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.
Nice way of showing the following equality?
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calculus
combinatorics
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0These are the Legendre polynomials, and the expression in terms of the differential is known as the Rodrigues formula; a bit of digging using either of those phrases will probably turn up useful info for you. – 2011-11-03
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0As 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