Talk:Lambert W function

Example 2
Details to solve $$x^x = z$$. The trick is to express $$x$$ as $$e^{\ln(x)}$$.

$$x^x = z $$ $$(e^{\ln(x)})^{(e^{\ln(x)})} = z $$ $$e^{\ln(x) e^{\ln(x)}} = z $$ $$\ln(x) e ^ {\ln(x)} = \ln(z) $$ $$\ln(x) = W(\ln(z)) $$ $$x = e ^ {W(\ln(z))} $$ $$x = \ln(z)/W(\ln(z)) $$

Corwinjoy (talk) 19:48, 6 August 2010 (UTC)corwinjoy
 * Is it similarly possible to solve  $$x^{x+1} = 2$$   ?   (x = 1.344070114603687...)   Is there an explicit expression for x? -- MathLine (talk) 15:57, 11 June 2013 (UTC)

Request
Can we get an image of a fractal related to the Lambert W fun? If anyone has one (or can construct one), this article would benefit from its inclusion. (I know that fractals are not why Lambert W is important, and that Lambert W has more benefit as an equation implicitly defined by an elementary equation though it, itself is not an elementary equation, however a picture (of a fractal), and its ascetic beauty cultivate further interest in this most interesting function). -- LinuxDude 17:20, 8 January 2007 (UTC)

Followup: Rob Corless has one (here: http://www.apmaths.uwo.ca/~rcorless/frames/PAPERS/LambertW/expal2.1200.jpg), with accompanying explanation here: (http://www.apmaths.uwo.ca/~rcorless/frames/PAPERS/LambertW/) maybe we should ask him? -- Tryptographer Thu Jan 8 04:39:47 EST 2009  —Preceding undated comment was added at 09:40, 8 January 2009 (UTC).

Python Algorithm
Can someone include input and print statements to get output from that Python code? —Preceding unsigned comment added by Charlesrkiss (talk • contribs) 17:58, 5 April 2008 (UTC)

Generalization of Lambert W function
The standard Lambert W function expresses exact solutions to what is called ``transcendental algebraic'' equations of the form:

exp(-c*x) = a_o*(x-r) where a_o, c, and r are real constants. (1)

The solution is: x = r + W(c*exp(-c*r)/a_o)

--> There has been a generalization of the Lambert W function[AAECC] within:

(i) a connection between gravity theory and quantum mechanics shown in the Journal of Classical and Quantum Gravity [gravity/QM] where the RHS of (1) is now a quadratic polynomial:

exp(-c*x) = b_o*(x-r_1)*(x-r_2)                              (2)

where b_o, c, and r_1 and r_2 are real constants.

When r_1 = r_2, both sides of (2) can be factored and reduced to (1) and thus the solution reduces to that of the standard W function.

(ii) analytical solutions for the eigenenergies of a special case of the quantum 3-body problem, namely the hydrogen molecular ion. Here the RHS of (2) is now an infinite order polynomial:

exp(-c*x) = b_o* product_{i=1)^{infinity} (x-r_i)           (3)

where b_o, c, and r_i for all 'i' are real constants.

Thus, it turns out that the Lambert W is of even greater fundamental importance than anybody realized.

--> I submit the current page on the Lambert function is INCOMPLETE and NEEDS TO BE UPDATED.

Generalization
The new material looks good. But I am wondering (as I was after reading Scott et al) what exactly the generalization actually is. Is it a function of three arguments (x, r1, r2)? The standard function has W(x)*exp(W(x)) = x. Is there a similar defining identity for the generalized version? Yours in confusion, Robinh (talk) 08:00, 11 December 2008 (UTC)

Re: generalization.

Thanks. I hope the following clarifications answers your question(s):

Actually, it's one argument 'x'. The coefficients a_o, r_1, r_2 are parameters. This is similar to the hypergeometric and Meijer G functions which have only one argument but many parameters (as many as you like). This is implied by eq. (14) of the AAECC paper. I admit that the work of Scott et al. does not address everything. In particular the variation and dependencies on these parameters needs to be explored.

The paper by Scott et al. identifies the generalization and its governing equation(s) and provides exact solutions to some special cases but did not examine e.g. what happens when two distinct parameters are allowed to vary until they are no longer distinct. Nor did they examine this function in the complex plane. For eq. (2) both tasks were studied by Byers-Brown in the 1970s in a quantum chemistry journal for the the double well Dirac delta potential model. The resulting branch structure in the complex plane for eq. (2) is by no means trivial - harder IMHO than the average case for the hypergeometric functions. There were controversies about the convergences of the series not resolved until the 1990s by Scott, Dalgarno, Babb and Morgan. BTW, this body of work is cited in the references by Scott et al in the AAECC paper.

