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 $p_n$?
Higher derivatives of an exponential function
1
$\begingroup$
calculus
real-analysis
derivatives
-
0Related: http://math.stackexchange.com/questions/4700/improve-my-proof-about-this-c-infty-function/ – 2010-11-26
2 Answers
6
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.
-
0There's the [Rodrigues formula](http://functions.wolfram.com/05.01.07.0003.01) among other things... ;) – 2010-11-26
-
0@J.M. Rodrigues formula? I clicked on your link and I didn't find it. – 2010-11-26
-
0@TCL: *That* is the Rodrigues formula for the Hermite polynomials. – 2010-11-26
-
1@J. M.: That was a cruel joke. :) – 2010-11-26
1
$p_n$ 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.