Exponential growth and decay (Part 6): Paying off credit-card debt via recurrence relations

The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’).

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

$A_{n+1} = r A_n - k$

The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line

In yesterday’s post, I demonstrated that the solution of this recurrence relation is

$A_n = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$ .

Let’s now study when the credit card debt will actually reach $0. To do this, we see $A_n = 0$ and solve for $n$ :

$0 = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$

$0 = r^n \left(P + \displaystyle \frac{k}{1-r} \right) - \displaystyle \frac{k}{1-r}$

$0 = r^n \left( P[1-r] + k \right) - k$

$k = r^n \left( P[1-r] + k \right)$

$\displaystyle \frac{k}{P[1-r] + k} = r^n$

$\displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) = n \ln r$

$\displaystyle \frac{ \displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) }{ \ln r} = n$

That’s certainly a mouthful. However, this calculation should be accessible to a talented student in Precalculus.

Let’s try it out for $k = 50$ , $P = 2000$ , and $r = 1 + \displaystyle \frac{0.25}{12}$ :

Remembering that each compounding period is one month long, this corresponds to $86.897/12 \approx 7.24$ years, which is nearly equal to the value of $4\ln 6 \approx 7.17$ years when we solved this problem using differential equations under the assumption of continuous compound interest (as opposed to interest that’s compoounded monthly).

Exponential growth and decay (Part 5): Paying off credit-card debt via recurrence relations

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

$A_{n+1} = r A_n - k$

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. However, this particular difference equation can be solved in a straightforward fashion that should be accessible to talented Precalculus students. Let’s use the above recurrence relation to try to find a pattern. For $n = 1$ , we find

$A_1 = r A_0 - k = r P - k$ .

For $n = 2$ , we find

$A_2 = r A_1 - k$

$A_2 = r (rP - k) - k$

$A_2 = r^2P - rk - k$

$A_2 = r^2 P - k (1 + r)$

For $n = 3$ , we find

$A_3 = r A_2 - k$

$A_3 = r \left[ r^2 P - k(1+r) \right] - k$

$A_3 = r^3 - rk(1+r) - k$

$A_3 = r^3 P - rk - r^2k - k$

$A_3 = r^2 P - k \left( 1 + r+r^2 \right)$

At this point, we can probably guess a pattern:

$A_n = r^n P - k \left( 1 + r + r^2 + \dots + r^{n-1} \right)$

Using the formula for a finite geometric series, this simplifies as

$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$ .

Indeed, though I won’t do it here, this can be formally proven using mathematical induction.

Exponential growth and decay (Part 4): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In the previous two posts, I presented the general formula

$A = \displaystyle \frac{k}{r} - \left( \frac{k}{r} - P \right) e^{rt}$

which can be obtained by solving a certain differential equation. So, if $r = 0.25$ , $k = 600$ , and $P = 2000$ , the amount left on the credit card after $t$ years is

$A(t) = 2400 - 400 e^{0.25t}$ .

On the other hand, if the debtor pays $1200 per year, the equation becomes

$A(t) = 4800 - 2800 e^{0.25t}$

Today, I’ll give some pedagogical thoughts about how this problem, and other similar problems inspired by financial considerations, could fit into a Precalculus course… and hopefully improve the financial literacy of high school students.

green line Under the theory that a picture is worth a thousand words, let’s take a look at the graphs of both of these functions:

Students should have no trouble distinguishing which curve is which. Clearly, by paying $1200 per year instead of $600 per year, the credit card debt is paid off considerably quicker.

There’s another immediate take-away from these graphs — especially the graph for $k = 600$ , when the debt is being paid off over 7 years. Notice that the debt is being paid off very slowly in the initial years. Only in the latter years does the pace of paying off the loan pick up. So the moral of the story is: if you can afford to pay extra in the early years of a debt (credit card, mortgage, etc.), it’s much more important to pay off an extra amount in the early years than in the later years.

I believe this to be an important lesson for students to learn before they bury themselves deeply in debt as young adults… and Precalculus provides a natural vehicle for teaching this lesson.

