The number $e$ and compound interest

Introduction

Investments left in an account that accrues interest at a regular rate grow exponentially. This document explores the effect of “compounding” that interest more regularly has on the effective rate of growth. This turns out to require a good understanding of exponential functions and the natural exponential base $e$. As such, this topic illustrates a nice application for beginning calculus students.

The basics of compound interest

The basic question that motivates compound interest is - how much money will a bank account hold if we invest a certain amount at a given interest rate and let it grow year by year? To be more precise, suppose that

$P$ denotes our initial investment (often called the principal),
$r$ denotes the annual interest rate, and
$t$ denotes the length of time that we leave the account alone.

We’d like a function $A(t)$ that tells us the amount that our account holds as a function of time $t$. For the time being we’ll assume that $t$ takes discrete steps in numbers of years.

Analysis

Let’s be clear about the annual interest rate. This is simply the factor by which the money grows from one year to the next.

At time $t=0$, we’ve just put the money in the bank. We have $A(0)=P$.
After $t=1$ year, $A(1) = P + rP = P(1+r)$.
After $t=2$ years, $A(2) = P(1+r) + rP(1+r) = P(1+r)(1+r) = P(1+r)^2$.
$\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$
After $t$ years, $A(t) = P(1+r)^t$.

Note that the interest that we make each year rolls into the investment the next year. This is the defining characteristic of compound interest. The effect is that your investment grows exponentially! Generally, though, the base $1+r$ is just a little larger than $1$.

Example: Suppose we invest $\$1000$ at an annual interest rate of $4\%$. How much will we have after

1 year?
10 years?

Solution:

After 1 year, we have $A(1) = 1000\times1.04 = 1040$.
After 10 years, we have $A(10) = 1000\times(1.04)^{10} = 1480.24$.

Frequency of compound interest

In the previous example, the interest is compounded annually. What if we want more frequent access to our funds, though? The typical approach is to compound the funds more frequently; we divide our rate, though, by the number of times we compound annually.

Let’s suppose that our annual rate is again $4\%$. Now, though, we wish to compound the interest quarterly, that is $4$ times throughout the year. The standard way to accomplish this is to use a quarterly rate of $1\%$. Then, after one year, the interest will have compounded four times. After ten years, the interest will have compounded 40 times. We’ll find that

After 1 year, we have $A(1) = 1000\times(1.01)^4 = 1040.60$.
After 10 years, we have $A(10) = 1000\times(1.01)^{40} = 1488.86$.

Not surprisingly, we make a bit more money as time goes on. The change is not huge, though; the main advantage is more frequent access.

Of course, we might want even more frequent access. We might want to compound monthly, or daily, or minutely, or anything really. We really should write down a general formula in terms of a parameter $n$ representing the number of times per year that we wish to compound the interest. This is not hard:

If we compound the interest $n$ times per year at an annual rate $r$, then

the rate per period is $r/n$,
the number of times we compound the interest over $n$ years is $nt$, so
\[A(t) = P(1+r/n)^{nt}. \tag{1}\]

Example: Suppose that on Jan 1 we invest $\$500$ at an annual interest rate of $6\%$ compounded daily. How much will we have on July 5 of a non leap year?

Solution: Our rate per period will be $0.06/365$ and July 5 is the $185^{\text{th}}$ day of the year or $184$ days after Jan 1. Thus we’ll have \[ 500\times (1+0.06/365)^{184} = \$515.35. \]

Example: Suppose that we invest $\$1$ on Jan 1 at an annual interest rate of $5\%$ compounded 12 times per year or approximately monthly. How much will we have on Jan 1 of the next year?

Solution: Our rate per period will be $0.05/12$ and we’ll compound 12 times. Thus we’ll have \[ (1+0.05/12)^{12} = \$1.05116. \] Again, we get a little more than the $5 that we’d get if we’d compounded just once. In fact, the effect is as if our interest were $5.116\%$. This is called the effective rate of interest.

Interest compounded continuously

What if we want constant access to our money? Compounding by the minute is insufficient; we want our money now!

The solution is to take the limit as $n\to\infty$ in Equation 1. Here’s that computation:

\[ \begin{aligned} \lim_{n\to\infty} P\left(1+\frac{r}{n}\right)^{nt} &= \lim_{n\to\infty} P\left(1+\frac{1}{n/r}\right)^{\frac{n}{r}rt} \\ &= \lim_{n\to\infty} P\left(\left(1+\frac{1}{n/r}\right)^{n/r}\right)^{rt} = Pe^{rt}. \end{aligned} \]

What a beautifully simple and pert little formula! To get there we used the fact that \[ \left(1+\frac{1}{n/r}\right)^{n/r} \to e \text{ as } n\to\infty. \] That’s just the definition of $e$, since $\displaystyle \frac{n}{r} \to \infty$ as $n\to\infty$.

Example: Suppose that $1000 was automatically deposited in a new account as the clock struck midnight on New Year’s Eve going into 2009. The account accrues interest at an annual rate of 6.5% compounded continuously. How much would the account hold at 9:17 AM on July 5?

Solution:

Of course, you can do this on a calculator. This example is a bit more complicated, though, so here’s how I’d approach it in code: