# Lambert $W$-Function Branch Identities

Henri Cohen

###### Abstract

After defining in detail the Lambert $W$-function branches, we give a large
number of _exact_ identities involving (infinite) symmetric functions of these
branches, as well as geometrically convergent series for all the branches. In
doing so, we introduce a family of polynomials which may be of independent
interest.

##  1 Introduction: the Lambert $W$-Function Branches

We begin by recalling the definition of the complex branches of the Lambert
$W$-function. Originally, the Lambert $W$-function was defined as the solution
of the implicit equation $we^{w}=z$ with $z\in{\mathbb{R}}$. It is immediate
to see that for $z\geq 0$ there is a single solution, denoted $W_{0}(z)$, with
a power series expansion given by Lagrange inversion

$W_{0}(z)=\sum_{n\geq 1}(-1)^{n-1}\dfrac{n^{n-1}}{n!}z^{n}\;,$

This power series converges for $|z|\leq 1/e$ (here and in the sequel
$e=\exp(1)$), and provides an analytic continuation of $W_{0}(z)$ to the whole
interval $[-1/e,\infty[$. When $-1/e\leq z<0$ there exists a _second_ solution
$W_{-1}(z)$ characterized by $W_{-1}(z)\leq-1$. Note that
$W_{0}(-1/e)=W_{-1}(-1/e)=-1$ and that $W_{-1}(z)\to-\infty$ as $z\to 0^{-}$.
Finally, if $z<-1/e$ there is no real solution.

For completeness and comparison with later formulas, note that Lagrange
inversion allows us to prove additional expansions such as

$\displaystyle W_{0}(z)^{k}$ $\displaystyle=\sum_{n\geq
k}(-1)^{n-k}k\dfrac{n^{n-k-1}}{(n-k)!}z^{n}\text{\quad for all
$k\in{\mathbb{Z}}\setminus\\{0\\}$\;,\ and}$
$\displaystyle\dfrac{1}{W_{0}(z)+1}$ $\displaystyle=\sum_{n\geq
0}(-1)^{n}\dfrac{n^{n}}{n!}z^{n}\;.$

More generally, if $F$ if any power series, we have

$[z^{n}]F(W_{0}(z))=[t^{n}]F(t)(1+t)e^{-nt}\;,$

where $[z^{n}]$ extracts the coefficient of $z^{n}$, see [2].

It is natural to extend these definitions to all complex numbers. Taking
logarithms in the initial implicit equation, we have $w+\log(w)-\log(z)=0$, so
for $k\in{\mathbb{Z}}$ we would like to denote by $w_{k}=W_{k}(z)$ the
solution of the equation $(E_{k})$:

$w_{k}+\log(w_{k})=\log(z)+2k\pi i\;,$

where $\log$ denotes the principal determination of the logarithm. Note that
this is _not_ the same as $w_{k}+\log(w_{k}/x)=2k\pi i$.

Because of the branching properties of the logarithm we will see that this
equation may have $0$, $1$, or $2$ solutions, so we must be more careful.
Although the definitions are in the literature, for instance in [2], we prefer
doing everything from scratch. The precise result is as follows:111We
evidently use the much more sensible French notation $]a,b[$ and similar for
intervals, instead of the dreadful $(a,b)$ which has so many other different
meanings.

###### Proposition 1.1

Let $z\in{\mathbb{C}}\setminus\\{0\\}$, $k\in{\mathbb{Z}}$, and denote by
$(E_{k})$ the equation $w+\log(w)=\log(z)+2k\pi i$.

  1. 1.

If $z\notin]-\infty,0[$, equation $(E_{k})$ has a unique complex solution
denoted $W_{k}(z)$, and we have
$W_{-k}(z)=\overline{\vphantom{T}W_{k}(\bar{z})}$.

  2. 2.

If $z\in]-\infty,0[$ and $k\neq 0,-1$, or $z\in]-\infty,-1/e[$ and $k=0$ or
$k=-1$, equation $(E_{k})$ has a unique complex solution denoted $W_{k}(z)$,
and we have $W_{-k}(z)=\overline{\vphantom{T}W_{k-1}(z)}$.

  3. 3.

If $z\in[-1/e,0[$ and $k=-1$, equation $(E_{k})$ has no solution (this is the
only case in which there is no solution).

  4. 4.

If $z\in[-1/e,0[$ and $k=0$, equation $(E_{k})$ has two real solutions denoted
$W_{-1}(z)$ and $W_{0}(z)$ characterized by $W_{-1}(z)\leq-1\leq W_{0}(z)$,
which coincide if $z=-1/e$.

We give the detailed elementary proof, since it is sometimes obscured in the
literature. It can also be proved using the principle of the argument, but the
proof is essentially the same, see [4].

Proof. As usual denoting by $\log(y)$ the principal branch of the logarithm,
we can write $\log(z)=a+\lambda\pi i$ with $a\in{\mathbb{R}}$ and
$-1<\lambda\leq 1$. Equation $(E_{k})$ is $w+\log(w)=a+(2k+\lambda)\pi i$, so
writing $w=\rho e^{i\theta}$ with $-\pi<\theta\leq\pi$ gives the system
$S_{2k+\lambda}(a)$, where $S_{\mu}(a)$ is

$\rho\sin(\theta)+\theta=\mu\pi\text{\quad
and\quad}\rho\cos(\theta)+\log(\rho)=a\;.$

Let us first find the solutions with $\theta\in]0,\pi[$. The first equation
has the unique solution $\rho=(\mu\pi-\theta)/\sin(\theta)$, and since we have
$\rho>0$, we must have $\theta<\mu\pi$. Thus if $\mu\leq 0$ there are no
solutions with $\theta\in]0,\pi[$, so assume $\mu>0$. Replacing in the second
equation gives

$F_{\mu}(\theta):=(\mu\pi-\theta)\operatorname{cotan}(\theta)+\log(\mu\pi-\theta)-\log(\sin(\theta))=a\;.$

We compute that

$F^{\prime}_{\mu}(\theta)=-\dfrac{\mu\pi-\theta-\sin(2\theta)}{\sin^{2}(\theta)}-\dfrac{1}{\mu\pi-\theta}\;,$

and a tedious computation shows that $F^{\prime}_{\mu}(\theta)<0$ for all
$\theta\in]0,\pi[$ such that $\theta<\mu\pi$. It follows that $F_{\mu}$ is a
strictly decreasing function of $\theta$.

When $\theta\to 0^{+}$ we have $F_{\mu}(\theta)\to+\infty$. If $\mu>1$, when
$\theta\to\pi^{-}$ we have $F_{\mu}(\theta)\to-\infty$, so $F_{\mu}$ has a
single root, and it follows that $S_{\mu}(a)$ has a single solution. If
$\mu<1$, when $\theta\to\mu\pi^{-}$ we again have $F_{\mu}(\theta)\to-\infty$,
so again $S_{\mu}(a)$ has a single solution. Finally, if $\mu=1$, when
$\theta\to\pi^{-}$ we compute that $F_{1}(\theta)\to-1$. It follows that if
$a>-1$, $S_{1}(a)$ has a single solution, while if $a\leq-1$ it has no
solution.

To summarize, $S_{\mu}(a)$ has a solution with $\theta\in]0,\pi[$ if and only
if $\mu>0$ and either $\mu\neq 1$ or $\mu=1$ and $a>-1$, and that solution is
unique.

Since changing $\theta$ into $-\theta$ in the system simply changes $\mu$ into
$-\mu$, we deduce that $S_{\mu}(a)$ has a solution with $\theta\in]-\pi,0[$ if
and only if $\mu<0$ and either $\mu\neq-1$ or $\mu=-1$ and $a>-1$, and that
solution is unique.

A solution of $S_{\mu}(a)$ with $\theta=0$ is possible if and only if $\mu=0$
and $\rho+\log(\rho)=a$, and since this is an increasing function of $\rho$
and tends to $-\infty$ when $\rho\to 0^{+}$ and to $+\infty$ when
$\rho\to\infty$, we deduce that there always exists a solution with $\theta=0$
when $\mu=0$, and it is unique (and evidently equal to $W_{0}(e^{a})$).

A solution of $S_{\mu}(a)$ with $\theta=\pi$ is possible if and only if
$\mu=1$ and $-\rho+\log(\rho)=a$. This is a concave function of $\rho$ tending
to $-\infty$ when $\rho\to 0^{+}$ and $\rho\to\infty$, with a maximum at
$\rho=1$ equal to $-1$, so there exists a solution with $\theta=\pi$ if and
only if $\mu=1$ and $a\leq-1$, and there are two solutions if $a<-1$
(evidently equal to $W_{0}(-e^{a})$ and $W_{-1}(-e^{a})$), and a double
solution if $a=-1$.

