Derivation of Continuous Compound Interest Formula without Calculus

Jacob Bernoulli, 1654-1705
Jacob Bernoulli, 1654-1705

My students, like most people, like money and find the topic of compound interest interesting. After completing a unit on simple, compound and continuous compound interest, one of my students told me that math is useful and interesting after all.

This post will discuss the derivation of the formula for the future value of an investment when interest is compounded continuously, FV = Pert. No prior understanding of the limit concept in calculus is required. I will be using the limit concept, but I will give an informal intuitive explanation of the limit concept as it comes up in the discussion. A recent post discussed an approach for deriving an equation that models exponential growth/decay. Problem (2) in that post showed the derivation of the compound interest formula FV = P(1 + r/k)kt where FV = the future value of the investment account, P = principle or one time lump-sum investment, r = annual percent rate of return expressed as a decimal, k = the number of times per year interest is compounded, and time t = the number of years the principal is invested.

Before I can get to the derivation of the equation FV = Pert, I need to explain what continuous compound interest means. Let’s consider an investment where P = $10,000, average annual rate of return = 7% = 0.07, and the investment collects interest over a period of 20 years. I adopted the standard banking convention rule that 1 year = 360 days. (Whether we use 365 or 360 days in a year makes no significant difference. Apparently banks like 30-day months.) The text box below shows how increasing the number of times per year interest is compounded affects the future value of an investment.


Students immediately notice that there is a point where it makes no difference how often interest is compounded, and they completely understand the difference between simple interest and compound interest. I tell them that the future value of the $10,000 investment, $40,552.00 in this example, represents the upper limit of one’s greed. When interest is compounded more times per year (k approaches infinity), and interest is compounded over smaller and smaller time intervals; say every second, every microsecond, or continuously. No matter what the principal is or the annual interest rate, there is always an upper limit of the future value of an investment, and the upper limit is reached when interest is compounded continuously.

In 1683 in the course of his study of continuous compound interest, Jacob Bernoulli (1654-1705) wanted to find the number that was the limiting value of the expression (1+1/n)^n as n approaches infinity. This is the first time that a number is defined as the limiting value of an expression. Bernoulli determined that this special number is bounded and lies between 2 and 3. In 1748 Leonard Euler (pronounced Oil-er) (1707-1783) published a document in which he named this special number e. He showed that e is the limiting value of the expression (1 + 1/n)n as n approaches infinity, and is approximately equal to 2.718281828459045235. He also gave another definition of e as the limiting value of the infinite sum 1 + 1/1! + 1/2! + 1/3! + . . . . Euler is generally given credit as the first to prove e is an irrational number.

To help you better understand the definition of the irrational number e, I will start by comparing the graphs of functions of the form y = (1 + 1/k)x where k is a fixed constant and the graph of the function y = (1 + 1/x)x. Refer to graphs (A) and (B) and the companion text box below. A quantity that approaches infinity means the quantity gets bigger and bigger without any upper boundary. A quantity that approaches a fixed constant means the quantity gets infinitely close to the fixed constant.





The purpose of the above graphs and the comments in the text box is to demonstrate that a subtle difference in the expressions (1 + 1/k)x and (1 + 1/x)x results in far different limiting values as x approaches ∞. The key result needed in the derivation of the continuous compound interest formula is the fact that e = limiting value of (1 + 1/x)x as x approaches ∞ when x is any positive real number. Considering that the expression (1 + 1/n)n is a rational number for every positive integer n, it is astonishing that the expression (1 + 1/n)n approaches an irrational number as n approaches ∞. I can now show you the derivation of the continuous compound interest formula FV = Pert.



• When I did the calculations for compounding every minute and compounding every second with my graphing calculator, I got results that were slightly different than the expected results. When I used double floating point precision real numbers in a computer program, program output agreed with the expected results. We need to constantly remind ourselves that calculator or computer calculations of expressions that involve very large numbers, or require a large number of iterations to arrive at a solution, results may be slightly different than the expected or theoretical value.

• Using problems similar to the examples in this post, I show my students how compound interest works and what continuous compounding of interest means. I have them enter the expressions into their graphing calculator as the lesson progresses. This gives them practice using their calculator and they gain a better understanding and appreciation of what compound interest is all about. They are astonished when I show them $10,000*e.07*20 = $40,552.00.

• For a class of curious or advanced students, it’s not wasted class time to show the derivation of the continuous compound interest formula. Less advanced students are usually content with learning how to use the formula. My handout, Basic Financial Formulas, provides an overview of useful financial formulas that you can use in your classroom.