So a paper fleshing out e.g. series and asymptotic series expansions requires a whole body of work and paper in itself, even if some aspects have been worked out already...

But we're following the history of the standard W function. Firstly, one identifies the function or its structure (somehow). In this case, the governing equations are transcendental-algebraic equations which are now easy to solve numerically on most computers. What's missing from a practical point of view would be e.g. asymptotic series expansions for large argument 'x'. For the standard W function, that came relatively late in its history. Feedback is important. Wikipedia plays a role in this regard by helping this body of work connect to other people 'discovering' or re-discovering the Lambert W function in its generality.

PS: Since the generalization is there, shouldn't we edit out the part of the talk discussion which shows the generalization? —Preceding unsigned comment added by 171.71.55.135 (talk) 20:48, 11 December 2008 (UTC)

Need help simplifying
Am not sure if this is the right place to make queries about productlog usage but perhaps someone can help?!

W(5.e^5)=5

Is there a simplification for:

W(-5.e^-5)

Where W denotes the product log function?

Neil Parker (talk) 19:15, 20 May 2009 (UTC)


 * See the equation $$ Y = X e ^ X \; \Longleftrightarrow \; X = W(Y) $$ in the article: $$Y=-5e^{-5}$$ means $$W(Y)=-5$$. Shreevatsa (talk) 22:04, 20 May 2009 (UTC)

Thanks but apparently there is a second solution. Try:

http://www58.wolframalpha.com and input ProductLog[-5*e^-5]

I am trying to understand where the second solution comes from.

Neil Parker (talk) 07:28, 21 May 2009 (UTC)


 * Please consult the reference desk for questions like this. &mdash; Tobias Bergemann (talk) 18:43, 23 May 2009 (UTC)

p^(ax+b)=cx+d
Is d=/=0 really required here? I do not see where that requirement comes from. —Preceding unsigned comment added by 77.4.43.178 (talk) 23:49, 22 June 2009 (UTC)

Reference
From the article:
 * However the inverse of wew was first described by Pólya and Szegö in 1925.

This could only refer to "Problems and Theorems in Analysis", originally published as G Pólya and G Szegö, Aufgaben und Lehrsätze aus der Analysis, Springer 1925. But I haven't read it so I'm not going to add this as a reference.

CRGreathouse (t | c) 19:18, 17 August 2009 (UTC)

DESY Lambert W
Now I understand why they did not yet find supersymmetry at DESY - desy_lambert_W is an extremely rough approximation to LambertW ! In x=1 (where W(1) = 0.5...), they differ by -0.06... this may seem little for everyday use, but for a scientific purpose, that's not serious ! In addition, it really does not make sense to declare it as "double precision". &mdash; MFH:Talk 16:36, 13 November 2009 (UTC)

DLMF notation
I suggest reverting, although the new notation could be noted as well. I have only seen this notation used in the DLMF, and while the DLMF can be considered authoritative, I think the notation of Corless et al. should be used until/if the DLMF notation becomes more widespread. One problem with p/m or +/- is that it is limited to only two branches, but the numerical subscript allows indexing all branches of the Lambert W function (this article fails to note this). The "old" notation is also used in computer algebra systems, which are the most important domain of use for the Lambert W function.

In fact, a part of the article says "characteristic equation $$\lambda=a e^{-\lambda}$$, leading to $$\lambda=W_k(a)$$ and $$y(t)=e^{W_k(a)t}$$, where $$k$$ is the branch index. If $$a$$ is real, only $$W_0(a)$$ need be considered" which is unintelligible right now when $$W_0$$ (and the general branches) aren't defined. Fredrik Johansson 04:39, 6 June 2010 (UTC)

I agree with the reversion to the old notation. Corless et al. is the standard reference for the Lambert W function, and uses the subscript k notation, which is more appropriate given the existence of the complex branches. I have never even seen the DLMF notation before - is it standardly used anywhere but there? In any case, I have switched the notation back to the Corless et al. notation in the meantime, but made note of the DLMF notation. Debate on the subject can be continued here. 71.201.198.158 (talk) 21:50, 6 July 2010 (UTC)

