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This is probably a lame question, but what is the general approach to solving $\log z +z \sigma +1=0$ for $z$? Wolfram Alpha obtains a Lambert W-function, but I don't quite see how.

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up vote 2 down vote accepted

Assuming $\sigma\neq 0$, you need something like Lambert's W function.

Taking exponentials of both sides, we have $$\begin{align*} e^{\log z + z\sigma + 1} &= 1\\ e^{\log z}e^{\sigma z} e&= 1\\ ze^{\sigma z}&= \frac{1}{e}\\ \sigma ze^{\sigma z} &= \frac{\sigma}{e} \end{align*}$$ Now let $x=\sigma z$. Then the equation is equivalent to $$xe^x = \frac{\sigma}{e},$$ which means that $x=W(\frac{\sigma}{e})$, hence $$z = \frac{x}{\sigma} = \frac{1}{\sigma} W\left(\frac{\sigma}{e}\right).$$

(If $\sigma=0$, then you just have $\log z + 1 = 0$, or $z=e^{-1}$.)

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This is kinda hard, is there any intuition behind W-function? At least of what order is it? – sigma.z.1980 Jun 11 '11 at 6:15
The $W$ function is, in a sense, a "magic" function. We know that $g(x) = xe^x$ is one-to-one when $x\geq e^{-1}$, so we know it has an inverse; $W$ is that inverse. As "magical", in a way, as the $\arcsin$ function. I'm not sure what you mean by "order" of the function. – Arturo Magidin Jun 11 '11 at 6:18
@sigma: I don't think that's a valid equation for the exponential (though I guess it works very near zero...); anyway, the Taylor series for $W$ around $0$ is $$W(x) = \sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}x^n.$$ – Arturo Magidin Jun 11 '11 at 18:21
yes you are right. I mean $e^x=1+x+O(x^2)$ – sigma.z.1980 Jun 12 '11 at 0:09

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