E in the Modern WorldDate: 01/06/98 at 13:41:33 From: Benjamin Vena Subject: e (logs and natural logs) I am looking for examples of how e exists in the modern world. For instance, the spread of epidemics, population growth, element decay, drug absorption, savings account interest. My teacher gave these examples and I need to find out how e affects the situation directly. Thank you for your time! Benjamin Vena Forest Hills Sr. High Sidman, PA Date: 01/07/98 at 08:08:02 From: Doctor Jerry Subject: Re: e (logs and natural logs) Hi Benjamin, I can't tell whether you are taking calculus or not. I'll give two responses, one calculus-based and the other less calculus-based. It is known from experiment that the rate of decay of a radioactive material present at any time t is proportional to the amount present. So, if A(t) is the amount of a given piece of material, A'(t)=k*A(t). The solution to this differential equation is A(t)=c*e^{k*t}. That's how e comes in, quite naturally. You might be interested in the savings account connection with e. If you start with 1 dollar and invest it a r percent per year, with the interest compounded k times per year (k=1 or 2 or 3 or ...), then after n/k years you will have (1+r/(100k))^n dollars in the bank. We write this as A(n*k) = (1+r/(100k))^n. Letting n*k=t, A(t) = (1+r/(100k))^{t/k}=[(1+r/(100k))^{1/k}]^{t} We're thinking about t as the time, which we regard as arbitrary but fixed. Now we ask what happens as k increases. So, we're going from compounding once a year to quarterly, daily, every second, etc. With t fixed, the only thing changing is k. So we examine (1+r/(100k))^{1/k}. The r is in the way, so we let r/(100k)=1/s. Making this change gives (1+r/(100k))^{1/k} = (1+1/s)^{r*s/100} = [(1+1/s)^s]^{r/100}. Now you need to whip out your calculator and examine what happens to (1+1/s)^s as s becomes very large. Here are some results from my calculator: s (1+1/s)^s 10 2.5937 100 2.7048 1000 2.7169 Sooner or later the calculator will start going wrong but you can see that the limit appears to be e = 2.718281828... So, 1$ invested for t years at r percent with "continuous compounding" will yield A(t) = e^{r*t/100}. In 1 year and 8%, A(1) = 1.083287... Not a lot better than the 1.08 you'd get at once a year compounding, but if you had more years or more dollars, or both, well... -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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