A common question I get is "What is e?" Here we mean the number, also called Euler's number or Euler's constant.

The easiest but most uninteresting answer

e is approximately 3. More accurately (but not much more) is that $$ e\approx 2.71828 $$

One step harder, but one step more interesting

e can be definied by a limiting sequence. To start the sequence, take the number \( (1+1/1)^1 \) which reduces to just 2. This is our first approximation of e. As a second approximation, take \( (1+1/2)^2 \) which is 2.25 and that's our second approximation. Our third approximation is \((1+1/3)^3\) and so on. Each approximation is closer and closer to the value of e.

Why is this interesting?

We can arrive at this sequence by way of a natural process! That is to say, we can connect this to something meaningful. Namely, we can relate this to a banking process, even more namely, compound interest. Recall that if you invest $100 at an interest rate of, say, 3% per year, then at the end of the year you have $103. But if you let it go another year at the same rate you then get $106.09.

Notice that the bump from the first year to the second was $3 but the bump from the second to the third was a slightly bigger $3.09. Well every time you do this, you "compound" the interest. The bigger your money is, the more you earn from the interest. And each year it gets bigger, and therefore it grows faster.

Now let's see this in a formula. The first year the money is $100, the second it's \(\$100(1+0.03)\). If this expression doesn't make sense, write a comment asking me to explain it and I'll make a separate blog post on that. Anyway, the third year, the money is now \(\$100(1+0.03)(1+0.03)\) for basically the same reason as last time, just applied to the new amount of money. And in general after n years, the bank total will be $$ \$100(1+0.03)^{n-1} $$ You can check that this makes sense, because when you plug in n=1 you get $100, and when you plug in n=2 you get $103, and so on.

In this explanation we compounded once per year. What if we compounded twice per year? Then the formula would be $$ \$100 \left(1+\frac{0.03}{2}\right)^{2(n-1)} $$ And if we computed thrice per year? $$ \$100 \left(1+\frac{0.03}{3}\right)^{3(n-1)} $$ What if we want to find out what happens as we compound infinitely often? Like one compounding per second, and then per milisecond and so on? Well, let's ignore the $100 and just take this to be 1, for simplicity. We can always multiply by whatever the true initial balance is lager if we want to change it to anything else, anyway. And let's set the rate to just 1, again for simplicity. We can change that too later on if it becomes interesting to do so. And let's just say that rather than n-1 we just call it n. (Again, if any of that is confusing, write a comment and I'll say more.) Then a formula for infinitely compounded interest would be $$ (1+1/n)^n $$ where we let n go to infinity. This is the same thing that we computed earlier.

In effect, that is what e is. e is the result of picking simple numbers and letting a compounding process go to infinity.

Harder but more interesting answers

There are still more interesting answers that come from calculus, but I'll save those for another day.