# Exponential growth and decay: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my recently completed series on various applications of exponential growth and decay.

Part 1: Introduction: continuous compound interest and the phrasing of homework questions

Paying off credit-card debt

Part 2: Solution using a differential equation.

Part 3: Teaching basic principles of financial literacy.

Part 4: More on financial literacy.

Part 5: Solution using a difference equation.

Part 6: Comparison of the two solutions (difference equation vs. differential equation).

Part 7: An alternative solution of the difference equation that can be derived by Precalculus students.

Part 8: Verifying the solution of the difference equation using a spreadsheet.

Part 9: Amortization tables.

Half-life

Part 10: Derivation of the formula for exponential decay using a differential equation.

Part 11: Rewriting the solution of the differential equation into the half-life formula.

Newton’s Law of Cooling

Part 12: Derivation of the formula using a differential equation.

Part 13: Classroom demonstrations of Newton’s Law of Cooling.

Logistic Growth Model

Part 14: A simple classroom demonstration of the logistic growth model.

Part 15: The governing differential equation for the logistic growth model.

Part 16: Tips on graphing the logistic growth function.

# Exponential growth and decay (Part 16): Logistic growth model

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications, radioactive decay, and Newton’s Law of Cooling.

Today, I discuss the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. In yesterday’s post, I described an in-class demonstration that engages students while also making the following formula believable:

$A(t) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-rt}}$.

I’d like to discuss some observations about this somewhat complicated function that will make producing its graph easier. The first two observations are within reach of Precalculus students.

1. Let’s figure out the $y-$intercept:

$A(0) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-r \cdot 0}} = \displaystyle \frac{Ly_0}{y_0+ L-y_0} = y_0$.

In other words, the number $y_0$ represents the initial number of people who have the infection.

2. Let’s figure out the limiting value as $t$ gets large:

$\displaystyle \lim_{t \to \infty} A(t) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0) \cdot 0} = \displaystyle \frac{Ly_0}{y_0} = L$.

As expected, all $L$ people will get the infection eventually. (Of course, Precalculus students won’t be familiar with the $\displaystyle \lim$ notation, but they should understand that $e^{-rt}$ decays to zero as $t$ gets large.

3. Let’s now figure out the point of inflection. Ordinarily, points of inflection are found by setting the second derivative equal to zero. Though this can be done for the function $A(t)$ above, it would be a somewhat daunting exercise!

The good news is that the points of inflection can be found quite simply using the governing differential equation, which is

$A' = r A [ L - A] = r L A - r A^2$

Let’s take the derivative of both sides, remembering that $r$ and $L$ are constants:

$A'' = r L A' - 2 r A A'$

$A'' = A' (r L - 2 r A)$

So the second derivative is equal to zero when either $A' = 0$ or else $r L - 2 r A = 0$. The first case corresponds to the trivial cases $A(t) \equiv 0$ and $A(t) \equiv L$; these constants are called the equilibrium solutions. The second case is the more interesting one:

$r L - 2 r A = 0$

$r L = 2 r A$

$\displaystyle \frac{L}{2} = A$

This suggests that, as the infection spreads throughout a population, the curve changes concavity at the time that half of the population becomes infected. In other words, the infection spreads fastest throughout the population at the time when half of the population has been infected.

The time at which the point of inflection occurs can be found by setting $A(t) = \displaystyle \frac{L}{2}$ and solving for $t$:

$\displaystyle \frac{L}{2} = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-rt}}$.

$\displaystyle \frac{1}{2} = \displaystyle \frac{y_0}{y_0+ (L-y_0)e^{-rt}}$.

$y_0 + (L-y_0) e^{-rt} = 2y_0$

$(L-y_0) e^{-rt} = y_0$

$e^{-rt} = \displaystyle \frac{y_0}{L-y_0}$

$-rt = \displaystyle \ln \left( \frac{y_0}{L-y_0} \right)$

$t = \displaystyle - \frac{1}{r} \ln \left( \frac{y_0}{L-y_0} \right)$

This technique for finding the points of inflection directly from the differential equation is possible whenever the differential equation is autonomous, which loosely means that the independent variable does not appear on the right-hand side.

# Exponential growth and decay (Part 15): Logistic growth model

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications, radioactive decay, and Newton’s Law of Cooling.

Today, I introduce the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. For example:

or

In yesterday’s post, I described an in-class demonstration that engages students while also making the following formula believable:

$A(t) = \displaystyle \frac{L}{1 + (L-1)e^{-rt}}$.

