home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.physics
- Path: sparky!uunet!gumby!destroyer!news.iastate.edu!pv343f.vincent.iastate.edu!abian
- From: abian@iastate.edu (Alexander Abian)
- Subject: Re: TIME HAS INERTIA. ABIAN's PROOF OF THE FUND. THM. OF ALGEBRA
- Message-ID: <abian.722124717@pv343f.vincent.iastate.edu>
- Keywords: CORRECTION FUNDAMENTAL THM. OF ALGEBRA
- Sender: news@news.iastate.edu (USENET News System)
- Organization: Iowa State University, Ames IA
- References: <abian.721281610@pv343f.vincent.iastate.edu> <abian.721303213@pv343f.vincent.iastate.edu>
- Date: Wed, 18 Nov 1992 22:11:57 GMT
- Lines: 87
-
- In <abian.721303213@pv343f.vincent.iastate.edu> abian@iastate.edu (Alexander Abian) writes:
-
- > CORRECTION
- >
- > PLEASE SUBSTITUTE THIS FOR PREVIOUS VERSION
-
- >Dear Mr. C.G
- >
- > You asked me to give a proof of the Fundamental Theorem of
- >Algebra by e-mail. I post only in TIME HAS INERTIA. Here is my own
- >proof (I am almost sure that I have published it somewhere).
- >
-
- > THEOREM. Every nonconstant polynomial over complex numbers has a
- > root.
-
- > PROOF. Assume on the contrary that a + bz +...+ kz^n with nonzero k,
- >
- >has no root, i.e., never vanishes. Thus, REWRITING THE POLYNOMIAL
-
- >IN DESCENDING POWERS kz^n + ... + bz + a it follows that the
-
- >quotient (since by our assumption never vanishes)
-
- >(1) 1 / (kz^n + ...+ bz + a)
- >
- >has a derivative at every point of the z-plane.
- >
- >Hence, in particular,
-
- >(2) 0 is an analytic point of 1/(kz^n + ... + bz + a), and there-
-
- >fore the latter has a Laurent expansion in powers of z valid in the entire
- >plane (since we are over complex numbers - this is not true over real
- >numbers).
- >
- > However, the Laurent expansion can be obtained by LONG DIVISION of (1).
- >
- >Performing the long division of (1), we obtain
-
- >(3) 1/(kz^n + ...+ bz + a) = (1/k) z^(-n) + ........
- >
- >But (3) shows that 0 is an essential isolated singularity or a pole of
- >
- > 1/(kz^n + ...+ bz + a)
- >
- >contradicting (2). Hence our assumption is false and the Theorem is
- >proved.
-
- >P.S. Note that the proof will fail if instead of a polynomial we con-
- > sider an entire function (i.e., a power series, i.e., so to speak
- >"an infinite polynomial"). The reason is that one cannot rewrite the
- >power series in descending order. Indeed, there are nonvanishing power
- >series such as the power series of e^z.
- > With best regards, Alexander ABIAN
- >
-
- >P.S. I typed the previous version and I thought that I pressed "i"
- > to check the spelling and the mistakes. Apparently I pressed
- > "s" because I could no longer edit the text.
- > Also, I know all the standard proofs e.g., by Liouville's Thm.
-
-
-
-
-
-
-
-
-
-
-
- >>--
- >> The tendency of maintaining the status-quo, Reaction to provocation and
- >> The tendency of maintaining again a status-quo.
- >> TIME HAS INERTIA and some energy is lost to move Time forward
- >> E = mcc (Einstein) must be replaced by E = m(0) exp(-At) (Abian)
- >--
- > The tendency of maintaining the status-quo, Reaction to provocation and
- > The tendency of maintaining again a status-quo.
- > TIME HAS INERTIA and some energy is lost to move Time forward
- > E = mcc (Einstein) must be replaced by E = m(0) exp(-At) (Abian)
- --
- The tendency of maintaining the status-quo, Reaction to provocation and
- The tendency of maintaining again a status-quo.
- TIME HAS INERTIA and some energy is lost to move Time forward
- E = mcc (Einstein) must be replaced by E = m(0) exp(-At) (Abian)
-