Moving to Lambert W relation
Is it just me, or does anyone else realize that the Lambert W function is not a function?! —Preceding unsigned comment added by 74.111.112.90 (talk) 19:31, 29 August 2010 (UTC)
 * It is common to call something a function when its branch cuts are functions. There is no confusion when it is used this way. Most people know for instance that the square root is a relation, but they know also that by convention $$ \sqrt{x} $$ usually only refers to the principal branch and it is also generally called a function.--75.80.43.80 (talk) 14:51, 3 March 2011 (UTC)

Why is this?
At it says that W(2)i (4/π)i is the Definite integral over a half-period of i^x. Perhaps this shold be in this artical? Robo37 (talk) 22:30, 18 August 2011 (UTC)

Deleted a line from special values
I took out this line:
 * $$exp(-W\left(0\right))(-ln(3^{1/3}) = 2.47805268028830\dots$$, the solution of y of $$ \sqrt[x]{x}= \sqrt[y]{y}$$ when x=3 and x≠y.

because (1) the math is badly formatted and it's unclear what is meant: the parentheses don't match up. (2) W(0)=0, so why this is in the expression is unclear. (3) What's so special about x=3 or $$\sqrt[x]{x}= \sqrt[y]{y}$$ that this should be listed in this article? If you disagree with the removal, please discuss here. Thanks!&mdash;24.85.177.92 (talk) 14:57, 23 August 2011 (UTC)
 * Well the number is listed at oeis and other websites, and the 3 just arises because we have 4 as the solution to x=2 and 2 as the solution to x=4, so 3 just kind of sits nicely in the middle, though I'm not sure whether that makes it notable enough or not. Maybe $$-2W({1 \over 2})$$ which is the solution to ex=x2, -iW(i) which is the solution to ex=xi, $$\ln{2} \over 2$$ which is the solution to W(x)=2x and/or $$ \sqrt {W(2) \over 2}$$ which is the solution to W(x)=x2 should be included as well/instead, as there representations are simpler? Robo37 (talk) 19:18, 26 August 2011 (UTC)

Meaning of "a library search"?
The statement "Corless and developers of the Maple Computer algebra system made a library search to find that this function was in fact ubiquitous to nature" isn't clear because "library" could refer to a conventional library of printed books or it could refer to software libraries. I don't know how a search of either could prove something about Nature. I think the statement should be clarified or omitted.

Tashiro (talk) 14:59, 23 October 2011 (UTC)


 * I find nothing wrong with the statement. Given that dates of their papers (early 1990s) proficiency with the web was not that well established back then.   Wikipedia came later.  Clearly these were bonafide "library" searches as in e.g. University library searches.  Their initial article with appeared in MapleTech showed that the Lambert W function appeared in a number of applications demonstrating the universality of the Lambert W function.  What is not clear about this? TonyMath (talk) 18:00, 14 September 2012 (UTC)


 * I think I understand in part the nature of your objection. It's because the notion that the work of Corless et al being apparently "reduced" to a library search seems inglorious and lacking in respect.  A couple of points should be made: (i) that search was absolutely essential in demonstrating that the W function was fundamental in nature - we needed to know that - and (ii) their results were really not so original after all.  In particular long before they had examined the W function in the complex plane, mathematical physicists like Byers-Brown  had already sussed that out in the context of the  Double well Dirac Delta function model in itself a 1-D representation of the Hydrogen molecular ion for a generalized case back in the 1970s!  It stands to reason: the Lambert W function is centuries old and thus it becomes virtually impossible for any of us to have a full assessment of all the developments involving that function.  We must remain objective when examining things with hindsight. TonyMath (talk) 23:29, 30 September 2012 (UTC)


 * I agree with the above comment regarding "ubiquitous in nature". The paper cited reads "it has appeared in the solution of enough problems to have earned the right to have a name of its own". A more precise citation is needed for the statement in the article. Isheden (talk) 12:16, 1 October 2012 (UTC)


 * Done. Corless' main paper does indeed talk about the many many applications of the Lambert W function though his paper cites only a few.TonyMath (talk) 15:55, 2 October 2012 (UTC)
 * I should also add that the wikipedia article itself with some 8 examples + generalizations also makes the case of the wide range of applications of the W function. Even Corless' paper cannot do full justice to that notion.  I think the W function could use a counterpart of the Fibonacci quaterly but for Lambert's function TonyMath (talk) 15:59, 2 October 2012 (UTC)

Question about the range of one of the special values
In the section special values, it says $$W\left(-\frac{\ln a}{a}\right)= -\ln a \quad            \left(\frac{1}{e}\le a\le e\right)$$

I believe this relation holds true for $$0<a\le e$$, can anyone confirm this?