Where does this formula come from? Suppose that a disease is spreading in a population of size $L$. It stands to reason that the rate at which the disease spreads is proportional to the number of possible contacts between those who have the disease and those who don’t. If $A(t)$ is the number of people who have the disease, then $L-A(t)$ is the number of people who don’t have the disease. Therefore, the product $A(t) [ L - A(t) ]$ is the number of possible contacts between those who have the disease and those who don’t. This leads to the governing differential equation

$A'(t) = c A(t) [ L - A(t) ]$,

where $c$ is the constant of proportionality. This is often rewritten by letting $c = \displaystyle \frac{r}{L}$, or $r = cL$:

$A'(t) = \displaystyle \frac{r}{L} A(t) [ L - A(t) ]$

$A'(t) = r A(t) \displaystyle \left[1 - \frac{A(t)}{L} \right]$

The good news is that this differential equation can be solved using separation of variables, just like the governing differential equations for continuous compound interest, paying off credit card debt, radioactive decay, and Newton’s Law of Cooling. The bad news is that it’s a lot harder to calculate the required integrals! After all, the right-hand side, after distributing, has a term containing $A^2$, which makes this differential equation non-linear.

Solving this differential equation is a bit tedious, and I don’t feel particularly obligated to re-invent the wheel since it can be found several places on the Internet. Suffice it to say that integration by partial fractions and some very tricky algebra is necessary to solve for $A(t)$ and obtain the solution above. Among several different sources (which likely use different letters than the ones I’m using here):

# Exponential growth and decay (Part 14): Logistic growth model

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications, radioactive decay, and Newton’s Law of Cooling.

Today, I introduce the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. Before I actually present the formula to my students, I usually perform a 8- to 10-minute demonstration to convince students that the formula actually works. This demonstration works well with between 15 and 45 students; I have personally not attempted this demo with a class larger than 45.

I wish I could take credit for the idea behind this demonstration. I’m afraid I can’t remember who told me the idea behind this demo about 15 years ago, but I’m thankful to him or her for this idea, as I’ve used it with great success over the years when teaching Precalculus and even when teaching Differential Equations.

Here’s the demonstration:

1. The class period before the demo, I ask my students to bring their calculators to class.

2. On the day of the demo, I prepare slips of paper with the numbers 1, 2, 3, etc. I hand these to my students as they take their seats before class starts (and, as needed, to students who arrive late).

3. I tell the class that we’re going to model how a rumor gets spread. On the chalkboard, I write down the numbers 0, 1, 2, …, up to $N$, the number of students in the class that day. Invariably, I get asked, “What’s the rumor?” In response, I’ll playfully point to someone in the front row and say, “The rumor is about him.”

4. I point out that, at time 0, only one person has heard the rumor…. me. I’m person number 0 (confirming the popular belief of my students). So I’ll cross out the 0 on the board and mark on a table that only one person has heard the rumor so far. (Here’s the spreadsheet that I’ve used to keep track of this information while simultaneously making a graph of the data: logisitic).

5. I begin to spread the rumor. To spread the rumor, I use my calculator to get a random number between $0$ and $N$. This can be done by just using the built-in random number generator found on many calculators and then multiplying by $N+1$. (After all, there are $L= N+1$ people in the room: $N$ students plus one instructor.) The part after the decimal point is not important; the number before the decimal point represents the next person to hear the rumor.

For example, in the figure below, I would tell that person #35 of my class of 37 students was the next to hear the rumor. (If my random number is 0, I’ll privately cheat and get until I get a random number other than 0. I only permit the possibility of cheating on the first step so that the data fits the predicted curve as accurately as possible.)

6. At this point, I’ll X out the number of the next person to hear the rumor (in this case, 35), and I then ask how many people have heard the rumor. Obviously, two people have heard the rumor. So I’ll note on the table that two people have heard the rumor after one step.

7. Now we repeat the process. I get a new random number, and I ask the first student to pull out his/her calculator to get a random number too. But there’s an important rule: if you get a number that’s already been called, that’s OK. This models what really happens when a rumor (or disease) spreads in a population — it’s perfectly possible to hear the rumor twice.

8. We repeat the process — X’ing out numbers that have been previously called and students calling out the next person to hear the rumor — until the entire class hears the rumor. At some point, it becomes easiest to ask students to only call out if they get a number that hasn’t been called yet. Invariably, there’s always one person at the end who hasn’t heard the rumor yet, and this student is often the subject of some good-natured ribbing. Eventually, a chart like the following is produced.

