## Preliminary

In Section 3.1 we used the formula for compound interest $$A = P \left( 1 + \frac{r}{n} \right)^{n t}$$ , where $$A$$ is the amount owed, $$P$$ is the original amount borrowed, $$r$$ is the annual interest rate and $$n$$ is the number of compounding intervals per year.

The base of this exponential function is $$b = \left( 1 + \frac{r}{n} \right)^n$$. Let us consider the special case when $$r = 100\% = 1$$. That is, let $$b = \left( 1 + \frac{1}{n} \right)^n$$.

## Exercise 3.2.1

Compute to six decimal places the value of $$b$$ when $$n = 1, n = 10, n = 100, n = 1000, n = 1,000,000$$.

See Solution

## The natural base $$e$$

If we let n increase without bound, then b will approach a limiting value of approximately $$2.718281828$$. This limiting number is sufficiently important in mathematics that it is given a special symbol: $$e$$. It is referred to as the natural base. Scientific calculators are capable of computing $$y=e^x$$ for any input number $$x$$.

Using methods from calculus it can be shown that the limiting value of $$b = \left( 1 + \frac{r}{n} \right)^n$$ when $$n$$ increases without bound is equal to $$e^r$$. In the formula for compound interest, allowing $$n$$ to increase without bound corresponds to compounding the interest continuously.

The formula for computing the amount owed when interest compounds continously at an annual rate of r is

$A = P e^{r t}$

## Exercise 3.2.2

Find the amount owed on a loan of $$\1000$$ compounded continuously if the annual rate is $$15\%$$.

See Solution

## Exercise 3.2.3

Sketch the graph of $$f ( x ) = e^x$$ for input values between $$- 1$$ and $$1$$.

See Solution