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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.

  • 0
    Put $y=1$ in the question I linked to. The answer is that $x=W(1)$ where $W$ is a certain nonelementary function. You won't be able to write down any explicit expression for $x$ in terms of elementary functions, if that's what you're after.2011-10-28

3 Answers 3

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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: $x\approx 0.567143$

graph

  • 0
    Perhaps one should add: sometimes $y$ou can find a solution if you e$x$pand the expression under study as a powerseries and invert this formally. It is possible that you find one which you know from other context (or which is composed of others) or that it approximates satisfactory (say exp(x) can also only be given as approximation if not left as algebraic symbol but we know we *could* compute it to arbitrary precision). Now the powerseries for x*exp(x) can formally be inverted ... however unfortunately ... in this case you were again lost, because it does not converge at x=12011-10-28
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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:

Table of x_k

and this plot of $f(x)$:

Plot of f(x)

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    Thank you for this different approach. That was not at all what I was looking for, but +1 anyway, because I think it's cool and gives a good approximation of values.2011-10-31
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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...