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How could I possibly solve for $x$ in the following equation using the $W$ function?

$$\frac{1-x}xe^x = C$$ for some constant number, $C$?

I prepared the left-hand term as in the following equation hoping that I am closer to formulate it using the $W$ function, but ended up with more $xs$ on the right-hand side.

$$xe^x= C \cdot \frac{x^2}{1-x}$$

Any help appreciated. Thank you!

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    $\begingroup$ I tried $C=5$ with Wolfram alpha and received only a numerical solution, not one using the Lambert-w-function. Indication what we actually need numerical methods here. $\endgroup$ – Peter Aug 30 '17 at 12:14
  • $\begingroup$ Maybe you can use the series expansion for $e^x\cdot \frac{1-x}{x}$. Using the first two summands of the series expansion you´ll get an quadratic equation. $\endgroup$ – callculus Aug 30 '17 at 12:43
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As said in comments, there is no solution(s) in terms of the usual Lambert function.

However, writing the equation as $$e^{-x}=\frac{1-x}{Cx}$$ solutions could be expressed in terms of the generalized Lambert function.

You could be, at least out of curiosity, be interested by this recent paper.

Numerical methods would be required (Newton being the simplest).

However, if $C$ is not too large, we could use Pade approximants built around $x=1$ and get $$\frac{ (1-x)}{x}e^x =-e (x-1)\frac {1+\frac{18}{35} (x-1)+\frac{13}{140} (x-1)^2 } {1+\frac{18 }{35}(x-1)-\frac{57}{140} (x-1)^2+\frac{8}{105} (x-1)^3 }$$ which would reduce the problem to a cubic equation in $y=x-1$. $$(32 C+39 e) y^3-(171 C-216 e) y^2+(216 C+420 e) y+420 C=0$$ which has only one real root if $C>0$.

For illustration purposes, I produced below the approximate solution (from the cubic) as well as the exact solution for a few values of $C$.

$$\left( \begin{array}{ccc} C & \text{approximate} &\text{exact} \\ 1 & 0.659045 & 0.659046 \\ 2 & 0.436051 & 0.436086 \\ 3 & 0.313007 & 0.313142 \\ 4 & 0.241166 & 0.241428 \\ 5 & 0.195250 & 0.195635 \\ 6 & 0.163666 & 0.164159 \\ 7 & 0.140703 & 0.141290 \\ 8 & 0.123290 & 0.123957 \\ 9 & 0.109647 & 0.110382 \\ 10 & 0.098677 & 0.099471 \end{array} \right)$$

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