9. Students immediately see that this is a different type of function than pure exponential growth. It actually does start off looking like exponential growth, but at some point the curve levels off. This makes sense because there’s a limiting value of $L=N+1$ (in this case, 38), which can’t happen for a model like $A = A_0 e^{rt}$

10. The punchline is that the spreadsheet secretly computes the actual curve predicted by the logistic growth model. The numbers are actually located in column C, which is conveniently hidden beneath the graph. The function is

$A = \displaystyle \frac{L}{1 + (L-1)e^{-rt}}$,

where $r = \ln 2$. (Had each person the rumor to two different people at each step, then $r$ would have been equal to $r = \ln 3$.) Here’s the graph, superimposed upon the data collected from class. I can do this pretty quickly in class because the curve is actually already drawn in the figure above… but it’s drawn in gray, the same color as the background. By changing the color to black, the graph becomes clear:

I never expect the curve to exactly fit the data, but it should come pretty close. After this fairly dramatic revelation, my students are completely sold that the mathematics that I’m about to show them actually works.

# Exponential growth and decay (Part 13): Newton’s Law of Cooling

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss Newton’s Law of Cooling, which describes how quickly a hot object cools in a room at constant temperature. (This law is not to be confused with Newton’s Three Laws of Motion.)

While Newton’s Law of Cooling is easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that the rate at which a hot object cools is proportional to the difference between its current temperature and the surrounding temperature. After solving the appropriate differential equation, the temperature $T(t)$ of the object is found to be

$T = S + (T_0 - S) e^{-kt}$,

where $t$ is the time, $S$ is the constant surrounding temperature, and $k$ is a constant that depends on the object.

Of course, students in Algebra II or Precalculus (or high school physics) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

This is the third application of exponential functions considered in this series; the previous two were continuous compound interest and radioactive decay. Unlike these previous two applications, Newton’s Law of Cooling can actually be demonstrated in class to engage students. All that’s required is the appropriate classroom technology and a standard-issue temperature probe.This stands in sharp contrast to the previous applications of exponential functions considered in this series. While students can easily envision making money via compound interest, no one will actually give them the money during class. And certainly I don’t encourage performing a real demonstration of radioactive decay with, say uranium-235, during class time! (There are ways of simulating radioactive decay using M&Ms or other manipulatives, however.)

A simple Google search yields thousands of webpages describes multiple classroom activities for Newton’s Law of Cooling. Some activities merely require collecting data and performing a regression fit to an exponential curve; such an activity would be appropriate for middle-school students. Other activities are more explicit about using Newton’s Law of Cooling. Here’s a sampling:

These links are aimed at a students at a variety of levels. Indeed, it’s possible to use a graphing calculator to plot the numerical derivative $T'(t)$ as a function of time, use linear regression to solve for the constant $k$, and then produce the exponential curve using this value of $k$. Several years ago, I saw an effective demonstration of this idea at the Joint Mathematics Meetings in which the presenters covered these aspects of Newton’s Law of Cooling in less than 10 minutes. (Naturally, additional time is needed when students perform these activities for themselves.)

# Exponential growth and decay (Part 12): Newton’s Law of Cooling

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss Newton’s Law of Cooling, which describes how quickly a hot object cools in a room at constant temperature. (This law is not to be confused with Newton’s Three Laws of Motion.)

While Newton’s Law of Cooling is easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that the rate at which a hot object cools is proportional to the difference between its current temperature and the surrounding temperature. In other words, in a room at a temperature of $70^\circ$F, a very hot object at $270^\circ$F will cool twice as fast than when its temperature drops to $170^\circ$F.

The above paragraph can be rewritten as a differential equation. Let $T(t)$ be the temperature of the object at time $t$, and let $S$ be the (constant) surrounding temperature that surrounds the object. Since the rate at which the substance decays is $dT/dt$, we find that

$\displaystyle \frac{dT}{dt} = - k (T - S)$,

where $k$ is a constant that depends on the object. The negative sign on the right-hand side isn’t strictly necessary, but it’s a reminder that amount present decreases as time increases.

This differential equation can be solved in several ways, including separation of variables (below, I’ll be sloppy with the constant of integration for the sake of simplicity):

$\displaystyle \frac{dT}{T-S} = -k$

$\displaystyle \int \frac{dT}{T-S} = - \displaystyle \int k \, dt$

$ln |T-S| = -k t + C$

$|T-S| = e^{-kt+C}$

$|T-S| = e^{-kt} e^C$

$T-S = \pm e^C e^{-kt}$

$T-S = C e^{-kt}$

$T = S + C e^{-kt}$

To solve for the constant, we usually use the initial condition $T(0) = T_0$, a number that must be given in the problem:

$T(0) = S + C e^{-k \cdot 0}$

$T_0 =S + C \cdot 1$

$T_0 - S = C$

Plugging back in, we obtain the final answer

$T = S + (T_0 - S) e^{-kt}$

Of course, students in Algebra II or Precalculus (or high school physics) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

The careful reader will notice that this derivation essentially parallels the previous derivations for continuous compound interest and for radioactive decay. In other words, these phenomena from apparently different realms of life have similar solutions because the governing differential equations are very similar.

