e as a Series and a Limit

Date: 03/30/98 at 11:58:33
From: CMLee
Subject: Definition of e

Why does e = 1 + 1/2! + 1/3! + 1/4! + ... 
and  lim (1 + 1/n) ^ n   n --> infinity? 

What is the relation between these two expressions (if any), i.e. how 
do they connect? 

How do you exactly define e and why/how does the definition work?
Please help. Thank you.

Date: 03/30/98 at 22:12:20
From: Doctor Rob
Subject: Re: Definition of e

The relation is found in calculus.  e is defined as the unique real 
number such that d(log_e(x))/dx = 1/x. That implies that e is given by 
the limit you gave. This comes from looking at the derivative as the 
limit of the difference quotient:

   1/x = lim (log_e[x+d]-log_e[x])/d  as d -> 0
   1/x = lim (log_e[(x+d)/x])/d  as d -> 0
   1/x = lim (log_e[1+d/x])/d  as d -> 0
   1 = lim log_e([1+d/x]^[x/d])  as d -> 0
   1 = log_e(lim [1+d/x]^[x/d])  as d -> 0
   e = lim (1+d/x)^(x/d)  as d -> 0
   e = lim (1+1/n)^n  as n -> infinity

by setting n = x/d, or d = x/n.

Now this implies that d(e^x)/dx = e^x: Let y = e^x, so x = log_e(y).
Now differentiate using the Chain Law: 1 = (1/y)*dy/dx, so dy/dx = y,
and d(e^x)/dx = e^x.

Now the Maclaurin series for any function f(x) infinitely 
differentiable at x = 0 is given by:

   f(x) = f(0) + f'(0)*x/1! + f''(0)*x^2/2! + f'''(0)*x^3/3! +
            f''''(0)*x^4/4! + ...

If we use:

     f(x) = e^x

then:

     f(x) = f'(x) = f''(x) = f'''(x) = f''''(x) = ..., so

     1 = f(0) = f'(0) = f''(0) = f'''(0) = f''''(0) = ..., so

     e^x = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ...

Now set x = 1.

By the way, you left off the first term from your infinite series 
for e.

You may also be able to establish the infinite series equation by 
using the binomial theorem on (1+1/n)^n, but I haven't seen this 
argument.

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