The Lambert W-function, i.e. the multivalued inverse of $z=we^w$, has countably many complex-valued branches $W_k(z)$. The relations between the branches are a bit involved and are summarized here. We will consider the behavior of the $k=0,-1$ branches for $x<0$. Using Mathematica, we obtain the following plots of $W_k(x)$ along the negative real axis:

enter image description here

Setting aside the green line for the moment, the two plots give the real and imaginary parts respectively of $W_0(x)$ (blue) and $W_{-1}(x)$ (orange). From this, we see that both branches are real for $x\in (-1/e,0)$. (The $k=0$ branch is additionally real for positive real $x$; no other branches obtain real values along the real line.) This is not surprising, as these branches correspond to the two real-valued inverses of $z=we^w$ along the real line.

What's perhaps surprising, though, is that (according to this plot) $\overline{W_{-1}(x)}=W_{0}(x)$ for all $x\in (-\infty,-1/e)$. (A derivation of this fact may be found in the Q&A linked in the comments below.) From this, we conclude that the average of these two branches, $\frac{1}{2}(W_0(x)+W{-1}(x))$, is real for all negative real $x$. This is the green line plotted above, and from the first plot we further ascertain that $\frac12 (W_0+W_{-1})$ is smooth across the point $x=-1/e$. (Further plotting in Mathematica suggests that $\frac12 (W_0+W_{-1})$ is analytic for all $x\neq 0$ such that $-\pi < \text{arg }x \leq \pi$.)

for all $x<0$ but not holomorphic across the real line.) By contrast, the two branches have square-root branching at $x=-1/e$.

This last property of $\frac{1}{2}(W_0+W_{-1})$ remains mysterious to me, so my question is:

Why is $\frac{1}{2}(W_0(x)+W{-1}(x))$ a smooth function for all real $x<0$?

  • 2
    $\begingroup$ I am surprised that nobody answered your very interesting question. It is a nice observation. I suppose that you noticed the zero for $x=-\frac \pi 2$ $\endgroup$ Aug 3, 2017 at 14:59
  • $\begingroup$ @ClaudeLeibovici Thanks! I hope someone will provide an equally interesting answer. As for the zero---no, I had entirely missed that. In retrospect, it's not so strange: the equation $-\pi/2=we^w$ has the purely imaginary solutions $w=\pm i \pi/2$, and these evidently correspond to the two main branches. $\endgroup$ Aug 3, 2017 at 15:40
  • 1
    $\begingroup$ my question here seems to be related: math.stackexchange.com/questions/1652745/… $\endgroup$
    – tired
    Aug 3, 2017 at 21:25
  • $\begingroup$ @tired Indeed, I think that the answer to your question resolves the reality part of my question. (That it's smooth across $z=-1/e$, however, still seems mysterious.) $\endgroup$ Aug 3, 2017 at 21:28
  • $\begingroup$ For $\Im(z)\geq 0$ and near the branch point, you can express the average as a convergent expansion using the known series expansions of $W_0$ and $W_{-1}$ near the branch point. The terms involving the square roots of $e z+1$ cancel and you end up with a power series that is convergent around $z=1/e$. This should explain the analyticity at the branch point. $\endgroup$
    – Gary
    Nov 30, 2017 at 11:59

2 Answers 2


It’s not a full answer, I only want to show why $\,\displaystyle\frac{W_0(x)+W_{-1}(x)}{2}\,$ is real for $\,x<0\,$ .

We can parameterize $\,\displaystyle W_0(x)=-\ln ((1+\frac{1}{t})^t) \,$ and $\,\displaystyle W_{-1}(x)=-\ln ((1+\frac{1}{t})^{t+1})\,$

(where $\,t\,$ is complex) and it follows $\,\displaystyle \frac{W_0(x)+W_{-1}(x)}{2}=-\ln ((1+\frac{1}{t})^{t+\frac{1}{2}})\,$ .

Using e.g. $\,\displaystyle t:=\frac{1}{e^{-i\alpha}-1}\,$ with real $\,\alpha\,$ we can express the complex line of $\,W_0(x)\,$ by $\,\displaystyle \frac{i\alpha}{e^{-i\alpha}-1}\,$

and $\,W_{-1}(x)\,$ by $\,\displaystyle \frac{-i\alpha}{e^{i\alpha}-1}\,$ so that we see $\,\overline{W_{-1}(x)}=W_0(x)\,$ .

