How to Taylor expand $\ln{(1-\exp(-i_t))}$ around $i$? My question here is how to Taylor expand around $i$ the function,
$$\ln{(1-\exp{(-i_t)})}.$$
How do I expand this to first order?
(Note: The variable $i_t$ is a time series variable and $i$ is its steady state.)
 A: This is the log1mexp function discussed in Mächler (2012), and it has a rather nasty general form for its derivatives.  For brevity, I will denote the function as:
$$L(x) = \ln (1-\exp(-x)) \quad \quad \quad \text{for } x > 0.$$
With a bit of algebraic work, the $n$th derivative of this function can be shown to be:
$$L^{(n)}(x) = \frac{(-1)^{n-1}}{(1-\exp(-x))^n} \sum_{k=1}^{n} \Big\langle \begin{matrix} n-1 \\ k-1 \end{matrix} \Big\rangle \exp(-kx).$$
where the values $\langle \ \cdots \ \rangle$ are the Eulerian numbers.  (Note that if $n>1$ then the last term in the summation is zero, due to the properties of the Eulerian numbers.)  You can confirm this form using a proof by induction using the recursive properties of the Eulerian numbers (which I will leave as an exercise for you).
Now, using this general form for the derivatives, we can obtain the Taylor expansion of the function $L$ around the point $a > 0$.  Taking a Taylor approximation up to order $M$ gives:
$$\begin{equation} \begin{aligned}
L(x) 
&\approx L(a) + \sum_{n=1}^M \frac{L^{(n)}(a)}{n!} \cdot (x-a)^n \\[6pt] 
&= L(a) + \sum_{n=1}^M \frac{(-1)^{n-1}}{n!} \cdot \Big( \frac{x-a}{1-\exp(-x)} \Big)^n \cdot \sum_{k=1}^{n} \Big\langle \begin{matrix} n-1 \\ k-1 \end{matrix} \Big\rangle \exp(-kx) \\[6pt] 
&= L(a) - \sum_{k=1}^{M} \exp(-kx) \sum_{n=k}^{M}  \frac{1}{n!} \cdot \Big\langle \begin{matrix} n-1 \\ k-1 \end{matrix} \Big\rangle \cdot \Big( \frac{a-x}{1-\exp(-x)} \Big)^n. \\[6pt] 
\end{aligned} \end{equation}$$

When $M=1$ we obtain the first-order approximation:
$$L(x) \approx L(a) - (a-x) \cdot \frac{\exp(-x)}{1-\exp(-x)}.$$
When $M=2$ we obtain the second-order approximation:
$$L(x) \approx L(a) - \exp(-x) \cdot \frac{a-x}{1-\exp(-x)}
- \exp(-x) \cdot \frac{1}{2} \cdot \Big( \frac{a-x}{1-\exp(-x)} \Big)^2.$$
