home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.physics
- Path: sparky!uunet!mcsun!sun4nl!relay.philips.nl!prle!dacosta
- From: dacosta@prl.philips.nl (Paulo da Costa 42147)
- Subject: Re: Abian and the Fundamental Theorem of Algebra
- Message-ID: <1992Nov22.131202.16201@prl.philips.nl>
- Sender: news@prl.philips.nl (USENET News System)
- Organization: none
- References: <1992Nov19.161552.1055@oracorp.com>
- Date: Sun, 22 Nov 1992 13:12:02 GMT
- Lines: 71
-
- In article <1992Nov19.161552.1055@oracorp.com> daryl@oracorp.com (Daryl McCullough) writes:
- >dacosta@prl.philips.nl (Paulo da Costa 42147) 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).
- >
- >Look, I am certainly not going to defend Abian's physics, but when it
- >comes to algebra and complex analysis, he knows what he is talking
- >about, and you don't. [...]
-
- Yes, I do. Maybe I went a little too fast.
-
- [...] The fact that f(0) = 1/a does *not* prove that 0
- >is not an essential singularity at 0. [...]
-
- It does, because f is a rational function, and therefore can only have
- a finite number of isolated singularities (at most n of them,
- corresponding to the zeros of the denominator). This does not depend on
- the FTA or anything of the sort, just on the fact that f is
- differentiable at any point where its denominator doesn't vanish (I
- trust that you know how to calculate the derivative of the inverse of a
- polynomial...). If f(0) is defined _at all_, then f is also
- differentiable (and therefore continuous) at that point, whence my
- assertion about the crappiness of his talk.
-
- [...]For example, consider the
- >function:
- >
- > f(z) = z sin(1/z) (for z nonzero)
- > f(0) = 0
- >
- >f has an essential singularity at 0, and it is also equal to 0 there.
- >To say that f has an essential singularity at a point z is a statement
- >about what happens to f in the *neighborhood* of the point, not *at*
- >the point.
-
- Correct, but irrelevant (sorry).
-
- >>What a crackpot.
- >
- >I don't like to call people crackpots; I prefer simply to call them
- >mistaken. In my opinion, Abian is mistaken about physics, and you
- >are mistaken about complex analysis.
-
- I hope you now see that's not the case. And I _do_ call someone a
- crackpot when he thinks he has found "gorgeous" solutions that went
- overlooked for centuries and goes on pushing them no matter how many
- holes people point out in them.
-
- >Daryl McCullough
- >ORA Corp.
- >Ithaca, NY
-
- --
- -- Paulo da Costa /\/\ Minha terra tem palmeiras /\/\
- -- dacosta@prl.philips.nl \/\/ Onde canta o sabia'... \/\/
-