Different definitions of e (Part 2): 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:

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 $A = Pe^{rt}$ .
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.

In yesterday’s post, I used a numerical example to justify the compound interest formula $A = \displaystyle P \left(1 + \frac{r}{n} \right)^{nt}$ when interest is compounded $n$ times a year. In a couple of days, I’ll discuss how the above formula naturally leads to the formula $A = P e^{rt}$ when interest is continuously compounded. Today, I’d like to give some pedagogical thoughts about both formulas.

The mathematics in yesterday’s post was pretty straightforward: apply the simple interest formula $I = P r t$ a few times and see if a pattern can be developed. However, my observation is that college students have no memory of being taught how the compound interest formula $A = P \displaystyle \left( 1 + \frac{r}{n} \right)^{nt}$ can be seen as a natural consequence of the simple interest formula. In other words, they’d just use the compound interest formula without having any conceptual understanding of where the formula came from.

For my math majors who aspire to become secondary teachers in the future, I’ll make my observation that there’s absolutely no reason why students couldn’t discover this formula on their own similar to the outline above. Doubtlessly, it would take more time that I use in my college class… I can usually cover the points in yesterday’s post in about 10 minutes or less, even allowing for students to pause and interject the next step of the calculation. So while the pace would be slower for a class of high school students, the mathematical ideas are simple enough to be understood by high school students.

Different definitions of e (Part 1): Discrete compound interest

In the previous series of posts, I consider how two different definitions of a logarithmic function are actually related to each other. 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:

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 $A = Pe^{rt}$ .
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

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

I begin this series of posts with a justification for the compound interest formula $A = \displaystyle P \left(1 + \frac{r}{n} \right)^{nt}$ when interest is compounded $n$ times a year. My experience is that most math majors are familiar with this formula from their high school experience but have absolutely no idea about why it is true, and so the presentation below fills in a major hole in their preparation to become secondary teachers themselves.

In the near future, I’ll discuss how the above formula naturally leads to the formula $A = P e^{rt}$ when interest is continuously compounded.

I start with a sequence of numerical examples.

A. Suppose that you invest $1,000 at 4% interest for 2 years. (At the time of this writing, a fixed interest rate of 4% is almost mythological, but let’s leave that aside for the sake of the problem.) How much money do you have if the money is compounded annually? Here are the brute force steps. (To make the presentation less dry, I make sure that my students are volunteering each answer before proceeding to the next step.)

Amount of interest earned in Year 1 = $\$1000(0.04) = \$40$ .
Total amount of money after Year 1 = $\$1000 + \$40 = \$1040$ .
Amount of interest earned in Year 2 = $\$1040(0.04) = \$41.60$ .
Total amount of money earned in Year 2 = $\$1040 + \$41.60 = \$1081.60$ .

B. Let’s repeat the above problem, except this time the 4% interest is compounded twice a year. In other words, 2% interest is applied every six months.

Amount of interest earned in first six months = $\$1000(0.02) = \$20$ .
Total amount of money after first six months = $\$1000 + \$20 = \$1020$ .
Amount of interest earned in second six months = $\$1020(0.02) = \$20.40$ .
Total amount of money earned after second six months = $\$1020 + \$20.40 = \$1040.40$ .

At this point, I’ll make a big production about how much work this is, and we’re only halfway done with this calculation! So, I’ll rhetorically ask my class, is there an easier way to do this? Let’s take a look back at the first calculation, adding some observations.

Amount of interest earned in Year 1 = $\$1000(0.04)$ .
Total amount of money after Year 1 = $\$1000 + \$1000(0.04) = \$1000(1 + 0.04) = \$1040$ .
Amount of interest earned in Year 2 = $\$1040(0.04)$ .
Total amount of money earned in Year 2 = $\$1040 + \$1040(0.04) = \$1040(1 + 0.04) = \$1000 (1+0.04)(1+0.04)$

$= \$1000(1+0.04)^2$ .

Then I’ll check with a calculator to confirm that $\$1000(1+0.04)^2$ is indeed equal to $\$1081.60$ .

Let’s now return to the problem when the 4% interest is compounded twice a year. We’re only halfway through the calculation, but let’s recapitulate what we’ve done so far. Since this is very similar to the above work, students usually can produce the logic very quickly.

Amount of interest earned in first six months = $\$1000(0.02)$ .
Total amount of money after first six months = $\$1000 + \$1000(0.02) = \$1000(1+0.02) = \$1020$ .
Amount of interest earned in second six months = $\$1020(0.02)$ .
Total amount of money earned after second six months = $\$1020 + \$1020(0.02) = \$1020(1 + 0.02) = \$1000(1+0.02)(1+0.02)$ .

$= \$1000(1+0.02)^2$ .