Exponential growth and decay (Part 3): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In yesterday’s post, I showed that the answer to this question was about 7.2 years. To obtain this answer, I started with the differential equation

$\displaystyle \frac{dA}{dt} = 0.25 A - 600$

which, given the initial condition $A(0) = 2000$ , has solution

$A(t) = 2400 - 400 e^{0.25t}$ .

I’ve read many Precalculus books; not many of them include applying exponential functions to the paying off of credit-card debt (or a mortgage on a house or car). Of course, yesterday’s derivation was well above the comprehension level of students in Precalculus. However, there’s no reason why Precalculus students couldn’t be given the general formula

$A = \displaystyle \frac{k}{r} - \left( \frac{k}{r} - P \right) e^{rt}$ ,

where $P$ is the initial amount, $r$ is the relative rate of growth, and $k$ is the amount paid per year. In other words, students could be given the formula without the full explanation of where it comes from. After all, many Precalculus textbooks give the formula for Newton’s Law of Cooling (the subject of a future post) with neither derivation nor explanation (though its derivation is nearly identical to the work of yesterday’s post), So I don’t see why also giving students the above formula for paying off credit-card debt isn’t more common.

Plugging in $k = 600$ , $r = 0.25$ , and $P = 2000$ into this equation again yields the function

$A(t) = 2400 - 400 e^{0.25t}$ ,

from which we find that it will take $t = 4\ln 6 \approx 7.2$ years to pay off the debt.

A natural follow-up question is “How much money actually was spent to pay off this debt?” By this point, the answer is quite easy: the lender paid $\$600$ per year for $4\ln 6$ years, and so the amount spent is

$\$600 \times 4 \ln 6 = \$2400 \ln 6 \approx \$4300$ .

When I teach this topic in differential equations, I let that answer sink in for a while. The original debt was only \$2000, but ultimately \$4300 needs to be paid over 7.2 years in order to pay off the debt.

The natural question is, “Why did it take so long?” Of course, the answer is that the debtor only paid the minimal amount — $50 per month, or $600 per year. It stands to reason that if extra money was paid each month, then the debt will be paid off faster at lesser expense.

To give one example, let’s repeat the calculation if the debtor paid twice as much ($100 per month, or $1200 per year). Then the amount owed as a function of time would be

$A(t) = \displaystyle \frac{1200}{0.25} - \left( \frac{1200}{0.25} - 2000 \right) e^{0.25t} = 4800 - 2800 e^{0.25t}$

To find when the credit card will be paid off, we set $A(t) = 0$ :

$0 = 4800 - 2800 e^{0.25t}$

$2800 e^{0.25t} = 4800)$

$e^{0.25t} = \displaystyle \frac{12}{7}$

$0.25t = \displaystyle \ln \left( \frac{12}{7} \right)$

$t = \displaystyle 4 \ln \left( \frac{12}{7} \right)$

$t \approx 2.16$

That’s certainly a lot faster! Also, the amount that’s spent over that time is also considerably less:

$\displaystyle 1000 \times 4 \ln \left( \frac{12}{7} \right) = 4000 \ln \left( \frac{12}{7} \right) \approx \$2156$ .

So, along with being a good way to practice proficiency with exponential and logarithmic functions, this problem lends itself for students discovering some basic principles of financial literacy.

Exponential growth and decay (Part 2): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In this post, I present the actual solution of this problem. In tomorrow’s post, I’ll give some pedagogical thoughts about how this problem, and other similar problems inspired by financial considerations, could fit into a Precalculus course.

Let’s treat this problem as a differential equation (though it could also be considered as a first-order difference equation… more on that later). Let $A(t)$ be the amount of money on the credit card after $t$ years. Then there are two competing forces on the amount of money that will be owed in the future:

The effect of compound interest, which will increase the amount owed by $0.25 A(t)$ per year.
The amount that’s paid off each year, which will decrease the amount owed by $\$600$ per year.

