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### E in the Modern World

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Date: 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.

Benjamin Vena
Forest Hills Sr. High
Sidman, PA
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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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Associated Topics:
High School Exponents

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