Summarizing, we have proved the following:

###### Lemma 1.2

The solutions to $S_{\mu}(a)$ are as follows:

  1. 1.

If $\mu>1$, or $0<\mu<1$, a unique solution with $\theta\in]0,\pi[$, if
$\mu<-1$ or $-1<\mu<0$ a unique solution with $\theta\in]-\pi,0[$, and in both
cases no solution with $\theta=0$ or $\theta=\pi$.

  2. 2.

If $\mu=-1$ and $a>-1$, a unique solution with $\theta\in]-\pi,0[$, and if
$\mu=-1$ and $a\leq-1$ no solution.

  3. 3.

If $\mu=1$ and $a>-1$, a unique solution with $\theta\in]0,\pi[$, and if
$\mu=1$ and $a\leq-1$ two solutions with $\theta=\pi$ which coincide for
$a=-1$.

  4. 4.

If $\mu=0$, a unique solution with $\theta=0$.

Since $\mu=2k+\lambda$ with $-1<\lambda\leq 1$ and $k\in{\mathbb{Z}}$, $\mu>1$
is equivalent to $k\geq 1$, $\mu<-1$ is equivalent to $k\leq-1$, $-1<\mu<1$
and $\mu\neq 0$ is equivalent to $k=0$ and $\lambda\neq 0$, $\mu=0$ is
equivalent to $k=\lambda=0$, $\mu=1$ is equivalent to $k=0$ and $\lambda=1$,
and $\mu=-1$ is equivalent to $k=-1$ and $\lambda=1$, so it is clear that this
lemma is equivalent to the proposition.  $\sqcap$$\sqcup$

###### Corollary 1.3

Keep the same notation.

  1. 1.

If $z\notin]-\infty,0[$, or $z\in]-\infty,0[$ and $k\neq 0,-1$, or
$z\in]-\infty,-1/e[$ and $k=0$ or $k=-1$, $W_{k}(z)$ is the unique solution of
$(E_{k})$, and $W_{-k}(z)=\overline{\vphantom{T}W_{k}(\bar{z})}$ if
$z\notin]-\infty,0[$, and $W_{-k}(z)=\overline{\vphantom{T}W_{k-1}(z)}$
otherwise.

  2. 2.

If $z\in[-1/e,0[$, $W_{-1}(z)$ is the unique real solution of $E_{0}$ _(not of
$E_{-1}$)_ less than or equal to $-1$, and $W_{0}(z)$ is the unique real
solution of $E_{0}$ greater or equal to $-1$.

To give a precise idea of what these results mean, let us give the example of
$W_{k}(x)$ for $-2\leq k\leq 1$, and $x$ in each of the three real intervals
$]-\infty,-1/e[$, $[-1/e,0[$, and $]0,\infty[$, for instance $x=-1$, $x=-0.1$,
and $x=1$:

$\displaystyle W_{-2}(-1)$ $\displaystyle=-2.06227...-7.58863...\quad
W_{-2}(-0.1)=-4.44909...-7.30706...$ $\displaystyle W_{-1}(-1)$
$\displaystyle=-0.31813...-1.33723...\quad W_{-1}(-0.1)=-3.57715...$
$\displaystyle W_{0}(-1)$ $\displaystyle=-0.31813...+1.33723...\quad\
\,W_{0}(-0.1)=-0.11183...$ $\displaystyle W_{1}(-1)$
$\displaystyle=-2.06227...+7.58863...\quad\
\,W_{1}(-0.1)=-4.44909...+7.30706...$

In both of these cases, $W_{-k}(z)=\overline{\vphantom{T}W_{k-1}(z)}$, except
in the second case for $k=0$ and $1$.

$\displaystyle W_{-2}(1)$ $\displaystyle=-2.40158...-10.77629...$
$\displaystyle W_{-1}(1)$ $\displaystyle=-1.53391...-4.37518...$
$\displaystyle W_{0}(1)$ $\displaystyle=\phantom{-}0.56714...$ $\displaystyle
W_{1}(1)$ $\displaystyle=-1.53391...+4.37518...$

In this case, $W_{-k}(z)=\overline{\vphantom{T}W_{k}(z)}$.

Remark. It is possible to define $W_{k}(z)$ for _any_ $k\in{\mathbb{C}}$ (not
only for $k\in{\mathbb{Z}}$) by the defining equation
$W_{k}(z)+\log(W_{k}(z))=\log(z)+2k\pi i$. Of course, if
$k\notin{\mathbb{Z}}$, this will not anymore correspond to a solution of
$ye^{y}=z$. Furthermore, it is clear that we can find $\theta$ such that
$-1/2<\theta\leq 1/2$ and such that
$W_{k}(z)+\log(W_{k}(z))=\log(ze^{2\theta\pi i})+2m\pi i$, where
$m\in{\mathbb{Z}}$ is such that $|m-k|<1$, so if for instance
$z\notin]-\infty,0]$, the solution to this equation will simply be
$W_{m}(ze^{2\theta\pi i})$. A similar remark applies to the equation
$w+\log(w/x)=2k\pi i$ whose solution is $W_{k+\varepsilon}(x)$ for some
$\varepsilon=-1$, $0$, or $1$ depending on $k$ and $x$.

##  2 The Fundamental Identity

We begin by the following lemma:

###### Lemma 2.1

Let $x\in{\mathbb{C}}\setminus\\{0\\}$ be fixed. We have

$\lim_{t\to\pm\infty}\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)-t}=\mp\dfrac{1}{2}\;,$

where here and below a sum or product for $k\in{\mathbb{Z}}$ is understood as
the limit as $K\to\infty$ of the sum or product for $-K\leq k\leq K$.

Proof. We will see below the trivial fact that $W_{k}(x)=2k\pi i+O(\log(k))$
when $|k|\to\infty$, so the series indeed converges if we sum symmetrically.
Denoting it by $f(t)$, we thus have $f(t)=f_{1}(t)+f_{2}(t)$ with

$f_{1}(t)=\sum_{k\in{\mathbb{Z}}}\dfrac{1}{2k\pi i-t}\text{\quad
and\quad}f_{2}(t)=\sum_{k\in{\mathbb{Z}}}\dfrac{2k\pi
i-W_{k}(x)}{(W_{k}(x)-t)(2k\pi i+t)}\;.$

Since $2k\pi i-W_{k}(x)=O(\log(k))$ the series for $f_{2}(t)$ converges
uniformly in $t$, hence $\lim_{t\to\pm\infty}f_{2}(t)=0$. On the other hand it
is classical that $f_{1}(t)=-\coth(t/2)/2$, so the result follows.
$\sqcap$$\sqcup$

###### Theorem 2.2

For $x\in{\mathbb{C}}\setminus\\{0\\}$ we have

$\prod_{k\in{\mathbb{Z}}}\left(1-\dfrac{t}{W_{k}(x)}\right)=e^{-t/2}-\dfrac{t}{x}e^{t/2}\;.$

Proof. The variable $x$ being fixed, consider the function
$F(t)=e^{-t/2}-(t/x)e^{t/2}$. It has order $1$, has no pole, and its zeros are
the roots of $te^{t}=x$, so by definition they are the $W_{k}(x)$. By the
Hadamard factorization theorem it thus has a Hadamard product which we can
write in the form

$F(t)=Ae^{Bt}\prod_{k\in{\mathbb{Z}}}(1-t/W_{k}(x))\;.$

Note that the standard way of writing the product would be
$\prod_{k\in{\mathbb{Z}}}(1-t/W_{k}(x))e^{t/W_{k}(x)}$, but since we know that
the product written as above converges if we interpret it as the limit of the
product for $|k|\leq K$, it is preferable to write it as we have done.

We trivially have $A=F(0)=1$. To determine $B$, we compute the logarithmic
derivative:

$\dfrac{F^{\prime}(t)}{F(t)}=B-\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)-t}\;,$

We immediately compute that $\lim_{t\to\pm\infty}F^{\prime}(t)/F(t)=\pm 1/2$,
so making $t\to\infty$ or $t\to-\infty$ and using the lemma gives $B=0$,
proving the theorem.  $\sqcap$$\sqcup$

Remarks.

  1. 1.