At this point, I’ll ask my class what they think how much money will accumulate after two years. Invariably, they guess the correct answer of $latex $1000(1+0.02)^4$. If my class seems to get it, I usually will just accept this as the correct answer without explicitly running through steps 5 through 8 to get to the end of the fourth six weeks.

C. What is the money makes 4% interest compounded four times a year for 2 years? By this point, my students can usually guess the answer: $A = 1000(1.01)^8$ . The $0.01$ comes from dividing the 4% into four parts. The 8 comes from the number of compounding periods over 2 years.

D. By this point, we have pretty much arrived at the compound interest formula: $A = P \displaystyle \left(1 + \frac{r}{n} \right)^{nt}$ . The above argument justifies the formula; the actual proof of the formula is very similar to the above numerical examples, and so I don’t use class time to formally prove it.

Tomorrow, I’ll give some pedagogical thoughts about these computations.

Different definitions of logarithm (Part 7)

There are two apparently different definitions of a logarithm that appear in the secondary mathematics curriculum:

From Algebra II and Precalculus: If $a > 0$ and $a \ne 1$ , then $f(x) = \log_a x$ is the inverse function of $g(x) = a^x$ .
From Calculus: for $x > 0$ , we define $\ln x = \displaystyle \int_1^x \frac{1}{t} dt$ .

The connection between these two apparently different ideas begins with the following theorem, which was proven in the few previous posts.

Theorem. Let $a \in \mathbb{R}^+ \setminus \{1\}$ . Suppose that $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ has the following four properties:

$f(1) = 0$

$f(a) = 1$

$f(xy) = f(x) + f(y)$ for all $x, y \in \mathbb{R}^+$

$f$ is continuous

Then $f(x) = \log_a x$ for all $x \in \mathbb{R}^+$ .

At this point, we have provided enough groundwork to make the connection between these two different ways of viewing a logarithm.

Let’s define the function (for $x > 0$ )

$A(x) = \displaystyle \int_1^x \frac{1}{t} dt$ .

I’ll illustrate this with the appropriate area under the hyperbola $y = \frac{1}{x}$ . (Please forgive the crudeness of this drawing; I’m only using Microsoft Paint.)

So if $x$ is the right-hand limit, then $A(x)$ is just the shaded area under the curve.

Often, someone will interject, “Hey, I know how to do that… it’s just the natural logarithm of $x$ .” To which I will respond, “Yes, that’s true. But why is it the natural logarithm of $x$ ?” I have yet to encounter a student who can immediately answer this question (which, of course, is the whole point of me presenting this in class). In other words, I want my students to realize that, many semesters ago, they pretty much accepted on faith that the above integral is equal to $\ln x$ , but they were never told the reason why. And now — several semesters after completing the calculus sequence — we’re finally going over the reason why.

To start, I’ll say, “OK, $A$ is defined as an integral. That means that it must have…” Someone will usually volunteer, “A derivative.” I’ll respond, “That’s right. The Fundamental Theorem of Calculus says that this function is differentiable. So, if something is differentiable, then it also must be…” Someone will usually volunteer, “Continuous.” My response: “That’s right. So $A$ must be continuous. So that’s Property 4: this function is continuous.” I’ll continue: “Let’s see if we can get the other properties.”

I’ll next move to Property 1, as it’s the next easiest. I’ll ask the class, “Can you prove to me that $A(1) = 0$ ?” After a moment of thought, someone will notice that

$A(1) = \displaystyle \int_1^1 \frac{1}{t} dt$ .

must be equal to $0$ since the left and right endpoints of the integral are the same.

Then I’ll skip over to Property 3, which requires a little more thought. To begin, we can write

$A(xy) = \displaystyle \int_1^{xy} \frac{1}{t} dt = \displaystyle \int_1^{x} \frac{1}{t} dt + \int_x^{xy} \frac{1}{t} dt$ .

Before proceeding, I’ll ask my class why the above line has to be true. After a couple moments, someone will volunteer something like “The area from $1$ to $x$ plus the area from $x$ to $xy$ has to be equal to the area from $1$ to $xy$ .”

I’ll then say something like, “We can simplify one of the integrals on the right-hand side right away. Which one?” Students quickly see that the first integral on the right, $\displaystyle \int_1^{x} \frac{1}{t} dt$ , is of course equal to $A(x)$ . So then I’ll ask, “So what do I want the last integral to be equal to?” Students look back at Property 3 and answer, “That should be $A(y)$ .

