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I'm doing the $n$th derivate of $e^{-x^2}$ and I can't see how to generalize it:

$\frac{d^0 y}{dx^0}=e^{-x^2}$

$\frac{dy}{dx}=e^{-x^2}(-2x)$

$\frac{d^2y}{dx^2}=-e^{-x^2}(4x^2-2)$

$\frac{d^3y}{dx^3}=-e^{-x^2}(8x^3-12x)$

$\frac{d^4y}{dx^4}=-e^{-x^2}(16x^4-48x^2+12)$

I just see that some numbers of the polynomial are multiplying between each other in each iteration, but I can't see how to generalize it.

Gary
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Iker Sebastián Bali Elizalde
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    See the physicist's Hermite polynomials here: https://en.wikipedia.org/wiki/Hermite_polynomials#Definition – Gary Mar 02 '24 at 05:53
  • $y=f(x)e^{-x^2}\implies y'= f'(x)e^{-x^2}-2xf(x)e^{-x^2}=[f'(x)-2xf(x)]e^{-x^2} $. So $f_{n+1}(x)=[f'_n(x)-2xf_n(x)]$. $f_n(x)=\sum_{k=0}^\infty c_{nk}x^k $ This implies the coefficients of the derivative is some linear combination of the coefficients for the previous expression. – TurlocTheRed Mar 02 '24 at 06:34
  • Look up Ladder Operators for Hermite Polynomials – TurlocTheRed Mar 03 '24 at 01:12

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