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Legendre polynomial integration

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    int x^m*P_n(x) dx=0 where integration is from (-1) to (+1).Given m<n

    2. Relevant equations



    3. The attempt at a solution

    I took integrand F(x) and saw that F(-x)=(-1)^(m+n)*F(x)
    should that help anyway?
     
  2. jcsd
  3. May 6, 2007 #2
    um....the integration is from -1 to 1, so first thing come to my mind is to make good use of the orthogonality property of Legendre polynomials.
    then...may be you can try Mathematical Induction for both odd and even polynomails with leading terms x^m , m< n

    you may have a look on
    http://en.wikipedia.org/wiki/Legendre_polynomial

    it may not be a fast and smart method..
    but i think it may works

    can anyone think of any better way??:smile:
     
  4. May 6, 2007 #3
    Regarding the orthogonality,it appears that x^m needs to be equal to some P_m(x).It would be too much particular case.Isn't it?

    Boas suggests to go for integration by parts.But it is so much horrible that I am stuck after 1st step.Can anyone please help?
     
  5. May 6, 2007 #4
    OK,you were correct.Because,
    the Legendre polynomials are complete on [-1,1], so:

    x^m=a linear combination of P_0,P_1,P_2,...P_m
    where m<n

    Hence the result follow directly from orthogonality.
     
  6. May 6, 2007 #5
    Yes.
    you got the point.
    just rewrite x^m in terms of Pn(x) and you may neglect those constants
    in this case.
     
  7. May 6, 2007 #6
    No need to neglect the constants.They simply go outside the integrals.
     
  8. Oct 20, 2007 #7
    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?
     
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