So, if we can show the final integral is equal to $A(y)$ , we have established Property 3. To this end, I will perform a somewhat unusual looking $u-$ substitution:

$t = ux$

In this formula, I encourage my students to think of $t$ as the old variable of integration, $u$ as the new variable of integration, and $x$ as an unknown number that is constant. So I’ll say parenthetically, “If $t = 5u$ , how do we find $dt$ ?” Students of course answer, “ $dt$ must be $5 du$ .” So I’ll follow up: “If $t = xu$ , how do we find $dt$ ?” Students get the idea:

$t = x \, du$

So to complete the $u-$ substitution, we must adjust the limits of integration. For the lower limit,

$t = x \Longrightarrow u = \displaystyle \frac{x}{x} = 1$

For the upper limit,

$t = xy \Longrightarrow u = \displaystyle \frac{xy}{x} = y$

So we can now complete the $u-$ substitution of the second integral:

$A(xy) = \displaystyle \int_1^{xy} \frac{1}{t} dt = \displaystyle \int_1^{x} \frac{1}{t} dt + \int_x^{xy} \frac{1}{t} dt$ .

$A(xy) = \displaystyle \int_1^{x} \frac{1}{t} dt + \int_1^{y} \frac{1}{xu} x \, du$ .

$A(xy) = \displaystyle \int_1^{x} \frac{1}{t} dt + \int_1^{y} \frac{1}{u} \, du$ .

Students recognize that, except for the variable of integration, the last integral is just $A(y)$ , which leads to the punch line

$A(xy) = A(x) + A(y)$

In other words, we have established that the function $A$ satisfies Property 3.

So the only property left is Property 2. To that end, let’s define the number $e$ so that the area in green above is equal to 1. There’s no other way to describe this number…. we just increase $x$ far enough along the $x-$ axis until the area under the hyperbola is equal to 1. Wherever this happens, that’s the number that we’ll call $e$ . So, by definition, $A(e) = 1$ .

Therefore, by the above theorem, we conclude that $A(x) = \log_e x$ , written more simply as $\ln x$ .

To summarize: using the above theorem, we are able to establish that the integral $\displaystyle \int_1^x \frac{1}{t} dt$ has all of the properties of a logarithm and therefore must be a logarithmic function. The only catch is that we had to define $e$ to be the base of this logarithm through an unusual definition concerning the area under a hyperbola.

Of course, this is not the “standard” definition of $e$ that is usually encountered in a Precalculus class. More on these different definitions in a future series of posts.

One more pedagogical note: My experience is that I can cover the content of the first 7 posts of this series in a single 50-minute lecture and still keep my students’ attention. Naturally, I’ll recapitulate the highlights of this logical development at the start of the next lecture by way of review, as this is an awful lot to absorb at once.

A surprising appearance of e

Here’s a simple probability problem that should be accessible to high school students who have learned the Multiplication Rule:

Suppose that you play the lottery every day for about 20 years. Each time you play, the chance that you win is $1$ chance in $1000$ . What is the probability that, after playing $1000$ times, you never win?

This is a straightforward application of the Multiplication Rule from probability. The chance of not winning on any one play is $0.999$ . Therefore, the chance of not winning $1000$ consecutive times is $(0.999)^{1000}$ , which we can approximate with a calculator.

Well, that was easy enough. Now, just for the fun of it, let’s find the reciprocal of this answer.

Hmmm. Two point seven one. Where have I seen that before? Hmmm… Nah, it couldn’t be that.

What if we changed the number $1000$ in the above problem to $1,000,000$ ? Then the probability would be $(0.999999)^{1000000}$ .

There’s no denying it now… it looks like the reciprocal is approximately $e$ , so that the probability of never winning for both problems is approximately $1/e$ .

Why is this happening? I offer a thought bubble if you’d like to think about this before proceeding to the answer.

The above calculations are numerical examples that demonstrate the limit

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

In particular, for the special case when $n = -1$ , we find

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

The first limit can be proved using L’Hopital’s Rule. By continuity of the function $f(x) = \ln x$ , we have

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{x}{n}\right)^n \right]$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} n \ln \left(1 + \frac{x}{n}\right)$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \ln \left(1 + \frac{x}{n}\right)}{\displaystyle \frac{1}{n}}$

The right-hand side has the form $\infty/\infty$ as $n \to \infty$ , and so we may use L’Hopital’s rule, differentiating both the numerator and the denominator with respect to $n$ .

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \frac{1}{1 + \frac{x}{n}} \cdot \frac{-x}{n^2} }{\displaystyle \frac{-1}{n^2}}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \displaystyle \frac{x}{1 + \frac{x}{n}}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \frac{x}{1 + 0}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = x$

Applying the exponential function to both sides, we conclude that

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

In an undergraduate probability class, the problem can be viewed as a special case of a Poisson distribution approximating a binomial distribution if there’s a large number of trials and a small probability of success.

The above calculation also justifies (in Algebra II and Precalculus) how the formula for continuous compound interest $A = Pe^{rt}$ can be derived from the formula for discrete compound interest $A = P \displaystyle \left( 1 + \frac{r}{n} \right)^{nt}$

All this to say, Euler knew what he was doing when he decided that $e$ was so important that it deserved to be named.