From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/3705 Path: news.gmane.org!not-for-mail From: Paul Taylor Newsgroups: gmane.science.mathematics.categories Subject: functions not polynomials Date: Wed, 21 Mar 2007 16:39:55 +0000 Message-ID: NNTP-Posting-Host: main.gmane.org X-Trace: ger.gmane.org 1241019473 9897 80.91.229.2 (29 Apr 2009 15:37:53 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:37:53 +0000 (UTC) To: categories@mta.ca Original-X-From: rrosebru@mta.ca Wed Mar 21 19:03:15 2007 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Wed, 21 Mar 2007 19:03:15 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1HU8m1-0003fR-LL for categories-list@mta.ca; Wed, 21 Mar 2007 18:55:13 -0300 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 59 Original-Lines: 51 Xref: news.gmane.org gmane.science.mathematics.categories:3705 Archived-At: Let f: Z x Z -> Z be a binary FUNCTION (in the sense of sets) on the integers, with the property that - for each x:Z, f(x,-) : Z -> Z is a (agrees with a unique) POLYNOMIAL, whose coefficients are functions of x; and similarly - for each y:Z, f(-,y) : Z -> Z is also a polynomial. Then f(x,y) was itself a polynomial in two variables. "You just use Newton's difference method to find the co..." .... unterexample. Taking N for simplicity first, consider the binomial coefficient (x,n) as an nth degree polynomial in x, ie (x,0) = 1 (x,1) = x (x,2) = x(x-1)/2 (x,3) = x(x-1)(x-2)/3! and so on. Newton's finite difference method provides the coefficients of these generating polynomials by taking successive differences (the finite analogue of successive derivatives). But then sum_n (x,n)(y,n) is a function NxN->N that is a polynomial in x for each y and vice versa but isn't itself a polynomial. None of the web pages that I found about this method was especially clear, so I'm not going to recommend any, but they are there if you want to look for them yourself. The more general method (as described on these pages, which is why I didn't think they were that good) takes differences from any sequence of points, not just 0, 1, 2, .... Hence, given any enumeration of a countable field K (this probably works for Z too) we can generalise both the method of fitting polynomials and the counterexample. Of course, since I'm a categorist, it was the categorical formulation that interests me. Any thoughts on that would be appreciated, despite the above failure of the simple example. In fact, I'd quite like to hear from any students of universal algebra out there who also know how monads encode algebraic theories, and might be willing to help as sparring partners for my present mad scheme. Paul Taylor