We continue the discussion that was started in the previous section: Simple versus 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.

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.
The following are equivalent:

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 ‘$\,\text{e}\,$’ 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:

Letting $\,n\,$ go to infinity in the Compound Interest Formula

As $\,n\,$ gets large, $\displaystyle\,\frac{r}{n}\,$ gets small.
That is, as $\displaystyle\,n\rightarrow\infty\,$, we have $\displaystyle\,\frac{r}{n}\rightarrow 0\,$.

As $\displaystyle\,\frac{r}{n}\,$ approaches zero, $\displaystyle\,1 + \frac{r}{n}\,$ approaches $\,1\,$.
That is, as $\,\displaystyle\frac rn\rightarrow 0\,$, we have $\,\displaystyle 1 + \frac rn \rightarrow 1\,$.

As $\,n\,$ gets big, $\,nt\,$ gets big.
That is, as $\,n\rightarrow\infty\,$, we have $\,nt\rightarrow\infty\,$.

So, as we add interest in more and more frequently (let $\,n\rightarrow\infty\,$), here's what happens in the Compound Interest Formula: $$A=P{ \bigl( \underbrace{1+\frac{r}{n}}_{\text{base approaches $1$}} \bigr) }^{ \overbrace{nt}^{\text{exponent approaches $\infty$}} }$$

We're looking at a ‘$\,1^\infty\,$’ form—which is just a shorthand for saying that the base approaches $\,1\,$, and the exponent approaches $\,\infty\,$.
But, here's the important question:

As the base and exponent travel together along their journeys to $\,1\,$ and $\,\infty\,$, 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:

Depending on precisely how the base approaches $\,1\,$, and how the exponent approaches infinity, the form $\,1^\infty\,$ can approach different numbers!
These three different $\,1^\infty\,$ forms approach (respectively) $\,\text{e}\,$, $\,\infty\,$, and $\,1\,$:

‘$\,1^\infty\,$’ is an example of what is called (in calculus) an indeterminate form.
An indeterminate form is when you can't figure out what is happening without further analysis.
That is, each occurence of ‘$\,1^\infty\,$’ must be investigated separately.

For us, the investigation will involve noting the similarity of our situation to the definition of the irrational number $\,\text{e}\,$.
Compare these: $$\lim_{n\rightarrow\infty} P(1 + \frac rn)^{nt} = \ ? \quad \quad \text{ and } \quad \quad \lim_{n\rightarrow\infty} (1+\frac 1n)^n := \text{e}$$ There are some pesky differences:

We can deal with these differences—keep reading!

Derivation of the Continuous Compounding Formula

So, you want $\displaystyle\,\frac rn\,$ to ‘look like’ $\displaystyle\,\frac 1n\,$?     Make it happen!
Define a new variable $\,m\,$ by this relationship: $$\text{Let}\ \frac 1m := \frac rn\,.\quad \quad \quad \quad (*)$$ Recall that ‘ $:=$ ’ means ‘equal, by definition’.
(By the way, this is the tricky part of the derivation.)

Cross-multiplication in equation (*) gives $\,n = mr\,$.

Now, rewrite the Compound Interest Formula in terms of $\,m\,$ instead of $\,n\,$, and do a bit of algebra: $$ P(1 + \frac rn)^{nt} \quad = \quad P(1 + \frac 1m)^{mrt} \quad = \quad P\left[(1 + \frac 1m)^m\right]^{rt} $$

Taking reciprocals in equation (*), we have $\displaystyle\,m = \frac nr\,$.
So, as $\,n\,$ goes to infinity, $\,m\,$ also goes to infinity.
And (here's the punch line!) as $\,m\,$ goes to infinity, $\displaystyle P\left[(1 + \frac 1m)^m\right]^{rt} \rightarrow P{\text{e}}^{rt}\,$ !!

CONTINUOUS COMPOUNDING letting $\,n\rightarrow\infty\,$ in the Compound Interest Formula
Letting $\,n\rightarrow\infty\,$ in the Compound Interest Formula, $\,\displaystyle A = P(1 + \frac rn)^{nt}\,$, yields the Continuous Compounding Formula: $$A = P{\text{e}}^{rt}$$ Roughly, continuous compounding describes interest being added in the instant it is earned.


Suppose that \$1000 is invested at $3\%$ annual interest.
What is the accumulation after ten years if compounded monthly, daily, and continuously?

Compounded monthly:     $\displaystyle A = P(1 + \frac rn)^{nt} = 1000(1 + \frac {0.03}{12})^{12\cdot 10} = \$1,349.35$

Compounded daily:     $ \displaystyle A = 1000(1 + \frac {0.03}{365})^{365\cdot 10} = \$1,349.84$

Compounded continuously:     $\displaystyle A = P{\text{e}}^{rt} = 1000{\text{e}}^{0.03\cdot10} = \$1,349.86$

Not much difference! You won't get rich if your bank decides to compound continuously!

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
logarithmic functions: review and additional properties
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
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(MAX is 16; there are 16 different problem types.)