Tjajab (talk) 12:22, 30 December 2012 (UTC)

New example?
This example could possibly replace example 2, as it is similar but I believe more interesting:


 * $$x^y=y^x\,$$


 * $$\Rightarrow y\ln x = x\ln y\,$$


 * $$\Rightarrow \ln y = y\frac{\ln x}{x}\,$$


 * $$\Rightarrow y = e^{y\frac{\ln x}{x}}\,$$


 * $$\Rightarrow -y\frac{\ln x}{x} = -\frac{\ln x}{x}e^{y\frac{\ln x}{x}}\,$$


 * $$\Rightarrow -\frac{\ln x}{x} = -y\frac{\ln x}{x}e^{-y\frac{\ln x}{x}}\,$$


 * $$\Rightarrow -y\frac{\ln x}{x} = W(-\frac{\ln x}{x})\,$$

Looking at the definition


 * $$z = W(z) e^{W(z)}\,$$

we could multiply both sides with $$e^{W(-\frac{\ln x}{x})}$$, getting


 * $$-y\frac{\ln x}{x}e^{W(-\frac{\ln x}{x})} = W(-\frac{\ln x}{x})e^{W(-\frac{\ln x}{x})}\,$$


 * $$\Rightarrow -y\frac{\ln x}{x}e^{W(-\frac{\ln x}{x})} = -\frac{\ln x}{x}\,$$


 * $$\Rightarrow y=e^{-W(-\frac{\ln x}{x})},$$

The interesting thing about this equation is that it is symmetric in x and y, implying that y=x must be one solution. This gives that $$y=-W(-\frac{\ln x}{x})$$ has the simplification $$y=\ln x$$ for at least one branch of W for any given x. Also, the symmetry gives that this function is its own inverse (as we could switch y and x in the first equation). From this one could also draw some conclusions about $$W(-\frac{\ln x}{x})$$, namely that it must equal $$-\ln x$$ in at least one branch of W for any given x.

I would very much appreciate some feedback on this.

Tjajab (talk) 13:37, 30 December 2012 (UTC)

Fixed point of the exponential function
-W(-1) is an fixed point of the exponential function. Here's the proof:
 * $$-1=W(-1) e^{W(-1)}$$
 * $$-W(-1)^{-1}= e^{W(-1)}$$
 * $$-W(-1) = e^{-W(-1)}$$

So $$z=-W(-1)$$ is z which satisfies $$e^z=z$$, and so is its complex conjugate. --31.147.15.123 (talk) 11:32, 8 October 2013 (UTC)

Product rule for Omega function
Product rule

ω(a)*ω(b)*exp[ω(a)+ω(b)]=a*b=ω(a*b)*exp[ω(a+b)]

Solve for ω(a*b)

ω(a)*ω(b)=c*exp(c) exp[ω(a)+ω(b)]=d*exp(d)

c=ω[ω(a)*ω(b)] d=ω[exp[ω(a)+ω(b)]]

ω(a*b)=c*d=ω[ω(a)*ω(b)]*ω[exp[ω(a)+ω(b)]]

For instance:

ω(2*3)=ω[ω(2)*ω(3)]*ω[exp[ω(2)+ω(3)]]

4.49.117.146 (talk) 22:14, 13 February 2014 (UTC)

W0 Taylor series expansion is wrong
Second term (n=2), gives denominator of 2 via the summation which is not what is in the expansion. Cross-checking with Wolfram Mathworld shows a difference. --Paddy (talk) 07:28, 30 August 2014 (UTC)

Request for more explicit notation to help non-mathematicians
I'm just a user trying to understand W well enough to use it. The equation at the beginning of the Applications section is clearer than those in the introduction, but one thing would really help I think: It would be much easier to picture for those unused to dealing with functions defined only in terms of the inverse of some other function if this equation could be restated with no implicit functional dependences. I.e. don't use F to represent F(x).