In the proofs of the lemma and the theorem, we have implicitly used the
evident fact that for $t$ real with $|t|$ sufficiently large, for all
$k\in{\mathbb{Z}}$ we have $t\neq W_{k}(x)$ and $t\neq 2k\pi i$.

  2. 2.

Since $W_{k}(x)$ is close to $2k\pi i$, the above theorem should be compared
with the identity (equivalent to the one we used above for $\coth(t/2)$):

$-t\prod_{k\in{\mathbb{Z}}\setminus\\{0\\}}\left(1-\dfrac{t}{2k\pi
i}\right)=e^{-t/2}-e^{t/2}\;.$

The following corollary gives the usual form of the Hadamard product:

###### Corollary 2.3

We have

$\prod_{k\in{\mathbb{Z}}}\left(1-\dfrac{t}{W_{k}(x)}\right)e^{t/W_{k}(x)}=e^{t/x}\left(1-\dfrac{t}{x}e^{t}\right)\;.$

Proof. Immediate from the theorem since identification of the coefficients of
$t^{1}$ gives $\sum_{k\in{\mathbb{Z}}}1/W_{k}(x)=1/2+1/x$.  $\sqcap$$\sqcup$

Using the theorem, differentiating with respect to $t$ or $x$, and/or
specializing, we easily find a large number of identities which we regroup in
the following corollary:

###### Corollary 2.4

We have the following identities:

$\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)-t}=\dfrac{1}{2}+\dfrac{1+t}{xe^{-t}-t}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)+1}=\dfrac{1}{2}\;,$
$\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)}=\dfrac{1}{2}+\dfrac{1}{x}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)^{2}}=\dfrac{2x+1}{x^{2}}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)^{3}}=\dfrac{3x^{2}+6x+2}{2x^{3}}\;.$
$\prod_{k\in{\mathbb{Z}}}\left(1-\dfrac{t}{W_{k}(x)}\right)=e^{-t/2}-\dfrac{t}{x}e^{t/2}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{(W_{k}(x)+1)(W_{k}(x)-t)}=\dfrac{1}{xe^{-t}-t}\;.$
$\sum_{k\in{\mathbb{Z}}}\dfrac{1}{(W_{k}(x)-t)^{2}}=\dfrac{xe^{-t}(2+t)+1}{(xe^{-t}-t)^{2}}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{(W_{k}(x)+1)^{2}}=\dfrac{1}{xe+1}\;.$
$\sum_{k\in{\mathbb{Z}}}\dfrac{1}{(W_{k}(x)+1)(W_{k}(x)-t)^{2}}=\dfrac{xe^{-t}+1}{(xe^{-t}-t)^{2}}\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{1}{(W_{k}(x)+1)^{3}}=\dfrac{1}{xe+1}\;.$
$\sum_{k\in{\mathbb{Z}}}\dfrac{W^{\prime}_{k}(x)}{W_{k}(x)-t}=\dfrac{1}{x}\left(\dfrac{1}{2}+\dfrac{t}{xe^{-t}-t}\right)\;,\quad\sum_{k\in{\mathbb{Z}}}\dfrac{W^{\prime}_{k}(x)}{W_{k}(x)}=\dfrac{1}{2x}\;.$

It is clear that if $F$ is any rational function, by decomposing into partial
fractions, any symmetrically convergent series of the form
$\sum_{k\in{\mathbb{Z}}}F(W_{k}(x))$ can be evaluated explicitly. For
instance,

$\displaystyle\sum_{k\in{\mathbb{Z}}}\dfrac{1}{W_{k}(x)^{2}+t^{2}}$
$\displaystyle=\dfrac{x(\cos(t)+\sin(t)/t)+1}{x^{2}+2xt\sin(t)+t^{2}}\text{\quad
and}$
$\displaystyle\sum_{k\in{\mathbb{Z}}}\dfrac{W_{k}(x)}{W_{k}(x)^{2}+t^{2}}$
$\displaystyle=\dfrac{1}{2}+\dfrac{x(\cos(t)-t\sin(t))-t^{2}}{x^{2}+2xt\sin(t)+t^{2}}\;.$

Acknowledgment: I am grateful to I. Pinelis for having given the identity
$\sum_{k\in{\mathbb{Z}}}1/(W_{k}(x)+1)=1/2$ in the mathoverflow forum, which
led to the present section of this paper, but his proof is completely
different, using an integral representation of this sum coming from the
residue theorem.

##  3 Asymptotics and Sum and Product Formulas

In this section we are going to show that even nonconvergent symmetric sums or
products of the $W_{k}$ can be evaluated explicitly after removing the
asymptotic main term. First note the following easy proposition:

###### Proposition 3.1

The variable $x$ being fixed, as $|k|\to\infty$ we have

$W_{k}(x)=2k\pi i-\log(2k\pi i)+\log(x)+\dfrac{\log(2k\pi i)-\log(x)}{2k\pi
i}+O\left(\dfrac{\log(|k|)^{2}}{k^{2}}\right)\;.$

Proof. For notational simplicity set $K=2k\pi i$, $L=\log(2k\pi i)=\log(K)$,
$X=\log(x)$, and $M=L-X=\log(2k\pi i)-\log(x)$. By successive approximation,
we find that if we set
$w_{k}=K-M+M/K+\varepsilon=K(1-M/K+M/K^{2}+\varepsilon/K)$ we have
$\log(w_{k})=L-M/K+O(L^{2}/K^{2})$, hence
$w_{k}+\log(w_{k})=K+X+\varepsilon+O(L^{2}/K^{2})$, and using that
$w_{k}+\log(w_{k})=\log(x)+2k\pi i$ we see that $\varepsilon=O(L^{2}/K^{2})$.
$\sqcap$$\sqcup$

Note that we will give below the complete asymptotic expansion.

###### Corollary 3.2

Let the variable $x$ be fixed.

  1. 1.

For $k>0$ sufficiently large we have $(2k-1)\pi<|W_{k}(x)|<(2k+1)\pi$.

  2. 2.

If $x$ is positive real, then for $K$ sufficiently large, the solutions to the
equation $we^{w}=x$ with $|w|<(2K+1)\pi$ are the $W_{k}(x)$ for $-K\leq k\leq
K$.

Proof. Immediate from the proposition and left to the reader.
$\sqcap$$\sqcup$

Note that if $x$ is not a positive real, there may be an additional solution
for $k=\pm(K+1)$.

###### Theorem 3.3

We have

$\lim_{K\to\infty}\left(\sum_{-K\leq k\leq
K}W_{k}(x)+\log(\pi^{2K}(2K)!)-(2K+1/2)\log(x)\right)=\dfrac{\log(2)}{2}\;.$

Proof. We will prove this theorem in two stages. First we will prove that the
left-hand side is a constant independent of $x$. Second, the appendix due to
Letong Hong and Shengtong Zhang proves that this constant is equal to
$\log(2)/2$, which I conjectured in the first version of this paper. Note that
their proof also shows that the left-hand side is independent of $x$, but we
have chosen to give the present proof first since it handles more easily the
case where $x$ is not a positive real.

For $k>0$ we have

$W_{k}(x)+W_{-k}(x)=-2\log(2k\pi)+2\log(x)+1/(2k)+O(\log(k)^{2}/k^{2})\;.$

It follows that

$\sum_{-K\leq k\leq
K}W_{k}(x)=-2\log(K!)+2K\log(x/(2\pi))+\log(K)/2+F(x)+o(1)$

for some function $F$. By Stirling’s formula, we have

$\displaystyle-2\log(K!)-$ $\displaystyle
2K\log(2\pi)+\log(K)/2=-(2K+1/2)\log(2K)+2K$
$\displaystyle\phantom{=}-2K\log(\pi)-\log(4\pi)+o(1)$
$\displaystyle=-\log((2K)!)-2K\log(\pi)-\log(4\pi)+\log(2\pi)/2+o(1)\;,$

so

$\sum_{-K\leq k\leq K}W_{k}(x)=-\log((\pi/x)^{2K}(2K)!)+A(x)+o(1)$

for some other function $A(x)$. To find $A(x)$, we note that222This is not
quite rigorous, but one can prove that the derivative of the $o(1)$ is still a
$o(1)$.

