Legendre polynomials formula

[inferi]

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 until n=7,I did that because i wanted to find something that i can build my formula but i could not.
anyone can help?

 

[arildno]

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

For a function h(x)=f(x)*g(x), we have:
$$\frac{d^{n}h}{dx^{n}}=\sum_{i=0}^{n}\binom{n}{i}f^{i}g^{n-i}$$
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.

 

[therector24]

sorry for bothering but can you please make it again with using the n'th derivative