PS: Thanks to the Wiki-mathematicians for this and other helpful articles Ma-Ma-Max Headroom (talk) 17:45, 5 September 2014 (UTC)


 * The simplest definition of W is:
 * If $$y=xe^x$$  then  $$x=\text{W}(y)$$
 * There is a small complication because there may be more than one value of x that satisfies $$y=xe^x$$  for a given value of y, so this definition does not quite define a function in the strict mathematical sense. But it is really no more complicated than
 * If $$y=x^2$$  then  $$x=\sqrt{y}$$
 * Gandalf61 (talk) 07:29, 6 September 2014 (UTC)
 * Here you use different notations x and W for the argument of the function to be inverted and the inverse function respectively. But the lead of our article uses the same letter W for both meanings. Perhaps this is also unnecessarily confusing? —David Eppstein (talk) 07:36, 6 September 2014 (UTC)
 * While I agree that this article is rather technical, I feel like elementary math students rarely encounter it.--Jasper Deng (talk) 07:33, 6 September 2014 (UTC)

This function is also of use in traffic queuing theory. I first came across y(x) = xe^x several years ago when investigating ALOHA. I tried to find the inverse function x(y). It is not possible since its an example of a one way function and has no inverse. Hence x(y) = W(y). (I am an engineer not a mathematician)

Request for help to solve an equation
Hi, I want to solve the following equation: (x^2-4*x+6)*exp(x)=y (1) It looks a bit like the following equation: x*exp(x)=y (2) The solution of equation (2) is: x=LambertW(y) I think the solution of equation (1) uses the function LambertW. Can somebody solve the equation (1)? Kindest regards, Jacob Safra — Preceding unsigned comment added by Jacobsafra (talk • contribs) 01:06, 7 October 2014 (UTC)


 * (1) is simply exp(-x) = (1/y)(x^2-4*x+6). This is a special case of eq.(2) of the generalized Lambert W function, the right-hand-side being a quadratic polynomial but with complex roots i.e. 2 +- sqrt(-2).   TonyMath (talk) 01:27, 12 October 2014 (UTC)

Dear TonyMath, First of all, thank you for your answer. The equation I am trying to solve is of the form exp(-c*x) = b_o*(x-r_1)*(x-r_2) (2) with c = 1 ; b_o = 1/y ; r_1 = 2-i*sqrt(2) and r_2 = 2+i*sqrt(2) thanks to you. However I don't know how do you proceed to find the value of x? Best, Jacob Safra — Preceding unsigned comment added by Jacobsafra (talk • contribs) 20:24, 20 October 2014 (UTC)
 * In general there will be no closed form solution. After all, this is a transcendental-algebraic equation. Using methods currently developed by a colleague, I can find some exact solutions.  E.g. x=1-i, y=(2+2*i)*e^(1-i) and x=1+i, y=(2-2*i)*e^(1+i).  These solutions are in the complex plane however. There may be other exact solutions. Does this originate from a physics or engineering problem? TonyMath (talk) 11:49, 26 October 2014 (UTC)

Thank you again for answering me. I will try to better explain what I want. This equation I found is part of the field of fluid mechanics and in particular provides a starting value for the skin friction coefficient of a turbulent flow. I’m going to study the following function: For all x>0; f(x)=(x^2-4*x+6)*exp(x) For all x>0; f’(x)=(x^2-2*x+2)*exp(x) For all x>0; x^2-2*x+2≥1 and exp(x)≥1 Therefore, for all x>0; f’(x)≥1>0 The function f is strictly increasing on the interval ]0; +∞[. Furthermore, the function f is continuous. Therefore, for all x>0, there is a unique y>6 such that f(x)=y. I know the value of y and I know how to solve the equation f(x)=y numerically. For example: y=100 000; x=7.905419368254814 y=100 000 000; x=15.506081342140432 Do you know how to find the function g such that g(y)=x (g is the inverse function of f, i.e. g=f-1). This would provide a general formula for y in terms of x without having to solve the equation numerically. Best, Jacob Safra — Preceding unsigned comment added by Jacobsafra (talk • contribs) 15:36, 26 October 2014 (UTC)
 * Getting interesting. Unless I misunderstand, your 'g' is just (or is related to) the inverse generalized Lambert W function and 'x' is just the solution.  I think Mathematica has an inverse for the standard Lambert W function but nobody, as far as I know, has defined an inverse for the generalized Lambert W function.  The bottom line, I think, is that you are looking for exact solutions.  I passed on your problem to that colleague.  Maybe she can find something.  I would like to know more about the physical problem.  Is there a reference for it? or is this a physical model you have been developing on your own?  At any rate, the SIGSAM papers in the references could provide (i) series solutions to this problem and (ii) a numerical method for getting ALL the roots.  Unfortunately they are not yet implemented on a general computer algebra system. TonyMath (talk) 03:10, 28 October 2014 (UTC)
 * So far, my colleague managed to prove that the solution to your problem is unique, but no closed form solution so far. TonyMath (talk) 03:02, 14 November 2014 (UTC)

