Approximation to the Lambert W function 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?
 A: 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 $a \ll x$.
A: First, let's consider $x=y+\log(y)$.
Assuming that $y$ is big, we can assume that $\log(y)$ is smaller than $y$. Therefore, we take a first estimate of $y=x$ and apply Newton's method to $f(y)=y+\log(y)-x$
$$
\begin{align}
y_{\text{next}}
&=y-\frac{f(y)}{f'(y)}\\
&=x-\frac{x+\log(x)-x}{1+\frac1x}\\
&=x\left(1-\frac{\log(x)}{x+1}\right)\tag{1}
\end{align}
$$
For $x\ge1$, this is a decent approximation for $\mathrm{W}(e^x)$.
Now consider that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{W}(e^x)=\frac{\mathrm{W}(e^x)}{\mathrm{W}(e^x)+1}\tag{2}
$$
Thus, for small $a$,
$$
\begin{align}
\mathrm{W}(e^{x{+}a})
&\approx\mathrm{W}(e^x)+a\frac{\mathrm{W}(e^x)}{\mathrm{W}(e^x)+1}\\
&=\mathrm{W}(e^x)\left(1+\frac{a}{\mathrm{W}(e^x)+1}\right)\\
&\approx\mathrm{W}(e^x)\left(1+\frac{a}{x+1}\right)\tag{3}
\end{align}
$$
Combining the approximations in $(1)$ and $(3)$, we get
$$
\mathrm{W}(e^{x{+}a})\approx x\left(1-\frac{\log(x)-a}{x+1}\right)\tag{4}
$$
