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janvdl · #3 May 15th, 2008, 07:25 AM

Quote:

Originally Posted by mr fantastic

Let $F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)$

$\Rightarrow F(x) = \Pr \left( -\sqrt{\frac{2x}{M}} < V < \sqrt{\frac{2x}{M}} \right)$

$= \frac{1}{\sqrt{2 \pi}} \, \int_{-\sqrt{\frac{2x}{M}} }^{\sqrt{\frac{2x}{M}} } e^{-u^2/2} \, du$

$= \frac{2}{\sqrt{2 \pi}} \, \int_{0}^{\sqrt{\frac{2x}{M}} } e^{-u^2/2} \, du$ .

Therefore $f(x) = \frac{dF}{dx} = \frac{2}{\sqrt{2 \pi}} \, e^{-x/M} \, \left (\frac{1}{\sqrt{2Mx}} \right) \,$ , $x > 0$
where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative

$= e^{-x/M} \, \frac{1}{\sqrt{\pi M x}} \,$ , $x > 0$

and f(x) = 0 for x < 0.

Thank you Mr F, I will work through this now and try to understand