Solving a ratio of Lambert W branches: $y=\frac{W_{-1}(x)}{W_0(x)},$ where y is positive and real.

Question

How might I go about solving a ratio of Lambert W branches? Namely,

$y=\frac{W_{-1}(x)}{W_0(x)},$ where y is positive and real.

Motivation:

I want to define a Cauchy filter for edge detection given octave bandwidth (log2 frequency bandwidth). I have equation for octave bandwidth of a cauchy filter:

$BW=\frac{\log\frac{W_{-1}(c)}{W_0(c)}}{\log{2}},$

where $c=-\frac{2^{-(1/a)}}{e}$, and $a=\omega_0\sigma$. Here $\omega_0$ is peak tuning frequency, $\sigma$ specifies the filter envelope (from Boukerroui, Noble, Brady. On the Choice of Band-Pass Quadrature Filters). What I want specifically then is an equation that returns $a$ in terms of $BW$.

That said, I'm interested in negative values for $x$ ranging from $(-1/e,0)$. I know this simplifies the problem because things get complex (pun intended) outside this range.

In practice, I can easily estimate what I want, but I think this is an interesting question the inverse of $W$ is so straightforward, but this problem is not.

Answer 1

Searching on the site, I found the exact solution. If you look here, for the range of interest to you, the two real branches of Lambert function can be parameterized as $$W_0(x)=\frac{\alpha \log (\alpha )}{1-\alpha }\qquad \text{and} \qquad W_{-1}(x)=\frac{ \log (\alpha )}{1-\alpha }$$ with $$x=\alpha ^{\frac{\alpha }{1-\alpha }} \log \left(\alpha ^{\frac{\alpha }{1-\alpha }}\right)$$

So, the solution is explicit since $y=\frac 1 \alpha$ $$\boxed{\color{blue}{\large x=y^{\frac{1}{1-y}}\,\frac{\log(y)}{y-1}}}$$

It is interesting to plot on the same graph this function and $x_0=-\frac{W(y)}{y}$

Answer 2

Expressing the inverse of $$y=\frac{W_{-1}(x)}{W_0(x)}$$seems to be of very little hope.

However, using the very first term of each series expansion, for a very first estimate $$y \sim \frac {\log(-x)} x \qquad\implies\qquad x_0=-\frac{W(y)}{y}$$

Consider that you look for the zero of $$f(x)=\frac{W_{-1}(x)}{W_0(x)}-y$$ $$f'(x)=\frac 1 x\,\frac{W_{-1}(x)}{ W(x)} \left(\frac{1}{W_{-1}(x)+1}-\frac{1}{W(x)+1}\right)$$ and use one single iteration of Newton method $$x_1=x_0- \frac {f(x_0)} {f'(x_0)}$$ is fully explicit (even if the formula is quite messy).

What is interesting to notice is that, for any $y \geq 5$ $$f(x_0) \times f''(x_0)~~>~0$$ So, by Darboux theorem, there will not be any overshoot of the solution during Newton iterations.

Trying for $y=10^{k}$, a few results

$$\left( \begin{array}{cccc} k & x_0 & x_1 & \text{solution} \\ 1 & -1.74553\times 10^{-1} & -1.94877\times 10^{-1} & -1.98090\times 10^{-1} \\ 2 & -3.38563\times 10^{-2} & -4.16932\times 10^{-2} & -4.44026\times 10^{-2} \\ 3 & -5.24960\times 10^{-3} & -6.46451\times 10^{-3} & -6.86702\times 10^{-3} \\ 4 & -7.23185\times 10^{-4} & -8.76201\times 10^{-4} & -9.20278\times 10^{-4} \\ 5 & -9.28457\times 10^{-5} & -1.10651\times 10^{-4} & -1.15117\times 10^{-4} \\ 6 & -1.13834\times 10^{-5} & -1.33739\times 10^{-5} & -1.38153\times 10^{-5} \\ 7 & -1.35143\times 10^{-6} & -1.56861\times 10^{-6} & -1.61181\times 10^{-6} \\ 8 & -1.56690\times 10^{-7} & -1.79998\times 10^{-7} & -1.84207\times 10^{-7} \\ 9 & -1.78417\times 10^{-8} & -2.03139\times 10^{-8} & -2.07233\times 10^{-8} \\ 10 & -2.00287\times 10^{-9} & -2.26281\times 10^{-9} & -2.30259\times 10^{-9} \\ \end{array} \right)$$

The second iteration of Newton method would give a more than decent result.

To still obtain explicit, more and more accurate approximations, use one single step of Halley or Householder (or even higher order methods which do not bear any name).

Below are the same results as above for Halley method

$$\left( \begin{array}{cccc} k & x_0 & x_1 & \text{solution} \\ 1 & -1.74553\times 10^{-1} & -1.98060\times 10^{-1} & -1.98090\times 10^{-1} \\ 2 & -3.38563\times 10^{-2} & -4.43629\times 10^{-2} & -4.44026\times 10^{-2} \\ 3 & -5.24960\times 10^{-3} & -6.86329\times 10^{-3} & -6.86702\times 10^{-3} \\ 4 & -7.23185\times 10^{-4} & -9.20015\times 10^{-4} & -9.20278\times 10^{-4} \\ 5 & -9.28457\times 10^{-5} & -1.15099\times 10^{-4} & -1.15117\times 10^{-4} \\ 6 & -1.13834\times 10^{-5} & -1.38140\times 10^{-5} & -1.38153\times 10^{-5} \\ 7 & -1.35143\times 10^{-6} & -1.61171\times 10^{-6} & -1.61181\times 10^{-6} \\ 8 & -1.56690\times 10^{-7} & -1.84199\times 10^{-7} & -1.84207\times 10^{-7} \\ 9 & -1.78417\times 10^{-8} & -2.07227\times 10^{-8} & -2.07233\times 10^{-8} \\ 10 & -2.00287\times 10^{-9} & -2.30254\times 10^{-9} & -2.30259\times 10^{-9} \\ \end{array} \right)$$