Approximate solutions of $W_{-1}(x)-W(x)=k\left(W_{-1}(x)+W(x) \right)$

Question

I am trying to find good approximations of the zero of function $$f(x)=\frac{W_{-1}(x)-W(x)}{W_{-1}(x)+W(x)}-k \quad \text{with} \quad 0 \leq k \leq 1\quad \text{and} \quad -\frac 1e \leq x \leq 0^-$$ From a numerical point of view, there are no major problem as long as $k$ is not too close to $1$.

$f(x)$ was developed as a series using $$W_{0,-1}(x)=\sum_{n=0}^\infty a_n\, y^n_\pm(x)\qquad \text{where} \qquad y_\pm(x)=\pm \sqrt{2(1+ex)}$$ (have a look here), the $a_n$'s forming he sequence $$\left\{-1,1,-\frac{1}{3},\frac{11}{72},-\frac{43}{540},\frac{769}{17280},-\frac{221} {8505},\frac{680863}{43545600},-\frac{1963}{204120},\frac{226287557}{37623398400} ,\cdots\right\}$$ This makes $$\frac{W_{-1}(x)-W(x)}{W_{-1}(x)+W(x)}=\sum_{n=0}^\infty b_n\, y_+^{2n+1}$$ where the $b_n$'s form the sequence $$\left\{1,-\frac{13}{72},\frac{433}{17280},-\frac{188297}{43545600},\frac{4028449}{75 24679680},-\frac{8293593643}{69528040243200},\cdots\right\}$$

Using series reversion, the result is $$y=\sqrt{2(1+ex)}=\sum_{n=0}^\infty c_n\, k^{2n+1}$$ where the $c_n$'s form the sequence $$\left\{1,\frac{13}{72},\frac{419}{5760},\frac{187553}{4838400},\frac{20011721}{836 075520},\frac{625581272639}{38626689024000},\frac{602467562104201}{516494013235 20000},\cdots\right\}$$

As shown below, the approximation is quite good up to $k \sim 0.85$ but it deteriorate quite quickly above.

$$\left( \begin{array}{ccc} k & \text{approximation} & \text{solution} \\ 0.00 & -0.367879 & -0.367879 \\ 0.05 & -0.367419 & -0.367419 \\ 0.10 & -0.366033 & -0.366033 \\ 0.15 & -0.363707 & -0.363707 \\ 0.20 & -0.360413 & -0.360413 \\ 0.25 & -0.356115 & -0.356115 \\ 0.30 & -0.350762 & -0.350762 \\ 0.35 & -0.344285 & -0.344285 \\ 0.40 & -0.336601 & -0.336601 \\ 0.45 & -0.327599 & -0.327599 \\ 0.50 & -0.317142 & -0.317142 \\ 0.55 & -0.305053 & -0.305053 \\ 0.60 & -0.291104 & -0.291104 \\ 0.65 & -0.274994 & -0.274994 \\ 0.70 & -0.256318 & -0.256318 \\ 0.75 & -0.234519 & -0.234519 \\ 0.80 & -0.208805 & -0.208691 \\ 0.85 & -0.178007 & -0.177601 \\ 0.90 & -0.140347 & -0.138902 \\ 0.95 & -0.093020 & -0.087549 \end{array} \right)$$

Question

How could I improve the approximation for the range $0.85 \leq k \leq 1$ ? Do I need to precise that this only art for art's sake ?

Edit

@Simply Beautiful Art provided an elegant and beautiful solution. As a tribute, I give below the result of the approximation for $k=1-10^{-p}$ $$\left( \begin{array}{ccc} p & \text{approximation} & \text{solution} \\ 1 & -0.13889384802843830195 & -0.13889550354096674629 \\ 2 & -0.02602863214400574022 & -0.02602863210501531541 \\ 3 & -0.00378956220505078207 & -0.00378956220504845816 \\ 4 & -0.00049497621618462337 & -0.00049497621618462337 \\ 5 & -0.00006102722387405542 & -0.00006102722387405542 \end{array} \right)$$

@user698573. Could you go ahead ? Remember that $-\frac 1e \leq x \leq 0^-$ — Claude Leibovici, Aug 09 '20 at 07:36
Yes you're right Claude but I don't want overlook nothing in your problem :-).Moreover and it's a negative point there is no nice antiderivative of the function related to the fraction . I continue for the fun ! — Miss and Mister cassoulet char, Aug 09 '20 at 09:15
Dividing by $W_0(x)$ see here https://www.wolframalpha.com/input/?i=Power+series+%28ProductLog%5B-1%2Cx%5D%2FProductLog%5B0%2Cx%5D%29 and https://www.wolframalpha.com/input/?i=Power+series+%28x-1%29%2F%28x%2B1%29 — Miss and Mister cassoulet char, Aug 09 '20 at 09:18
Ah I understand now you use it https://www.wolframalpha.com/input/?i=Power+series+%28ProductLog%5B-1%2Cx%5D%2FProductLog%5B0%2Cx%5D%29+at+x%3D-1%2Fe .Vous êtes malin Claude ! — Miss and Mister cassoulet char, Aug 09 '20 at 09:22
@user698573. Dividing by $W_0(x)$ does not change anything (imho). — Claude Leibovici, Aug 09 '20 at 09:31
Well I don't think so with my little transformation we can find an inverse function see here https://www.wolframalpha.com/input/?i=Inverse+function+ln%28%28%28ProductLog%5B-1%2Cx%5D%2FProductLog%5B0%2Cx%5D%29-1%29%29 .Obviously it's a partial inverse but now I think we can do someting .Are you agree ? Ps:Your question is really interesting ! — Miss and Mister cassoulet char, Aug 09 '20 at 10:08
$k$ close to one is like $x$ close to zero, and $W_{-1}(x)$ close to $-\infty$. The function $ze^{z}$ has an essential singularity at $z = \infty$ (in that $z^{-1} e^{z^{-1}}$ has an essential singularity at $z = 0$). So isn't it a bit like you wanting a Taylors' series for $g(x) = \frac{\log(x)+x}{log(x)-x}$ near $x = 0$? — Stephen Montgomery-Smith, Aug 09 '20 at 15:03
(But I should add that I am not an expert by any means, so maybe my thoughts are naive.) — Stephen Montgomery-Smith, Aug 09 '20 at 15:04
@StephenMontgomery-Smith There isn't a Laurent series about $x=0$ since you have a logarithmic singularity. — Simply Beautiful Art, Aug 09 '20 at 15:05
@StephenMontgomery-Smith But you suggested we find the Taylor series of $\frac{\log(x)+x}{\log(x)-x}$ about $x=0$? — Simply Beautiful Art, Aug 09 '20 at 15:11
@StephenMontgomery-Smith It's not at all clear to me what you are trying to say. — Simply Beautiful Art, Aug 09 '20 at 15:13

