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- Newsgroups: sci.math
- Path: sparky!uunet!newsgate.watson.ibm.com!yktnews!admin!yktnews!victor
- From: victor@watson.ibm.com (Victor Miller)
- Subject: Re: An interesting limit problem.
- Sender: news@watson.ibm.com (NNTP News Poster)
- Message-ID: <VICTOR.92Jul31103607@terse4.watson.ibm.com>
- In-Reply-To: elkies@ramanujan.harvard.edu's message of 30 Jul 92 19:06:54 GMT
- Date: Fri, 31 Jul 1992 14:36:07 GMT
- Reply-To: victor@watson.ibm.com
- Disclaimer: This posting represents the poster's views, not necessarily those of IBM
- References: <1992Jul29.000223.27339@massey.ac.nz> <1992Jul29.115656.23253@gdr.bath.ac.uk>
- <MARTIN.92Jul29125203@lyra.cis.umassd.edu>
- <1992Jul30.150656.14324@husc3.harvard.edu>
- Nntp-Posting-Host: terse4.watson.ibm.com
- Organization: IBM, T.J. Watson Research Center
- Lines: 74
-
- >>>>> On 30 Jul 92 19:06:54 GMT, elkies@ramanujan.harvard.edu (Noam Elkies) said:
-
- Noam> In article <MARTIN.92Jul29125203@lyra.cis.umassd.edu>
- Noam> martin@lyra.cis.umassd.edu (Gary Martin) writes:
- >And have we even had a proof that the limit is 1 that didn't use a machine?
-
- Noam> The proof I posted only used the machine to substitute one power series
- Noam> into another, which is just straightforward arithmetic --- it could have
- Noam> been done easily by hand in a few minutes, but why bother?
-
- >There was one posting that made the substitution y=tan(sin x) (or something
- >similar) and claimed that the result was the reciprocal of the original
- >expression, but I don't think that claim was correct.
-
- Noam> Me neither --- and even if it was, one would still have to show that the
- Noam> limit exists and does not equal -1. But that posting did contain the kernel
- Noam> of an idea that does provide a valid proof along the lines indicated in
- Noam> V.Miller's recent posting. Basically we are comparing the commutators
- Noam> [f,g] and [F,G] in the infinite-dimensional Lie group of power series
- Noam> a1*x+a2*x^2+... (a1<>0) under composition, with F,G the inverse functions
- Noam> of f,g; these commutators either both reduce to the identity (the power
- Noam> series x) or are both of the form x+an*x^n+... with the same n. This
- Noam> also leads to an explanation of the fact that in our case f=sin,g=tan
- Noam> (with both f,g odd functions with leading coefficient 1) the difference
- Noam> f(g)-g(f) [and so also F(G)-G(F)] vanishes to 7th order at the origin.
-
- > 2 2
- > sin sin
- > --- x - --- x
- > cos cos 0
- >----------------------- = --- = 0. (Though they might leave out the
- > 2 2 2 2 0 equal signs, too. :( )
- >Arc sin Arc sin
- >------- x - ------- x
- > 2 2
- > cos cos
-
- Noam> 5 2
- Noam> Ouch. Do they automatically write 2 9 = 2592 too? ;-)
-
- Noam> --Noam D. Elkies (elkies@zariski.harvard.edu)
- Noam> Department of Mathematics, Harvard University
-
- Well, I decided to work this out last night. It turned out to be
- pretty simple. This brought up a more fundamental question: why was
- this seemingly arcane fact well-known. Does something like this arise
- in physics, for example?
-
-
- Let k be field, and F and G in R = k[[x]] be distinct power
- series starting x + O(x^2). Suppose further, that p(x) is a
- polynomial of degree n-1 such that F(x) = p(x) + a x^n + O(x^{n+1}),
- G(x) = p(x) + b x^n + O(x^{n+1}), with a \ne b. Let q(x) be the
- unique polynomial of degree n-1 such that p(q(x)) = x mod x^n. In
- particular there is a c in k such that p(q(x)) = x + c x^n +
- O(x^{n+1}).
-
- We have, for e in k, F(q(x) + e x^n) = F(q(x)) + F'(q(x)) e x^n +
- O(x^{n+1}) = p(q(x)) + a q(x)^n + e x^n + O(x^{n+1}) =
- x + (c+a+e) x^n + O(x^{n+1}). Thus
-
- F^{-1}(x) = q(x) - (c+a) x^n + O(x^{n+1}).
-
- Similarly
-
- G^{-1}(x) = q(x) - (c+b) x^n + O(x^{n+1}).
-
- so (F(x) - G(x))/(F^{-1}(x) - G^{-1}(x)) = -1 + O(x^2).
-
- --
- Victor S. Miller
- Vnet and Bitnet: VICTOR at WATSON
- Internet: victor@watson.ibm.com
- IBM, TJ Watson Research Center
-