Simplification of a logarithm in the context of a network I am reading Mark Newman's Networks textbook and equation 11.14 is puzzling me.
$$
\ln{u} = (n-1)\ln{\Big[ 1 - \frac{c}{n-1} (1-u) \Big]} \\
 \approx -(n-1)\frac{c}{n-1}(1-u)
$$
With my understanding of logarithm rules, I would have simplified the expression like this:
$$
\ln{u} = -(n-1)\ln{\Big[\frac{c}{n-1} (1-u) \Big]}
$$
So I don't understand how we can drop the ln on the right hand side of the equation.
The assumptions here are that n is a large integer, u is a fraction between 0 and 1 and c is a positive real number.
Thank you
 A: This makes use of the Taylor expansion of $\log(1+x)$ near $x=0$, which is given by $\log(1+x) = x - x^2 + x^3 - \cdots$ (see this related question: taylor series of $\ln(1+x)$?).
For convenience, let's set $\varepsilon_n = \frac{c}{n-1}(1-u)$. Notice that this is a positive number, since $u <1$ and $c > 0$ are both assumed to be true.
In the equation you wrote down, the fact that $n$ is a large integer means that we can assume $\varepsilon_n \ll 1$. Why? Well, if $u$ is between $0$ and $1$, then $(1-u)$ is as well, so $\varepsilon_n \leq \frac{c}{n-1}$. Since we're being told that $n$ is large, we should take that to mean $c \ll n$; this means that $c \ll n-1$ as well. Then our inequality $\varepsilon_n \ll 1$ follows.
For very small numbers (like $\varepsilon_n$), we can treat powers of them as approximately zero: for example, if $\varepsilon_n = 0.1$, then $\varepsilon_n^2 = 0.01$, $\varepsilon_n^3 = 0.001$, etc., and only $\varepsilon_n$ is large enough to bother paying attention to. This lets us simplify our Taylor expansion to $\log(1+\varepsilon_n) \approx \varepsilon_n$, which gives the simplification you cited from the textbook.