Simply Beautiful Art · Accepted Answer · 2020-08-09T16:00:04.937

Dividing through by $W_{-1}(x)$, we get an equation in $q(x)=W_0(x)/W_{-1}(x)=\exp(W_{-1}(x)-W_0(x))$, which may be solved.

$$\frac{1-q(x)}{1+q(x)}=k$$

$$q(x)=\frac{1-k}{1+k}$$

For simplicity let $\kappa=\frac{1-k}{1+k}$ and $L=\ln\kappa$.

Unfortunately $q$ does not admit a nice expansion at $0$. We may write $q$ in its exponential form, with the fixed-point iteration

$$W_{-1}(x_{n+1})-W_0(x_n)=L$$

$$x_{n+1}=\kappa(L+W_0(x_n))\exp(W_0(x_n))$$

Using the initial approximation $x_0=0$, we get $x_1=\kappa L$. Three more iterations gives us the expansion

$$x_2=\kappa L+\kappa^2L(1+L)+\tilde{\mathcal O}(\kappa^3)$$

$$x_3=\kappa L+\kappa^2L(1+L)+\kappa^3L\left(1+2L+\frac12L^2\right)+\tilde{\mathcal O}(\kappa^4)$$

$$x_4=\kappa L+\kappa^2L(1+L)+\kappa^3L\left(1+2L+\frac12L^2\right)+\kappa^4L\left(1+3L+\frac32L^2+\frac16L^3\right)+\tilde{\mathcal O}(\kappa^5)$$

where $\tilde{\mathcal O}$ is big O up to a poly-logarithmic factor.

Truncating $x_4$ gives us the approximations

\begin{array}{c|c|c} k&\text{approximation}&\text{solution}\\\hline 0.50&-0.320426&−0.317142\\ 0.55&-0.306505&−0.305053\\ 0.60&-0.291622&−0.291104\\ 0.65&-0.275102&−0.274994\\ 0.70&-0.256284&−0.256318\\ 0.75&-0.234445&−0.234519\\ 0.80&-0.208665&−0.208691\\ 0.85&-0.177591&−0.177601\\ 0.90&-0.138894&−0.138902\\ 0.95&−0.087549&−0.087549 \end{array}

which works decently well, and better than the provided approximation for $0.8\le k\le1.0$.

This is for sure a fantastic work which solves the problem. It works perfect very close to $k=1^-$. I thank you very much. Cheers :-) — Claude Leibovici, Aug 10 '20 at 06:15
Paraphrasing Droopy : "You know what ?" . You did not make my day but my week, my month and more ! — Claude Leibovici, Aug 10 '20 at 14:00

score 1 · Answer 2 · answered Aug 29 '20 at 17:41

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\a{\alpha}$

\begin{align}\Wm(x)-\Wp(x)-k(\Wm(x)+\Wp(x)) &=0\tag{1}\label{1}\end{align}

An exact solution by means of Parametric representation of the real branches of the Lambert $\W$ function.

Let \begin{align} \Wp(x)&=\frac{\a\ln \a}{1-\a} \tag{2}\label{2} ,\\ \Wm(x)&=\frac{\ln \a}{1-\a} \tag{3}\label{3} \\ \text{for }\quad \a\in(0,1) . \end{align}

Then \eqref{1} becomes

\begin{align} \frac{\ln \a}{1-\a} - \frac{\a\ln \a}{1-\a} - k\,\left( \frac{\ln \a}{1-\a} + \frac{\a\ln \a}{1-\a} \right) &=0 \tag{4}\label{4} , \end{align}

\begin{align} \a&=\frac{1-k}{1+k} \tag{5}\label{5} \end{align}

and the solution is

\begin{align} x&= \a^{\tfrac \a{1-\a}} \ln\Big(\a^{\tfrac \a{1-\a}}\Big) = \a^{\tfrac1{1-\a}} \ln\Big(\a^{\tfrac1{1-\a}}\Big) \tag{6}\label{6} . \end{align}

$\endgroup$

Thank you very much. This is superb ! – Claude Leibovici Aug 30 '20 at 02:36 — Claude Leibovici, Aug 30 '20 at 02:36

Approximate solutions of $W_{-1}(x)-W(x)=k\left(W_{-1}(x)+W(x) \right)$

2 Answers2