$A^{\prime}(x)+\dfrac{2K}{x}+o(1)=\sum_{-K\leq k\leq K}W^{\prime}_{k}(x)\;.$

Now since
$W^{\prime}_{k}(x)=(1/x)W_{k}(x)/(W_{k}(x)+1)=(1/x)(1-1/(W_{k}(x)+1))$, we
have by Pinelis’s formula

$\sum_{-K\leq k\leq K}W^{\prime}_{k}(x)=\dfrac{2K+1/2}{x}+o(1)\;,$

so $A^{\prime}(x)=1/(2x)$, hence $A(x)=\log(x)/2+A$ for some constant $A$. The
fact that $A$ is real follows by choosing $x>0$ and using that
$W_{-k}(x)=\overline{\vphantom{T}W_{k}(x)}$, proving that the left-hand side
is indeed a real constant. We refer to the appendix for the proof that
$A=\log(2)/2$ and a second proof of the fact that the left-hand side is
constant.  $\sqcap$$\sqcup$

The proof shows that the implicit $o(1)$ term is a $O(\log(K)^{2}/K)$.

###### Corollary 3.4

We have

$\lim_{K\to\infty}\dfrac{\prod_{-K\leq k\leq
K}W_{k}(x)}{\pi^{2K}(2K)!}=\sqrt{x/2}\;,$

and more generally

$\lim_{K\to\infty}\dfrac{\prod_{-K\leq k\leq
K}(W_{k}(x)-t)}{\pi^{2K}(2K)!}=\sqrt{x/2}(e^{-t/2}-(t/x)e^{t/2})\;,$

Proof. By definition, we have $\log(W_{k}(x))=\log(x)+2k\pi i-W_{k}(x)$. It
follows from the theorem that

$\displaystyle\sum_{-K\leq k\leq K}\log(W_{k}(x))$
$\displaystyle=(2K+1)\log(x)-\sum_{-K\leq k\leq K}W_{k}(x)$
$\displaystyle=\log(\pi^{2K}(2K)!)+\log(x)/2-\log(2)/2+o(1)\;,$

so the result follows for $t=0$, and the more general result from the formula
for $\prod_{k}(1-t/W_{k}(x))$ given above.  $\sqcap$$\sqcup$

##  4 A Family of Polynomials

Before going further, we are now going to study in great detail a family of
polynomials important for our work. These polynomials date back at least to
Comtet [1].

###### Definition 4.1

Let $A(X)$ be a given polynomial such that $A(0)=0$.

  1. 1.

We define the Lambert family of polynomials $P_{n}(X)=P_{n,A}(X)$ associated
to $A$ by $P_{0}(X)=A(X)$ and for $n\geq 0$ by the recursion

$P_{n+1}(X)=-P_{n}(X)+n\int_{0}^{X}P_{n}(t)\,dt\;.$

  2. 2.

We denote by $F_{A}(T,X)=\sum_{n\geq 0}P_{n,A}(X)T^{n}$ the generating
function of the $P_{n}(X)$.

###### Lemma 4.2

$P_{n}$ is the Lambert family associated to $A$ if and only if

$(1+T)\dfrac{\partial F_{A}}{\partial X}=A^{\prime}(X)+T^{2}\dfrac{\partial
F_{A}}{\partial T}$

and $P_{n}(0)=0$ for all $n$.

Proof. Since $\partial F_{A}/\partial X=\sum_{n\geq 0}P^{\prime}_{n}(X)T^{n}$
we have

$\displaystyle(1+T)\partial F_{A}/\partial X$
$\displaystyle=dP_{0}/dX+\sum_{n\geq
1}(P^{\prime}_{n}(X)+P^{\prime}_{n-1}(X))T^{n}$
$\displaystyle=A^{\prime}(X)+\sum_{n\geq
1}(n-1)P_{n-1}(X)T^{n}=A^{\prime}(X)+T^{2}\partial F_{A}/\partial T\;.$

Conversely, if this is satisfied we have $P^{\prime}_{0}(X)=A^{\prime}(X)$ and
$P^{\prime}_{n}(X)+P^{\prime}_{n-1}(X)=(n-1)P_{n-1}(X)$, and since we assume
in addition that $P_{n}(0)=0$ we have $P_{0}(X)=A(X)$ and
$P_{n}(X)+P_{n-1}(X)=(n-1)\int_{0}^{X}P_{n-1}(t)\,dt$.  $\sqcap$$\sqcup$

###### Lemma 4.3

If $P_{n}$ is the Lambert family associated to $A(X)$ then
$\int_{0}^{X}P_{n}(t)\,dt$ is the Lambert family associated to
$\int_{0}^{X}A(t)\,dt$.

Proof. If $Q_{n}(X)=\int_{0}^{X}P_{n}(t)\,dt$ it is clear that
$Q^{\prime}_{n+1}(X)+Q^{\prime}_{n}(X)=nQ_{n}(X)$, so the result follows since
$Q_{n}(0)=0$.  $\sqcap$$\sqcup$

###### Theorem 4.4

Let $L_{n}$ be the Lambert family associated to $A(X)=-X$, $F(T,X)=\sum_{n\geq
0}L_{n}(X)T^{n}$ its generating function, and $k\geq 1$. By abuse of notation,
write $F^{k}(T,X)=\sum_{n\geq 0}L_{n,k}(X)T^{n}$ (including for $k\leq-1$), so
that $L_{n,1}(X)=L_{n}(X)$.

  1. 1.

For $n\geq 1$ we have

$L_{n,2}(X)=-2(n+1)\int_{0}^{X}L_{n}(t)\,dt=-\dfrac{2(n+1)}{n}(L_{n+1}(X)+L_{n}(X))\;.$

Equivalently, $L_{n,2}(X)/(n+1)$ is the Lambert family associated to
$A(X)=X^{2}$.

  2. 2.

The function $F$ satisfies the differential equation in $T$

$TFF^{\prime}+(1+T)F^{\prime}+F=0\;.$

  3. 3.

The function $F$ satisfies the implicit equation

$\log(1+TF)+F=-X\;.$

Proof. (1). By Lemma 4.2 applied to $A(X)=-X$ we have

$\displaystyle(1+T)\partial(F^{k})/\partial X$
$\displaystyle=kF^{k-1}(1+T)\partial F/\partial X$
$\displaystyle=kF^{k-1}(A^{\prime}(X)+T^{2}\partial F/\partial
T)=-kF^{k-1}+T^{2}\partial(F^{k})/\partial T\;.$

Identifying coefficients of $T^{n}$, it follows that

$L^{\prime}_{n,k}(X)+L^{\prime}_{n-1,k}(X)=(n-1)L_{n-1,k}(X)-kL_{n,k-1}(X)\;.$

We prove (1) by induction. Since
$L_{0,2}(X)=(-X)^{2}=(-2)\int_{0}^{X}(-t)\,dt$, it is true for $n=0$. Assume
that it is true for $n-1$: we thus have

$L^{\prime}_{n,2}(X)=2nL_{n-1,1}(X)-2(n-1)n\int_{0}^{X}L_{n-1,1}(t)\,dt-2L_{n,1}(X)$

and since $(n-1)\int_{0}^{X}L_{n-1,1}(t)\,dt=L_{n,1}(X)+L_{n-1,1}(X)$ we
obtain $L^{\prime}_{n,2}(X)=-2(n+1)L_{n,1}(X)$, proving our induction
hypothesis since $L_{n,k}(0)=0$ for all $n$, $k$. It follows from Lemma 4.3
that $L_{n,2}(X)/(n+1)$ is the Lambert family associated to $X^{2}$.

(2). For $n\geq 1$ the coefficient of $T^{n}$ in
$TFF^{\prime}+(1+T)F^{\prime}+F=(T/2)(F^{2})^{\prime}+F^{\prime}+TF^{\prime}+F$
is equal to

$\displaystyle\phantom{=}(n/2)L_{n,2}(X)+(n+1)L_{n+1,1}+nL_{n,1}+L_{n,1}$
$\displaystyle=-n(n+1)\int_{0}^{X}L_{n,1}(t)\,dt+(n+1)n\int_{0}^{X}L_{n,1}(t)\,dt=0\;,$

and the coefficient of $T^{0}$ is $L_{0,1}(X)+L_{1,1}(X)=-X+X=0$, proving the
differential equation.

(3). If we set $G=\log(1+TF)+F$, we have

$\displaystyle G^{\prime}$ $\displaystyle=(TF^{\prime}+F)/(1+TF)+F^{\prime}$
$\displaystyle=((1+T)F^{\prime}+F+TFF^{\prime})/(1+TF)=0$

by (2), so $G(T)=G(0)=F(0)=-X$.  $\sqcap$$\sqcup$

Remarks.

  1. 1.

The proof of (3) shows that the general solution to the differential equation
in (2) is a solution to the implicit equation $\log(1+TF)+F=A(X)$, where
$A(X)=F(0)$.

