348 CHAPTER 5 Exponential and Logarithmic Functions Now suppose that the number n of times that the interest is compounded per year gets larger and larger; that is, suppose that →∞ n . Then = →∞ h n r , and the expression in brackets in equation (3) equals e. That is, →A Pe .r Table 8 compares ( ) + r n 1 , n for large values of n, to er for = = r r 0.05, 0.10, = r 0.15, and = r 1. As n becomes larger, the closer ( ) + r n 1 n gets to e .r No matter how frequent the compounding, the amount after 1 year has the upper bound Pe .r Need to Review? The number e is defined on page 302. Using Continuous Compounding The amount A that results from investing a principal P of $1000 at an annual rate r of 10% compounded continuously for a time t of 1 year is = = ⋅ = A e $1000 $1000 1.10517 $1105.17 0.10 EXAMPLE 3 THEOREM Continuous Compounding The amount A after t years due to a principal P invested at an annual interest rate r compounded continuously is = A Pert (4) = 100 n = 1000 n = 10,000 n er = r 0.05 1.0512580 1.0512698 1.051271 1.0512711 = r 0.10 1.1051157 1.1051654 1.1051704 1.1051709 = r 0.15 1.1617037 1.1618212 1.1618329 1.1618342 = r 1 2.7048138 2.7169239 2.7181459 2.7182818 Table 8 ( ) + 1 r n n When interest is compounded so that the amount after 1 year is Pe ,r the interest is said to be compounded continuously . Now Work PROBLEM 13 2 Calculate Effective Rates of Return Suppose that you have $1000 to invest and a bank offers to pay you 3 percent annual interest compounded monthly. What simple interest rate is needed to earn an equal amount after one year? To answer this question, first determine the value after one year of the $1000 investment that earns 3 percent compounded monthly. ( ) = + = A $1000 1 0.03 12 $1030.42 12 Use ( ) = + A P r n 1 n with = = = P r n $1000, 0.03, 12. So the interest earned is $30.42. Using = I Prt with = = t I 1, $30.42, and = P $1000, the annual simple interest rate is = 0.03042 3.042%. This interest rate is known as the effective rate of interest .
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