I have an expression and i am almost sure what it equals:
$ e^{-W_{-1}\left(-\frac{log\left(x\right)}{x}\right)} $
I only need a simplified version of this expression for $x\geq e$.
I assume:
$\forall x\in\mathbb{R}: \left(x\geq e\right)\Rightarrow \left(e^{-W_{-1}\left(-\frac{log\left(x\right)}{x}\right)}=x\right)$
I tested it with some values, but i did not find a general way to prove this. Maybe someone has an idea?
Thank you,
Regards Kevin
Let $X=-\frac{\ln(x)}{x}$
and $y=e^{-W(X)}$ hense $W(X)=-\ln(y)$
By definition of the $W$ function : $W(X) e^{W(X)}=X$
$-\ln(y) e^{-\ln(y)}=X$
$-\frac{\ln(y)}{y}=X=-\frac{\ln(x)}{x}$
$y=x$
$x=y=e^{-W(X)}= e^{-W\left(-\frac{\ln(x)}{x}\right)}$