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Proofs of e

Date: 03/21/2002 at 21:37:08
From: Christine
Subject: Proof of number e

I was asked by my algebra teacher to find the proof of e. I found 
out that the definition of e is: e = lim, as n approaches infinity 
(1+ 1/n)^n. Is there a proof for this or for 1+2(e-1/n)?

Thank you so much ahead of time for your help!

Date: 03/22/2002 at 09:27:16
From: Doctor Pete
Subject: Re: Proof of number e

Hi Christine,

If we define the number e to equal the limit of (1 + 1/n)^n as n -> 
Infinity, then we do not need to prove that this limit is equal to e, 
because it is true by definition.

However, there are several "definitions" for the number e.  For 
instance, it is the unique real number greater than 1 such that the 
area under the curve y = 1/x from x = 1 to x = e is equal to 1. Or we 
may define e to be the value of the infinite series

  e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ... .

And of course, we may define e to be the limit you describe. All of 
these are equivalent; however, the challenge is in proving that these 
definitions are equivalent. To do so rigorously requires basic 
calculus, and concepts from real analysis.

We can show that the series definition implies that e is also equal to 
the limit formula.  Let

  s[n] = 1/0! + 1/1! + ... + 1/n!,
  t[n] = (1 + 1/n)^n,

for n >= 0.  Then by the Binomial Theorem,

  t[n] = 1 + 1 + (1 - 1/n)/2! + (1 - 1/n)(1 - 2/n)/3! + ...
           + (1 - 1/n)(1 - 2/n)(...)(1 - (n-1)/n)/n!.

Therefore, t[n] <= s[n], so that

  lim sup t[n] <= e.
  n -> oo

Now, if n >= m, then

  t[n] >= 1 + 1 + (1 - 1/n)/2! + ... + (1 - 1/n)(...)(1 - (m-1)/n)/m!.

Let n -> infinity, keeping m fixed.  Hence

  lim inf t[n] >= 1 + 1 + 1/2! + ... + 1/m!,
  n -> oo

so we also find

  s[m] <= lim inf t[n].
          n -> oo

Now let m -> infinity:

  e <= lim inf t[n].
       n -> oo

Therefore, e <= lim inf t[n] <= lim sup t[n] <= e, and it follows that

  lim t[n] = e.

To see how the limit definition implies the integral (area under the 
curve) definition, we will first show that the derivative of the 
natural logarithm is given by

  d/dx [ln(x)] = 1/x.

(Heretofore we have defined the function ln(x) to be the inverse of 
the function y = e^x with e defined as a limit; thus by definition ln 
e = 1.)

By definition of the derivative, we find

  d/dx [ln(x)] = lim (ln(x+h)-ln(x))/h,
               = lim (1/h)ln((x+h)/x)
               = lim ln((1 + h/x)^(1/h))

as h -> 0. Since x is a fixed positive real, let u = x/h; then the 
limit becomes

  lim ln((1 + 1/u)^(u/x)) = lim (1/x)(ln(1 + 1/u)^u)
                          = 1/x lim ln(1 + 1/u)^u,

as u -> infinity. But this latter limit is precisely ln e = 1, so the 
result follows. Therefore by the Fundamental Theorem of Calculus, 
there exists a real number t > 1 such that

  1 = Integral[1/x dx, {1, t}] = ln(t) - ln(1) = ln(t),

therefore ln(t) = 1, or t = e.

That wraps up the equivalence of the three major definitions for e.  
It may be beyond the scope of what your instructor asked of you (or it 
may not), but the general point is that in mathematics, definitions do 
not need proof; they are true by assumption. Occasionally, however, 
some definitions lead to such powerful and often-used theorems in of 
themselves that the role of theorem and definition become reversed.  
For another example of this, consider the formula for the dot product, 
or the definition of the circular trigonometric functions.

- Doctor Pete, The Math Forum
Associated Topics:
College Analysis
College Calculus
High School Analysis
High School Calculus
High School Transcendental Numbers

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