I have an equation involving the $W(x)$ function.

Question

This is the equation:

$e^{-x}=W\left(\left(xe^{x}\right)^{-1}\right)$

I was trying to find similarities between $\exp(x)$ and $\ln(x)$. Then I starting working on this equation:

$-\exp(-x)=\ln(x)$

By algebra I came across that big fat equation on the extreme top.

You can see these 2 equations are the same, I tried Wolfram Alpha and got this value:

$x ≈ 0.56714329...$.

But anyway it would be helpful if anybody can help me give a intuitive way to derive the value for $x$.

Answer 1

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align}\e^{-x}&=\W\left(\left(x\e^{x}\right)^{-1}\right)\tag{1}\label{1}\end{align}

\begin{align} \e^{-x}\exp(\e^{-x}) &= \W\left(\left(x\e^{x}\right)^{-1}\right) \exp\left(\W\left(\left(x\e^{x}\right)^{-1}\right)\right) \tag{2}\label{2} ,\\ \e^{-x}\exp(\e^{-x}) &= \left(x\e^{x}\right)^{-1} \tag{3}\label{3} ,\\ \exp(\e^{-x}) &= x^{-1} \tag{4}\label{4} ,\\ \e^{-x} &= -\ln x \tag{5}\label{5} ,\\ -x\e^{-x} &= x\ln x \tag{6}\label{6} . \end{align}

Both sides of \eqref{6} are in the form $u\exp u$, so we can apply $\W$ function to simplify it: \begin{align} \W(-x\e^{-x}) &= \W(x\ln x) \tag{7}\label{7} ,\\ -x &= \ln x \tag{8}\label{8} . \end{align} And \eqref{8} can be easily rearranged to $u\exp u$ form as well:

\begin{align} \exp(-x) &= x \tag{9}\label{9} ,\\ x\exp(x)&=1 \tag{10}\label{10} \end{align}

hence we can apply $\W$ again to get the answer:

\begin{align} \W\left(x\exp(x)\right)&=\W(1) \tag{11}\label{11} ,\\ x&=\W(1)=\Omega \tag{12}\label{12} . \end{align}

For the $\Omega$ constant see A030178.

$\endgroup$

Answer 2

There are a few main ways to find the answer. The first way is without using the W-Lambert function nesting functions proven in this graph. Assume that x=k in the left hand side:

$$\mathrm{-e^{-x}=ln(x)\implies x=-ln(-ln(…-ln(k)…))=e^{-e^{…^{-k}}}= 0.5671432904097838729999686622103555497538157871865125081351310792230457930866}$$

One can essentially “undo” recursion to find a formula:

For the logarithm answer:

$$\mathrm{-e^{-x}=ln(x)\implies e^{-x}=-ln(x)\implies -x=ln(-ln(x))\implies x=-ln(-ln(x))}$$ This just isolated x in the left hand side. If we take x=-ln(x) then we can plug in -ln(x) for x or just x=-ln(-ln(x)). Since we can do this, we can undo the recursion to get x=-ln(x) and solve as seen above from there:

$$\mathrm{x=-ln(-ln(x))\implies x=-ln(x)\implies e^x=\frac{1}{x}\implies xe^x=1\implies x=W(1)=Ω}$$

From the recursive exponential answer: $$\mathrm{x=e^{-e^{-x}}\implies e^{-x}=x\implies 1=xe^x\implies x=W(1)=Ω}$$

The answer is that $x=\Omega=W(1)$. This is called the Omega constant. Please correct me and give me feedback!