I am looking for a good approximation for the $W_0$ branch of the Lambert $W$ function. I am looking for values $0 < x < e$ only, so I expect something simpler than the general Taylor expansion. Thanks.
Approximation – Lambert W Function Near Zero
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Related Solutions
Here is a post that I made on sci.math a while ago, regarding a method I have used.
Analysis of $we^w$
For $w\gt0$, $we^w$ increases monotonically from $0$ to $\infty$. When $w\lt0$, $we^w$ is negative.
Thus, for $x\gt0$, $\mathrm{W}(x)$ is positive and well-defined and increases monotonically.
For $w\lt0$, $we^w$ reaches a minimum of $-1/e$ at $w=-1$. On $(-1,0)$, $w e^w$ increases monotonically from $-1/e$ to $0$. On $(-\infty,-1)$, $w e^w$ decreases monotonically from $0$ to $-1/e$. Thus, on $(-1/e,0)$, $\mathrm{W}(x)$ can have one of two values, one in $(-1,0)$ and another in $(-\infty,-1)$. The value in $(-1,0)$ is called the principal value.
The iteration
Using Newton's method to solve $we^w$ yields the following iterative step for finding $\mathrm{W}(x)$: $$ w_{\text{new}}=\frac{xe^{-w}+w^2}{w+1} $$
Initial values of $w$
For the principal value, when $-1/e\le x\lt0$, and when $0\le x\le10$, use $w=0$. When $x\gt10$, use $w=\log(x)-\log(\log(x))$.
For the non-principal value, when $x$ is in $[-1/e,-.1]$, use $w=-2$; and if $x$ is in $(-.1,0)$, use $w=\log(-x)-\log(-\log(-x))$.
This says that for the branch you want, use the iteration with an initial value of $w=0$.
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
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.