Combining, we obtain the differential equation

$\displaystyle \frac{dA}{dt} = 0.25 A - 600$

There are a variety of techniques by which this differential equation can be solved. One technique is separation of variables, thus pretending that $dA/dt$ is actually a fraction. (In the derivation below, I will be a little sloppy with the arbitrary constant of integration for the sake of simplicity.)

$\displaystyle \frac{dA}{0.25 A - 600} = dt$

$\displaystyle \int \frac{dA}{0.25 A - 600} = \displaystyle \int dt$

$\displaystyle 4 \int \frac{0.25 dA}{0.25 A - 600} = \displaystyle \int dt$

$4 \ln |0.25A - 600| = t + C$

$\ln |0.25A - 600| = 0.25 t + C$

$|0.25A - 600| = e^{0.25 t + C}$

$|0.25 A - 600| = C e^{0.25t}$

$0.25A - 600 = C e^{0.25t}$

$0.25 A = 600 + C e^{0.25t}$

$A = 2400 + C e^{0.25t}$

To solve for the missing constant $C$ , we use the initial condition $A(0) = 2000$ :

$A(0) = 2400 + C e^0$

$2000 = 2400 + C$

$-400 = C$

We thus conclude that the amount of money owed after $t$ years is

$A(t) = 2400 - 400 e^{0.25t}$

To determine when the amount of the credit card will be reduced to $0, we see $A(t) = 0$ and solve for $t$ :

$0 = 2400 - 400 e^{0.25 t}$

$400 e^{0.25 t} = 2400$

$e^{0.25t} = 6$

$0.25t = \ln 6$

$t = 4 \ln 6$

$t \approx 7.2 \hbox{~years}$

In tomorrow’s post, I’ll give some pedagogical thoughts about this problem and similar problems.

Exponential growth and decay (Part 1): Phrasing of homework questions

I just completed a series of posts concerning the different definitions of the number $e$ . As part of this series, we considered the formula for continuous compound interest

$A = Pe^{rt}$

Indeed, this formula can be applied to other phenomena besides the accumulation of money. Unfortunately, as they appear in Precalculus textbooks, the wording of questions involving exponential growth or decay can be either really awkward or mathematically imprecise (or both). Here’s a sampling of problems that I’ve collected from various sources:

One thousand bacteria on a petri dish are placed in an incubator, encouraging a relative rate of growth of 10% per hour. How many bacteria will there be in two days?

This is mathematically precise, as it relates to the differential equation $A'(t) = r A(t)$ with solution $A = P e^{rt}$ . The meaning of the value of $r$ is clear from dimensional analysis: the units of $A'(t)$ are $\hbox{bacteria}/ \hbox{hour}$ , while the units of $A(t)$ are $\hbox{bacteria}$ . Therefore, the units of $r$ must be $\hbox{hour}^{-1}$ . So saying that there’s a “relative rate of growth of 10% per hour” makes total sense.

Of course, when Precalculus students are solving this problem, they have no idea about what a differential equation is, making the word relative seem superfluous to the problem.

A sum of $5000 is invested at an interest rate of 9% per year. Find the time required for the money to double if the interest is compounded continuously.

What the problem is trying to say is “Let $r = 0.09$ .” But this is a horrible way to write this in ordinary English! After all, if we plug $r = 0.09$ and $t = 1$ into the formula, we obtain

$A = P e^{0.09 \times 1} \approx 1.09417P$

So it would appear that the interest rate after one year is about 9.417%, and not 9%.

Indeed, if we read the problem at face value that the interest rate is 9% per year, then it stands to reason that, after one year, we have

$P(1.09) = P e^{r \cdot 1}$

$1.09 = e^r$

$\ln 1.09 = r$

In a nutshell, saying that there is “an interest rate of 9% per year” can easily be interpreted to mean that the annual percentage rate is 9% year, and this can be a conceptual barrier for literally-minded students.

I don’t have a good solution for this impasse between ordinary English and giving clear directions to students about what numbers should be used in the formula. But I do think that it’s important for teachers to be aware of this possible misunderstanding as students read their homework questions.

