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$p = \lambda e^{ -\lambda }$

If $p$ and $e$ are known. How can I calculate $\lambda$?

I tried on log on both side but it did not help. Any suggestions?

1 Answers 1

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You say that $e$ is known -- does that mean you're using $e$ as a variable name rather than to denote Euler's number? That would be rather confusing.

In either case, what you need is the Lambert W function. Using the defining relation

$$z=W(z)\mathrm e^{W(z)}$$

(where $\mathrm e$ as usual denotes Euler's number) and

$$-p=-\lambda\mathrm e^{-\lambda}\;,$$

you get

$$\lambda =-W(-p)\;.$$

Note that this is a) multivalued and b) only defined for $p\le1/e$.

If you really did intend to use $e$ as a variable name, we can replace it by $a$ to keep things clear and then write

$$ \begin{eqnarray} p &=& \lambda a^{-\lambda} \;, \\ p &=& \lambda \mathrm e^{-\lambda\log a} \;, \\ -p\log a &=& -\lambda\log a \mathrm e^{-\lambda\log a} \;, \\ \lambda &=& -\frac{W(-p\log a)}{\log a} \;. \end{eqnarray}$$

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    I think the LHS of your third line is $-p \log a$ and that $\lambda = -W(-p\log a)/\log a$.2011-12-07
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    @Patrick: Thanks -- already corrected; you were faster :-)2011-12-07