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Let $$p_n(x)e^{-x^2}$$ be the $n$th derivative of $$e^{-x^2}.$$ Find a formula for $p_n(x)$. We have $p_1(x)=-2x, p_2(x)=4x^2-2$, etc. But what is the general formula for $p_n$?

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Related: math.stackexchange.com/questions/4700/… –  Aryabhata Nov 26 '10 at 3:05
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up vote 6 down vote accepted

That depends on what you mean by "general formula." These are (up to some normalization) the Hermite polynomials. They satisfy a nice recurrence and have a nice generating function. You could torture some kind of general formula out of the generating function but I really don't see the point.

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There's the Rodrigues formula among other things... ;) –  J. M. Nov 26 '10 at 3:13
    
@J.M. Rodrigues formula? I clicked on your link and I didn't find it. –  TCL Nov 26 '10 at 3:24
    
@TCL: That is the Rodrigues formula for the Hermite polynomials. –  J. M. Nov 26 '10 at 3:26
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@J. M.: That was a cruel joke. :) –  Hans Lundmark Nov 26 '10 at 8:22
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$p_n$ satisfies the recurrence, $p_{n+1}(x) = p_n'(x) - 2xp_n(x)$ with $p_0(x) = 1$. This looks to me like it might give you some Tchebyshev polynomial. Something like $p_n(x) = 2 (-1)^n T_n(x)$ .

EDIT: I was wrong with the Tchebyshev polynomial thing. Sorry.

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