  2. 2.

It is clear that the theorem can be proved by using Lagrange inversion, but I
have preferred to give the above direct proof.

It is natural to call the $L_{n}(X)$ the _Lambert polynomials_.

###### Proposition 4.5

In addition to the partial differential equation $(1+T)\partial F/\partial
X=-1+T^{2}\partial F/\partial T$ of Lemma 4.2, the function $F$ also satisfies
the following ones:

$\dfrac{\partial F}{\partial X}=-\dfrac{TF+1}{TF+T+1}\text{\quad
and\quad}F\dfrac{\partial F}{\partial X}=(TF+1)\dfrac{\partial F}{\partial
T}=-T\dfrac{\partial F}{\partial T}-F\;.$

Proof. By Lemma 4.2 and Theorem 4.4 (2) we have

$\displaystyle\dfrac{\partial F}{\partial X}$
$\displaystyle=\dfrac{1}{1+T}\left(-1+T^{2}\dfrac{\partial F}{\partial
T}\right)$
$\displaystyle=-\dfrac{1}{1+T}\left(1+T^{2}\dfrac{F}{TF+T+1}\right)=-\dfrac{TF+1}{TF+T+1}\;.$

Thus, by Theorem 4.4 once again we have $TF+T+1=-F/F^{\prime}$, hence
$\partial F/\partial X=(TF+1)(\partial F/\partial T)/F$, and the last formula
again follows from the theorem.  $\sqcap$$\sqcup$

###### Corollary 4.6

We have the identity

$\sum_{n\geq
1}\dfrac{L^{\prime}_{n}(X)}{n}T^{n}=-\log\left(\dfrac{F(T,X)}{-X}\right)\;.$

Proof. Indeed, if we differentiate with respect to $T$, the derivative of the
left hand side is $\sum_{n\geq 1}L^{\prime}_{n}(X)T^{n-1}=(\partial F/\partial
X+1)/T$, while that of the right-hand side is $-(\partial F/\partial T)/F$,
and equality follows from the last formula given by the proposition. The
identity follows since both sides vanish for $T=0$,  $\sqcap$$\sqcup$

###### Corollary 4.7

  1. 1.

Generalizing (1) of the theorem, for $k\geq 2$ we have

$\displaystyle L_{n,k}(X)$
$\displaystyle=-\dfrac{k}{k-1}(n+k-1)\int_{0}^{X}L_{n,k-1}(t)\,dt$
$\displaystyle=-\dfrac{k}{k-1}\dfrac{(n+1)L_{n+1,k-1}(X)+(n+k-1)L_{n,k-1}(X)}{n}\;,$

and $L_{n,k}(X)/\binom{n+k-1}{n}$ is the Lambert family associated to
$A(X)=(-X)^{k}$.

  2. 2.

For all $k\geq 2$ we have

$\displaystyle L_{n,k}(X)$
$\displaystyle=(-1)^{k-1}k(k-1)\binom{n+k-1}{n}\int_{0}^{X}(X-t)^{k-2}L_{n}(t)\,dt$
$\displaystyle=(-1)^{k-1}k\binom{n+k-1}{n}\int_{0}^{X}(X-t)^{k-1}L^{\prime}_{n}(t)\,dt\;.$

  3. 3.

For any polynomial $A(X)$ such that $A(0)=0$, the Lambert family associated to
$A$ is given by

$\displaystyle P_{n,A}(X)$
$\displaystyle=-\int_{0}^{X}A^{\prime}(X-t)L^{\prime}_{n}(t)\,dt$
$\displaystyle=-A^{\prime}(0)L_{n}(X)-\int_{0}^{X}A^{\prime\prime}(X-t)L_{n}(t)\,dt\;.$

Proof. (1). Multiplying by $F^{k-2}$ the second identity of the proposition we
obtain $F^{k-1}\partial F/\partial X=-TF^{k-2}\partial F/\partial T-F^{k-1}$,
and identifying the coefficients of $T^{n}$ gives
$L^{\prime}_{n,k}/k=-((n+k-1)/(k-1))L_{n,k-1}$, proving the first formula
since $L_{n,k}(0)=0$.

It is clear that $(n!/(n+k-1)!)L_{n,k}(X)$ is a Lambert family for $k=1$.
Assume that this is the case for $k-1$ with $k\geq 2$. Using the formula just
proved and using the induction hypothesis, we have

$\displaystyle\phantom{=}\dfrac{n!}{(n+k-1)!}L^{\prime}_{n,k}(X)+\dfrac{(n-1)!}{(n+k-2)!}L^{\prime}_{n-1,k}(X)$
$\displaystyle=-\dfrac{k}{k-1}\left(\dfrac{n!}{(n+k-2)!}L_{n,k-1}(X)+\dfrac{(n-1)!}{(n+k-3)!}L_{n-1,k-1}(X)\right)$
$\displaystyle=-\dfrac{k}{k-1}(n-1)\dfrac{(n-1)!}{(n+k-3)!}\int_{0}^{X}L_{n-1,k-1}(t)\,dt$
$\displaystyle=(n-1)\dfrac{(n-1)!}{(n+k-2)!}L_{n-1,k}(X)\;,$

proving that $(n!/(n+k-1)!)L_{n,k}(X)$ is a Lambert family, and it is
associated to $L_{0,k}(X)/(k-1)!=(-X)^{k}/(k-1)!$. Using this and replacing in
the first formula proves the second, proving (1).

(2). This follows immediately from the formula giving iterated integrals.

(3). By linearity, it is clear that

$P_{n,A}(X)=-A^{\prime}(0)L_{n}(X)-\int_{0}^{X}A^{\prime\prime}(X-t)L_{n}(t)\,dt\;,$

so the result follows by integrating by parts.  $\sqcap$$\sqcup$

The polynomials $L_{n,k}$ can be given explicitly:

###### Proposition 4.8

We have

$L_{n,k}(X)=(-1)^{n+k}k!\binom{n+k-1}{n}\sum_{j=1}^{n}s(n,n+1-j)\dfrac{X^{k-1+j}}{(k-1+j)!}\;,$

where the $s(n,m)$ are the Stirling numbers of the first kind. In particular,

$\displaystyle L_{n}(X)$
$\displaystyle=(-1)^{n+1}\sum_{j=1}^{n}s(n,n+1-j)\dfrac{X^{j}}{j!}\;,\text{\quad
or equivalently,}$ $\displaystyle\int_{0}^{\infty}e^{-Xt}L_{n}(t)\,dt$
$\displaystyle=X^{-(n+1)}\prod_{j=1}^{n-1}(j-X)\;.$

Proof. The result for $k=1$ follows immediately from the recursion for
Stirling numbers $s(n+1,k)=s(n,k-1)-ns(n,k)$, the general result by successive
integration, and the integral formula from the definition of Stirling numbers.
$\sqcap$$\sqcup$

Remark. In view of the above results, it would seem that the most natural
polynomials are the $L^{\prime}_{n}(X)$ and not the $L_{n}(X)$: they are
monic, have constant term equal to $(-1)^{n-1}$, satisfy the same Lambert-type
recursion (but they are not a Lambert family since $L^{\prime}_{n}(0)\neq 0$),
but it is immediate to see that the equations involving them are more
complicated, so we have preferred to take the $L_{n}(X)$ themselves as
fundamental building blocks. We will see in the next section that the higher
derivatives of $L_{n}$ enter in the expansion of $F^{k}$ for $k\leq-1$.

The first few $L_{n}(X)$ are the following:

$\displaystyle L_{0}(X)$ $\displaystyle=-X\;,\ L_{1}(X)=X\;,\
L_{2}(X)=X^{2}/2-X\;,$ $\displaystyle L_{3}(X)$
$\displaystyle=X^{3}/3-(3/2)X^{2}+X\;,\
L_{4}(X)=X^{4}/4-(11/6)X^{3}+3X^{2}-X\;.$

##  5 Further Results on Lambert Polynomials

The preceding section dealt essentially with the explicit computation of the
polynomials $L_{n,k}$ when $k\geq 1$. The present section deals with the case
$k\leq-1$.

