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The correct link to Alain Connes paper is http://www.alainconnes.org/docs/Companion.pdfSun, 01 Nov 2015 17:38:50 +0000
Does anyone know if it's still possible to buy one of these?Tue, 13 Mar 2012 18:21:16 +0000
But how do you know that the vectors v, tv, t2v, t3v, ...., tnv are all distinct to begin with?Fri, 06 Jan 2012 01:30:26 +0000
[...] Those of us involved in mathematics education often whinge about … [...]Wed, 13 Apr 2011 12:07:16 +0000
As I see it, the set of residues of integers mod p^k, k a positive integer, meet the wikipedia criteria to form a commutative ring under residue addition and multiplication mod p^k. I want to know if this set is a field. For a ring to be a field, Wikipedia defines the requirement that every non zero element be invertible. In this set, every element that is not a factor of zero is invertible, so it is slightly different from the wikipedia requirement.Thu, 27 Jan 2011 17:15:26 +0000
Forget that question Steve. It looks like the binomial theorem for (a+b)^p will handle any type of value ( that I can think of) for a and b, including (cosx+sinx). I am thefore taking it that I can create my factorisation when a+b add to zero.Thu, 27 Jan 2011 13:50:15 +0000
After a marriage and a major house refurbishment I'm back to finish off my problem. Having claimed victory over the devil, I find myself amusingly playing devil's advocate to try and find a flaw in my paper. The binomial theorem works for negative integers. Does it work when one of the integers is 0? My problem is to analyse the sum of two pth powers of integers, p an odd prime. We will do it with residues of these integers modulo p^k. The integers :a:b: are prime to p, but their sum has a factor p, and possibly a facot with higher powers of p. One stratagem generally employed is to express the sum as ((a+b)-b)^p +b^p and expand out the first term with the binomial theorem. The b^p terms cancel out leaving a factored expression (a+b)() and you prove the () factor has a single factor p. Now I can prove what a and b must be for all the various combinations, but my factorisation is of the form (a+b)() where () does not have the factor p. The devil points this out, but I reply I can take my solution :a:b: mod p^k and exponentiate it using the stratagem and the result will be a residue expression mod p^k of (a+b)p(). Is this also valid if a+b=0 mod p^k, i.e. the stratagem forms (0)p() mod p^k? If not, I need to add another section.Fri, 23 Apr 2010 00:14:42 +0000
What about if you want to display polynomial long division in modular.., for example, I am trying to long divide a polynomial in mod7Thu, 22 Apr 2010 15:51:34 +0000
Ignore the last paragraph above. What remains is my argument. (a+b) can only be zero. I don't want to get into Dr. Faustus and the devil, but in dealing with the devil in mathematics he has two very powerful weapons, infinity and zero. I managed to control infinity, but zero almost beat me. Thanks for being able to bounce ideas off you. Over and out.Fri, 16 Apr 2010 07:18:17 +0000
Needless to say I cannot say a+b is zero mod p^3, so that is not the answer. My last effort is as follows. I form a+b residues mod p^2 on a 1:1 basis from other residues mod p^2. I end up with 10 say mod p^2. 5 of them don't add to zero and can be rejected. The other 5 add to zero and are formed by adding two residues mod p^2. I can factor these as p.(a+b) where a+b is one of the rejected a+b pairs. Since I am not factorising an integer, but a residues sum, I contend that the (a+b) factor in p.(a+b) is a residue mod p. I cannot make it an (a+b) residue mod p^2, and therefore I maintain my 1:1 relationship with a+b mod p^2.