Compounding Interest and e

Date: 11/11/98 at 11:52:55
From: John Radway
Subject: How in the world did they come up with e?

I know that e is equal to 2.718281828..., and that it is the base of 
natural logarithms, but what does that number mean? How is it 
calculated? How did they come up with it?

Date: 11/11/98 at 14:11:24
From: Doctor Anthony
Subject: Re: How in the world did they come up with e?

In the 1730s 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 is added twice a year, we replace r by r/2 and t by 2t. So 
the formula becomes:

   A = P(1 + r/(2 * 100))^(2t)   = amount after t years.

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

   A = P(1 + r/(3 * 100))^(3t)

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

   A = P[1 + r/(N * 100)]^(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 the denominators of 1/n^2, 1/n^3, and so on, and 
distribute one n to each of the terms n, n-1, n-2, etc in the 
numerator, getting for each arbitrary term of the sequence:

   1 * (1-1/n) * (1-2/n) * ...  

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
  http://mathforum.org/dr.math/