• The derivation of the continuous compound interest formula is a great opportunity to expose advanced high school algebra, college algebra and pre-calculus students to the limit concept in calculus.

• As mentioned earlier, very term of the sequence an = (1 + 1/n)n is a rational number, but the sequence itself converges to the irrational number e. Most calculus students find this very counterintuitive. What a great opportunity to launch a discussion of any number of related math concepts!

• The constants 0, 1, π, e, and i where i2 = -1 are the five most important constants in mathematics because they are widely used in equations that describe relationships in all branches of mathematics and science. The equation eπi + 1 = 0, which is due to Leonhard Euler, is one of the most interesting and intriguing equations in mathematics. Euler used the symbol e for the irrational constant, and in his honor, e is named Euler’s number.

• Both Bernoulli and Euler were prolific mathematical giants. Much of what is routinely used in mathematics and science can be traced back to the work of these two great men. L’Hospital’s Rule in calculus is due to Bernoulli, not L’Hospital. L’Hospital published the rule, but Bernoulli discovered the rule and gave it, for a fee, to L’Hospital.

Because of limits on food, living space, disease, existing technology, war, and other factors, most populations have limited growth as opposed to unlimited exponential growth which is modeled by the classic exponential growth equation P = P0bt/k. A limited growth population starts growing almost exponentially, but it reaches a critical point in time where its growth rate slows, and the population starts to exponentially and asymptotically approach an upper limit. There are several models that are used to describe limited growth of a population. In my next post, I will discuss the logistic function which was used by the Belgium mathematician Pierre Francois Verhulst (1804-1849) to study limited population growth. The logistic function also has applications in artificial neural networks, biology, chemistry, demography, ecology, economics, biomathematics, geoscience, mathematical psychology, sociology, political science, probability, and statistics.

Exponential Growth and Decay from a Data-Pairs Approach

header_expgrwthdcyMy last two posts discussed the mathematics of linear growth and decay. If you have not read those posts, you might find it helpful to read them before continuing. This post focuses on finding an exponential equation that expresses a relationship between two variables by first constructing a table of data-pairs to better understand the relationship and see the pattern in the relationship.

Most exponential growth/decay relationships involve a time variable t and the amount A of some quantity at time t. Amount could be the current value of an investment account, population of a city, remaining kilograms of radioactive material, assessed value of a truck, etc. The text box and observations below explain how and why the basic fundamental exponential growth/decay formula A = A0*bt/k works, and the role that the parameters A0, b, and k play in the equation. Periodic growth factor is another way to think of the base multiplier b.


Some observations about A = A0*bt/k where b > 0:
• The point (0, A0) is the intercept on the vertical axis of the graph.
• Base multiplier b is a periodic growth or decay factor.
• If 0 < b < 1, the equation models exponential decay.
• If b > 1, the equation models exponential growth.
• Exponential growth/decay is about repeated multiplication by growth/decay factor b.
• A0 and any other point on the graph determines a unique exponential equation.
• If A0 is positive, the graph is above and asymptotic to the horizontal axis.
• If A0 is negative, the graph is below and asymptotic to the horizontal axis.

Most discussions about finding the equation of an exponential relationship don’t start by looking at data-pairs in a table. After only a couple of demonstrations of how to apply the data-pairs approach, students quickly develop the ability to find the three key parameters of an exponential growth/decay relationship. Exponential equations of the form A = A0*bt/k where base b is a rational number are much easier to comprehend than equations of the form A = A0*ekt where e is the irrational math constant = 2.718281828459045 . . .  . I will use four familiar math problems that involve an exponential relationship to illustrate the table data-pairs approach. In the comments section of this post, you will find an example that further clarifies my reason for expressing most exponential growth/decay equations as A = A0*bt/k where b is a rational number, and the reason that the solutions of population and radioactive growth/decay problems tend to be expressed in terms of base e only.

Problem 1: Consider a population of bacteria that is growing exponentially 50% every 4 hours and the current population is 60 bacteria. Let t = number of hours in the future and N = the number of bacteria after t hours.
(a) Find an equation that expresses N as a function of t.
(b) Find the population after 10 hours and 45 minutes ago.
(c) Express N as a function of t if the population is increasing 5% every 15 minutes.

