Legendre polynomials and binomial series

LANS
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Homework Statement


Where P_n(x) is the nth legendre polynomial, find f(n) such that
\int_{0}^{1} P_n(x)dx = f(n) {1/2 \choose k} + g(n)

Homework Equations



Legendre generating function:
(1 - 2xh - h^2)^{-1/2} = \sum_{n = 0}^{\infty} P_n(x)h^n

The Attempt at a Solution



I'm not sure if that g(n) term is necessary.

First I integrate both sides of the generating function on 0->1. I can then replace the (1+h^2)^1/2 term with a binomial series. I'm not sure how to cancel out the rest of the factors to solve for P_n. Any help would be appreciated, thanks.
 
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LANS said:

Homework Statement


Where P_n(x) is the nth legendre polynomial, find f(n) such that
\int_{0}^{1} P_n(x)dx = f(n) {1/2 \choose k} + g(n)


Homework Equations



Legendre generating function:
(1 - 2xh - h^2)^{-1/2} = \sum_{n = 0}^{\infty} P_n(x)h^n

The Attempt at a Solution



I'm not sure if that g(n) term is necessary.

First I integrate both sides of the generating function on 0->1. I can then replace the (1+h^2)^1/2 term with a binomial series. I'm not sure how to cancel out the rest of the factors to solve for P_n. Any help would be appreciated, thanks.

The generating function is
\frac{1}{\sqrt{1-2xh + h^2}},
not what you wrote (see, eg., http://en.wikipedia.org/wiki/Legendre_polynomials ).

Anyway, why try to solve for P_n? When you integrate you just need the coefficient of h^n on both sides.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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