Different definitions of e (Part 12): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$ .2. We have the limits

$e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ .

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$ . From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed:

$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$ :

$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$

In yesterday’s post, I showed that using the original definition (in terms of an area under a hyperbola) does not lend itself well to numerically approximating $e$ . Let’s now look at the other two methods.

2. The limit $e = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ gives a somewhat more tractable way of approximating $e$ , at least with a modern calculator. However, you can probably imagine the fun of trying to use this formula without a calculator.

3. The best way to compute $e$ (or, in general, $e^x$ ) is with Taylor series. The fractions $\frac{1}{n!}$ get very small very quickly, leading to rapid convergence. Indeed, with only terms up to $1/6!$ , this approximation beats the above approximation with $n = 1000$ . Adding just two extra terms comes close to matching the accuracy of the above limit when $n = 1,000,000$ .

More about approximating $e^x$ via Taylor series can be found in my previous post.

Different definitions of e (Part 11): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$ .2. We have the limits

$e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ .

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$ . From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed:

$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$ :

$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$

Let’s now consider how the decimal expansion of $e$ might be obtained from these three different methods.

1. Finding $e$ using only the original definition is a genuine pain in the neck. The only way to approximate $e$ is by trapping the value of $e$ using various approximation. For example, consider the picture below, showing the curve $y = 1/x$ and trapezoidal approximations on the intervals $[1,1.8]$ and $[1.8,2.6]$ . (Because I need a good picture, I used Mathematica and not Microsoft Paint.)

Each trapezoid has a (horizontal) height of $h = 0.8$ . Furthermore, the bases of the first trapezoids have length $\displaystyle \frac{1}{1} = 1$ and $\displaystyle \frac{1}{1.8}$ , while the bases of the second trapezoid of length $\displaystyle \frac{1}{1.8}$ and $\displaystyle \frac{1}{2.6}$ . Notice that the trapezoids extend above the hyperbola, so that

$\displaystyle \int_1^{2.6} \frac{dx}{x} < \displaystyle \frac{0.8}{2} \left( 1 + \frac{1}{1.8} \right) + \frac{0.8}{2} \left( \frac{1}{1.8} + \frac{1}{2.6} \right)$

$\displaystyle \int_1^{2.6} \frac{dx}{x} < 0.9983 < 1$

However, the number $e$ is defined to be the place where the area under the curve is exactly equal to $1$ , and so

$\displaystyle \int_1^{2.6} \frac{dx}{x} < \displaystyle \int_1^{e} \frac{dx}{x}$

In other words, we know that the area between $1$ and $2.6$ is strictly less than $1$ , and therefore a number larger than $2.6$ must be needed to obtain an area equal to $1$ .

Great, so $e > 2.6$ . Can we do better? Sadly, with two equal-sized trapezoids, we can’t do much better. If the height of the trapezoids was $h$ and not $0.8$ , then the sum of the areas of the two trapezoids would be

$\displaystyle \frac{h}{2} \left( 1 + \frac{2}{1+h} + \frac{1}{1+2h} \right)$

By either guessing and checking — or with the help of Mathematica — it can be determined that this function of $h$ is equal to 1 at approximately $h \approx 0.8019$ , thus establishing that $e > 1 + 2h \approx 2.6039$ .

We can try to better with additional trapezoids. With four trapezoids, we can establish that $e > 2.6845$ .

With 100 trapezoids, we can show that $e > 2.71822$ .

However, trapezoids can only provide a lower bound on $e$ because the original trapezoids all extend over the hyperbola.

To obtain an upper bound on $e$ , we will use a lower Riemann sum to approximate the area under the curve. For example, notice the following picture of 19 rectangles of width $h = 0.1$ ranging from $x =1$ to $x = 2.9$ .

