$n$th derivative of $e^{-x^2}$

Question

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.

Answer 1

Hint: If the recurrence relation holds then the $n$-th derivative is a telescopic sum

Answer 2

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.

Answer 3

The question is answered here:

• Hermite polynomials recurrence relation

More informatiion at:

• Hermite polynomial

The keywords are "Hermite polynomial" and "Rodrigues formula".