🚀 go-pugleaf

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. The fact that f(0) = 1/a does *not* prove that 0
is not an essential singularity at 0. 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.

>> PS.  I know all the standard proofs e.g., by Liouville's Thm.
>>       But no proof is as gorgeous as the above. [...]

>That's because those are right, the above is wrong.

>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.

Daryl McCullough
ORA Corp.
Ithaca, NY