Fixed point of a method

Question

Is there any way to compute the fixed point of this equation : $xe^{-x} = e^{-3}$.

If it is a quadratic equation then i can use $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ but i don't know how can i directly compute a fixed point for such equations.

I need this to compute the convergence factor which is given by $F^\prime(x^\prime)$ where $x^\prime$ is the fixed point.

Any help would be great!

Answer 1

Assuming you rearranged the equation as

$$x=e^{x-3}=:f(x),$$

(magenta curve) you have

$$f'(x')=e^{x'-3}=x'.$$

Now notice that

$$0<e^{0-3}$$ and $$1>e^{1-3}$$ and there must be a root $x'\in(0,1)$. Hence the fixed-point is stable.

Note that there is another solution in $(4,5)$ (as $e<4$ and $e^2>5$), corresponding to divergent iterations.

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The situation is reversed with $f(x):=\ln x+3$, that yields the same solutions (blue curve), but the stable fixed point in $(4,5)$.

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Additional comment:

It doesn't make much sense to solve the equation to discuss convergence, as the fixed-point method will most probably be used for the purpose of finding that root. It is more appropriate to determine a gross bracketing.

Answer 2

Thanks to Lambert function you can rewrite

$$ -(x)e^{-x}=-e^{-3} $$ which is

$$ x=-W_0(-e^{-3}) $$ or

$$x=-W_{-1}(e^{-3})$$ Hopefully well defined because $-e^{-3} \in [-e^{-1},+\infty[$