The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
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 
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!   
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

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.