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- Newsgroups: sci.physics
- Path: sparky!uunet!mcsun!sun4nl!relay.philips.nl!prle!dacosta
- From: dacosta@prl.philips.nl (Paulo da Costa 42147)
- Subject: Re: TIME HAS INERTIA. Att: PRATT FUND. THM. OF ALGEBRA
- Message-ID: <1992Nov18.175601.18259@prl.philips.nl>
- Keywords: CORRECTION FUNDAMENTAL THM. OF ALGEBRA
- Sender: news@prl.philips.nl (USENET News System)
- Organization: none
- References: <abian.721281610@pv343f.vincent.iastate.edu> <abian.721303213@pv343f.vincent.iastate.edu> <abian.722071648@pv343f.vincent.iastate.edu>
- Date: Wed, 18 Nov 1992 17:56:01 GMT
- Lines: 28
-
- In article <abian.722071648@pv343f.vincent.iastate.edu> abian@iastate.edu (Alexander Abian) writes:
- >In <abian.721303213@pv343f.vincent.iastate.edu> abian@iastate.edu (Alexander Abian) writes:
- [...]
- >>(3) 1/(kz^n + ...+ bz + a) = (1/k) z^(-n) + ........
- >>
- >>But (3) shows that 0 is an essential isolated singularity of
- >>
- >> 1/(kz^n + ...+ bz + a)
- >>
- >>contradicting (2). Hence our assumption is false and the Theorem is
- >>proved.
-
- This step fails miserably. Reality check: You must have a<>0 (otherwise
- your polynomial would have a root at z=0, contrary to your assumption).
- In this case, the inverse of the polynomial at z=0 is just 1/a. Your
- talk of "essential isolated singularity" is total crap, the function is
- even CONTINUOUS there (i.e., your expansion under (3) is plain wrong --
- "long division" is not what you did there).
-
- >> PS. I know all the standard proofs e.g., by Liouville's Thm.
- > But no proof is as gorgeous as the above. [...]
-
- That's because those are right, the above is wrong.
-
- What a crackpot.
- --
- -- Paulo da Costa /\/\ Minha terra tem palmeiras /\/\
- -- dacosta@prl.philips.nl \/\/ Onde canta o sabia'... \/\/
-