The Natural Exponential Function and Continuous Compounding

Increasing Compoundings and Negligibly Changing Account Values

In the previous topic/lesson, we gave a formula/function that allows one to compute the value of an account after \(t\) years based on a given yearly rate, \(r\) (as a decimal), a number of compoundings per year, \(k\), and an initial account value called the principal, \(P\). See below.

$$Value(t)=P\left(1+\frac{r}{k}\right)^{k\cdot t}$$

We also briefly discussed how the number of compoundings per year are essentially how many times the account builds on itself, noting that the more compoundings the account undergoes, the greater the account value is in the same amount of time. At the same time though, increasing the number of compoundings from, say, 2 to 365 per year results in only a moderate increase in value; similarly too if we increased our compoundings from daily to every minute (525600 minutes in a year). Given an investment of $100,000 in an account yielding an annual rate of 4% per year, the amount the account is worth after one year given these three different compounding amounts is given in the dropdown box below.

If one continues to increase the number of compoundings per year, such as was done in the table above, one will notice that as the number of compoundings become sufficiently large, there is not as much of an increase in the account’s value (see hourly versus every minute compounding above). One has to wonder whether there is an upper bound for how high an account’s value can go if we keep increasing the number of compoundings forever. The answer to this curiosity is “yep! There’s an upper bound.”

Continuous Compounding

The formula above comes from the compounding formula discussed above and in the previous lesson, but by letting the number of compoundings increase without bound. This is expressed in the limit below:

$$\lim_{k\rightarrow \infty} P\left(1+\frac{r}{k}\right)^{k\cdot t}=Pe^{rt}$$

The limit above essentially is asking for the value of the compound interest function when the number of compoundings gets “infinitely large.”

In the quick example below, we demonstrate how compounding continuously “matches up” with what happened when we increased the number of compoundings for our $100,000 example above.

The number \(e\) and the Natural Exponential Function

The continuous compounding formula is a slightly more complicated (i.e. transformed) version of what is called the natural exponential function,

$$f(x)=e^x.$$

where \(e=2.71828\) is an irrational number (i.e. can’t be written as a fraction of integers).

The natural exponential function has a panoply of applications in math, science, and finance. One of its most important features is that (loosely speaking) the rate at which the function increases (as x values increase) is the same as the speed the rate of change changes.

That’s confusing. So to put it another way, suppose you are driving your car down the road, and after 30 miles of driving you see that you are going 30 miles an hour, and are also accelerating at 30mph per hour (i.e. your speed is increasing by 30 miles per hour every hour). Your position, speed, and acceleration all match. With the natural exponential function, no matter where you are in terms of position, your speed and acceleration match your position. So, if you are at the 31-mile mark, you’re going 31 miles per hour, and increasing your speed at a rate of 31 miles per hour every hour. If you are at the 32-mile mark, you’re going 32 miles per hour, and increasing your speed at a rate of 32 miles per hour every hour… and so on and so forth.

The function(s) \(f(x)=ce^x\) where \(c\in \mathbb{R}\) are the only functions in all of mathematics with the aforementioned rate-of-change property (i.e. a constant times the natural exponential function). This makes the natural exponential function very special indeed.