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Math Help - Using the generating function for the Legendre polynomials

  1. #1
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    Using the generating function for the Legendre polynomials

    Using the identity:

    \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} } with:

    x \in [-1,1], r \in (-1,1)

    Derive an expression for P_{2}(x) the legendre polynomial of degree 2.

    Any help here will be amazing as I absolutely haven't got a clue!

    Also from this it also asks to evaluate P_{3}'(0) but i take that if you can find P_{2}(x) then you can also find P_{2}'(0) and use the same method for P_{3}(x)

    Thank you
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  2. #2
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    Quote Originally Posted by garunas View Post
    Using the identity:

    \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} } with:

    x \in [-1,1], r \in (-1,1)

    Derive an expression for P_{2}(x) the legendre polynomial of degree 2.

    Any help here will be amazing as I absolutely haven't got a clue!

    Also from this it also asks to evaluate P_{3}'(0) but i take that if you can find P_{2}(x) then you can also find P_{2}'(0) and use the same method for P_{3}(x)

    Thank you
    you need to expand the right hand side in a Taylor series around zero

    so let

    f(r)=(1-2xr+r^2)^{-\frac{1}{2}}

    Now you get

    1+f'(0)r+\frac{f''(0)}{2!}r^2+\frac{f'''(0)}{3!}r^  3+...

    Now just equate coefficients this gives

    P_2(x)r^2=\frac{f''(0)}{2!}r^2 \implies P_2(x)=\frac{f''(0)}{2!}

    Can you finish from here?
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  3. #3
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    ...i can sure finish from here. Thanks a lot!
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  4. #4
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    Quote Originally Posted by garunas View Post
    Using the identity:

    \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} } with:

    x \in [-1,1], r \in (-1,1)

    Derive an expression for P_{2}(x) the legendre polynomial of degree 2.

    Any help here will be amazing as I absolutely haven't got a clue!

    Also from this it also asks to evaluate P_{3}'(0) but i take that if you can find P_{2}(x) then you can also find P_{2}'(0) and use the same method for P_{3}(x)
    \frac{1}{\sqrt {1-2xr+r^2} } = \bigl(1-(2xr-r^2)\bigr)^{-1/2}. Use the first few terms of the binomial expansion of (1-t)^{-1/2} (with t=2xr-r^2) to find the coefficient of r^2.

    You could get P_3'(0) by the same method. Alternatively, it might be slightly easier to start by differentiating \frac{1}{\sqrt {1-2xr+r^2} } with respect to x, then put x=0, and finally use the binomial series to find the coefficient of r^3.
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