The rectangles all lie below the hyperbola. The width of each one is $h = 0.1$ , and the heights vary from $\frac{1}{1.1}$ to $\frac{1}{2.9}$ . Therefore,

$\displaystyle \int_1^{2.9} \frac{dx}{x} > \displaystyle 0.1 \left( \frac{1}{1.1}+ \frac{1}{1.2} + \dots + \frac{1}{2.9} \right)$

$\displaystyle \int_1^{2.9} \frac{dx}{x} > 1.0326 > 1$

In other words, we know that the area between $1$ and $2.9$ is strictly greater than $1$ , and therefore a number smaller than $2.9$ must be needed to obtain an area equal to $1$ . So, in a nutshell, we’ve shown that $e < 2.9$ .

Once again, additional rectangles can provide better and better upper bounds on $e$ . However, since rectangles do not approximate the hyperbola as well as trapezoids, we expect the convergence to be much slower. For example, with 100 rectangles of width $h$ , the sum of the areas of the rectangles would be

$h \displaystyle \left( \frac{1}{1+h} + \frac{1}{1+2h} + \dots + \frac{1}{1+100h} \right)$

It then becomes necessary to plug in numbers for $h$ until we find something that’s decently close to $1$ yet greater than $1$ . Or we can have Mathematica do the work for us:

So with 100 rectangles, we can establish that $e < 2.7333$ . With 1000 rectangles, we can establish that $e < 2.71977$ .

Clearly, this is a lot of work for approximating $e$ . With all of the work shown in this post, we have shown that $e = 2.71\dots$ , but we’re not yet sure if the next digit is $8$ or $9$ .

In the next post, we’ll explore the other two ways of thinking about the number $e$ as well as their computational tractability.

Different definitions of e (Part 10): Connecting the two definitions

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.

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of $e$ and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

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

Proof #2. Let’s write the left-hand side as

$L = \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt}$ .

Let’s take the natural logarithm of both sides:

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

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

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

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

$\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + nt \ln \left( 1 + \frac{r}{n} \right)\right]$

$\ln L = \ln P + \displaystyle \lim_{n \to \infty} \frac{t \displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

The limit on the right-hand side follows the indeterminate form $0/0$ , as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator with respect to $n$ , we find

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \frac{1}{1 + \frac{r}{n}} \cdot \frac{-r}{n^2}}{\displaystyle \frac{-1}{n^2}}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty}\frac{r}{1 + \frac{r}{n}}$

$\ln L = \ln P + \displaystyle t \frac{r}{1 + 0}$

$\ln L = rt + \ln P$

We now solve for the original limit $L$ :

$L = e^{rt + \ln P}$

$L = e^{rt} e^{\ln P}$

$L = Pe^{rt}$

Different definitions of e (Part 9): Connecting the two definitions

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

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

We begin with the second definition, which is usually considered the true definition of $e$ . From this definition, I have shown in a previous post that we can derive the differentiation formulas

$\displaystyle \frac{d}{dx} (\ln x) = \frac{1}{x} \qquad$ and $\qquad \displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

beginning with this definition of the number $e$ .

Theorem. $\displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = Pe^{rt}$ .

Proof #1.In an earlier post in this series, I showed that

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \ln \left(1 + \frac{h}{x} \right)^{1/h}$

Let’s now replace $h$ with $1/n$ . Also, replace $x$ with $1/r$ . Then we obtain

$r = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{1/n}{1/r} \right)^{n}$

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

Multiply both sides by $t$ :

$rt = \displaystyle t \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{n}$

$rt = \displaystyle \lim_{n \to \infty} t \ln \left(1 + \frac{r}{n} \right)^{n}$

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

$rt = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{nt}$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

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

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

Finally, we multiply both sides by $P$ :

$P e^{rt} = \displaystyle \lim_{n \to \infty} P \left(1 + \frac{r}{n} \right)^{nt}$

(A second proof of this theorem, using L’Hopital’s Rule, will be presented in tomorrow’s post.)

This firmly established, at last, the connection between the continuous compound interest formula and the area under the hyperbola. I’ve noted that my students feel a certain sense of accomplishment after reaching this point of the exposition.