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Legendre polynomial of zero

  1. Mar 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Using the Generating function for Legendre polynomials, show that:
    ##P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}##

    2. Relevant equations
    Generating function: ##(1-2xt+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(x) t^n##

    3. The attempt at a solution
    I put ##x=0## in the generating function and got ##(1+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(0) t^n## Now, I think I should expand ##(1+t^2)^{-1/2}## using the binomial theorem, then I get an expression that's only valid for even values of ##n##, but it's not the expression stated in the question. So, how can I solve this?
     
  2. jcsd
  3. Mar 18, 2016 #2

    Ray Vickson

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    So, what does that tell you about ##P_n(0)## for odd ##n##?
     
  4. Mar 22, 2016 #3

    vela

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    "Valid" isn't the right word here. The expansion only contains terms with even exponents, so what does that tell you about the coefficients of the odd terms?
     
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