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I observed that $f^{(n)}(x)= \begin{cases} e^{-x^2} & \text{if $n=0$}\\ -2xe^{-x^2} & \text{if $n=1$}\\ f^{(n-1)}(x)-f^{(n-2)}(x) & \text{otherwise.} \end{cases}$

How to get the closed form?
Edit: This recurrence does not hold.

k5f
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    How did you find this recurrence relation ? – lmsteffan Sep 06 '14 at 12:12
  • I guessed it and proved by induction. Is it wrong? – k5f Sep 06 '14 at 12:18
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    I think it is wrong. $f^{(2)}(x)=-2e^{-x^2}+(-2x)^2e^{-x^2}$. – Bart Michels Sep 06 '14 at 12:20
  • Did you try to validate it by applying it to $f^{(2)}$ ? I think it is not equal to $f^{(1)} - f^{(0)} $ – lmsteffan Sep 06 '14 at 12:21
  • I don't think that recurrence is true, since that would mean that $f^{(n)}(x)=(a_nx+b_n))f(x)$ for some $a_n,b_n$, and that is not the case- $f^{(2)}(x)=(4x^2-2)e^{-x^2}$. – Thomas Andrews Sep 06 '14 at 12:28
  • To obtain the right recurrence, write $f^{(n)}(x)=e^{-x^2}P_n(x)$ where $P_n$ is a sequence of polynomials.

    We get $f^{n+1}(x)=e^{-x^2}(P_n'(x)-2xP_n(x))$, hence $P_{n+1}(x)=P_n'(x)-2xP_n(x)$. There may be a nice closed form for the coefficients of $P_n$, consider the similar question http://math.stackexchange.com/questions/193702. (I just deleted my answer because I thought it would better be a comment.)

     – Bart Michels Sep 06 '14 at 12:35
  • @barto You could've left it as it was. I'm trying to find the sequence, it seems that the coefficent of the highest order of $n$th polynomial is $2^n$, except the first one. – k5f Sep 06 '14 at 12:43
  • Check this. – Mhenni Benghorbal Sep 06 '14 at 14:52

3 Answers3

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Hint: the recursion $f_n=f_{n-1}-f_{n-2}$ always has a solution of the form

$$ C_1(1/2+i\sqrt{3}/2)^n + C_2(1/2-i\sqrt{3}/2)^n, $$

with $C_1,C_2$ constants to be determined from the initial values $f_0,f_1$.

Edit: Your recurrence does not hold.

Kim Fierens
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  • He mentioned that the $n$-th derivatives obey a certain recurrence formula, with $f_n = D^n(e^{-x^2})$. Thus $f_n$ would be a legitimate sequence (of functions). But the point is moot anyway, since the recurrence he derived turns out to be incorrect. – Kim Fierens Sep 06 '14 at 12:55
  • Why not? The recurrence holds for every $x$ separately, and only the initial values explicitly depend on $x$. Example: $f_n(x)=2f_{n-1}(x)$, with $f_0(x)=x$. Then $f_n=2^nx=f_n(x)$. – Kim Fierens Sep 06 '14 at 13:05
  • I mean you cannot solve it that way unless you have had enough caffeine. – copper.hat Sep 06 '14 at 13:14
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Hint: If the recurrence relation holds then the $n$-th derivative is a telescopic sum

Belgi
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The question is answered here:

More informatiion at:

The keywords are "Hermite polynomial" and "Rodrigues formula".
Community
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Han de Bruijn
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