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

### Euler's Number

Date: 01/28/98 at 16:29:49
From: Jeffery Holcomb
Subject: Euler's number

During a recent math assignment I came in contact with Euler's number.
I also heard it referred to as the natural base. When I asked why it
was called this my teacher said that he did not know. Since then I
have looked in several different places and have not been able to find
an answer. If you could please tell me why Euler's number is referred
to as the natural base I would greatly appreciate it.

Date: 01/28/98 at 18:13:01
From: Doctor Anthony
Subject: Re: Euler's number

The base of logs that has most application in mathematics is 'e'.
Here is a note on how this number arises naturally from the laws of
compound interest growth.

In the 1730's Euler investigated the result of compounding interest
continuously when a sum of money, say, is invested at compound
interest.

If interest is added once a year we have the usual formula for the
amount, A,  with principal P, rate of interest r percent per annum,
and t the time in years:

A = P(1 + r/100)^t

If interest were added twice a year, then we replace r by r/2 and we
replace t by 2t. So formula becomes

A = P(1 + r/(2x100))^(2t)   = amount after t years.

If 3 times a year then A at the end of t years would be:

A = P(1 + r/(3x100))^(3t)

and if we added interest N times a year, then after t years the amount
A would be

A = P[1 + r/(Nx100)]^(Nt)

Now to simplify the working we put r/(100N) = 1/n  so  N = nr/100

and  A = P[1 + 1/n]^(nrt/100)

A = P[(1 + 1/n)^n]^(rt/100)

We now let n -> infinity and we must see what happens to the
expression

(1 + 1/n)^n  as n tends to infinity.

Expanding by the binomial theorem

(1 + 1/n)^n = 1 + n(1/n) + n(n-1)/2! (1/n)^2 + n(n-1)(n-2)/3! (1/n)^3
+ ...

now take the n's in 1/n^2,  1/n^3 etc, in the denominators and
distribute one n to each of the terms n, n-1, n-2, etc. in the
numerator, getting

1 x (1-1/n) x (1-2/n) x .....  so we now have

(1 + 1/n)^n = 1 + 1 +  1(1-1/n)/2! + 1(1-1/n)(1-2/n)/3! + .....

Now let n -> infinity and the terms 1/n, 2/n etc. all go to zero,
giving

(1 + 1/n)^n = 1 + 1 + 1/2! + 1/3! + 1/4! + .....

and this series converges to the value we now know as e.

If you consider e^x you get

(1 + 1/n)^(nx) and expanding this by the binomial theorem you have

(1 + 1/n)^(nx) = 1 + (nx)(1/n) + nx(nx-1)/2! (1/n)^2 + ....

and carrying through the same process of putting the n's in the
denominator into each term in the numerator as described above you
obtain

e^x = 1 + x + x^2/2! + x^3/3! + ....

and differentiating this we get

d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + ...

=     1 + x + x^2/2! + x^3/3! + ....

=   e^x

Reverting to our original problem of compounding interest
continuously, the formula for the amount becomes

A = P.e^(rt/100)

You might like to compare the difference between this and compounding
annually.

If P = 5000,   r = 8,   t = 12 years

Annual compounding gives  A = 5000(1.08)^12  = 12590.85

Continuous compounding gives  A = 5000.e^(96/100)  = 13058.48

The difference is not as great as might be expected.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/

Associated Topics:
High School Exponents
High School Interest

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