From: victor@watson.ibm.com (Victor Miller)
Subject: Re: An interesting limit problem.
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
	<MARTIN.92Jul29125203@lyra.cis.umassd.edu>
	<1992Jul30.150656.14324@husc3.harvard.edu>
Organization: IBM, T.J. Watson Research Center

>>>>> 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
Newsgroups: sci.math