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.
Best Answer
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 $x_0=-\frac{W(y)}{y}$