Associated Topics || Dr. Math Home || Search Dr. Math

### The Nature of e

```
Date: 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

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/
```
Associated Topics:
High School Interest
High School Transcendental Numbers

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search