It follows $\,\displaystyle \frac{W_0(x)+W_{-1}(x)}{2}=-\ln ((1+\frac{1}{t})^{t+\frac{1}{2}})= -\frac{\alpha}{2}\cot\frac{\alpha}{2}\,$ which is real (for $\,x<0\,$).

  • $\begingroup$ +1. Do you have a reference where such a parameterization is introduced/mentioned? $\endgroup$
    – g.kov
    Sep 14, 2020 at 5:59
  • $\begingroup$ @g.kov : (Sorry for the delay, for a while I am not active here.) I don't know where such a parametrization is introduced, I only used the fact $\,{x_1}^{1/x_1}={x_2}^{1/x_2}\,$ where $\,x_1\neq x_2\,$ for $\,x_1:=(1+\frac{1}{t})^ t\,$ and $\,x_2:=(1+\frac{1}{t})^{1+t}\,$ . $\endgroup$
    – user90369
    Sep 28, 2020 at 14:57

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

Using the parametric representation, of $\Wp(x),\,\Wm(x)$ we arrive at

\begin{align} \Wp(\alpha) &=-\tfrac\alpha2\,\cot\tfrac\alpha2-\tfrac\alpha2\cdot\i \tag{1}\label{1} ,\\ \Wm(\alpha) &=-\tfrac\alpha2\,\cot\tfrac\alpha2+\tfrac\alpha2\cdot\i \tag{2}\label{2} ,\\ x(\alpha)&= -\frac{\alpha}{2\sin\tfrac\alpha2} \cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2) \tag{3}\label{3} \end{align}

for $x\le-\tfrac1\e$, $\alpha\ge0$, since

\begin{align} \lim_{\alpha\to0}x(\alpha)=-\tfrac1\e \tag{4}\label{4} . \end{align}

Combining this with the parametric representation of $\Wp(x),\,\Wm(x)$ we have \begin{align} \Wp(a)&=\frac{a\ln a}{1-a} \tag{5}\label{5} ,\\ \Wm(a)&=\frac{\ln a}{1-a} \tag{6}\label{6} ,\\ x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}} \tag{7}\label{7} \end{align}
for $x\in[-\tfrac1\e,0]$, $a\in[0,1]$.

The function of interest

\begin{align} f(x)&=\tfrac12\,(\Wp(x)+\Wm(x)) \tag{8}\label{8} \end{align}
for $x<0$ is then represented as two pieces \begin{align} f_1(\alpha)&=-\tfrac\alpha2\,\cot\tfrac\alpha2 \tag{9}\label{9} ,\\ x(\alpha)&= -\frac{\alpha}{2\sin\tfrac\alpha2} \cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2) ,\quad \alpha>0 \tag{10}\label{10} ,\\ \text{and }\quad f_2(a)&=\tfrac12\,\ln a\cdot\frac{1+a}{1-a} \tag{11}\label{11} ,\\ x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}} ,\quad a\in[0,1] \tag{12}\label{12} ,\\ \lim_{\alpha=0}x(\alpha)&= \lim_{a=1}x(a)= -\tfrac1\e \tag{13}\label{13} ,\\ \lim_{\alpha=0}f_1(\alpha)&= \lim_{a=1}f_2(a)=-1 \tag{14}\label{14} . \end{align}

Omitting details of calculation, this leads to the desired result,

\begin{align} \left. \frac{d f_1(\alpha)/d\alpha}{dx(\alpha)/d\alpha} \right|_{\alpha=0} &= \left. \frac{d f_2(a)/da}{dx(a)/da} \right|_{a=1} =-\tfrac23\e \tag{15}\label{15} ,\\ \left. \frac{d^2 f_1(\alpha)/d\alpha^2}{d^2x(\alpha)/d\alpha^2} \right|_{\alpha=0} &= \left. \frac{d^2 f_2(a)/da^2}{d^2x(a)/da^2} \right|_{a=1} =-\tfrac23\e \tag{16}\label{16} , \end{align}

that means that $f(x)$ is indeed $C^2$-continuous at $x=-\tfrac1\e$.

Surprisingly, the value of first and second derivation at this point is the same.



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .