LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?

Question

I just found by numerical heuristics for some systematic numbers $q(x)_\text{heuristical}$ depending on $x=1,2,3,4,\ldots$ using WolframAlpha the suggested interpretation in terms of the LambertW-function, such that the suggestive formula for my various examples of $q(x)$ was: $$ q(x)_\text{LambertW} = {-x \over W(-x \exp(-x))} \tag 1$$ Here $-x$ was not equal to $q(x)$ and thus the value and the formula as well was an interesting item, so I wrote that formula for the most likely analytic explanation of my heuristical values $q$ in my current small treatize.
Here is a short table $$\small \begin{array} {} x & q(x)_\text{heuristical} & q(x)_\text{LambertW} & \text{err} \\ \hline 0.250000 & 0.096649 & 1.00000 & -0.903350 \\ 0.500000 & 0.284668 & 1.00000 & -0.715332 \\ 0.750000 & 0.576834 & 1.00000 & -0.423166 \\ \hline 1.00000 & 1.00000 & 1.00000 & -1.47621E-21 \\ 1.25000 & 1.59076 & 1.59076 & \epsilon \\ 1.50000 & 2.39700 & 2.39700 & \epsilon \\ 1.75000 & 3.48066 & 3.48066 & \epsilon \\ 2.00000 & 4.92155 & 4.92155 & \epsilon \\ 2.25000 & 6.82233 & 6.82233 & \epsilon \\ 2.50000 & 9.31487 & 9.31487 & \epsilon \\ 2.75000 & 12.5686 & 12.5686 & \epsilon \\ 3.00000 & 16.8010 & 16.8010 & \epsilon \\ 3.25000 & 22.2915 & 22.2915 & \epsilon \\ 3.50000 & 29.3986 & 29.3986 & \epsilon \\ 3.75000 & 38.5828 & 38.5828 & \epsilon \end{array} $$ ("$\epsilon$" being machine/software-epsilon; for the limit and a graph of the function see W/A)

After a second read I got now aware that the denominator clearly should equal $-x$ by the definition of the LambertW, and so it should $q(x) = 1$ for all cases. Urrgh...
Well, LambertW is multivalued, so I looked in the exact definition at mathworld and tried then negative $x$ to make the argument of the LambertW a positive value, and of course for $x \lt 1$ we have indeed the identity $q=-x$ in (1) but $x=1$ is a special point.

So I think I should "polish" my written text around the formula for that positive $x \gt 1$. But also using the notation for the first branch does not recover $q=-x$ and on the other hand the $q \ne -x$ for $x \gt 1$ seem to be completely natural evaluations.

So I'm a bit disorientated and my question is ...

Q: ... how should I formulate this for the reader so that no irritation would occur?

[Added]: In mathworld there is a close relative to the LambertW shown as $$h(z)=-{W(-\ln z)\over \ln z}$$ and I recall a discussion somewhere that in some cases the $h(z)$ -function might be preferable over the expression by the LambertW - but do not exactly remember the scope of that proposal.

Answer 1

Q: ... how should I formulate this for the reader so that no irritation would occur?

Hallo Gottfried,

it's a bit tricky to get such identities to work, without knowing the exact range of $W$. The trick is twofold: The main functional identity (for the inverse) is:

$$W_k(z)\cdot\exp(W_k(z))=z, k\in\mathbb{Z}$$

and is valid throughout and for all branches, while the reverse identity (for the inverse) is valid only within the strip-domain ranges of $W_k$. That is, the identities:

$$W_k(z\cdot\exp(z))=z,k\in\mathbb{Z}(*)$$

are only valid for $z\in D_k$, where $D_k$ are the corresponding strip-domains in the Quadratrix of Hippias, which is shown below, which are labeled with $k\in\mathbb{Z}$:

.

The exact delimitations of the strips where these identities hold are given as curves $D_k$ and the corresponding identities are given in Lemma 2.6 here.

Within the unique strip-domain $D_0$, we have $W_0(z\cdot \exp(z))=z$, but if you switch strip-domains you have to switch identities, as per (*).

