The Nature of eDate: 02/23/97 at 17:49:09 From: Kelvin Chen Subject: About natural logs Dear Dr. Math, I have a project question that I cannot answer: Explain in cogent prose what is "natural" about e. I know that most of the problems involve with natural growth use e. But what exactly is "e"? Thank you for your time. Sincerely, Kelvin C. Date: 02/24/97 at 08:39:11 From: Doctor Jerry Subject: Re: About natural logs Hi Kelvin, Like pi, which could be said to be "naturally" associated with circles, e occurs in several different "growth" areas of mathematics and applied mathematics. If an amount A of money is invested at an APR of R percent and the interest is compounded k times a year, how much money P(t) will you have after t years? You probably know that this amount is: P(t) = A(1+R/(100k))^(k*t) If A = $1000, R = 6, and k = 12, for example, P(10) = $1819.40 If the money is compounded 365 times a year (daily compounding), so that k = 365 (A and R stay the same): P(10) = $1822.03 If k = 365*24 (your money is compounded every hour): P(10)= $1822.12 Some banks advertise that they offer "continuous compounding." This means that in the formula for P(t), k becomes infinite. What does P(t) approach? I'll come back to this in a moment. First, we look at the sequence: (1+1/N)^N, N = 1,2,3,... If you calculate several of these numbers, you will find that they approach the number e = 2.718281828 as N becomes very large. For N = 1000, for example, which you can do on your calculator, (1+1/1000)^(1000) = 2.7169, approximately. Write P(t) as: P(t) = A{[(1+1/(100k/R)]^(100k/R)}^(Rt/100) Let 100k/R = N (I know that 100k/R may not be an integer, but we'll ignore this - it doesn't matter). Then: P(t) = A[(1+1/N)^N]^(Rt/100) So, as k becomes large, so does N, and P(t) approaches: P(t) = A*e^(Rt/100) This is the formula for continuous compounding. It's just like the biological growth/radioactive deacay formula. Try it out with A = 1000, R = 6, and t = 10: P(t) = 1000*e^(6*10/100) = $1822.12 Did you notice how e came in "naturally"? One other thing. In calculus, we use radians so that the derivative of sin(x) is cos(x). If x were measured in degrees, we wouldn't get this simple answer. If Log means base 10 logs and ln means base e logs (that is, natural logarithms), the derivative of ln(x) is 1/x. The derivative of Log(x) is not quite as simple. This makes e the "natural" base to use. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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