Help understanding this approximation In a paper that I'm reading, the authors write:-

$$N_e \approx \frac{3}{4} (e^{-y}+y)-1.04. \tag{4.31}$$
Now, an analytic approximation can be obtained by using the expansion
  with respect to the inverse number of "e-foldings" ($N_e$ is the
  number of "e-foldings"). For instance, eq. $(4.31)$ yields:-
$$e^y = \dfrac{3}{4N_e} - \dfrac{9\ln(N_e)}{16(N_e)^2} -
 \dfrac{0.94}{(N_e)^2} + O\left(\dfrac{\ln^2(N_e)}{(N_e)^3}\right)$$

Can anyone tell me how this approximation is done? I'm not familiar with the $O$ notation either. What does it mean? How do the authors arrive at that expression?
If anyone should require it, the original paper can be found here: https://arxiv.org/pdf/1001.5118.pdf?origin=publication_detail
 A: The assumption seems to be that $y \ll -1$ and $N_\epsilon \gg 1$.  For brevity let's set $x = e^y$ and $N_\epsilon = N$ so that the equation becomes
$$
N = \frac{3}{4 x} + \frac{3}{4}\ln x - 1.04. \tag{1}
$$
The condition $y \ll -1$ now corresponds to $0 < x \ll 1$ for the new variable.  In this case the term involving $1/x$ is dominant on the right-hand side, and since the other two terms are negative we have
$$
N < \frac{3}{4 x} \quad \Longleftrightarrow \quad 0 < x < \frac{3}{4 N}.
$$
This is our initial estimate.  Since $N \gg 1$ this implies that
$$
\frac{3}{4}\ln x - 1.04 = O(\ln N),
$$
and hence from $(1)$ that
$$
\frac{3}{4x} = N + O(\ln N).
$$
Solving for $x$ yields
$$
\begin{align}
x &= \frac{3}{4(N+O(\ln N))} \\
&= \frac{3}{4N} \cdot \frac{1}{1+O\left(\frac{\ln N}{N}\right)} \\
&= \frac{3}{4N} \left(1+O\left(\frac{\ln N}{N}\right)\right) \tag{2.1} \\
&= \frac{3}{4N} + O\left(\frac{\ln N}{N^2}\right). \tag{2.2}
\end{align}
$$
This matches the first term of the desired estimate.  Achieving more terms is simply a matter of repeating this process.
Substituting $(2.1)$ for $x$ into the sub-dominant terms of $(1)$, namely $\frac{3}{4}\ln x - 1.04$, yields
$$
\begin{align}
\frac{3}{4}\ln x - 1.04 &= \frac{3}{4} \ln\left[\frac{3}{4N} \left(1+O\left(\frac{\ln N}{N}\right)\right)\right] - 1.04 \\
&= \frac{3}{4}\ln \frac{3}{4} - \frac{3}{4}\ln N + \frac{3}{4}\ln\left(1+O\left(\frac{\ln N}{N}\right)\right) - 1.04 \\
&= -\frac{3}{4}\ln N + \frac{3}{4}\ln \frac{3}{4} - 1.04 + O\left(\frac{\ln N}{N}\right),
\end{align}
$$
so that $(1)$ becomes
$$
N = \frac{3}{4x} - \frac{3}{4}\ln N + \frac{3}{4}\ln \frac{3}{4} - 1.04 + O\left(\frac{\ln N}{N}\right).
$$
Solving for $x$ as above (though with a bit more algebra this time) yields the approximation
$$
x = \frac{3}{4N} - \frac{9}{16} \frac{\ln N}{N^2} + \frac{C}{N^2} + \frac{27}{64} \frac{(\ln N)^2}{N^3} + O\left(\frac{\ln N}{N^3}\right), \tag{3}
$$
where
$$
C = \frac{9}{16} \ln \frac{3}{4} - \frac{3}{4} \cdot 1.04 \approx -0.94.
$$
Using the original variables $x = e^y$ and $N = N_\epsilon$, $(3)$ is

$$
e^y = \frac{3}{4N_\epsilon} - \frac{9}{16} \frac{\ln N_\epsilon}{N_\epsilon^2} + \frac{C}{N_\epsilon^2} + \frac{27}{64} \frac{(\ln N_\epsilon)^2}{N_\epsilon^3} + O\left(\frac{\ln N_\epsilon}{N_\epsilon^3}\right), \tag{4}
$$

as desired.
