How to solve $x\cdot\mathrm e^x=1$?

Question

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?

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edit

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

Answer 1

You can easily verify that there is only one solution:

1. if $x\leq0$ then $x\cdot\mathrm e^x\leq0<1$;

2. if $x>0$ then $x\cdot\mathrm e^x=1$ iff $\mathrm e^x = \frac1x$ (see the graph below); indeed, $\mathrm e^x$ increases and $\frac1x$ decreases on the set $\{x>0\}$ so there is no more than one solution. The solution exists since $\mathrm e^{0.1}<10$ but on the other hand $\mathrm e^1>1$ and hence by Intermediate Value Theorem there is a point $x\in (0.1,1)$ such that $\mathrm e^x = \frac1x$. This point you can easily find numerically: $x\approx 0.567143$

Answer 2

There are infinitely many complex numbers that satisfy your equation. Put another way, the Lambert function, which is the inverse function of $x\cdot\exp\,x$, has many branches. The unique real solution is $W(1)\approx0.567143290409783873$; the other complex solutions include $W_{-1}(1)\approx -1.5339133197935745079 - 4.3751851530618983855\,\, i$ and $W_1(1)\approx -1.5339133197935745079 + 4.3751851530618983855\,\,i$, among others...

Answer 3

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 $f(x)$: