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 \(b = \left( 1 + \frac{1}{n} \right)^n\).
Compute to six decimal places the value of \(b\) when \(n = 1, n = 10, n = 100, n = 1000, n = 1,000,000\).
See SolutionIf 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
\[A = P e^{r t}\]
Find the amount owed on a loan of \(\$1000\) compounded continuously if the annual rate is \(15\%\).
See SolutionSketch the graph of \(f ( x ) = e^x\) for input values between \(- 1\) and \(1\).
See Solution