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I would appreciate if someone can help me answer the following questions.

Although I read several papers and documents on the Lambert W function, I could not assess on what set is this function (or at least its principle branch) analytic (or holomorphic)? Hence, on what set can one apply the identity theorem on it? I know that the principal branch of the function admits a convergent Taylor series at $0$ with a positive radius of convergence $1/{\rm e}$ given by $$ W_0(z)=\sum\limits_{n=0}^\infty \frac{(-n)^{n-1}}{n!}z^n $$ Hence, $W_0(z)$ is analytic at $0$. However, is it also analytic elsewhere in the complex plane?

According to Wikipedia, "The function defined by this series can be extended to a holomorphic function defined on all complex numbers with a branch cut along the interval $(−\infty, −1/\text{e}]$; this holomorphic function defines the principal branch of the Lambert W function". I don't quite understand this statement. How can the extended function of this set be defined on all complex numbers away from the branch cut, if it only converges for $-\frac{1}{\text{e}}<z<\frac{1}{\text{e}}$ and diverges for all other $z$.

From my understanding, a function is said to be analytic on an open set $D$, if the function converges to its Taylor series in a neighborhood of every point in the set $D$. If the Taylor series of the function $W_0(z)$ at an arbitrary $z_0$ is not known to have a closed-form, does this mean that this function is not analytic at $z_0$? or could it be analytic without a known Taylor series expansion for arbitrary $z_0$? And if the Taylor series around an arbitrary point $z_0$ is not known in closed-form, how can one obtain the radius of convergence of the series? Does the radius of convergence play any role in applying the identity theorem?

In How to derive the Lambert W function series expansion?, there is an example showing how to write the Taylor series expansion of the Lambert W function around $\text{e}$. However, it is not clear to me if this applies to an arbitrary point $z_0$ (oher than $0$ and $\text{e}$) and how can one obtain the radius of converge of this non-closed form series and whether can one claim that the function is analytic at $z_0$?

If we extend the function to the complex domain, it is known that the lambert W function has the derivative $\frac{\text{d}W}{\text{d}z}=\frac{W(z)}{z+\text{e}^{W(z)}}=\frac{W(z)}{z(W(z)+1)}$ for $z\neq \{0,-1/\text{e}\}$. If we differentiate this infinite number of times, and the derivative exists at $z_0$, then the function is infinitely differentiable $z_0$. In this case, it only remains to prove that the function is equal its own Taylor series at a neighborhood of every point of its domain (or some open set) for the function to be holomorphic, right? Can this be shown for arbitrary $z_0$? And hence for some open set?

I tried to apply the Lagrange inversion theorem to get the Taylor series of $W_0(z)$ at arbitrary $z_0$, but I could not converge to a closed-form.

In Short, in my problem, i need to use the identity theorem on the Lambert W function. However, i need to check first on what set this theorem applies. In other words, on what set is the Lambert W function analytic?

Any help is appreciated. Thanks a lot in advance.

LARA
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  • The Wikipedia page doesn't say this---it says that the extension is to $\mathbb C$ with a branch cut. In other words, not all of $\mathbb C$ (which, as you rightly say, would make it an entire function, and hence it couldn't have branch points). – Richard Martin Nov 19 '18 at 12:01
  • Thanks for your reply. Right, it says with a branch cut along the interval $(-\infty,-1/\text{e})$. But does it mean that the Lambert W function is analytic for example on $(0,\infty)$? or does this apply only for this series (around $0$) with a radius of convergence of $1/\text{e}$, i.e., on $(-1/\text{e}<z<1/\text{e})$? – LARA Nov 19 '18 at 12:11
  • OK, so the radius of convergence is only $1/e$ so you can only use the power series as far as there. Outside that disk you may or may not be able to continue the function, and it's saying that you can do so along the +ve real axis. For example, by means of expansion around $z=1/2$ which would get you a little further than $1/e$. – Richard Martin Nov 19 '18 at 12:17
  • Do you mean that for example the further we move on the +ve real axis, then the Taylor series expansion will have a larger radius of convergence? So can we say that the function converges to its Taylor series in a neighborhood of every point on $(0,\infty)$? In other words, can we say the function is analytic on $(0,\infty)$ and apply the identity theorem?? – LARA Nov 19 '18 at 12:25
  • Yes, assuming that the Wikipedia assertion is correct. More precisely, if we expand around $x=r$ then the radius of cgce will be $r+1/e$. Obtaining the coefficients might be hard though ... – Richard Martin Nov 19 '18 at 12:32
  • Great. Thanks a lot for your precious time and help. I would appreciate if you
    1. Can give me a reference which can help one to obtain the set on which the LambertW function is analytic.
    2. Correct me if I am wrong in the following. Consider two functions $f(z)$ and $W_0(z)$ that agree on a set of points $\mathcal{S}$ $\subset \mathcal{R}^+$, having an accumulation point, i.e.,

    $$f(z)=W_0(z),, \forall z\in \mathcal{S}$$ Extending $z$ to the complex domain, and considering the domain $D$ defined by $\mathcal{Re}{z}>0$, then $W_0(z)$ is analytic over $D$. Hence, $f(z)=W_0(z)$ on $D$.

     – LARA Nov 19 '18 at 13:05
  • Unsure but 2) it is very important to remember that the accumulation point must be in the domain of differentiability. Thus $\sin 1/z$ and 0 agree on the set $\pi/n$, $n\in\mathbb N$, but the accumulation point (0) is outside the domain of differentiability of $\sin 1/z$.
    • – Richard Martin Nov 19 '18 at 13:11