The “Mathematical Constant” and Continuously Compouding Interest

One of the (many) aspects of the “Mathematical Constant” $e$ is that:

$\lim_{x\to\infty} (1+\frac{1}{x})^x = e$

This property makes $e$ very useful for working on compounding interest problems. How so?

Let’s start with the basic time value of money formula giving the relationship between the PV (present value) and FV (future value) given R (the rate of interest):

$FV = PV * (1 + R)^n$

In the formula above $n$ represents the number of compounding periods. Given normal conventions, this is typical represented as $m * Y$ where m is months (or other periods – it represents how many chunks we divide the year into for receiving interest payments) and Y is years. Also, the $R$ (or rate) above is normally quoted in years, so we should divide this by the months: $R/m$.

So now we have:

$FV = PV * (1 + \frac{R}{m})^{mY}$

We can make use of the $e$ formula above if we want continuously compounding interest – that is where $m$ is as high as possible, converging on $\infty$.

Let’s start by making $x=\frac{m}{R}$ and we would get:

$FV = PV * (1 + \frac{1}{x})^{x * YR}$

We can now drop this into our formula for $e$ given that we want m (or x, which is effectively our proxy for m) to approach $\infty$:

$FV = PV * [\lim_{x\to\infty} (1+\frac{1}{x})^{x}]^{RY}$

We can write this simply as:

$FV = PV * e^{RY}$

So if we invest $100 at 6.5% continuously compounding interest for a year, we will end up with:

$FV = 100 * e^{.065*1}$ $FV = \$106.72$