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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 $\exp(k_0t) \approx 1+ k_0t$

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  • $\begingroup$ Your question is mathematics. As far as I can see, it involves no physics. $\endgroup$ Nov 6 '16 at 23:50
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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..

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  • $\begingroup$ Thanks. But then they should better say that both functions agree in first order, what is something different then saying one of the functions is an approximation of the other. $\endgroup$
    – newandlost
    Nov 7 '16 at 13:49
  • $\begingroup$ Not really, it is similar (: $\endgroup$
    – JalfredP
    Nov 7 '16 at 14:42

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