$\pi$ is the most well-known special number, but there is also $e$. Whereas $\pi$ has the easy interpretation of being the ratio of a circle’s circumference to its diameter, and everyone knows what a circle is, $e$ arises in calculus with which many people have no familiarity.
Here’s something interesting…
Without calculus, the closest one gets to encountering $e$ is in finance. If you have $P$ dollars in an account that earns interest at an annual rate $r$ compounded $n$ times per year, you will have at the end of the year:
$$\displaystyle A = P\left( 1 + \frac{r}{n} \right)^n $$
Supposing a thoroughly unrealistic 100% interest, so that $r = 1$, let’s see how the amount $A$ at the end of the year depends on $n$ with a starting principal $P = 1$:
| Compounding interval | $n$ | $A$ |
| Yearly | 1 | 2 |
| Monthly | 12 | 2.613035290224676 |
| Weekly | 52 | 2.692596954437168 |
| Daily | 365 | 2.7145674820219727 |
| Second-ly | 31557600 | 2.7182817764125247 |
One might expect that as the compounding interval shrinks, the amount at the end of the year would grow without bound. But even with the interval shrunk to one second, we get hardly more than with an interval of one day. In fact, as $n$ grows larger $A$ approaches a finite limiting value – that number is $e = 2.718281828459045…$.
From the Calculus Side
What we have so far is the claim (I haven’t proven it to you, but it is true) that
$$\displaystyle e = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n $$
I would say that it’s not at all obvious that this limit even exists, that it approaches a finite number as $n\to\infty$, nor is it clear why anyone would be … interested … in it outside the narrow context of finance. In the calculus world, however, $e$ is everywhere in the form of the exponential function $e^x$. This function is unique in that it is its own derivative:
$$\displaystyle \lim_{h\to 0} \frac{e^{x+h} – e^x}{h} = e^x $$
This function shows up all the time when dealing with quantities whose rate of change is proportional to their current quantities, like bacteria multiplying in a petri dish or decaying radioactive substances.
But how do we know that $e^x$ is its own derivative (after all, “My calculus teacher said so” is not valid mathematics), and why is this strange number $e$ be the base of that exponential rather than some other number? This is the question we will now answer. We will show that a function which is its own derivative must be an exponential, and that its base is the same limit we arrived at before.
Derivation
Suppose $f(x)$ is a function that equals its own derivative:
$$\displaystyle f'(x) = \lim_{h\to 0} \frac{f(x+h) – f(x)}{h} = f(x) $$
Assuming for the moment that $f(x) \neq 0$, we may divide through by it:
$$\displaystyle \frac{f'(x)}{f(x)} = 1 $$
Consider the chain rule: $\frac{d}{dx} g(f(x)) = g'(f(x))f'(x)$. If we can find a $g(x)$ such that $g'(x) = 1/x$, we will have then $\frac{d}{dx} g(f(x)) = f'(x)/f(x) = 1$ implying that $g(x)$ will be the inverse function of the $f(x)$ we want to find. So our $g(x)$ must satisfy:
$$\displaystyle g'(x) = \lim_{h\to 0} \frac{g(x+h) – g(x)}{h} = \frac{1}{x} $$
Since $h$ is going to zero, at the cost of assuming $x\neq 0$ we may replace $h$ by $hx$ without changing the limit:
$$\displaystyle g'(x) = \lim_{h\to 0} \frac{g(x+xh) – g(x)}{xh} = \lim_{h\to 0} \frac{g(x(1+h)) – g(x)}{xh}= \frac{1}{x} $$
Now we observe that if $g(x)$ is a logarithm, i.e. $g(ab) = g(a) + g(b)$ from which it follows that $g(1) = 0$, then the limit becomes:
$$\displaystyle \begin{aligned} g'(x) &= \lim_{h\to 0} \frac{g(x(1+h)) – g(x)}{xh} \\ &= \lim_{h\to 0} \frac{g(x) + g(1+h) – g(x)}{xh} \\ &= \lim_{h\to 0} \frac{g(1+h) – g(1)}{xh} = \frac{g'(1)}{x} \end{aligned}$$
