🚀 go-pugleaf

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)