If:
$$x = y + \log(y) -a$$
Then the solution for $y$ using the Lambert W function is:
$$y(x) = W(e^{a+x})$$
In a paper I'm reading, I saw an approximation to this solution, due to "Borsch and Supan"(?):
$$y(x) = W(e^{a+x}) \approx x\left(1 – \frac{\log x – a}{1+x}\right)$$
Any idea how this approximation was derived?
Best Answer
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
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 $a \ll x$.