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


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

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

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..

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.