###### Proposition 5.1

We have the additional partial differential equation

$\dfrac{\partial^{2}F}{\partial
X^{2}}=-T^{2}\dfrac{\partial^{2}(1/F)}{\partial T^{2}}\;.$

Proof. By Proposition 4.5 we have $\partial F/\partial X=-(TF+1)/(TF+T+1)$.
Differentiating once more we deduce that

$\displaystyle\dfrac{\partial^{2}F}{\partial^{2}X}$
$\displaystyle=-\dfrac{T^{2}}{(TF+T+1)^{2}}\dfrac{\partial F}{\partial
X}=T^{2}\dfrac{TF+1}{(TF+T+1)^{3}}\;.$

On the other hand, writing for simplicity ′ instead of $\partial/{\partial
T}$, by Theorem 4.4 (2) we have
$(1/F)^{\prime}=-F^{\prime}/F^{2}=1/(F(TF+T+1))$. Now,

$\displaystyle(TF^{2}+TF+F)^{\prime}$
$\displaystyle=F^{2}+2TFF^{\prime}+F+TF^{\prime}+F^{\prime}=F^{2}+TFF^{\prime}$
$\displaystyle=F^{2}-TF^{2}/(TF+T+1)=F^{2}(TF+1)/(TF+T+1)\;,$

hence $(1/F)^{\prime\prime}=-(TF+1)/(TF+T+1)^{3}$, proving the proposition.
$\sqcap$$\sqcup$

###### Corollary 5.2

Recall that we have defined $L_{n,k}$ for all $k\in{\mathbb{Z}}$ by
$F^{k}(T,X)=\sum_{n\geq 0}L_{n,k}(X)T^{n}$.

  1. 1.

We have $L_{0,-1}(X)=L_{1,-1}(X)=-1/X$, and for $n\geq 2$

$L_{n,-1}(X)=-\dfrac{L_{n}^{\prime\prime}(X)}{n(n-1)}\;,\text{\quad in other
words}$ $\dfrac{1}{F(T,X)}=-1/X-(1/X)T-\sum_{n\geq
2}\dfrac{L_{n}^{\prime\prime}(X)}{n(n-1)}T^{n}\;.$

  2. 2.

More generally, let $k\leq-2$. We have

$(n+k)L_{n,k}(X)=-\dfrac{k}{(k+1)}L^{\prime}_{n,k+1}(X)\;,$

and for $n\geq 1-k$ we have

$L_{n,k}(X)=(-1)^{k-1}k\dfrac{(n+k-1)!}{n!}L^{(1-k)}_{n}(X)\;.$

Proof. (1). The formulas for $L_{0,-1}(X)$ and $L_{1,-1}(X)$ are obtained
directly, and the formula for $L_{n,-1}(X)$ is equivalent to the PDE of the
proposition. It can also be easily obtained by differentiating the formula of
Corollary 4.6.

(2). Multiplying by $F^{k-1}$ the last PDE of Proposition 4.5, we obtain
$F^{k}\partial F/\partial X=-TF^{k-1}\partial F/\partial T-F^{k}$, and
identifying the coefficients of $T^{n}$ gives
$L^{\prime}_{n,k+1}/(k+1)=-((n+k)/k)L_{n,k}$, proving the first formula, and
the second follows by induction.  $\sqcap$$\sqcup$

The above corollary does not give $L_{n,k}$ for $n\leq-k$. For this, we need
to introduce another family of polynomials, closely linked to $L_{n}$ as
follows:

###### Definition 5.3

For $n\geq 1$ we define a family $M_{n}$ of functions by $M_{1}(X)=-1/X$ and
$M_{n+1}(X)=M^{\prime}_{n}(X)/n-M_{n}(X)$.

Note that the $M_{n}$ are polynomials in $1/X$ (for instance
$M_{2}(X)=(X+1)/X^{2}$ and $M_{3}(X)=-(2X^{2}+3X+2)/(2X^{3})$), but the
derivative $M^{\prime}_{n}(X)$ is of course taken with respect to $X$, not
with respect to $1/X$.

###### Proposition 5.4

Let $k\leq-1$.

  1. 1.

We have

$L_{n,k}=\begin{cases}-\dfrac{k}{n}\dfrac{M_{n}^{(-k-n)}(X)}{(-k-n)!}\text{\quad
for $1\leq n\leq-k$\;,}\\\ (-X)^{k}\text{\quad for $n=0$.}\end{cases}$

  2. 2.

In other words, for $k\leq-1$ we have

$\displaystyle F(T,X)^{k}$ $\displaystyle=(-X)^{k}-k\sum_{1\leq
n\leq-k}\dfrac{M_{n}^{(-k-n)}(X)}{n(-k-n)!}T^{n}$
$\displaystyle\phantom{=}+(-1)^{k-1}k\sum_{n\geq
1-k}\dfrac{(n+k-1)!}{n!}L^{(1-k)}_{n}(X)T^{n}\;.$

Proof. In the proof of Theorem 4.4 (1) we have seen the recursion

$L^{\prime}_{n,k}(X)+L^{\prime}_{n-1,k}(X)=(n-1)L_{n-1,k}(X)-kL_{n,k-1}(X)\;,$

which is valid for all $k$ including negative ones. Thus, for $n\geq 1$ we
have

$L^{\prime}_{n+1,-n}(X)+L^{\prime}_{n,-n}(X)=nL_{n,-n}(X)+nL_{n+1,-n-1}(X)\;.$

On the other hand, the recursion given in (2) of the above corollary implies
that $L^{\prime}_{n+1,-n}(X)=0$ for $n\geq 1$. We thus have the recursion
$L_{n+1,-n-1}(X)=L^{\prime}_{n,-n}(X)/n-L_{n,-n}(X)$, which is the recursion
for $M_{n}$, proving the result for $n=-k$ since $L_{1,-1}(X)=-1/X=M_{1}(X)$.

The same recursion for $1\leq n<-k$ gives
$L_{n,k}(X)=(-k/((k+1)(k+n)))L^{\prime}_{n,k+1}(X)$, so by induction for
$1\leq j\leq-k-n$ we have, since $n\geq 1$ hence $j\leq-k-1$:

$L_{n,k}(X)=\dfrac{k}{k+j}\dfrac{(-k-j-n)!}{(-k-n)!}L^{(j)}_{n,j+k}\;,$

and choosing $j=-k-n$ and using what we just proved gives the result for
$1\leq n\leq-k$. The case $n=0$ is trivial directly since $L_{0,k}$ is the
coefficient of $T^{0}$ in $F^{k}$, hence equal to $(-X)^{k}$.
$\sqcap$$\sqcup$

The functions $M_{n}$ can also be given explicitly in terms of Stirling
numbers:

###### Proposition 5.5

$M_{n}(X)=\dfrac{1}{(n-1)!}\sum_{j=0}^{n}(-1)^{j-1}\dfrac{s(n,j+1)j!}{X^{j+1}}=\dfrac{(-1)^{n}}{(n-1)!}\int_{0}^{\infty}e^{-Xt}\prod_{j=1}^{n-1}(t+j)\,dt\;.$

Proof. Left to the reader.  $\sqcap$$\sqcup$

The coefficients of $M_{n}(X)$ are of course closely related to those of
$L_{n}(X)$, but I do not see any natural way of expressing $M_{n}(X)$ in terms
of the $L_{n}(X)$, except by the following formula coming from the definition
of the beta function:

###### Proposition 5.6

$M_{n}(X)=\dfrac{n(n+1)}{X^{n+1}}\int_{0}^{\infty}\dfrac{L_{n}(-Xt)}{(t+1)^{n+2}}\,dt\;.$

Proof. Simply use the expansions of $L_{n}(X)$ and $M_{n}(X)$ in terms of
Stirling numbers. In addition, one can integrate once or twice by parts and
obtain similar formulas involving $L^{\prime}_{n}$ and $L^{\prime\prime}_{n}$.
$\sqcap$$\sqcup$

##  6 Convergent Series for the Lambert Branches

The result of Proposition 3.1 is asymptotic, and can be pushed further if
desired. The result is that not only does it give an asymptotic expansion to
any desired number of terms, but in fact a _convergent series_ for $W_{k}(x)$
even for small $k$. We can even put these series in a slightly more general
setting as follows.

###### Proposition 6.1

Fix $x\in{\mathbb{C}}\setminus\\{0\\}$, let
$K\in{\mathbb{C}}\setminus\\{0\\}$, and define

$M=K+\log(K)-\log(x)-2k\pi i\;.$

  1. 1.

As $|K|\to\infty$ we have the asymptotic expansions

$W_{k}(x)=K+\sum_{n\geq 0}\dfrac{L_{n}(M)}{K^{n}}\text{\quad
and\quad}\log(W_{k}(x))=\log(K)-\sum_{n\geq 1}\dfrac{L_{n}(M)}{K^{n}}\;,$

where the $L_{n}(z)$ are as usual the Lambert polynomials.

