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/