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Is there a good way of showing that ${d^n\over dx^n}(x^2-1)^n|_{x=1}=2^nn!$? I have tried  binomial expanding the thing then differntiate term-by-term, which seems a bit clumsy. Perhaps there's a closed form for ${d^n\over dx^n}(x^2-1)^n$? Thanks.

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    As Steven Stadnicki says, http://en.wikipedia.org/wiki/Rodrigues%27_formula coupled with $P_n(1)=1$ gives the statement (no proof) you need.2011-11-03

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By applying the product rule repeatedly, we get $ \frac{\mathrm{d}^n}{\mathrm{d}x^n}(fg)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}g^{(k)} $ let $f=(x-1)^n$ and $g=(x+1)^n$, then at $x=1$, all terms but the $\binom{n}{0}f^{(n)}g^{(0)}$ term are annihilated by $f^{(n-k)}(1)$. Since $f^{(n)}(1)=n!$ and $g^{(0)}(1)=2^n$, we get that $\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x^2-1)^n=2^nn!$

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Let $f(t) = ((1+t)^2 - 1)^n$. You want $f^{(n)}(0)$. Now from the Maclaurin series $f(t) = \sum_{j=0}^\infty \frac{f^{(j)}(0)}{j!} t^j$, $f^{(n)}(0)/n!$ is the coefficient of $t^n$ in $f(t)$. But $f(t) = (2t+t^2)^n = (2 + t)^n t^n$, so the coefficient of $t^n$ there is $2^n$.

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    I was sure there had to be some Taylor-series way of doing it but just wasn't finding it. Nice!2011-11-03
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I'm not sure how to make it rigorous, but when you take the $n$ derivatives, you need to apply it to the $(x^2-1)^n$ term every time, not to the powers of $2x$ that come from differentiating the inside. Otherwise you will have at least one factor of $x^2-1$ left, which will give $0$. Each time you differentiate you get a factor $2x$ times the exponent. The exponent decreases from $n$ down to $1$, giving the $n!$ term. You wind up with $(2x)^n n! + (x^2-1)(stuff)$, which evaluated at $x=1$ gives $2^nn!$

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    I thi$n$k my answer makes your idea rigorous.2011-11-03