I am looking for a simple, inexpensive and very accurate approximation of the Lambert function ($W_0$ branch) ($-1/e < x < 0$).
Approximation – Lambert Function $W_0$ Branch Approximation
approximationlambert-w
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I don't know how simple you need it, and since you never said anything on how accurate you want your approximant to be (i.e., to how many correct decimal places should the approximant match the Lambert function?),
$$W_0(z)\approx\ln(1+z)\frac{1+\frac{123}{40}z+\frac{21}{10}z^2}{1+\frac{143}{40}z+\frac{713}{240}z^2}$$
should be good enough, which has a maximum error of around $1.6\times 10^{-4}$ for $z\in[0,e]$.
The rational portion here is a Padé approximant; probably one might do better with a minimax rational approximation, but I don't have the patience and inclination to derive it since your question's rather vague to begin with.
We can use a procedure known as "bootstrapping" to determine an approximation for the Lambert $W$ function. Let's go back to its definition.
For $x > 0$ the equation
$$ we^w = x $$
has exactly one positive solution $w = W(x)$ which increases with $x$. Note that $(w,x) = (1,e)$ is one such solution, so if $x > e$ then $w > 1$. By taking logarithms of both sides of the equation we get
$$ \log w + w = \log x $$
or
$$ w = \log x - \log w. \tag{1} $$
When $x > e$ we therefore have
$$ w = \log x - \log w < \log x. $$
In other words, our first approximation is that
$$ 1 < w < \log x \tag{2} $$
when $x > e$. We then have
$$ 0 < \log w < \log\log x, $$
and plugging this into $(1)$ yields
$$ \log x - \log \log x < w < \log x, \tag{3} $$
where the left side is positive for $x > 1$. Taking logarithms as before yields
$$ \log\log x + \log\left(1 - \frac{\log\log x}{\log x}\right) < \log w < \log\log x, $$
and upon substituting this back into $(1)$ we get
$$ \log x - \log\log x < w < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right). $$
Since $w = W(x)$ we have shown that
$$ \log x - \log\log x < W(x) < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right) \tag{4} $$ for $x > e$.
In your particular case we're interested in $W(e^{x+a})$, for which we have
$$ x+a - \log(x+a) < W(e^{x+a}) < x+a - \log(x+a) - \log\left(1 - \frac{\log(x+a)}{x+a}\right) $$
for $x+a > 1$. In this sense we have
$$ W(e^{x+a}) \approx x+a - \log(x+a) = x\left(1 - \frac{\log(x+a) - a}{x}\right) \tag{5} $$
when $x+a$ is large. Now by applying Taylor series a couple times we see that, for $x$ large and $a \ll x$,
$$ \begin{align} \frac{\log x - a}{x+1} &= \frac{\log x - a}{x} \cdot \frac{1}{1+\frac{1}{x}} \\ &\approx \frac{\log x - a}{x} \left(1-\frac{1}{x}\right) \\ &= \frac{\log x - a}{x} - \frac{\log x - a}{x^2} \\ &= \frac{\log(x+a-a) - a}{x} - \frac{\log x - a}{x^2} \\ &= \frac{\log(x+a) + \log\left(1-\frac{a}{x+a}\right) - a}{x} - \frac{\log x - a}{x^2} \\ &= \frac{\log(x+a) - a}{x} + \frac{\log\left(1-\frac{a}{x+a}\right)}{x} - \frac{\log x - a}{x^2} \\ &\approx \frac{\log(x+a) - a}{x} - \frac{a}{x(x+a)} - \frac{\log x - a}{x^2} \\ &\approx \frac{\log(x+a) - a}{x}. \end{align} $$
We may then conclude from $(5)$ that
$$ W(e^{x+a}) \approx x \left(1 - \frac{\log x - a}{x+1}\right) $$
for $x$ large and $a \ll x$.
Best Answer
Here is a post that I made on sci.math a while ago, regarding a method I have used.
This says that for the branch you want, use the iteration with an initial value of $w=0$.