Discussion of branches
The discussion on branches and what they mean seems a bit sparse and confusing (there are 2 or 3 lines in the summary). From other sources, it appears there exists a number of branches W_k. A more involved discussion on W_o and W_-1 and W_k would be great. If some one could add a section that discusses the branches, it would improve the article. I don't have the expertise to do so. Gsonnenf (talk) 02:22, 30 September 2015 (UTC)

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-1 branch of W function
Should we add a formula for the -1 branch of the Lambert W Function? I only see formulas for the 0 branch. 63.227.221.214 (talk) 20:14, 5 May 2021 (UTC)

Use table in Software section
I think it would be a good idea to create a table in the Software section of this article. It could make it easier to read. Somerandomuser (talk) 05:14, 17 September 2021 (UTC)

The case a=b
In the History section we are told that Euler derived the equation x^a - x^b = (a-b)cx^(a+b), and then considered the case a = b. But this makes no sense as it stands (since both sides vanish); either there is a mistake, or else an essential intermediate equation is missing, without which the sentence is more or less meaningless. JBritnell (talk) 11:57, 3 July 2022 (UTC)


 * What do you mean by the symbol "c" ?
 * Why should there be a problem with 0 = 0 making sense, exactly? 2601:200:C000:1A0:C0A5:7B4A:921E:7CA (talk) 01:11, 5 July 2022 (UTC)
 * I don't mean anything by c, I was quoting a formula from the article.
 * It is indeed true that 0=0, and I did not mean to dispute it. But that doesn't help me to extract much useful information from the paragraph. To follow it, I would need to know something about the form of the series solution, and something about the limits that are taken. (I imagine that the solution is a power series in c with coefficients involving a and b, and that Euler takes the limit as b - a → 0, but that is just a guess.)


 * (Edit: I've glanced at Euler's article, and so far as I can make out without much Latin, my guess appears to be on the right lines. Euler uses v rather than c.) JBritnell (talk) 01:50, 5 July 2022 (UTC)


 * If you cannot provide a meaning for c, then nobody can possibly understand your question. It is unclear why you would ask a question that you yourself do not understand. 2601:200:C000:1A0:C0A5:7B4A:921E:7CA (talk) 04:44, 5 July 2022 (UTC)
 * Why be rude? As I said above, the equation with c in it is given in the article, in a section that I am trying to understand. (I think I do understand it now, after looking at the original article, and I may attempt to improve the section, but I am a bit wary of doing so on the basis of my dubious understanding of Euler's Latin.) JBritnell (talk) 10:52, 5 July 2022 (UTC)

Riemann surface?
I notice that there is currently no mention of a Riemann surface in the article.

Any multivalued analytic function, like the inverse function f-1 of a nonlinear analytic function f, has an interesting Riemann surface — call it M — which may be completed around finite branch points (those like the inverse function of z^n, n ≥ 2). On that Riemann surface (call it M') the inverse function is a bona fide, single-valued function, which (if there are any poles) can be taken further to a meromorphic completion, where the codomain ℂ ∪ {∞} now includes infinity, taking the topology of the sphere S2. Call the twice-completed surface M^.

The "completed" inverse function — call it by the same name f-1 — then has become a single-valued holomorphic function from the Riemann surface to the sphere:


 * f-1 : M^ →  S2.

Of course the Riemann surface is the result of glueing all the branches together along their common edges. But it is also valuable to understand the Riemann surface explicitly. Especially because it has a topology and a conformal structure that are global properties, not merely local ones, that are essential invariants of the inverse function f-1 (now a bona fide single-valued meromorphic function).

If there is anyone reading this who is knowledgeable about the Riemann surface of the Lambert W function, I hope they will add it to the article. 2601:200:C000:1A0:C0A5:7B4A:921E:7CA (talk) 21:22, 4 July 2022 (UTC)