The solution is given in the text box below. Problem solvers should carefully read the problem, create a table of data-pairs, determine the equation parameters, and then write the equation that models the problem situation. A companion exponential growth graph with a series of slope/rate triangles is provided to show the role that the equation parameters play in the relationship. Of course, the problem solver should always check the solution by using a computer graphing program to graph the equation.



Problem 2: Suppose a person invests $10,000 in a CD that will earn interest at 6%/year and interest is compounded monthly. Let t = the number of years in the future and V = the value of the investment after t years.
(a) Express V as a function of t.
(b) Find the value of the investment after 10 years and 20 years.
(c) Express V as a function of t if interest is compounded 360 times per year.


Problem 3: The half-life of a radioactive substance equals the time it takes (20 days, 149 years, 5,700 years, etc.) for the substance to lose half its mass. Consider a radioactive substance with a half-life of 60 days that currently has a 100 kg mass. Let time t = the number of days in the future and A = the mass of the remaining substance in kg at time t. Refer to table and companion graph below.

(a) Find a formula that expresses A as a function of time t in days.
(b) Find the mass of the substance after 135 days.
(c) Find a formula for A(t) if the half-life = 6 hours instead of 60 days.


Problem 4: The two exponential growth/decay graphs along with key points on the graphs are shown below.
(a) For graph A: Write an equation that expresses y as a function of x.
(b) For graph B: Write an equation that expresses y as a function of x.



Here are four exercises that you can give to your students. The graphs are a mixture of linear and exponential growth/decay graphs. Using the points on the graph, find the equation of the graph. If you wish, remind them that they should first create a table of data-pairs. Let them do the exercises with a partner and then check their answers by using a computer to graph the equations. We want to create a save environment in which kids feel free to experiment and check their answers for understanding. It’s OK to make a mistake, just fix it. If the first attempt to fix a mistake fails, so what? Try again. This is how real people learn to do anything that is worthwhile. The solutions are given at the end of this post.


• Consider the two mathematically equivalent equations below that model the population growth of a small town where t equals the number of years after 2010.

P = 5,200(1.08)t/4  and  P = 5,200e0.019240260t

The first equation immediately tells us the population of the town was 5,200 in 2010, and the population is increasing 8% every 4 years. The second equation tells us the population of the town was 5,200 in 2010, but by just inspecting the second equation, only God can figure out that the population is increasing 8% every four years. (Increasing 8% every 4 years is slightly less than increasing 2% every year.)

• It’s a snap to find the derivative of functions of the form y = Aekt. To find the derivative of functions of the form y = Abx/k where base b is a rational number requires a little more work. I suspect this is the reason that the solutions of population and radioactive grow/decay problems tend to be expressed in terms of base e only. From my point of view, this is not a sufficient reason to do so because converting an exponential function from one base to another base is a simple procedure. My free handout Logarithmic Base Conversion shows how to do this.

• All modern physicists know that the equations they discovered can only give us an approximation of how nature’s laws work. The brilliant physicist Richard Feynman, over and over again, stated this fundamental fact in his lectures and talks. In reference to problem (3) above, if we conducted an experiment with a radioactive material by measuring the remaining mass of the material at various points in time, we would find a discrepancy between the experimental results and the predicted results. No matter how accurately we measure mass and time, the errors can’t be taken out of the experimental observations. We can only say that the remaining mass of radioactive material at time t lies in an area of uncertainly which is the area under a probability distribution curve. This is why least-squares regression equations are used to describe the relationship between two variables.

• The formula for calculating the future value of an account after t years when interest is compounded continuously is FV = Pert where P = the principal and r = the annual interest rate expressed as a decimal. It’s impossible to express this relationship with a base that is a rational number. In a future post, I will give a derivation of this formula in a manner that does not require an understanding of concepts in calculus.

• I have used the handout, Introduction to Exponential Growth and Decay, with college algebra students, pre-calculus students, and as a review for more advanced students. To download the free student and teacher versions of the handout, go to There are other free handouts on properties of exponents, properties of logarithms, solving exponential/logarithmic equations, and logarithmic base conversion.

• All graphs in this post were created with my software, Basic Trig Functions. I designed the software to help teachers quickly make custom content for their classrooms. This software allows you to easily copy any graphic and then import it directly into a document (e.g. lesson plan, class handout, test) or further manipulate it in various graphic processing programs.

My next post will show how to solve Newton’s Law of Cooling problems without understanding differential calculus.

Solutions to exercises:
Graph A: y = -2x + 40
Graph B: y = 40*0.5x/2.5
Graph C: y = 20*1.5x/5
Graph D: y = x + 10