct 20, 2007 #1
    There is a question where you should find a formula for P-n(0) using the Legendre polynomials:
    P-n(x)=1/(2^n*n!) d^n/dx^n(x^2-1)^n , n=0,1,2,3......

    I tried to derive seven times by only substituting the n untill n=7,I did that because i wanted to find something that i can build my formula but i could not.
    anyone can help?
     
  2. jcsd
  3. Oct 20, 2007 #2

    arildno

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    Well, we have:
    [tex]P_{(n+1)}(x)=\frac{1}{(n+1!)2^{(n+1)}}\frac{d^{n}}{dx^{n}}(\frac{d}{dx}(x^{2}-1)^{n+1}))=\frac{2(n+1)}{(n+1!)2^{(n+1)}}\frac{d^{n}}{dx^{n}}(x*(x^{2}-1)^{n})=\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x*(x^{2}-1)^{n})[/tex]

    For a function h(x)=f(x)*g(x), we have:
    [tex]\frac{d^{n}h}{dx^{n}}=\sum_{i=0}^{n}\binom{n}{i}f^{i}g^{n-i}[/tex]
    where the upper indices of f and g signifies the i'th and the (n-i)-th derivative, respectively.

    Using this relation with f(x)=x, and that all terms are zero beyond i=1, you may transform the expression for the (n+1)'th Legendre polynomial in term of a recurrence relation involving the n'th and (n-1)'th Legendre polynomials.
     
  4. Oct 20, 2007 #3
    sorry for bothering but can you please make it again with using the n'th derivative
     
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