Recurrence relation with derivatives

Question

I have to show the following:

\begin{equation} D^{n+1}e^{-x^2} = -2x D^n e^{-x^2}-2nD^{n-1}e^{-x^2} \end{equation}

where $D^{n}=\frac{d^n}{dx^n}$

Would you give me a hint where to start?

Answer 1

You can do it easily with induction.

The inductive step: let $D^{k+1}e^{-x^2} = -2x D^k e^{-x^2}-2kD^{k-1}e^{-x^2}$ for some non-negative integer $k$.

Then, differentiating both sides with respect to $x$ (applying product rule on the right hand side):

$D^{k+2}e^{-x^2} \\ = -2(D^ke^{-x^2} + xD^{k+1}e^{-x^2}) - 2(kD^ke^{-x^2}) \\ = -2xD^{k+1}e^{-x^2} - 2D^ke^{-x^2} - 2kD^ke^{-x^2}\\ = -2xD^{k+1}e^{-x^2} - 2(k+1)D^ke^{-x^2}$

which proves the inductive step.

All that remains is to show it holds for $k=0$, i.e. just find the first derivative of $e^{-x^2}$ and demonstrate trivially that it equals the right hand side (remember $D^0 e^{-x^2} = e^{-x^2}$) and you're done.