4.5 Properties of the Constant Number e

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

$A_{t} = P {(1 + \frac{r}{n})}^{n t}$ $A_{t} = P {(1 + \frac{r}{n})}^{n t}$

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.

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

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

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

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

$A_{t} = P {(1 + \frac{r}{n})}^{n t}$ $A_{t} = P {(1 + \frac{r}{n})}^{n t}$

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

$\begin{matrix} A_{t} & = lim n \to \infty P {(1 + \frac{r}{n})}^{n t} = P \cdot lim n \to \infty {(1 + \frac{r}{n})}^{n t} = P \cdot lim n \to \infty {[{(1 + \frac{r / r}{n / r})}^{\frac{n}{r} \cdot t}]}^{r} = P \cdot lim n \to \infty {[{(1 + \frac{1}{n / r})}^{\frac{n}{r}}]}^{r t} = P \cdot lim n \to \infty {[{(1 + \frac{1}{n / r})}^{\frac{n}{r}}]}^{r t} \end{matrix}$ $\begin{aligned} A_{t} & = lim_{n \to \infty} P {(1 + \frac{r}{n})}^{n t} \\ = P \cdot lim_{n \to \infty} {(1 + \frac{r}{n})}^{n t} \\ = P \cdot lim_{n \to \infty} {[{(1 + \frac{r / r}{n / r})}^{\frac{n}{r} \cdot t}]}^{r} \\ = P \cdot lim_{n \to \infty} {[{(1 + \frac{1}{n / r})}^{\frac{n}{r}}]}^{r t} \\ = P \cdot lim_{n \to \infty} {[{(1 + \frac{1}{n / r})}^{\frac{n}{r}}]}^{r t} \end{aligned}$

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

$\begin{matrix} A_{t} & = P \cdot lim m \to \infty {[{(1 + \frac{1}{m})}^{m}]}^{r t} = P e^{r t} \end{matrix}$ $\begin{aligned} A_{t} & = P \cdot lim_{m \to \infty} {[{(1 + \frac{1}{m})}^{m}]}^{r t} \\ = P e^{r t} \end{aligned}$

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

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

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

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

$10,000 = 5,000 e^{0.10 t}$ $10,000 = 5,000 e^{0.10 t}$

Divide both sides by 5,000 to get

$2 = e^{0.10 t}$ $2 = e^{0.10 t}$

Then, take the natural log on both sides:

$\begin{matrix} ln 2 & = ln e^{0.10 t} = 0.10 t \end{matrix}$ $\begin{aligned} \ln 2 & = \ln e^{0.10 t} \\ = 0.10 t \end{aligned}$

Finally, solve for $t$ to get

$\begin{aligned} t & = \frac{\ln 2}{0.10} \\ = 6.93 years \end{aligned}$

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