Find the inverse and the asymptotic expansion of a logarithmic equation $x(y)=y \ln(y)$

Question

Concerning the logarithmic equation $x(y)=y\ln(y)$, is there an explicit expression for its inverse function $y=y(x)$?
Mathematica returns $y=\frac{x}{\text{ProductLog}[x]}$, which should be wrong since its limit for $x\to 0^-$ is $1$ rather than the correct answer $0^+$.
I am specifically interested in the behavior $x\to 0^-$ and $y\to 0^+$, and looking for an asymptotic expansion. Implied by iteration, I guess that:
$y=\frac{x}{\ln(-x)}+o[\frac{x}{\ln^2(-x)}]$
Is this true? What exactly are next orders?

Edit

One more related question: how do Mathematica obtain the form $\frac{x}{\text{ProductLog}[x]}$ (here $\text{ProductLog}$ is Lambert-W function as some answers already pointed out)?
What I've tried is to use substitution $y=\exp(t)$ and obtained $y=\exp[W(x)]$. These two forms are the same. While I still would like to want to know how.

Answer 1

To understand what's going on, it helps to know a bit about Mathematica's ProductLog[k,x] function, known more commonly as the Lambert-W function. It arises from trying to invert the function $x(y)=ye^y$. The tricky thing is that, as a graph quickly shows, this function is not one-to-one for all real $y$, and hence there can be no inverse function valid for all $x$. Instead, one must split the domain as $(-1/e,\infty)$ and $(-\infty,-1/e)$. The function $x(y)=ye^y$ then has the inverse functions $y=W_0(x)$ and $y=W_{-1}(x)$. If we want to get values of $y\in(-1/e,\infty)$, we use the former; otherwise, we use the latter. These are denoted in Mathematica by ProductLog[0,x]=ProductLog[x] and ProductLog[-1,x]. (There are in fact an infinity of complex-valued branches, but they aren't relevant here.)

To see the relevance, a quick graph of $x(y)=y\ln y$ shows that there's a similar issue here: the function is only one-to-one for $y>0$. (To connect with your numerical observation, note that the line $x=0$ intersects with the graph of $x(y)=y\ln y$ at both $y=0$ and $y=1$.) As a consequence, there are actually two inverse functions corresponding to the two branches; evidently, you want the other one (i.e. replace ProductLog[x] with ProductLog[-1,x]). A quick use of Mathematica's Limit command confirms that $x/W_{-1}(x)\to 0$ from above as $x\to 0^{-}$.

Answer 2

$y\log{y}$ is not injective on $(0,1)$, so Mathematica is giving you one branch of the inverse, which starts at $x=-1/e$, $y=1/e$ and then increases monotonically through $y=1$ at $x=0$ and so on. You can access the other branch, which starts at the same point and decreases to $x=0$, $y=0$ by replacing ProductLog[x] with ProductLog[-1,x].

It's quite difficult to approximate this as $x \uparrow 0$: the DLMF offers $$Wm(x) = -\log{(-1/x)}-\log{\log{(-1/x)}}-\frac{\log{\log{(-1/x)}}}{\log{(-1/x)}} + \dotsb, $$ where $Wm$ is what Mathematica calls ProductLog[-1,#]&. One could then divide $x$ by this and use the Taylor expansion of $1/(1-z)$ to get the next couple of terms in the series you are interested in, but it doesn't seem like a good idea to complicate this expansion further!

Answer 3

$$x=y\log(y)\implies y=\frac{x}{W(x)}$$ Since you are looking at $x$ around $0^+$, use the expansion $$W(x)=x-x^2+\frac{3 x^3}{2}+O\left(x^4\right)$$ and long division will give $$y=\frac{x}{W(x)}=1+x-\frac{x^2}{2}+\frac{2 x^3}{3}+O\left(x^4\right)$$ If you want better approximations, consider Padé approximants such as $$\frac{3 x+2}{x+2}$$ $$\frac{101 x^2+174 x+60}{17 x^2+114 x+60}$$ $$\frac{18881 x^3+59724 x^2+48780 x+11400}{1927 x^3+28044 x^2+37380 x+11400}$$