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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.)

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  • $\begingroup$ The idea is the same as for the growth of compound interest when the period of compounding is taken to be smaller and smaller. $e$ "crops up" because $(1+\frac{r}{n})^{n}\to e^r$ when $n\to\infty$. $\endgroup$
    – Conifold
    Commented Feb 5, 2020 at 18:36

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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.

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  • $\begingroup$ Thanks very much. This is is a very clear answer. My only question is: how can I solve the equation $f'(t) = cf(t)$? I managed to verify your solution by substituting $f(t)$ for $ae^{ct}$, but I wouldn't know how to solve the equation from scratch. $\endgroup$
    – Joe
    Commented Feb 5, 2020 at 19:32
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    $\begingroup$ To solve the problem from scratch would require some machinery, but I can give you some direction. I think it's instructive to consider $y(x) = a\cdot b^{x}$ and try and calculate the derivative from the limit definition. We'll have $$y'(x) = \lim_{h\to 0}\frac{ab^{x+h}-ab^x}h = ab^x \lim_{h\to 0} \frac{b^h - 1}h.$$ Observe that the last limit is precisely $y'(0)$, so we can rewrite $y'(x) = y'(0)\cdot y(x)$. (continued) $\endgroup$
    – Fimpellizzeri
    Commented Feb 5, 2020 at 19:56
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    $\begingroup$ (continued) So in studying the derivative of a basic family of functions (exponential ones), we see that they satisfy a differential equation like $(*)$. Of course, this sweeps the existence of the limit $y'(0)$ under the rug, but this is large enough for a comment and the calculations get technical. Moreover, there's something to be said about the solutions of these equations being unique, which is also a foundational theorem in ordinary differential equations. $\endgroup$
    – Fimpellizzeri
    Commented Feb 5, 2020 at 19:56
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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.

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  • $\begingroup$ you could have used yoctoseconds ... $\endgroup$
    – user645636
    Commented Feb 5, 2020 at 18:41
  • $\begingroup$ @zugzug Thanks very much for this answer. I'd love to hear the answer using calculus, but I only understand the very basics. E.g. how to find the derivative of $y=ax^n$, $y=n^x$ (using the chain rule) and integrating polynomial functions, but that is about the limit of my knowledge. If that is enough to understand the explanation using calculus, then I'd love to hear it. $\endgroup$
    – Joe
    Commented Feb 5, 2020 at 18:57
  • $\begingroup$ I will downvote this answer, and I'll explain why; there's an obvious difference between the compound interest situation. If $r$ were the 'nominal' growth rate of the year, we'd have $r=2$ and and then $\lim_{n\to\infty}(1+r/n)^n = e^r = e^2 \approx 7.39$. Of course, the population after $1$ year was doubled, not multiplied by $7.39$. This would be a good example to show how compound interests with high compounding frequencies can blow the nominal interest rate out of the water, but falls a bit flat when modeling the population growth problem even though they are related. $\endgroup$
    – Fimpellizzeri
    Commented Feb 5, 2020 at 18:59
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    $\begingroup$ @Joe All that's needed from calculus for this problem is to understand the solution to $y' = c\cdot y$, which says that the growth rate of $y$ is proportional to $y$. The solution to this is $y(x) = Ke^{cx}$, which can be verified a posteriori via differentiation of the exponential, but is also one of the foundational results in calculus. $e$ has many equivalent definitions, and one of the definitions is that $e^x$ is the unique solution to $y' = y$ with $y(0) = 1$. $\endgroup$
    – Fimpellizzeri
    Commented Feb 5, 2020 at 19:04
  • $\begingroup$ In fact, this definition of $e^x = \exp(x)$ is often employed in other scenarios, like matrices or differential geometry. In the former case, we can still use other definitions that rely on addition and multiplication of elements being well defined (as they are for matrices), but when doing differential geometry in manifolds we may not have this luxury -- and yet, the definition of $\exp$ via differential equations is still applicable. $\endgroup$
    – Fimpellizzeri
    Commented Feb 5, 2020 at 19:09

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