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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.

Community
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SteeveDroz
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  • http://en.wikipedia.org/wiki/Lambert_W_function – Hans Lundmark Oct 28 '11 at 12:42
  • Thanks @Hans for the comparision, but I don't think it's the same question. The links you gave don't answer me. – SteeveDroz Oct 28 '11 at 12:53
  • 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. – Hans Lundmark Oct 28 '11 at 13:28

3 Answers3

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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

SBF
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    Alternatively, you can also try intersecting $x$ and $\exp(-x)$. :) – J. M. ain't a mathematician Oct 28 '11 at 12:44
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    Thanks, that's all very comprehensible. My question was: is there a way to solve the equation on paper, without appromiation. I'm not looking for the answer, I'm looking for the method. – SteeveDroz Oct 28 '11 at 12:51
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    @Oltarus: nice to be of help, though solve the equation on paper is a bit incorrect problem. If I tell you that the answer is $\sin {(16! - 2\sqrt{15})}$ is it a solution (quite close to the real one, btw)? As you can read at Wiki page which Hans has cited, $W$ is not expressible through elementary function, so any answer will be either $x = W(1)$ or $x \approx 0.567$ – SBF Oct 28 '11 at 12:57
  • @oltarus: "is there a way to solve the equation on paper, without approxmiation." - so for example, if I say the solution of $\exp(x)=2$ is $x=\log,2$, you'll object? – J. M. ain't a mathematician Oct 28 '11 at 12:59
  • Hehe, I don't object, I might laugh. Ok, I understand: the function is not expressible except by $x=W(1)$. Thanks guys! – SteeveDroz Oct 28 '11 at 13:07
  • Perhaps one should add: sometimes you can find a solution if you expand 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=1 – Gottfried Helms Oct 28 '11 at 13:51
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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)

GarouDan
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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. – SteeveDroz Oct 31 '11 at 07:35
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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...

J. M. ain't a mathematician
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