4.5 Properties of the Constant Number e

In Chapter 3, we introduced the constant number $e$, which is 2.71828182845904523 . . . This number $e$ is an irrational number, and it is the base of natural logarithms. How is $e$ derived? In order to understand $e$, we need to go back to the formula to find the future amount of money if the annual interest rate is $r$ and the number of compounding periods is $n$.

$A_t=P{(1+\frac{r}{n})}^{nt}$

Suppose that the interest rate is 100%, and you are depositing one dollar today for one year. This is how the future amount of savings changes with the increase in the compounding period.

$\begin{array}{rl}1\cdot {(1+\frac{1}{1})}^1& =2\\ 1\cdot {(1+\frac{1}{10})}^{10}& =\mathrm{2.593742...}\\ 1\cdot {(1+\frac{1}{100})}^{100}& =\mathrm{2.704814...}\\ 1\cdot {(1+\frac{1}{\mathrm{1,000}})}^{\mathrm{1,000}}& =\mathrm{2.716924...}\\ 1\cdot {(1+\frac{1}{\mathrm{10,000}})}^{\mathrm{10,000}}& =\mathrm{2.718146...}\\ & ...\\ & ...\\ & ...\\ 1\cdot {(1+\frac{1}{\mathrm{100,000,000}})}^{\mathrm{100,000,000}}& =\mathrm{2.718282...}\end{array}$

Based on this pattern, we find that the number $e$ is

$e=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n$

How can we apply the definition of $e$ to the following equation?

$A_t=P{(1+\frac{r}{n})}^{nt}$

We will find the limit of the equation as $n$ approaches infinity:

$\begin{array}{rl}{A}_t& =\underset{n\to \mathrm{\infty }}{lim}P{(1+\frac{r}{n})}^{nt}\\ & =P\cdot \underset{n\to \mathrm{\infty }}{lim}{(1+\frac{r}{n})}^{nt}\\ & =P\cdot \underset{n\to \mathrm{\infty }}{lim}{\left[{(1+\frac{r/r}{n/r})}^{\frac{n}{r}\cdot t}\right]}^r\\ & =P\cdot \underset{n\to \mathrm{\infty }}{lim}{\left[{(1+\frac{1}{n/r})}^{\frac{n}{r}}\right]}^{rt}\\ & =P\cdot \underset{n\to \mathrm{\infty }}{lim}{\left[{(1+\frac{1}{n/r})}^{\frac{n}{r}}\right]}^{rt}\end{array}$

Now to make it a little simpler, let $m=\frac{n}{r}$ . When $n$ goes to infinity, $m$ goes to infinity since $r$ is a positive constant. Therefore, we can rewrite the above expression as

$\begin{array}{rl}{A}_t& =P\cdot \underset{m\to \mathrm{\infty }}{lim}{\left[{(1+\frac{1}{m})}^m\right]}^{rt}\\ & =P{e}^{rt}\end{array}$

You may have noticed that this is the formula to find the future value $A_t$ of the amount $P$ saved today at the interest rate $r$ for $t$ years.

Just for your information, the number $e$ can also be defined in another way. The number $e$ is the limit

$e=\underset{x\to 0}{lim}{(1+x)}^{\frac{1}{x}}$

We can see that the two definitions are equivalent by substituting $x=\frac{1}{n}$ . As $n$ approaches infinity, $x$ goes to 0.

Even though we practiced continuous compounding earlier, let’s practice again to make sure that you can use the equation.

Suppose that you invest $5,000 in an account that gives you an annual interest rate of 10% compounded continuously. How many years will it take for the account to be worth double the amount of the initial investment? The following equation is our starting point:

\mathrm{10,000}=\mathrm{5,000}{e}^{0.10t}

Divide both sides by 5,000 to get

$2=e^{0.10t}$

Then, take the natural log on both sides:

$\begin{array}{rl}\ln 2& =\ln e^{0.10t}\\ & =0.10t\end{array}$

Finally, solve for $t$ to get

$\begin{array}{rl}t& =\frac{\ln 2}{0.10}\\ & =6.93\text{years}\end{array}$