User:Algebraic

You can call me Al.

Per-generation and instantaneous growth rates
Let $$n_i(t)$$ be the number of organisms of type $$i$$ at time $$t$$, and let $$R$$ be the per-capita reproductive rate per generation. If $$t$$ counts generations, then
 * $$n_i(t+1) = n_i(t)R\!$$

and
 * $$n_i(t) = n_i(0)R^t.\!$$

Now we wish to move to the case where $$t$$ is continuous and real-valued. As before,
 * $$n_i(t+1) = n_i(t)R\!$$

but now

where the last simplification follows from L'Hôpital's rule. Explicitly, let $$\epsilon=\Delta t$$. Then
 * align="right" |$$n_i(t+\Delta t)\!$$
 * $$=n_i(t)R^{\Delta t}\!$$
 * align="right" |$$n_i(t+\Delta t) - n_i(t)\!$$
 * $$= n_i(t)R^{\Delta t} - n_i(t)\!$$
 * align="right" |$$\frac{n_i(t+\Delta t) - n_i(t)}{\Delta t}$$
 * $$=\frac{n_i(t)R^{\Delta t} - n_i(t)}{\Delta t}$$
 * align="right" |$$\frac{n_i(t+\Delta t) - n_i(t)}{\Delta t}$$
 * $$=n_i(t) \frac{R^{\Delta t} - 1}{\Delta t}$$
 * align="right" |$$\lim_{\Delta t \to 0} \left[{n_i(t+\Delta t) - n_i(t) \over \Delta t}\right]$$
 * $$=\lim_{\Delta t \to 0} \left[ n_i(t) \frac{R^{\Delta t} - 1}{\Delta t}\right]$$
 * align="right" |$$\frac{d n_i(t)}{dt}$$
 * $$=n_i(t) \lim_{\Delta t \to 0} \left[\frac{R^{\Delta t} - 1}{\Delta t}\right]$$
 * align="right" |$$\frac{d n_i(t)}{dt}$$
 * $$=n_i(t) \ln R\!$$
 * }
 * align="right" |$$\frac{d n_i(t)}{dt}$$
 * $$=n_i(t) \lim_{\Delta t \to 0} \left[\frac{R^{\Delta t} - 1}{\Delta t}\right]$$
 * align="right" |$$\frac{d n_i(t)}{dt}$$
 * $$=n_i(t) \ln R\!$$
 * }
 * }


 * $$\lim_{\Delta t \to 0} \left[{R^{\Delta t} - 1 \over \Delta t}\right]$$
 * $$= \lim_{\epsilon \to 0} \left[\frac{R^{\epsilon} - 1}{\epsilon}\right]$$
 * $$=\lim_{\epsilon \to 0} \left[\frac{\frac{d}{d\epsilon}\left(R^{\epsilon} - 1\right)}{\frac{d}{d\epsilon}\epsilon}\right]$$
 * $$=\lim_{\epsilon \to 0} \left[\frac{R^{\epsilon}\ln R}{1}\right]$$
 * $$=\ln R \lim_{\epsilon \to 0} \left[R^{\epsilon}\right]$$
 * $$=\ln R\!$$
 * }
 * $$=\lim_{\epsilon \to 0} \left[\frac{R^{\epsilon}\ln R}{1}\right]$$
 * $$=\ln R \lim_{\epsilon \to 0} \left[R^{\epsilon}\right]$$
 * $$=\ln R\!$$
 * }
 * $$=\ln R \lim_{\epsilon \to 0} \left[R^{\epsilon}\right]$$
 * $$=\ln R\!$$
 * }
 * $$=\ln R\!$$
 * }

The solution to the equation
 * $$\frac{d n_i(t)}{dt} = n_i(t) \ln R$$

is
 * $$n_i(t) = n_i(0) e^{t\ln R} = n_i(0) R^{t}.\!$$

Note that the continuous case and the original discrete-generation case agree for all values of $$t$$. We can define the instantaneous rate of increase $$r = \ln R$$ for convenience.