# Thread: Leibniz's formula

1. ## Leibniz's formula

Use Leibniz's formula to establish that:
$\displaystyle Z_n = \frac{{d^n }}{{dx^n }}e^{ - \frac{{x^2 }}{2}}$
is a solution of the differential equation $\displaystyle \frac{{d^2 Z_n }}{{dx^2 }} + x\frac{{dZ_n }}{{dx}} + (n + 1)Z_n = 0$

I know that Leibniz's formula gives the nth derivative of a product of functions, but in this case $\displaystyle e^{ - \frac{{x^2 }}{2}}$ isn't a product.

So far I've got:
$\displaystyle \begin{array}{l} y = e^{ - \frac{{x^2 }}{2}} \\ \frac{{dy}}{{dx}} = - xe^{ - \frac{{x^2 }}{2}} \\ \frac{{d^2 y}}{{dx^2 }} = 3xe^{ - \frac{{x^2 }}{2}} - x^3 e^{ - \frac{{x^2 }}{2}} \\ \end{array}$
but I'm at a loss as to how to generalise to n.

Help would be hugely appreciated.

2. first of all your second derivative is wrong. The correct expression is:

$\displaystyle \frac{{d^2 }} {{dx^2 }}e^{ - \frac{{x^2 }} {2}} = \left( {x^2 - 1} \right)e^{ - \frac{{x^2 }} {2}}$

so after substituting into the differential equation we get:

$\displaystyle \frac{{d^n }} {{dx^n }}\left( {\left( {x^2 - 1} \right)e^{ - \frac{{x^2 }} {2}} } \right) - x\frac{{d^n }} {{dx^n }}xe^{ - \frac{{x^2 }} {2}} + (n + 1)\frac{{d^n }} {{dx^n }}e^{ - \frac{{x^2 }} {2}} = 0$

HINT: $\displaystyle {\left( {x^2 - 1} \right)}$ has only 2 derivatives before it vanishes. and x has only one....