How does population growth relate to $e$?

Question

Say that there is a population of $10,000$. From year $0$ to year $1$, we know that the population has grown to $20,000$. How might I model this growth?

It clearly isn't sensible to say that the population suddenly 'jumped' from $10,000$ to $20,000$ as the year drew to a close. A better model might be to split the year up into $365$ segments, where on each day the population is multiplied by $\sqrt[365]{2}$ or approximately $1.0019$. However, even this feels unsatisfying, as the population would have grown during that day, which would have (very slightly) affected the rate of growth. The discrete model used in this example does not take into account the fact that the population would be slightly higher in the afternoon compared to the morning, which (at least in theory) would have an effect on population growth. If I use a continuous model, then I expect to see $e$ crop up. However, I can't seem to find it. Why is this? And, more generally, how does continuous modelling work—is it premised on splitting the year up into $n$ segments, where $n \to \infty$? (An answer that does not use university-level maths would be appreciated, but I understand if this is not possible.)

Answer 1

Your expectation is not unfounded. We would model it as such: let $f(t)$ be the population at time $t$. Then $f(0) = 10000$ and $f(1) = 20000$.

We implicitly assume that the growth of the population is proportional to to the population, and moreover that this proportion is constant in time. This would translate symbolically into

$$f'(t) = cf(t) \tag{$*$},$$

where $c$ is the constant of proportionality, to be determined. The solution to the differential equation $(*)$ is

$$f(t) = ae^{ct},$$

where $a$ is a constant to be determined.
From $f(0) = a$, we find that $a = 10000$. For $c$, we compute

$$f(1) = 10000e^c = 20000 \implies e^c = 2 \implies c = \log 2.$$

Wrapping up, we find that the function that continuously describes the growth of of the population through time is

$${f(t) = 10000\cdot 2^t}.$$

$e$ need not show up in the final answer, but it's intrinsically tied to the solution of $(*)$ which is the heart of our problem's modeling.

Answer 2

EDIT: This is only an answer to how continuous compound growth works. In this problem, you would want a different approach.

There is a beautiful answer using calculus so I'll try to put it in not calculus terms.

Let the time period units be small (say, milliseconds). Then every time period, the population (assuming the simplistic compounding model that doesn't account for limit resources, etc.) growths by a factor of $(1+\frac{r}{n})$, where $r$ is the nominal growth rate for the year and $n$ is the number of time periods that fit into a year. In my example about milliseconds, $n$ will obviously be huge.

Imagine this compounding every time period. Since there are $n$ time periods per year, by the end of time $t$ years, the population will be $A_0 (1+\frac{r}{n})^{nt}$, where $A_0$ is the initial value.

What happens if instead of using milliseconds, we decide that we want a finer measurement? The finer the measurement, the larger $n$ becomes. It turns out in the "limit" (just means for $n$ really big), $(1+\frac{r}{n})^{nt} \approx e^{rt}$ and the population is then $\approx A_0 e^{rt}$.

Technically speaking, "compounding continuously" just means the limit of a bunch of small discrete compounds. It's often used when compounding occurs so often that it's not agreed upon how one would stop at discrete steps.