As you can see from the graph, point $z=-1$ is a little special, because it is a branch point which is shared by the principal branch and the branch $k=-1$. The ray $(-\infty,-1)$ as well as the lower part of the delimiting curve of the principal branch are part of the $k=-1$ branch.

So, for example, when you try with Maple:

W:=(k,z)->LambertW(k,z);
W(-1,-5*exp(-5));
-5

or:

W(-1,-1*exp(-1));
-1

and because $x=-1$ is a shared branch point between the $k=0$ and $k=-1$ branches, check also that:

W(0,-1*exp(-1));
-1

But for $x<-1$ it is $\mathit{not}$ true that $W_0(x\cdot \exp(x))=x$, as is dictated by the identities (*).

For your second question:

and I recall a discussion somewhere that in some cases the h(z) -function might be preferable over the expression by the LambertW..

The utility function $h(z)=\frac{W_k(-\ln(z))}{-\ln(z)}$ is used only because it provides a partial inverse to the power tower ${^\infty}z$, in terms of the $W_k$ function branches. Otherwise it has no relation with the $W$ function perse.

Gottfried,

for your second comment:

If you set your $q$ as:

$$q_k(z)=\frac{z}{W_k(z\cdot\exp(z))},z\in D_k$$

then you'd be getting the identity function, because in each strip-domain $D_k$, you'd have:

$$W_k(z\cdot\exp(z))=z\Rightarrow q_k(z)=1,\forall z\in D_k$$

Similarly if you set it as:

$$q_k(z)=\frac{-z}{W_k(-z\cdot\exp(-z))},z\in D_k$$

you'd again get the identity function, because the curves of the Quadratrix are given as:

$$f(y)=-y\cot(y)+yi\Rightarrow\\$$ $$f(-y)=y\cot(-y)-yi=-y\cot(y)-yi=\overline{f(y)}\Rightarrow$$ $$q_k(z)=1,\forall z\in D_k$$

Concluding, if $w=\pm z$, then

$$q_k(w)=\frac{w}{W_k(w\cdot\exp(w))}\Rightarrow q_k(w)=1,\forall w\in D_k$$

Which means this only gives you the identity function throughout $\mathbb{C}$, unless $w\notin D_k$, in which case your $q_k$ will be some sort of conjugation ratio of $w$ over what another branch gives for what should be $w$, but isn't. I have no idea how you can use that, as a continuation of sorts.

Note however that a "continuation" of $W_k$ already exists, by Taylor's Theorem. The principal branch of $W_0$ is analytic at the origin and has a Maclaurin series there. Therefore $W_k$'s "continuation" is the standard Taylor series around any wanted point $z_0$ on the principal branch (or any other branch) modulo the corresponding branch cut on the negative axis ($(-\infty,-1/e)$ for principal, $(-\infty,0)$ for others) and modulo your winding number around the origin.

You know $W_k^{(n)}(z)=\frac{d^n W_k(z)}{dz^n}$, therefore choose a path from 0 to $z_0\in D_0$ and go there and find the corresponding Taylor series expansion around $z_0$. It will be:

$$\sum_{n=0}^\infty \frac{W_k^{(n)}(z_0)}{n!}\cdot (z-z_0)^n$$

You just have to be a little careful and note when you cross branches while going to your $z_0$. If you repeat the process and wind your path around 0, you are winding up once and are passing to either branch 1 or -1, depending on your direction. So, the "standard" analytic continuation covers all the branches of $W$ (For more on this, here).

Answer 2

I think I've now sufficiently understood the source of the problem and think I can formulate the explanation in my text better, thanks to the answer of Ioannis.
Here is a graph of the curves along the formula, which generates the $q_n(x)_\text{emp}$ when $n$ is taken to the limit towards $\infty$, and the curve for the expression $q(x)_\text{LambertW}$ by the LambertW-reference:

[Update]: The treatize, where I deal with this in more length&width is here . The page where my question is related to is at page 8 to 9