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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: <1992Nov20.100104.17600@prl.philips.nl>
- Keywords: CORRECTION FUNDAMENTAL THM. OF ALGEBRA
- Sender: news@prl.philips.nl (USENET News System)
- Organization: none
- References: <abian.722071648@pv343f.vincent.iastate.edu> <1992Nov18.175601.18259@prl.philips.nl> <1992Nov19.074930.15845@CSD-NewsHost.Stanford.EDU>
- Date: Fri, 20 Nov 1992 10:01:04 GMT
- Lines: 49
-
- In article <1992Nov19.074930.15845@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
- >In article <1992Nov18.175601.18259@prl.philips.nl> dacosta@prl.philips.nl (Paulo da Costa 42147) writes:
- >>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.
- >
- >You've just argued that the step succeeds. The proof assumes the
- >negation of the FTA and works towards a contradiction. As soon as a
- >reality check (or anything else) produces a contradiction then the
- >proof is complete.
-
- No. He got his "contradiction" by making an invalid expansion. If you
- really try to do the "long division" the way he did you'll see that the
- remainder after each step contains steadily growing NEGATIVE powers of
- z, so the expansion is not valid in the vicinity of z=0, where he
- proceeded to use it (the expansion converges for z *outside* a radius
- of convergence that includes all zeros of his polynomial -- to prove
- that it converges _anywhere_, he first has to arrive at his
- contradiction...).
-
- >The gist of Abian's proof is that expansion (3) indicates a singularity
- >where, as both you and Abian point out, there can't be one.
-
- No, the gist of Abian's "proof" is that he introduced a falsehood while
- he was trying to get to a contradiction, after which it becomes really
- easy, doesn't it?
-
- >I was hoping someone would pronounce on the validity of (3), [...]
-
- That's exactly what I did. It's bullshit. Satisfied?
-
- >Vaughan Pratt A fallacy is worth a thousand steps.
-
- I don't think we should go on discussing this is sci.physics.
- --
- -- Paulo da Costa /\/\ Minha terra tem palmeiras /\/\
- -- dacosta@prl.philips.nl \/\/ Onde canta o sabia'... \/\/
-