TITLE: Double Exponential Function and Growth of the Human Population
QUESTION [6 upvotes]: I read a paper:
The Hyperexponential Growth of the Human Population
on a Macrohistorical Scale,Varfolomeyev SD, Gurevich KG, J Theor Biol. 2001 Oct 7;212(3):367-72 doi:10.1006/jtbi.2001.238
In the first part they show that the growth of the human population follows the differential equation: 
$$\frac{\mathrm dN}{\mathrm dt}=kN^2$$
The solution by simply integrating is 
$$N(t)=\frac{N_0}{1-kN_0t}$$
which is obviously divergent at $t = 1/kN_0$.
Because they are not happy with a divergent function, they look for a better solution and introduce an accelerating function $J(t)$:
\begin{equation}\frac{\mathrm dN}{\mathrm dt}=kJ(t)N(t)\tag 1\end{equation} 
Now they say that if we assume that $J(t)=J_0 e^{k_0t}$, the solution to 1 is
$$N(t)=N_0 \exp\left(\frac{k_\textrm{app}}{k_0}\exp(k_0t)\right)\tag 2$$
where $k_0$ is a kinetic parameter of the accelerating function and $k_\textrm{app}$ is a complex parameter including $J_0$.
Now they say that by Taylor approximation of $\exp(k_0t)$ to first order, they would get back the divergent function
$$N(T) = \frac{N_0}{1-(k_\textrm{app}/k_0)-k_\textrm{app}t}\tag 3$$
My questions:
1) I can solve the differential equation as well, but I do not get $k_\textrm{app}$ to be complex. 
2) I do not understand how they get from $(2)$ to $(3)$. How do they get to $(3)$ by approximating $\exp(k_0t) \approx 1+ k_0t$

REPLY [3 votes]: For the second part they are using the fact that for $x\sim 0$
$${1\over 1-x}\sim 1+x$$
so that
$$e^x \sim 1+x \sim{1\over 1-x}$$
Not sure about the complex $k_{app}$ though..