Leibniz's formula

Printable View

October 30th 2008, 01:45 PM
free_to_fly

Leibniz's formula

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

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

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

Help would be hugely appreciated.
October 30th 2008, 02:45 PM
Peritus

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

$\frac{{d^2 }}<br /> {{dx^2 }}e^{ - \frac{{x^2 }}<br /> {2}} = \left( {x^2 - 1} \right)e^{ - \frac{{x^2 }}<br /> {2}}$

so after substituting into the differential equation we get:

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

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

All times are GMT -8. The time now is 10:46 PM.

Copyright © 2005-2014 Math Help Forum. All rights reserved.

Copyright © 2005-2014 Math Help Forum. All rights reserved.