I want to show that the coefficient of $x^n$ in $P_n(x)$ is $(2n)!/ (2^n(n!)^2)$
my problem is that I cannot find the the $n$-th derivative of $(x^2-1)^n$ to be able to simplify Rodrigues' formula of pn! can someone give me a hint?
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WolframAlpha says $$\sum_{k=0}^{n} \sum_{j=0}^{k} \frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = \frac{\Gamma( a + n)}{\Gamma(a)}$ is the pochhammer symbol and $\Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way. – ViktorStein Jan 18 '19 at 12:35
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Maybe look here – ViktorStein Jan 18 '19 at 12:41
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Possible duplicate of Proving a property of Legendre Polynomials – ViktorStein Jan 18 '19 at 12:41
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According to Rodrigue's formula,
$$P_n(x)=\left(\frac{1}{2^n n!}\right) \frac{d^n}{dx^n}\left(x^2-1\right)^n$$
The degree of the polynomial $q_n(x)=\left(x^2-1\right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) \dots (n+1) = \frac{(2n)!}{n!}$.
Multiplying by the coefficient $\left(\frac{1}{2^n n!}\right)$ you get the desired result $\frac{(2n)!}{2^n (n!)^2}$
mathcounterexamples.net
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Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator! – user635977 Jan 18 '19 at 12:51