User talk:Magidin

From Wikipedia, the free encyclopedia

Contents

[edit] Quotient field or Field of fractions

I've proposed on Talk:Quotient field to rename the page. I'd be very grateful to hear your comments there on what you think - thanks! — ciphergoth 00:07, 26 February 2006 (UTC)

[edit] Mexican mathematician bios

Hi Arturo,

I see you did your undergrad at UNAM. I was wondering if you might want to contribute any articles (or even stubs) on Mexican mathematicians. It's a little embarrassing that there are only two articles in category:Mexican mathematicians (of course there could be some existing articles that just need categorization, but I could only find one, which is now one of the two). --Trovatore 18:06, 11 April 2006 (UTC)

[edit] Kaplansky

Do you happen to have any citation for Irving Kaplansky's death (I noticed your edit)? I tried calling the UC math dept, for a link to an online obituary, but no one answered. Maybe I should try MSRI. I assume it's true. If it isn't, this is REALY embarrasing. I note that his good friend George Mackey died a few months earlier. Some friends of Mackey who were also friends of the Kaplansky's had not heard the news of Kaplansky's death. --CSTAR 17:46, 28 June 2006 (UTC)

[edit] Proofs of Fermat's theorem on sums of two squares

I am translating your Euler's proof into french. The first four steps are OK for me, but I have difficulties with the last one. I don't see why Since the kth differences of the sequence 1^k, 2^k, 3^k,\dots are all equal to k!. Would you be kind enough to help me a little bit? Thanks Jean-Luc W a french contributor with no english account.

This is essentially elementary algebra. We want to prove that the nth differences of xn are all equal to n!. If you start with a polynomial of degree n, and consider the first difference then it is easy to see that the result is a polynomial in k of degree n − 1. So if we start with f(x) = xn, the first difference is a polynomial of degree n − 1, the second of degree n − 2, etc., and the nth difference will therefore be a constant polynomial (degree 0); the next difference will therefore have to be 0. Now start with f(x) = xk. The first difference has leading term equal to kxk − 1. The leading term of the second difference is k(k − 1)xk − 2. The leading term of the third difference is k(k-1)(k-2)x^{k-3},\ldots. The leading term for the (k − 1)st difference is k(k-1)(k-2)\cdots (2)x, and the leading term of the kth difference is just k(k-1)\cdots(2)(1)=k!. Since the kth differences are constant, all these differences are equal to k!. Magidin 21:06, 16 October 2006 (UTC)

PS: I think there are two slight lack of precision in step 4. In case you are interested to know, I will be happy to communicate about.81.249.39.156 19:58, 16 October 2006 (UTC)

I took the presentation from Harold M. Edwards Fermat's Last Theorem. A Genetic Introduction to Algebraic Number Theory, GTM 50, Springer Verlag, pp. 46-48. It is possible I missed something.Magidin 21:06, 16 October 2006 (UTC)
Great, I have got the point, thank's a lot. 81.249.72.33 10:34, 17 October 2006 (UTC) Jean-Luc W

[edit] Abel-Ruffini theorem

Greetings, I think we met on scimath and discussed S6 etc. Glad we straightened the misunderstanding out and parted well! I went to your site later and read up on capability which was fascinating. Right now I'm worried about the correctness of the proof in the Abel-Ruffini theorem article as of 12/17/06 and hoping you have the time and inclination to look it over.Regards,Rich 15:05, 17 December 2006 (UTC)

[edit] Deletion (2x) of link to short & direct FLT proof

I would appreciate it if you can give an explanation for you twice deleting a link to a published short and direct FLT proof (nov.2005 Acta Mathematica Univ. Bratislava: http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html pp 169-184). I hope it is not considered 'sacriledge' to add that link - other people than yourself may be interested. So unless you find fault with the proof, in which case I would like to learn about it, I would appreciate you leaving that link alone for others to access (I'll put it in the 'eternal link' section now. Thank you. Benschop 17 January 2007

Retrieved from "http://en.wikipedia.org../../../m/a/g/User_talk%7EMagidin_404f.html"