How do Mathematicians determine when to use the constant $e$?

Question

I'm only an undergrad so sorry if this is a dumb question, but I was studying Poisson distribution and it struck me that so many models involve "e". So it got me wondering; how/when/where/why do they decide to use it? I'm assuming they don't build a model and include "e" just because, and there must be some sort of fundamental intuition behind when its use is appropriate.

Answer 1

Actually, on the contrary - often you do use $e$ "just because". Or, well, almost. The thing is, any exponential expression can be written as a power of $e$ (for example $2^{3x - 2}$ is just $\frac{1}{4}e^{3x\ln{2}}$). And powers of $e$ are super convenient for all sorts of reasons - the first being that $\frac{d}{dx}e^x = e^x$ and $\int e^xdx = e^x + c$, so powers of $e$ are ridiculously easy to differentiate and integrate. What that means is that whenever a mathematician is building a model that involves some sort of exponential, they'll usually convert it into a power of $e$ unless they have a pressing reason not to.

EDIT: To walk through that example:

First, remember that $a^{b + c} = a^ba^c$, and that for any $a$ we have $a = e^{\ln{a}}$ (this isn't a fancy property of $e$, just logarithms in general). So

$$2^{3x - 2} = 2^{3x}2^{-2} = (e^{\ln{2}})^{3x}2^{-2}$$

Next, use the fact that $(a^b)^c = a^{bc}$, and that $2^{-2} = \frac{1}{4}$.

$$(e^{\ln{2}})^{3x}2^{-2} = \frac{1}{4}e^{(\ln{2}) \cdot 3x}$$

Finally, rewriting it to be a little more readable, we get $\frac{1}{4}e^{3x\ln{2}}$.

Answer 2

An example of daily life usage of $e$ and why it comes about might be in calculating interest.

For compound interest if we have an initial investment of $P_0$ which grows at an interest rate of $r$ which is compounded $n$ times per unit time (e.g. a year), then the total amount of money the investment is worth after $t$ amount of unit time would be given as:

$$P(t)=P_0(1+\frac{r}{n})^{nt}$$

For example, if we were to compound the interest only once per year it would be $P_0(1+r)^t$, if we did it twice per year, $P_0(1+\frac{r}{2})^{2t}$, if we were to do it 365 times per year it would be $P_0(1+\frac{r}{365})^{365t}$ etc...

Now, an astute observer might notice that if we keep increasing the number of times per year that the interest is compounded, we eventually reach a limit where there is no noticable change. Indeed, by taking the limit as $n$ approaches infinity we arrive at the idea of continuous interest.

For continuous interest if we have an initial investment of $P_0$ which grows at a continuous interest rate of $r$, then the total amount of money the investment is worth after $t$ amount of time would be given as:

$$P(t)=P_0e^{rt}$$

That is to say, $\lim\limits_{n\to\infty}P_0(1+\frac{r}{n})^{nt}=P_0e^{rt}$. This observation occurs as a direct result of how $e$ is defined.

As a result, most problems involving exponential growth and decay (e.g. radioactivity, population growth/decline, investments) will if modeled as having their growth/decay occurring continuously rather than happening at set intervals involve some function of $e$.

Answer 3

With respect to the Poisson distribution, you should read up on the connections between Poisson arrival processes, exponential interarrival time distributions and the Poisson distributions for the number of arrivals in a time period.

You could avoid the use of $e$ in this context, but only by adding additional constant factors that would make all of the expressions more complicated.