# Exponential growth and decay (Part 11): Half-life

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss radioactive decay and the half-life formula.

One way of writing the formula for how a radioactive substance (carbon-14, uranium-235, brain cells) decays is

$A(t) = A_0 e^{-kt}$

In yesterday’s post, I showed how this formula is a natural consequence of a certain differential equation. Of course, students in Algebra II or Precalculus (or high school chemistry) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

There is another correct way to write this formula in terms of half-life. Sadly, my experience is that many students are familiar with these two different formulas but are not aware of how the two formulas are connected. As we’ll see, while yesterday’s post using differential equations is inaccessible to Algebra II and Precalculus students, the derivation below is entirely elementary and can be understood by good students in these courses.

Let $h$ be the half-life of the substance. By definition, this means that

$A(h) = \displaystyle \frac{1}{2} A_0$

Substituting into the above formula, we find

$\displaystyle \frac{1}{2} A_0 = A_0 e^{-kh}$

Let’s now solve for the constant $k$ in terms of $h$:

$\displaystyle \frac{1}{2} = e^{-kh}$

$\displaystyle \ln \left( \frac{1}{2} \right) = - k h$

$\displaystyle -\frac{1}{h} \ln \left( \frac{1}{2} \right) = k$

Let’s now substitute this back into the original formula:

$A = A_0 e^{-kt}$

$A = A_0 e^{ -\left[ -\frac{1}{h} \ln \left( \frac{1}{2} \right) \right] t}$

$A = A_0 e^{\ln \left( \frac{1}{2} \right) \cdot \frac{t}{h}}$

$A = A_0 \left[e^{\ln \left( \frac{1}{2} \right)} \right]^{t/h}$

$A = A_0 \displaystyle \left( \frac{1}{2} \right)^{t/h}$

This is the usual half-life formula, where the previous base of $e$ has been replaced by a new base of $\displaystyle \frac{1}{2}$ . For most applications, a base of $e$ is preferred. However, for historical reasons, the base of $\displaystyle \frac{1}{2}$ is preferred for problems involving radioactive decay. For example,

$A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2h/h}$

$A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2}$

$A(2h) = \displaystyle \frac{1}{4} A_0$

In other words, after two half-life periods, only one-fourth (half of a half) of the substance remains.

# Exponential growth and decay (Part 10): Half-life

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss radioactive decay and the half-life formula.

While these formulas are easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that if amount $A$ of a radioactive substance (carbon-14, uranium-235, brain cells) is present, the rate at which the substance decays is proportional to the amount of the substance currently present. This can be rewritten as a differential equation, since the rate at which the substance decays is $dA/dt$. So we find that

$\displaystyle \frac{dA}{dt} = - k A$

The negative sign on the right-hand side isn’t strictly necessary, but it’s a reminder that amount present decreases as time increases.

This differential equation can be solved in several ways, including separation of variables (below, I’ll be sloppy with the constant of integration for the sake of simplicity):

$\displaystyle \frac{dA}{A} = -k$

$\displaystyle \int \frac{dA}{A} = - \displaystyle \int k \, dt$

$ln |A| = -k t + C$

$|A| = e^{-kt+C}$

$|A| = e^{-kt} e^C$

$A = \pm e^C e^{-kt}$

$A = C e^{-kt}$

To solve for the constant, we usually use the initial condition $A(0) = A_0$, a number that must be given in the problem:

$A(0) = C e^{-k \cdot 0}$

$A_0 = C \cdot 1$

$A_0 = C$

Plugging back in, we obtain the final answer

$A(t) = A_0 e^{-kt}$

Of course, students in Algebra II or Precalculus (or high school chemistry) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

# Exponential growth and decay (Part 9): Amortization tables

This post is inspired by one of the questions that I pose to our future high school math teachers during our Friday question-and-answer sessions. In these sessions, I play the role of a middle- or high-school student who’s asking a tricky question to his math teacher. Here’s the question:

A student asks, “My father bought a house for $200,000 at 12% interest. He told me that by the time he fi nishes paying for the house, it will have cost him more than$500,000. How is that possible? 12% of $200,000 is only$24,000.”

Without fail, these future teachers don’t have a good response to this question. Indeed, my experience is that most young adults (including college students) have never used an amortization table, which is the subject of today’s post.

