Compound interest (§6.7)

There are several applications of exponential functions. To solve for the input of one of these functions is then an application of logarithms.

If you invest (or borrow) an amount of money at a fixed rate of interest, then the amount of money that you have (or owe) at the end of a period of time is an exponential function of time. There are three basic formulas that you want to use:

In these formulas, the variables have the following meaning:

With simple interest, the interest is applied once, at the end of the time period; this is effectively compound interest where n = 1/t. With intermittent compound interest, the interest is applied several times and added to the original amount, so that interest can be charged on the interest later. With continuous compound interest, the interest is added to the original amount continuously; this is like compound interest where n is effectively infinite.

To illustrate how continuous compound interest comes about, suppose that P = $1000.00, r = 6% = 6/100 = 3/50 = 0.06 (per year), and t = 1 year. With simple interest, this gives A = P(1 + rt) = ($1000.00)(1 + (0.06)(1)) = $1060.00. With n = 1/t = 1/(1) = 1, that is compounded annually, the result is the same (although calculated differently): A = P(1 + r/n)nt = ($1000.00)(1 + (0.06)/(1))(1)(1) = $1060.00. If we compound n = 4 times per year (quarterly), then there is some interest on the interest, so the result is larger: A = P(1 + r/n)nt = ($1000.00)(1 + (0.06)/(4))(4)(1) = $1061.363550624 ≈ $1061.36. Here I've rounded off to the nearest cent, which I'll do from now on. If we compound n = 12 times per year (monthly), then there is more interest on the interest, so the result is even larger: A = P(1 + r/n)nt = ($1000.00)(1 + (0.06)/(12))(12)(1) ≈ $1061.68. If we compound 30 times more often than that, that is n = 360 times per year (daily by bankers' traditional counting methods), then the amount is even larger, although not very much larger: A = P(1 + r/n)nt = ($1000.00)(1 + (0.06)/(360))(360)(1) ≈ $1061.83. If you increase n still more, then in principle A will continue to increase, although in fact you won't notice it for a while due to rounding. Even if you use n = 1235, you still get A ≈ $1061.83. If you use n = 1236, then you finally get A ≈ $1061.84; after that … you'll always get A ≈ $1061.84, no matter how large you make n. And continuous compound interest gives you this limiting result directly: A = Pert = ($1000)e(0.06)(1) = $1000e0.06 ≈ $1061.84. No matter how often you compound the interest, the final result will never be larger than that!

Another way to see where the special number e comes into it is to use some properties of exponents to write the formula for intermittent compound interest as A = P(1 + r/n)nt = P(1 + r/n)(n/r)(rt) = P((1 + r/n)n/r)rt = P((1 + x)1/x)rt, where I've written x for r/n (so that n/r is its reciprocal, 1/x). With r fixed (0.06 for example), as n gets arbitrarily large, x = r/n will get arbitrary close to 0 (while remaining positive). So if (1 + x)1/x gets arbitrarily close to some number e as x gets arbitrarily close to 0, then A gets arbitrarily close to Pert as n gets arbitrarily large. And that's exactly what happens; you can approximate e as closely as you like by using a sufficiently small positive number x in the expression (1 + x)1/x. (Compare the book's definition of e on the top of page 431; the n in their definition is not my n but rather my 1/x.)

If you know the final amount A and want to find the principal P, then solve the equation for P. If you know both A and P and want to find the amount of time t, then you must take a logarithm to solve the equation. It's also possible to solve for r, but not for n (at least not with the operations that we use in this class).


Go back to the course homepage.
Valid HTML 4.01 Transitional

This web page was written between 2012 and 2016 by Toby Bartels, last edited on 2016 May 12. Toby reserves no legal rights to it.

The permanent URI of this web page is http://tobybartels.name/MATH-1150/2016SP/logapps/.