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Math Help - Leibniz's formula

  1. #1
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    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.
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    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....
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