In the past few posts, we have considered the solution of the following recurrence relation, which is often used to model the payment of a mortgage or of credit-card debt:

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

With this difference equation, the rate at which the principal is reduced can be simply computed using Microsoft Excel. This tool is called an amortization schedule or an amortization table; see E-How for the instructions of how to build one. Here’s a sample Excel spreadsheet that I’ll be illustrating below: Amortization schedule. My personal experience is that many math majors have never seen such a spreadsheet, even though they are familiar with compound interest problems and certainly have the mathematical tools to understand this spreadsheet.

Here’s a screen capture from the spreadsheet:

The terms of the loan are typed into Cells B1 (length of loan, in years), B2 (annual percentage rate), and B3 (initial principal). Cell B4 is computed from this information using the Microsoft Excel command $\hbox{PMT}$:

$=\hbox{PMT}(\hbox{B2}/12,\hbox{B1}*12,-\hbox{B3})$

This is the amount that must be paid every month in order to pay off the loan in the prescribed number of years. Of course, there is a formula for this:

$M = \displaystyle \frac{Pr}{12 \displaystyle \left[1 - \left( 1 + \frac{r}{12} \right)^{-12t} \right]}$

I won’t go into the derivation of this formula here, as it’s a bit complicated. Notice that this formula does not include escrow, points, closing costs, etc. This is strictly the amount of money that’s needed to pay down the principal.

The table, beginning in Row 8 of the above picture, shows how quickly the principal will be paid off. In row 8, the interest that’s paid for that month is  computed by

$=\hbox{B8} * \ \hbox{B}\\hbox{2}/12$

Therefore, the amount of the monthly payment that actually goes toward paying down the principal will be

$= \\hbox{B}\\hbox{4} - \hbox{C8}$

Column E provides an opportunity to pay something extra each month; more on this later. So, after taking into account the payments in columns D and E, the amount remaining on the loan is recorded in Cell F8:

$= \hbox{B8} - \hbox{D8} - \hbox{E8}$

This amount is then copied into Cell B9, and then the pattern can be filled down.

The yellow graph shows how quickly the balance of the loan is paid off over the length of the loan. A picture is worth a thousand words: in the initial years of the loan, most of the payments are gobbled up by the interest, and so the principal is paid off slowly. Only in the latter years of the loan is the principal paid off quickly.

So, it stands to reason that any extra payments in the initial months and years of the loan can do wonders for paying off the loan quickly. For example, here’s a screenshot of what happens if an extra $200/month is paid only in the first 12 months of the loan: A definite bend in the graph is evident in the initial 12 months until the normal payment is resumed in month 13. As a result of those extra payments, the curve now intersects the horizontal axis around 340. In other words, 20 fewer months are required to pay off the loan. Stated another way, the extra payments in the first year cost an extra $\200 \times 12 = \2400$. However, in the long run, those payments saved about $\536.82 \times 20 \approx \10,700$! # Exponential growth and decay (Part 8): 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. The solution of this difference equation is

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

A great advantage of using a difference equation to solve this problem is that the solution can be easily checked with a simple spreadsheet. (Indeed, pedagogically, I would recommend showing a spreadsheet like this before doing any of the calculations of the previous few posts, so that students can begin to wrap their heads around the notion of a difference equation before the solution is presented.)

To start the spreadsheet, I wrote “Step” in Cell A1 and “Amount” in Cell B1. Then I entered the initial conditions: $0$ in Cell A2 and $2000$ in Cell B2. (In the screenshot below, I changed the format of column B to show dollars and cents.) Next, I entered $=\hbox{A2}+1$ in Cell A3 and

$=\hbox{B2}*(1+0.25/12)-50$

in Cell B3. Finally, I copied the pattern in Cells A3 and B3 downward. Here’s the result:

After the formula the algebraic solution of the difference equation has been found, this can be added to the spreadsheet in a different column. For example, I added the header “Predicted Amount” in Cell D1. In Cell D2, I typed the formula

$=2000*\hbox{POWER}(1+0.25/12,\hbox{A2})-50*(1-\hbox{POWER}(1+0.25/12,\hbox{A2}))/(1-(1+0.25/12))$

Finally, I copied this pattern down the Column D. Here’s the result:

Invariably, when I perform a demonstration like this in class, I elicit a reaction of “Whoa…. it actually works!” Even for a class of math majors. Naturally, I tease them about this… they didn’t believe me when I used algebra, but now it has to be true because the computer says so.

Here’s the spreadsheet that I used to make the above pictures: CreditCardDebt.