home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!Sunburn.Stanford.EDU!pratt
- From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
- Subject: Re: need proof: (1 + 1/n)^n ==> e
- Message-ID: <1992Dec31.045851.26887@CSD-NewsHost.Stanford.EDU>
- Sender: news@CSD-NewsHost.Stanford.EDU
- Organization: Computer Science Department, Stanford University.
- References: <1hfv4sINNdlk@usenet.INS.CWRU.Edu> <1992Dec29.094842.3685@CSD-NewsHost.Stanford.EDU> <VICTOR.92Dec30104755@terse.watson.ibm.com>
- Date: Thu, 31 Dec 1992 04:58:51 GMT
- Lines: 69
-
- In article <VICTOR.92Dec30104755@terse.watson.ibm.com> victor@watson.ibm.com writes:
- >The following simple proof comes from "Inequalities" by P. Korovkin,
- >Blaidell Scientific paperbacks.
- >
- >set x_n = (1+1/n)^n, and y_n = (1+1/n)^{n+1}.
- >
- >We show that x_n is monotone increasing and y_n is monotone
- >descreasing. Since x_n < y_n we find that they both approach a common
- >limit, which we call e.
-
- This only shows that x_n and y_n each approach a limit, not necessarily
- the same one. For the purposes of defining e it suffices to take e to
- be the limit of x_n. However y_n is particularly convenient for
- showing that x_n and y_n do approach the same limit, y_n/x_n being
- simply 1+1/n.
-
- The use of the arithmetic-geometric mean inequality in Korovkin's
- argument is not only slick but shows that the whole sequence (1+1/n)^n
- is monotone, not just the n-a-power-of-2 part (Michael Somos pointed
- this argument out to me yesterday). However it makes the proof that
- much less elementary, not that the inequality is hard to prove but
- including the proof does add a few lines.
-
- Korovkin's y_n gives me an idea for shortening my elementary
- powers-of-2 proof that (1+1/N)^N has a limit. I'll continue to write N
- for powers of 2 and n for positive integers.
-
- 1 N
- Lemma 1. X = (1 + -) is monotone increasing.
- N N
-
- 1 1 2 1 N 1 2N
- Proof. 1 + - < ( 1 + -- ) , hence ( 1 + - ) < ( 1 + -- )
- N 2N N 2N
-
-
- Lemma 2. Ditto with - in place of + throughout, call this sequence Y .
- N
-
- Proposition. X has a limit.
- N
-
- 1 N 1 N
- Proof. X = (1 + -) < (1 + ---) = 1/Y < 1/Y . So X is bounded. QED
- N N N-1 N 1 N
-
- This ought to be getting close to the shortest possible elementary
- proof of this proposition.
-
-
- We can also show that Y has a limit, namely 1/e, by tying it to X thus:
- N+1 N
-
-
- n+1 n n n+1 n
- Lemma 3. X Y = ( --- ) ( --- ) = --- , for all positive n.
- n n+1 n n+1 n+1
-
- Note that Lemma 3 holds for all n, but as long as we only understand
- X_N we can only apply it to Y_{N+1}. This raise the following
-
- Problem. Give comparably elementary proofs of the stronger result that
- X_n (not just X_N) has a limit, and the yet stronger result that X_n is
- monotone increasing. Somos' ordered-sequences argument does not yield
- these. Korovkin's AGM argument seems the most elegant way to obtain
- the latter, with the former an immediate corollary. Arguments based on
- expanding (a+b)^n seem to entail unreasonably tedious calculation.
- --
- Vaughan Pratt There's safety in certain numbers.
-