NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17486

Text File | 1992-12-29 | 3.7 KB | 88 lines

Newsgroups: sci.math
Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!Sunburn.Stanford.EDU!pratt
From: [e-mail] (Vaughan R. Pratt)
Subject: Re: need proof:  (1 + 1/n)^n ==> e
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: Computer Science Department,  Stanford University.
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Tue, 29 Dec 1992 21:57:44 GMT
Lines: 76

In article <[e-mail]> [e-mail] (Vaughan R. Pratt) writes:
>In article <[e-mail]> [e-mail] (Michael Somos) writes:
>>The tone of the question suggests that a very elementary proof of
>>the limit (1+1/n)^n is requested. [...]
>>
>>[Very interesting method here]
>>
>[equally elementary "establishment" proof]

I spent some time thinking about the relationship between Michael
Somos' proof and the more conventional proof I gave.  First let me make
the following change to my

>Proposition.  If 1 < M < N are powers of 2 then
>
>    (1+1/M)^M < (1+1/N)^N < e < (1+1/N+1/N^2)^N < (1+1/M+1/M^2)^M

The corresponding part of Michael's proof reads:

>>if x < y , then  -x < -y  and P(x) < P(y) < 1/P(-y) < 1/P(-x)

Michael's upper bound is simpler and easier to work with than my
1+1/N+1/N^2, which should be extended out to 1+1/N+1/N^2+1/N^3+... = 1/(1-1/N)
to give:

         1 M          1 N                1 -N          1 -M
    (1 + -)   <  (1 + -)   <  e  <  (1 - -)    <  (1 - -)
         M            N                  N             M

(This is better and hence more likely to be a standard argument than
what I had before.)  The last inequality can now be proved as easily as
the first:

        1         2     1           1   2
    1 - -  <  1 - -- + ---  = ( 1 - -- )    etc., valid even for N=1
        N         2N   N^2          2N

and making for a closer parallel to Michael's proof.

                      2 -N    1/N
The pointwise ratio now becomes (1-1/N )   < B   , where B is the limit
of the upper sequence, so that part of my argument doesn't change
appreciably, other than to simplify slightly.

The absence of the term B^{1/N} from Michael's argument has me worried:
maybe I'm just missing something but it seems to me his argument that
S(-x^2) approaches 0 was a bit quick and if spelled out should involve
something akin to the B^{1/N} argument.

The sentence

>>Finally, consider the sequence of lists  [1] , [1/2,1/2] , ... ,
>>[1/n,1/n,...,1/n] , ...

is ambiguous: is the third list meant to be [1/3,1/3,1/3] or
[1/4,1/4,1/4,1/4]?  The preceding paragraph says "All we need now is to
show that an increasing sequence of lists ...  will converge to a
limit," suggesting the latter, since [1/2,1/2] < [1/3,1/3,1/3] does not
hold (the larger list has to be obtainable from the smaller by
splitting elements).  In that case Michael's proof establishes
convergence of (1+1/N)^N just for N restricted to powers of 2, as with
mine.  Extending the proof to show convergence of (1+1/n)^n for all n>0
seems to entail the same complications required to so extend my proof.

With these changes and remarks I think the relationship between the two
proofs can be characterized by saying that the latter is what one
obtains by tuning all arguments in Michael's proof to the case of
constant unit-sum lists of length a power of 2, i.e. [1/N,...,1/N]
where N, a power of 2, is the length of the list.

This raises the question of whether there are other applications of
these interesting partially ordered lists besides those involving just
constant unit-sum lists.  Hopefully so, since they seem to form a nice
structure, and one that is certainly new to me.
--
Vaughan Pratt                There's safety in certain numbers.