# Different definitions of e (Part 4): Continuous compound interest

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

1. Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$.
2. Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$, so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

I should say at the outset that the second definition is usually considered the true definition of $e$. However, compound interest usually appears earlier in the mathematics curriculum than definite integrals, and so an informal definition of $e$ is given at that stage of the curriculum.

At this point in the exposition, I have justified the formula $A = \displaystyle P \left(1 + \frac{r}{n} \right)^{nt}$ for computing the value of an investment when interest is compounded $n$ times a year. We have also seen that $A$ increases as $n$ increases, but that $A$ appears to level off as $n$ gets very large. This observation forms the basis for the continuous compound interest formula $A = P e^{rt}$.

To begin, let’s consider plug in variables to make the compound interest formula as simple as possible. Let’s start with 1 dollar (so that $P = 1$) that earns 100% interest (so that $r = 1$) for one year (so that $t = 1$). This isn’t financially realistic, of course, but let’s run with it. Then the compound interest formula becomes

$A = \displaystyle \left( 1 + \frac{1}{n} \right)^n$

As before, let’s see what happens as $n$ increases. As before, I’ll plug numbers into a calculator in real time, asking my students to use their calculators along with me.

1. If $n = 1$, then $A = (1+1)^1 = 2$. I’ll usually double-check with my class to make sure that they believe this answer… that $1 compounded once at 100% interest results in$2.
2. If $n = 2$, then $A = (1.5)^2 = 2.25$.
3. If $n = 4$, then $A = (1.25)^4 \approx 2.441$.
4. If $n = 10$, then $A = (1.1)^{10} \approx 2.593$.
5. If $n = 1000$, then $A = (1.001)^{1000} \approx 2.71692$
6. If $n = 1,000,000$, then $A = (1.000001)^{1000000} \approx 2.71828$

As before, the final amount $A$ appears to be increasing toward something. That something is defined to be the number $e$. So, as $n$ tends toward infinity, we’ll define the limiting value to be the number $e$.

In my experience, college students have no memory of learning how they first saw the number $e$ when they were in high school. They remember is it as coming out of nowhere, as a number in a formula or as a button on a calculator. It really shouldn’t be this way. The above calculation is a natural consequence of the discrete compound interest formula, which makes the appearance of the number $e$ to be a bit more natural.

Of course, this “definition” of the number $e$ is highly informal. What we’re really claiming is

$\displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n = e$.

At this point in the mathematical curriculum, students only have the haziest notion of what a limit actually means, let alone the more formal treatment that’s presented in calculus… not to mention a proper $\delta-\epsilon$ treatment of limit in an honors calculus class or in real analysis. So, mathematically speaking, the above argument should not be considered a proper definition of the number $e$, but a working definition so that high school students can get comfortable with the number $e$ before seeing it again in their future mathematical courses.

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