# Is the Lambert W function analytic? If not everywhere then on what set is it analytic?

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 $$-\frac{1}{\text{e}} and diverges for all other $$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.

• 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). Nov 19, 2018 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, 2018 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 $1/e$. Nov 19, 2018 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, 2018 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 ... Nov 19, 2018 at 12:32

To numerically prove that $$W_0(z)$$ is analytic for $$z>0$$, I implemented the Taylor series of $$W_0(z)$$ around some arbitrary point $$Z_0$$ using [1, eq. (8), (11), and Table I] and found I got a good fit between the exact lambertw() function of matlab and its Taylor series for a good radius of convergence. Moreover, the good fit is satisfied for many different values of $$Z_0$$, from which I can conjecture that the LambertW function is analytic at least on the set $$(0,\infty)$$, since the Taylor series of the function converges on a neighborhood of every point on $$(0,\infty)$$
I also contacted Prof. David Jeffrey (an expert on the Lambert W function and an author of  and ), where he confirmed that the principal branch of the LambertW function is analytic everywhere in the complex domain with the exception of the branch cut along the negative real axis, i.e., except on $$-\infty