  2. 2.

These series converge to $W_{k}(x)$ and $\log(W_{k}(x))$ when $2(1+|M|)<|K|$
(we will see below the exact domain of convergence in special cases).

  3. 3.

More generally, we have the following formulas:

$\displaystyle(W_{k}(x)-K)^{j}$ $\displaystyle=\sum_{n\geq
0}\dfrac{L_{n,j}(M)}{K^{n}}\text{\quad for all
$j\in{\mathbb{Z}}\setminus\\{0\\}$\;,}$ $\displaystyle\dfrac{-M}{W_{k}(x)-K}$
$\displaystyle=1+\dfrac{1}{K}+\sum_{n\geq
2}\dfrac{ML^{\prime\prime}_{n}(M)}{n(n-1)K^{n}}\;,$
$\displaystyle\dfrac{1}{1+W_{k}(x)}$ $\displaystyle=\sum_{n\geq
1}\dfrac{L^{\prime}_{n}(M)}{K^{n}}\;,\text{ and
}\log\left(\dfrac{W_{k}(x)-K}{-M}\right)=-\sum_{n\geq
1}\dfrac{L^{\prime}_{n}(M)}{nK^{n}}$

with the same convergence properties.

Proof. (1). Denote by $R$ the right-hand side of the expansion for $W_{k}(x)$,
and recall that $F(T,X)=\sum_{n\geq 0}L_{n}(X)T^{n}$. By Theorem 4.4, we have

$\log(R)=\log(K)+\log(1+(1/K)F(1/K,M))=\log(K)-M-F(1/K,M)\;.$

It follows that

$R+\log(R)=K+\log(K)-M=\log(x)+2k\pi i\;,$

which is exactly equation $(E_{k})$ defining the Lambert branches, and since
$W_{k}(x)$ is the solution of $(E_{k})$ when $k\neq 0,1$, we deduce that it
gives its asymptotic expansion. The expansion for $\log(W_{k}(x))$ follows
from $W_{k}(x)+\log(W_{k}(x))=\log(x)+K$.

(2). Let us study the convergence properties. By the recursion formula for
Stirling numbers, it is immediate to show that $|s(n,k)|\leq 2^{n}n!/k!$, so
replacing in the explicit formula for $L_{n}(X)$ we deduce that
$|L_{n}(M)|\leq 2^{n}(1+|M|)^{n}$, proving the convergence result. Using the
stronger bound $|s(n,n-k)|\leq(n/2)^{k}\binom{n-1}{k}$ conjectured by René Gy
and proved by Mike Earnest in the stackexchange forum, we can replace $2^{n}$
by $e^{n/2}$.

(3). These formulas follow from (1), the definition, and using
$W^{\prime}_{k}(x)=W_{k}(x)/(x(1+W_{k}(x)))$ and $L^{\prime}_{0}(X)=-1$.
$\sqcap$$\sqcup$

###### Corollary 6.2

  1. 1.

We have the asymptotic expansions

$W_{k}(x)=K+\sum_{n\geq 0}\dfrac{L_{n}(M)}{K^{n}}$

(as well as all the others given in the proposition) for instance for $K=2k\pi
i$ and $M=\log(K)-\log(x)$ (which gives an asymptotic expansion when
$|k|\to\infty$), or for $K=2k\pi i+\log(x)$ and $M=\log(K)$ (which gives
another asymptotic expansion when $|k|\to\infty$, and also when $|x|\to\infty$
or $x\to 0$).

  2. 2.

If $x\in[-1/e,0[$, we have the same expansions also for $k=-1$ by choosing
$K=\log(-x)$ and $M=\log(-K)$.

  3. 3.

All these expansions converge if and only if

$\log(|K|)+1-\Re(M)+\min_{m\in\\{-1,0\\}}\Re(W_{m}(-e^{M-1}))>0$

(note that this condition is automatically satisfied in case (2)).

Proof. (1) is clear, (2) is proved in exactly the same way as the proposition,
and (3) is proved in [2] and [3].  $\sqcap$$\sqcup$

Remarks.

  1. 1.

The series with $K=2k\pi i+\log(x)$ is the one usually given, for instance in
[1] and [2]. However, when $k\neq 0$ the series with $K=2k\pi i$ often
converges faster.

  2. 2.

We could choose $K=W_{k}(x)$, in which case $M=0$ by definition, so the
identity is a triviality since $L_{n}(0)=0$ for all $n\geq 0$.

  3. 3.

Even for $k=1$ (which is certainly not in the “asymptotic” regime) and $x=1$,
we can compute $W_{1}(1)$ to $38$ decimals using slightly more than $100$
terms of the series (of course the usual iterative methods to compute
$W_{k}(x)$ are much more efficient since they are based on Newton or Halley
iterations).

  4. 4.

We have remarked above that it is possible, although not very useful, to
define $W_{k}(x)$ even when $k\notin{\mathbb{Z}}$. It is clear that by
construction the above series are still valid in this more general context.

In [3], some alternate expansions are given for the principal branch $W_{0}$,
but they are trivially generalizable to all branches. For instance, instead of
choosing $K$ as main variable, we can choose $K_{1}=K+1$. Thus, we write

$\displaystyle W_{k}(x)$ $\displaystyle=K+F(1/K,M)=K_{1}-1+F(1/(K_{1}-1),M)$
$\displaystyle=K_{1}-1+\sum_{n\geq 0}L_{n}(M)/(K_{1}-1)^{n}$
$\displaystyle=K_{1}-1+\sum_{n\geq 0}L_{n}(M)/K_{1}^{n}\sum_{j\geq
0}\binom{n+j-1}{n-1}/K_{1}^{j}$ $\displaystyle=K_{1}-1+\sum_{N\geq
0}(1/K_{1}^{N})\sum_{n=0}^{N}\binom{N-1}{n-1}L_{n}(M)\;,$

so

$W_{k}(x)=K_{1}-1+\sum_{N\geq 0}\dfrac{P_{n}(M)}{K_{1}^{N}}\text{\quad
with\quad}P_{n}(X)=\sum_{n=0}^{N}\binom{N-1}{n-1}L_{n}(X)\;,$

and these new polynomials $P_{n}(X)$ can be expressed using another type of
Stirling numbers, and also have recursion properties coming from those of
$L_{n}(X)$, such as

$P_{n+1}(X)=\int_{0}^{X}(nP_{n}(t)-(n-1)P_{n-1}(t))\,dt\;,$

whose proof is left to the reader. The first few $P_{n}(X)$ are

$\displaystyle P_{0}(X)$ $\displaystyle=-X\;,\ P_{1}(X)=X\;,\
P_{2}(X)=X^{2}/2\;,\ P_{3}(X)=X^{3}/3-X^{2}/2\;,$ $\displaystyle P_{4}(X)$
$\displaystyle=X^{4}/4-(5/6)X^{3}\;,\
P_{5}(X)=X^{5}/5-(13/12)X^{4}+X^{3}/2\;.$

In particular, note that $P_{N}(X)$ is divisible by $X^{\lfloor
N/2\rfloor+1}$.

For the branch $k=0$, [3] mentions that the domain of convergence is much
larger than that of the original series, but for the other branches the
difference is not that large since $K+1$ is not so different from $K$.

##  7 Additional Properties and Conjectures

###### Conjecture 7.1

  1. 1.

The roots of $L_{n}(X)$ are all nonnegative real, and less than or equal to
$n+\log(n)+\gamma+O(1/n)$, where $\gamma$ is Euler’s constant.

  2. 2.

More generally, the $j$th root of $L_{n}(X)$ in decreasing order is asymptotic
to $n/j+\sum_{j\leq m\leq n}1/m$.

  3. 3.

The roots of $L_{n}(X)$ and $L_{n+1}(X)$ interlace.

  4. 4.

More generally (and more precisely), for $k\geq 1$ the roots of $L_{n,k}(X)$
are all nonnegative real, and their maximum has an asymptotic expansion

$n+\log(n)+\gamma+k-1+C_{0}(k)/n+C_{1}(k)/n^{2}+...\;,$

where $C_{m}(k)$ is a polynomial of degree $m$ in $k$, in particular
$C_{0}(k)$ is independent of $k$. This is still true for $k\leq 0$ if we
interpret $L_{n,k}(X)$ as $L_{n}^{(1-k)}(X)$.

Note that the fact that the real roots of $L_{n,k}(X)$ are nonnegative follows
immediately from Proposition 4.8 and the fact that Stirling numbers alternate
in sign, and the fact that the roots of $L_{n+1}(X)$ and $L_{n}(X)$ interlace
follows from the recursion once proved that all the roots of $L_{n}(X)$ are
real.

