Different definitions of e (Part 3): Discrete 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. In tomorrow’s post, I’ll discuss how the above formula naturally leads to the formula $A = P e^{rt}$ when interest is continuously compounded.

The bridge between these two formulas is considering increasing values of $n$. So far in the presentation, we have considered an investment of \$1000 making 4% interest for 2 years. In the first post of this series, we made the following computations:

1. If interest is compounded annually ($n = 1$), then $A = \1000(1.04)^2 = \1081.60$.

2. If interest i compounded semiannually ($n = 2$), then $A = \1000(1.02)^4 \approx \1082.43$.

3. If interest is compounded quarterly ($n = 4$), then $A = \1000(1.01)^8 \approx \1082.86$.

So I ask my class, “What happens to the final amount as interest is compounded more frequently?” They easily observe that the final amount increases somewhat. A natural question, then, is to find how much it can increase. So let’s make the compounding more frequent and let’s see what happens.

4. Daily: ($n = 365$). Then $A = \1000 \displaystyle \left( 1 + \frac{0.04}{365} \right)^{730} \approx \1083.28$.

5. About twice a minute ($n = 1,000,000$): Then $A = \1000 \displaystyle \left( 1 + \frac{0.04}{1,000,000} \right)^{2,000,000} \approx \1083.29$.

Of course, I perform all of these calculations in real time on a calculator so that students can follow along:

Students quickly observe that the final amount continues to increase as $n$ increases. However, the final amount appears to be leveling off… we can’t make the final amount arbitrarily large just by compounding the interest more frequently.

This provides a natural bridge to continuous compound interest, the topic of tomorrow’s post.

I’ll also note parenthetically that this is why financial institutions are required to disclose the annual percentage rate of a loan (among other things). Otherwise, banks could get away with declaring “Only 2% interest monthly!!” That sounds like 24% annual interest. However, $(1.02)^{12} \approx 1.26824$, and so the annual percentage rate would really be 26.824%.

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