Lambert Function Question

Question

Given $y$, for the Lambert $W$ function $y=xe^x\implies W(y)=x$, can you determine $x$? Specifically, how would you evaluate $e^{W(-1/3)}$ without a calculator?

Answer 1

Numerical methods are the only way if you don't have access to Lambert function. Newton iterative schemes converge quite fast.

Suppose that you want to solve $x e^x= k$. Provided a guess $x$, it will be updated by $$\frac{k e^{-x}+x^2}{x+1}$$ if you use Newton method.

For illustration purposes, let me consider the case where $k=100$ and admit that I am so lazy that I start iterating at $x=2$.

For Newton method, the successive iterates will then be $5.84451$, $5.03292$, $4.30675$, $3.74915$, $3.45533$, $3.38852$, $3.38564$, $3.38563$ which the solution.

However, if you write the equation as $x +\log x=\log k$, the updating formula will be $$\frac{x (\log (k)-\log (x)+1)}{x+1}$$ For the same example as before, the successive iterates will then be $3.27468$, $3.38520$, $3.38563$ which the solution.

Answer 2

$$x\cdot\exp(x)=a\iff x=a\cdot\exp(-x)\iff x=a\cdot\exp(-a\cdot\exp(-a\cdot\exp(\ldots)))$$