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- Newsgroups: sci.physics
- Path: sparky!uunet!cs.utexas.edu!sdd.hp.com!ux1.cso.uiuc.edu!news.iastate.edu!pv343f.vincent.iastate.edu!abian
- From: abian@iastate.edu (Alexander Abian)
- Subject: Re: TIME HAS INERTIA Attn: Mr. PRATT
- Message-ID: <abian.722318103@pv343f.vincent.iastate.edu>
- Keywords: FUNDAMENTAL THM. OF ALGEBRA
- Sender: news@news.iastate.edu (USENET News System)
- Organization: Iowa State University, Ames IA
- References: <1992Nov19.074930.15845@CSD-NewsHost.Stanford.EDU> <1992Nov20.100104.17600@prl.philips.nl> <1992Nov20.190744.6915@meteor.wisc.edu> <1992Nov20.230233.18271@CSD-NewsHost.Stanford.EDU>
- Date: Sat, 21 Nov 1992 03:55:03 GMT
- Lines: 81
-
-
- Dear Mr. Pratt: 11-20-92
-
- You write:
-
- > Now that I look again I don't see how it
- >follows that this sum of singularities need be a singularity. Unless
- >Abian can demonstrate this, I would say *that* was a real flaw in
- >his argument. Mr. Abian?
-
- I have given the demonstration (see the unabridged version that I think
- I specially posted for you). Now, you are asking for more details, e.g. that
- I have to prove that long division in descending powers gives the Laurent
- expansion. I can give that proof (and in my proof no use of FTA is made).
- But then you may ask me for the proof of something else, and so on.
- On the other hand, if you are accepting my wager then please show
- my (unabbridged)proof to a best (and impartial) specialist in the Theory
- of Analytic Functions at Stanford. If he finds my proof erroneous, let
- him point out the mistakes in my proof and then I will pay you $500.
- But if he finds no mistakes in my proof then you pay me $500.
-
- Perhaps before receiving your answer, I should not write the following.
- But I sense that you have a very keen mind and probably you will enjoy
- reading the following. (It is so beautiful that brings tears in my eyes).
-
- Consider the function F given by F(z) = 1/(2-3z+z^2)
-
- (1) Long divide 1 by 2-3z+z^2 (in ascending order) it yields
-
- 1/2 + (3/4)z + (7/8) z^2 + (15/16) z^3 + ...
-
- giving Laurent (in this case Taylor) expansion of F(z) valid for |z|<1
-
-
- (2) Long divide 1 by z^2-3z+2 (in descending order) it yields
-
- 1/(z^2) + 3/(z^3) + 7/(z^4) +...
-
- giving Laurent expansion of F(z) valid for |z| > 2
-
- Now, the dramatic part:
-
- (3) Long divide 1 by -3z+2+z^2 (with leading term -3z) by skillfully
- collating the like powers of z after very skillfully computerized
- programming it will yield (very good approximation) of
-
- -1/(z) -(1/2) -1/(z^2) -z/4 - 1/(z^3) -....
-
- giving Laurent expansion of F(z) valid for 1 < |z| < 2
-
-
- Now, I am cautioning you (3) is a complicated step and requires very
- skillfully computerized programming to collate the like powers of z even
- for an approximate expansion of F(z). Maybe it is not worth the ef-
- fort to do on the computer. But the theory is gorgeous.
-
-
- Now, many variations on the above theme are possible (i) is it
- advantageous in (3) to start with 1 long dividing by -3z+z^2+2 (in
- this order); (ii) how to reshuffle the order in subsequent steps, etc.
-
- Also, obviously this applies to any polynomial (of any degree).
-
- Also the above considerations together with Cauchy -Hadamard
- formula for the value of radius of convergence will be a method
- of estimating the absolute values of roots of any polynomial P(z) from
- the(computerized) values of the coefficients of various powers of z
- in Laurent expansions of P(z) in various annuli.
-
- It is close to midnight and I am very tired and exhausted. There
- may be some errors and typos in the above. But I had to get this out
- of my chest, otherwise I won't be able to sleep, eat , etc.
-
- With best wishes and regards,
-
- Alexander 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)
-