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.

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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}:

recurrencecreditcard

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

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 lineA 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

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

creditcardStudents 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

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 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}.

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.

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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

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 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.

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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:

  1. The effect of compound interest, which will increase the amount owed by 0.25 A(t) per year.
  2. 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}

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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 6): 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.

logareagreen line

In yesterday’s post, I presented an informal derivation of the continuous compound interest formula A = Pe^{rt} from the discrete compound interest formula A = \displaystyle P \left( 1 + \frac{r}{n} \right)^{nt}. In today’s post, I’d like to give the more formal derivation using calculus.

What does it mean for something to compound continuously? In a nutshell, the rate at which the money increases should be proportional to the amount currently present. In other words, \$10,000 should earn ten times as much interest as \$1,000. Since A'(t) is the rate at which the money increases and A(t) is the current amount, that means

A'(t) = r A(t)

for some constant of proportionality r. This is a differential equation which can be solved using standard techniques. We divide both sides by A(t) and then integrate:

\displaystyle \frac{A'(t)}{A(t)} = r

\displaystyle \int \frac{A'(t)}{A(t)} dt = \int r dt

\ln |A(t)| = r t + C

|A(t)| = e^{rt+C} = e^rt e^C = C_1 e^{rt}

A(t) = \pm C_1 e^{rt}

A(t) = C_2 e^{rt}

(Technically, a better solution would use an integrating factor [see also MathWorld], but I find that the above derivation is much more convincing to students who are a few semesters removed from a formal course in differential equations.) When presenting this in class, I’ll sometimes lazily write C in place of C_1, with the understanding that e to an arbitrary constant is just an arbitrary positive constant. Also, on the last line, plus or minus an arbitrary constant is just an arbitrary constant (which I’ll usually write as C instead of C_2).

To solve for the missing constant C_2, we use the initial condition A(0) = P:

A(0) = C_2 e^{r\cdot 0}

P = C_2 \cdot 1

P = C_2

Replacing C_2 by P, we have arrived at the continuous compound interest formula A(t) = Pe^{rt}.

Different definitions of e (Part 5): 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.

logareagreen lineAt 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 made the informal definition

\displaystyle \left(1 + \frac{1}{n} \right) \to e \qquad as \qquad n \to \infty.

We are now in position to give an informal derivation of the continuous compound interest formula. Though this derivation is informal, I have found it to be very convincing for my Precalculus students (as well as to my class of future high school teachers).

The basic idea is to rewrite the discrete compound interest formula so that it contains a term like \displaystyle \frac{1}{\hbox{something}} instead of \displaystyle \frac{r}{\hbox{something}}. In this way, we can think like an MIT freshman and reduce to previous work.

To this end, let n = mr. Then the discrete compound interest formula becomes

A = P\displaystyle \left( 1 + \frac{r}{n} \right)^{nt}

A = P\displaystyle \left( 1 + \frac{r}{rm} \right)^{rmt}

A = P\displaystyle \left( 1 + \frac{1}{m} \right)^{mrt}

A = P\displaystyle \left[ \left( 1 + \frac{1}{m} \right)^m \right]^{rt}

Inside of the brackets is our familiar friend \displaystyle \left( 1 + \frac{1}{m} \right)^m, except that the name of the variable has changed from n to m. But that’s no big deal: as n tends to infinity, then m does as well since m = n/r and both n and r are positive. Therefore, as interest is compounded more frequently, we have replace the thing inside the brackets with the number e. This leads us to the formula for continuous compound interest:

A =P e^{rt}

Again, my experience is that college students have no conceptual understanding of this formula or even a memory of seeing it derived once upon a time. 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 perhaps a harder sell to high school students that the other calculations that I’ve posted in this series, but I firmly believe that this explanation is within the grasp of good students at the time that they take Algebra II and Precalculus.
Of course, the above derivation is highly informal. For starters, it rests upon the limit

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

which cannot be formally proven using only the tools of Algebra II and Precalculus. Second, the above computation rests upon the continuity of the function A(x) = P x^{rt}, so that we can simply replace \displaystyle \left( 1 + \frac{1}{m} \right)^m with its limit e. My experience is that students are completely comfortable making this substitution, even though professional mathematicians realize that interchanging limits requires continuity.

So, mathematically speaking, the above argument should not be considered a proper derivation of the continuous compound interest formula. Still, I have found that the above argument to be quite convincing to Algebra II and Precalculus students, appropriate to their current level of mathematical development.

 

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.

logareagreen lineAt 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.

ecalculatorIn 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.

 

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.

logareagreen lineAt 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:

interestcalculator

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%.