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I want to understand how this step is performed. Can you tell me that how this value of Po is obtained from the first equation.!

$$P_o=\frac{1}{\sum_{k=0}^\infty\left(\frac{\alpha}{u}\right)^k\frac{1}{k!}}$$

$$P_o=e^{-\frac{\alpha}{u}}$$

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    $\begingroup$ It's rather unreadable. Could you type it? $\endgroup$
    – Arpan
    Commented Mar 31, 2015 at 15:20
  • $\begingroup$ Is it better now.? $\endgroup$
    – A Ahmed
    Commented Mar 31, 2015 at 15:21
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    $\begingroup$ It appears false to me. Can't see all that well, but it appears that where there is multiplication by $k!$, it would need to instead be division. $\endgroup$
    – 2'5 9'2
    Commented Mar 31, 2015 at 15:21
  • $\begingroup$ yes sorry, it is 1/k! $\endgroup$
    – A Ahmed
    Commented Mar 31, 2015 at 15:25
  • $\begingroup$ I've edited it and now it is 1/k!. $\endgroup$
    – user26486
    Commented Mar 31, 2015 at 15:30

1 Answer 1

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$$P_o=\frac{1}{\sum_{k=0}^\infty\left(\frac{\alpha}{u}\right)^k.\frac{1}{k!}}$$

Substitute $\frac{\alpha}{u}=x$.

The denominator is $$f(x)=1+x+\frac{x^2}{2!}+\dots$$

Notice that this is the Taylor expansion for $e^x$, and this yields the answer. $$f(x)=e^x$$ $$P_0= \frac{1}{f(x)} = e^{-\frac{\alpha}{u}}$$

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    $\begingroup$ Thank you so much Arpan. God bless you. $\endgroup$
    – A Ahmed
    Commented Mar 31, 2015 at 15:36
  • $\begingroup$ Well... Thanks, I guess :) $\endgroup$
    – Arpan
    Commented Mar 31, 2015 at 15:36

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