TITLE: How to solve $x\cdot\mathrm e^x=1$?
QUESTION [8 upvotes]: Possible Duplicate:
Inverse of $y=xe^x$ 

I would like to solve the equation $x \cdot\mathrm e^x=1$.
I know it has an answer, I could find it with a calculator, but I don't remember how to solve it on paper.
Any help?

edit
I know the answer is $x \approx 0.567143$. I don't want the answer, I want a method to find it.

REPLY [4 votes]: Newton-Rapshon Theorem:
$f(x)=xe^x-1$ (your function).
$f'(x)=e^x x+e^x$ (the derivative of your function).
After this, put in the Newton-Rapson formula:
$x_{k}=x_{k-1}-\frac{f(k-1)}{f'(k-1)}$
and define a first attempt, for example, $x_0=1$.
The final formula, for your equation will look like this:
$x_{k}=\frac{x_{k-1}^2+e^{-x_{k-1}}}{x_{k-1}+1}$, $\quad$
$x_0=1$.
Now you need to find this $x_k$ values because, the limit of $x_k$ when $k$ tends to infinity solves the original equation.
See this table:

and this plot of $f(x)$: