I'd like to know if the function $ y = f(x) : [0,1] \rightarrow [0,1]$ defined implicitly by the transcendental equation $$\displaystyle y = e^{(y-1)/x}$$ is "well known" (name, properties) or is related to some other well known function.
Implicit function $y = e^{(y-1)/x}$
2 Answers
It is solvable employing the Lambert W function as follows.
Your equation is $\rm\quad\quad\ Y \ \: =\ e^{\:Z-YZ}\ \: $ for $\rm\: \ Z\ =\: -\frac{1}X$
$\rm\quad\quad\iff\ \ Y\:Z\ e^{Y\:Z}\ \:=\:\ Z\ e^Z$
$\rm\quad\quad\iff\ \ Y\:Z\ \ \:=\:\ \ W(Z\ e^Z) $
$\rm\quad\quad\iff\ \ Y\: =\: -X\ W(-\frac{1}X e^{-1/X}) $
To give a more detailed solution, we start by rearranging the original equation as
$$y\exp\left(\frac{1-y}{x}\right)=1$$
and then
$$y\exp\left(\frac1{x}\right)\exp\left(-\frac{y}{x}\right)=1$$
$$-y\exp\left(-\frac{y}{x}\right)=-\exp\left(-\frac1{x}\right)$$
$$-\frac{y}{x}\exp\left(-\frac{y}{x}\right)=-\frac1{x}\exp\left(-\frac1{x}\right)$$
which we can now invert to
$$-\frac{y}{x}=W\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$
and finally we have
$$y=-xW\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$
Though the Lambert function has two real branches, only the "principal branch" gives a nontrivial solution; the solution corresponding to the $W_{-1}(\cdot)$ branch is $y=1$