3.3: Continuous Compounding
We continue the discussion that was started in Section 3.1: Simple and Compound Interest.
In a nutshell:
- You invest some money (the principal ) in a bank which pays interest.
- After a specified time (the compounding period ), you add in the interest that has been earned on the principal.
- After another compounding period, you add in the interest earned on the principal plus accumulated interest.
Repeat.
Compound interest , by definition, is interest calculated on the principal amount together with accumulated interest . Interest can be added in at different fixed intervals: annually, monthly, weekly, daily, and so on. Adding in interest more and more often produces a slightly better accumulation. (Most banks add in interest daily.)
What happens as you increase the number of times that interest is added in each year?
Your earnings will get bigger. How much bigger? Will your earnings increase without bound?
Equivalently, what happens as you add in interest over shorter and shorter periods of time?
In other words, what happens as you let the compounding period approach zero?
You'll end up with more money. How much more? Will you get rich?
The amount of money produced under these situations does NOT increase without bound! (Wishful thinking, but doesn't happen.)
Instead, the irrational number ‘e’ emerges to describe the limiting amount earned.
The resulting formula is called the Continuous Compounding Formula and is the subject of this section.
A quick review of the Compound Interest Formula
- Put \(P\) dollars (the principal ) in a bank.
- Assume the bank offers an annual interest rate \(r\). For example, 3% per year corresponds to \(r\)=0.03.
- Add in the earned interest \(n\) equally-spaced times each year. In other words, assume there are \(n\) compounding periods per year. For example, adding in interest monthly corresponds to \(n\)=12. Adding in interest daily corresponds to \(n\)=365
- Continue this process for \(t\) years.
- The accumulation (principal plus interest) is given by the variable \(A\) in the Compound Interest Formula: \(A=P\left(1+\dfrac{r}{n}\right)^{nt}\)
Letting \(n\) go to infinity in the Compound Interest Formula
As \(n\) gets large, \(\dfrac{r}{n}\) gets small. That is, as \(n \rightarrow \infty\), we have \(\dfrac{r}{n} \rightarrow 0\).
As \(\dfrac{r}{n}\) approaches zero, \(1+\dfrac{r}{n}\) approaches 1. That is, as \(\dfrac{r}{n} \rightarrow 0\), we have \(1+\dfrac{r}{n} \rightarrow 1\)
As \(n\) gets big, \(nt\) gets big. That is, as \(n \rightarrow \infty\), we have \(n t \rightarrow \infty\)
So, as we add interest in more and more frequently (let \(n \rightarrow \infty\)), here's what happens in the Compound Interest Formula:
\[A=P\left(1+\dfrac{r}{n}\right)^{n t} \nonumber \]
We're looking at a '1’ form—which is just a shorthand for saying that the base approaches 1, and the exponent approaches .
But, here's the important question:
As the base and exponent travel together along their journeys to 1 and , respectively, what number (if any) does the entire expression approach?
A discussion of the ‘\(1^{\infty}\)’ form
There are lots of things that make the ‘\(1^{\infty}\)' form both interesting and difficult to analyze. All of the following are true:
- 1 to any finite power equals 1. For example, \(1^{1,000,000,000}=1\)
- When a number a bit less than 1 is raised to higher and higher powers, it approaches zero. For example, \(\left(\dfrac{99}{100}\right)^{n}\) approaches 0 as \(n\) approaches infinity. (This is an exponential function with base between zero and one.)
- When a number a bit more than 1 is raised to higher and higher powers, it approaches infinity. For example, \(\left(\dfrac{101}{100}\right)^{n}\) approaches as \(n\) approaches infinity. (This is an exponential function with base greater than one.)
Depending on precisely how the base approaches 1, and how the exponent approaches infinity, the form \(1^{\infty}\) can approach different numbers!
‘\(1^{\infty}\)’ is an example of what is called (in calculus) an indeterminate form . An indeterminate form occurs when you can't figure out what is happening without further analysis . That is, each occurrence of must be investigated separately. For us, the investigation will involve the following definition of the irrational number \(e\)
\[\left(1+\dfrac{1}{n}\right)^{n} \rightarrow \rightarrow \rightarrow \rightarrow \rightarrow \rightarrow e \approx 2.718 \quad \text { as } n \rightarrow \infty \nonumber \]
Continuous Compounding
Letting \(n \rightarrow \infty\) in the Compound Interest Formula, \(A=P\left(1+\dfrac{r}{n}\right)^{n t}\) yields the Continuous
Compounding Formula:
\[A=P e^{r t} \nonumber \]
Roughly, continuous compounding describes interest being added in the instant it is earned.
Example \(\PageIndex{1}\)
Suppose that $1000 is invested at 3% annual interest. What is the accumulation after ten years if compounded monthly, daily, and continuously?
Solution
Compounded monthly:
\[A=P\left(1+\dfrac{r}{n}\right)^{n t}=1000\left(1+\dfrac{0.03}{12}\right)^{12 \cdot 10}=\$ 1,349.35 \nonumber\]
Compounded daily:
\[A=1000\left(1+\dfrac{0.03}{365}\right)^{365 \cdot 10}=\$ 1,349.84 \nonumber\]
Compounded continuously:
\[A=P e^{r t}=1000 e^{0.03 \cdot 10}=\$ 1349.86 \nonumber\]
Not much difference! You won't get rich if your bank decides to compound continuously!
Reference
- References (7)
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0