Numerically,

$C_{0}(k)=-4.3469909700078207218721533281574751800...$

The above conjecture is certainly intimately linked to the formula

$\int_{0}^{\infty}e^{-Xt}L_{n}(t)\,dt=X^{-(n+1)}\prod_{j=1}^{n-1}(j-X)$

given by Proposition 4.8, but I do not yet see a proof.

##  8 Integral Representation of the Branches

The following is part of a result given in [4] and is given for completeness:

###### Proposition 8.1

  1. 1.

Set $P=\pi/2+1$, $K=2k\pi i+\log(x)$, and $f(t)=t-\log(t)+K$. Assume that
either $k\neq 0,-1$, or that $k=0$ or $k=-1$ and $x\notin[-1/e,0]$. We have
$W_{k}(x)=N/D$, with

$\displaystyle N$
$\displaystyle=\dfrac{K}{K^{2}+P^{2}}+\int_{0}^{\infty}\dfrac{tdt}{(t^{2}+1)(f(t)^{2}+\pi^{2})}$
$\displaystyle D$
$\displaystyle=\dfrac{P}{K^{2}+P^{2}}-\int_{0}^{\infty}\dfrac{dt}{(t^{2}+1)(f(t)^{2}+\pi^{2})}\;.$

  2. 2.

Set $P=\pi/2-1$, $K=(2k-\operatorname{sign}(k))\pi i+\log(x)$, and
$f(t)=t+\log(t)-K$. Assume that $k\neq-1,0,1$. Then $W_{k}(x)=-N/D$ with the
same formulas as in (1).

Remarks

  1. 1.

I refer to [4] for an explicit but much more complicated formula for $k=0$ or
$k=-1$ and $x\in[-1/e,0[$.

  2. 2.

The conditions given by the author for the validity of these integral formulas
are not quite the same as those given here. I believe at least that the above
are correct.

  3. 3.

As the author mentions, the change of variable $t=\sinh(u)$ transforms the
integrals into exponentially decaying ones, which can thus be easily computed.
But in fact the doubly-exponential integration method does this automatically.

Appendix: Proof of Theorem 3.3

By Letong Hong and Shengtong Zhang

Without loss of generality assume that $x$ is a fixed positive real
number333This is purely for simplicity. The argument works for any
$x\in{\mathbb{C}}\backslash\\{0\\}$ with minor modifications., and set
$f(w)=we^{w}-x$. By Corollary 3.2, for $K$ sufficiently large the set of zeros
$z$ of $f$ with $\left|z\right|\leq(2K+1)\pi$ is precisely $\\{w_{k}:-K\leq
k\leq K\\}$, where we set $w_{k}=W_{k}(x)$. By Jensen’s formula applied to $f$
we have

$\displaystyle\log x$ $\displaystyle=\sum_{-K\leq k\leq
K}\log\left|w_{k}\right|-(2K+1)\log((2K+1)\pi)+\frac{I((2K+1)\pi)}{2\pi}\;,\text{
with}$ $\displaystyle I(\rho)$ $\displaystyle=\int_{0}^{2\pi}\log\left|f(\rho
e^{i\theta})\right|\,d\theta\;.$

Since $w_{k}e^{w_{k}}=x$, we have $\log\left|w_{k}\right|=\log x-\Re w_{k}$,
so summing over $k\in[-K,K]$ we obtain

$\sum_{-K\leq k\leq K}\Re w_{k}=2K\log
x-(2K+1)\log((2K+1)\pi)+\frac{I((2K+1)\pi)}{2\pi}\;.$ (1)

We now study the integral $I((2K+1)\pi)$ when $K$ is large, so we set
$\rho(K)=(2K+1)\pi$. The function $f(w)=we^{w}-x$ behaves like $we^{w}$ when
$\Re w\geq 0$ and is near $-x$ when $\Re w\leq-2\log\rho(K)$. We now handle
the intermediate case.

###### Lemma 8.2

For $K$ sufficiently large and any $w$ with $\left|w\right|=\rho(K)$, we have

$\left|f(w)\right|\gg 1\;.$

Proof. If $\left|we^{w}\right|\notin[x/2,2x]$ the result follows by the
triangle inequality. Otherwise, the real part of $w$ must be $O(\log
x+\log(\rho(K)))$, therefore the imaginary part of $w$ must be
$\pm\rho(K)+o(1)$. Since $\rho(K)$ is an odd multiple of $\pi$, it follows
that the argument of $we^{w}$ must be $\pm\frac{\pi}{2}+o(1)$. Since $x$ is
positive real, this implies that $\left|we^{w}-x\right|$ is also bounded from
below.  $\sqcap$$\sqcup$

We can now give a precise estimate for $I((2K+1)\pi)$:

###### Lemma 8.3

For $K$ sufficiently large, we have

$\displaystyle\int_{0}^{2\pi}\log\left|f(\rho(K)e^{i\theta})\right|d\theta=\pi\log\rho(K)+2\rho(K)+\pi\log
x+o(1)\;.$ (2)

Proof. For simplicity, write $\rho$ instead of $\rho(K)$, set $w=\rho
e^{i\theta}$, and let $\epsilon=\frac{3\log\rho}{\rho}$. We split the integral
into three terms corresponding to
$\theta\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},2\pi]$,
$\theta\in[\frac{\pi}{2},\frac{\pi}{2}+\epsilon]\cup[\frac{3\pi}{2}-\epsilon,\frac{3\pi}{2}]$,
and $\theta\in[\frac{\pi}{2}+\epsilon,\frac{3\pi}{2}-\epsilon]$. In the first
term, we have $\Re w\geq 0$, so $\left|we^{w}\right|=\rho e^{\Re w}\geq\rho$.
It follows that $\log\left|f(w)\right|=\log\left|we^{w}\right|+O(\rho^{-1})$,
hence

$\displaystyle\int_{\theta\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\pi]}\log\left|f(\rho
e^{i\theta})\right|$
$\displaystyle=\int_{\theta\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\pi]}\log\left|we^{w}\right|+O(\rho^{-1})$
$\displaystyle=\int_{\theta\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\pi]}(\log\rho+\Re
w)+O(\rho^{-1})$ $\displaystyle=\pi\log\rho+2\rho+O(\rho^{-1}).$

In the second term, since $w$ has negative real part, we have the upper bound
$\left|f(w)\right|\leq x+\left|we^{w}\right|\leq 2\rho$, and by the previous
lemma, it is bounded from below by a constant. It follows that
$\log\left|f(w)\right|\ll\log\rho$, so the second term tends to $0$.

Finally, in the third term we have $\Re w\leq-2\log\rho$, so that
$\left|we^{w}\right|=\rho e^{\Re w}\ll\rho^{-1}$, therefore we have
$\log\left|f(w)\right|=\log x+O(\rho^{-1})$. It follows that

$\int_{\theta\in[\frac{\pi}{2}+\epsilon,\frac{3\pi}{2}-\epsilon]}\log\left|f(\rho
e^{i\theta})\right|=\pi\log x+o(1).$

Summing the three terms proves the lemma.  $\sqcap$$\sqcup$

Plugging this estimate into (1), and using that for $x>0$ we have
$\Im(w_{-k})=-\Im(w_{k})$, we deduce that as $K\to\infty$:

$\sum_{-K\leq k\leq K}W_{k}(x)=(2K+\frac{1}{2})\log
x-(2K+\frac{1}{2})\log((2K+1)\pi)+(2K+1)+o(1).$

Using Stirling’s formula, it is immediate to deduce that

$\sum_{-K\leq k\leq K}W_{k}(x)=(2K+\frac{1}{2})\log
x-\log(\pi^{2K}(2K)!)+\dfrac{\log(2)}{2}+o(1)\;,$

proving Theorem 3.3.

## References

  * [1] L. Comtet, Advanced Combinatorics, Springer (1974).
  * [2] R. Corless, G. Gonnet, D. Hare, D. Jeffrey, and D. Knuth, On the Lambert W Function, Advances in Comp. Math. 5 (1996), pp. 329–359.
  * [3] G. Kalugin and D. Jeffrey, Convergence in ${\mathbb{C}}$ of series for the Lambert $W$ Function, arXiv:1208.0754.
  * [4] A. Kheyfits, Explicit solutions of transcendental equations and the Lambert $W$ function, Fractional Calculus and Applied Analysis, Jan. 2004, 12p.