Since we want $g(x)$ such that $g'(x) = 1/x$, we need $g(x)$ to be the logarithm whose derivative at $x=1$ is $1$. If $g(x)$ is a logarithm, then $f(x) = g^{-1}(x)$ is an exponential function. If $g'(1) = 1$ then $f'(0) = 1$ (note that $f(0) = 1$ matching this derivative, as we sought). Now knowing that $f(x)$ is an exponential, $f(x) = b^x$, we need to know only what is the value of $b$. Observe that $b = f(1)$, and since $f(x)$ is an exponential:
$$\displaystyle \begin{aligned} b = f(1) &= f\left(\frac12 + \frac12\right) = f\left(\frac12\right)^2 \\ &= f\left(\frac13 + \frac13 + \frac13\right) = f\left(\frac13\right)^3 \\ &= f\left(\frac{1}{n}\right)^n \end{aligned}$$
where $n$ can be any positive number. As $n$ increases, $1/n$ approaches $0$ and we can estimate $f(1/n)$ using $f(0)$ and $f'(0)$ which we know both equal 1. By definition of the derivative, we have that:
$$\displaystyle \lim_{n\to\infty} \frac{f(1/n) – 1}{1/n} = 1$$
Rephrasing this fact into an estimate for $f(1/n)$, we have that for any $\epsilon > 0$ there exists an $N_\epsilon$ such that for all $n > N_\epsilon$:
$$\displaystyle 1 + \frac{1-\epsilon}{n} < f\left(\frac{1}{n}\right) < 1 + \frac{1+\epsilon}{n}$$
From which follows:
$$\displaystyle \left(1 + \frac{1-\epsilon}{n}\right)^n < f\left(\frac{1}{n}\right)^n < \left(1 + \frac{1+\epsilon}{n}\right)^n$$
The term in the middle is the base we want to find:
$$\displaystyle \left(1 + \frac{1-\epsilon}{n}\right)^n < b < \left(1 + \frac{1+\epsilon}{n}\right)^n$$
And the terms on the left and right are clearly reminiscent of the interest-bearing limit from earlier, except for the $\epsilon$ terms. We may rewrite them with help of the binomial theorem. The one on the right:
$$\displaystyle \left(1 + \frac{1+\epsilon}{n}\right)^n = \left(1+\frac{1}{n}\right)^n + \sum_{k=1}^n \frac{n!}{k!(n-k)!} \left(1+\frac{1}{n}\right)^{n-k}\left(\frac{\epsilon}{n}\right)^k $$
On the left we get a similar expression with $\epsilon$ replaced by $-\epsilon$. We can bound the magnitude of the $\epsilon$ terms:
$$\displaystyle \begin{aligned} &\left| \sum_{k=1}^n \frac{n!}{k!(n-k)!} \left(1+\frac{1}{n}\right)^{n-k}\left(\frac{\pm\epsilon}{n}\right)^k \right| \\ &= \left(1+\frac{1}{n}\right)^n \left| \sum_{k=1}^n \frac{n!}{k!(n-k)!} \left( \frac{\pm\epsilon}{n+1} \right)^k \right| \\ &\le \left(1+\frac{1}{n}\right)^n \sum_{k=1}^n \epsilon^k \\ &= \left(1+\frac{1}{n}\right)^n \frac{\epsilon}{1-\epsilon}\end{aligned}$$
So we can bracket the base $b$ like so:
$$\displaystyle \left(1 + \frac{1}{n}\right)^n – \left(1+\frac{1}{n}\right)^n \frac{\epsilon}{1-\epsilon}< b < \left(1 + \frac{1}{n}\right)^n + \left(1+\frac{1}{n}\right)^n \frac{\epsilon}{1-\epsilon}$$
Since $\epsilon$ can be made arbitrarily small (just not zero) at the cost of $n$ needing to become arbitrarily large, the base must be equal to the quantity obtained by setting $\epsilon = 0$ and letting $n\to\infty$, giving us what we expect:
$$\displaystyle b = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n $$
Technically we have to first show that this limit exists before we can draw this conclusion – this is easily established by expanding the power using the binomial theorem and showing that the resulting series is monotonically increasing as a function of $n$ and also bounded. We get the boundedness portion for free here because we can see the limit is bounded above by the base $b$ times some constant factor to compensate for the $\epsilon$ fraction on the left-hand side. But I won’t go through the details here.