But I'm out of my depth in the world of holomorphic functions of two variables
See figures 3--6 in my note at https://www.dropbox.com/s/9vvl6l6ym1zkret/binomial.pdf showing various 3-D views of this ingeniously complicated function. I was inordinately proud of those 1/Beta plots, after RWG confessed to having been defeated by the challenge of producing them. My self-satisfaction was soon punctured on discovering that David Fowler had beaten me to it by 20-odd years, using vastly more primitive tools --- David Fowler The Binomial Coefficient Function The American Mathematical Monthly, 103, (1996), 1--17 <http://www.jstor.org/stable/2975209>. At least we both managed to produce essentially the same picture! [ HB --- what was appending " ...?dl=1 " intended to achieve --- FireFox downloads the file instead of opening in browser? ] WFL On 7/24/17, Dan Asimov <dasimov@earthlink.net> wrote:
P.S.
----- The equation is in the world of meromorphic functions. -----
Well, that was the intention, anyway. But I'm out of my depth in the world of holomorphic functions of two variables
Sorry, I was using a quaint notation common in complex variables where f(z,w) denotes the function f itself (and not in this case the value of f at the point (z,w)). The equation is in the world of meromorphic functions.
—Dan
From: Fred Lunnon <fred.lunnon@gmail.com>
f(z,w) = 1 / beta(w+1, z-w+1) (and by symmetry = 1 / beta(z-w+1, w+1) )
unless, of course, the denominator happens to vanish ... WFL
On 7/24/17, Dan Asimov <dasimov@earthlink.net> wrote: Needless to say: If f(z,w) is defined as
f(z,w) = gamma(z+1) / (gamma(w+1) * gamma(z-w+1))
then it's a ratio of meromorphic functions, therefore meromorphic.
But of course since
beta(z,w) = gamma(z) gamma(w) / (gamma(z+w)
then we have
f(z,w) = 1 / beta(w+1, z-w+1)
(and by symmetry
= 1 / beta(z-w+1, w+1) )
—Dan
----- Why stop at two? There are a denumerable number of singularities of
Gamma(n+1) / Gamma(k+1) Gamma(n-k+1) at negative integer n , and once you decide to start tinkering with approaching them from multiple randomly selected directions, a whole continuum of choices opens before you --- though admittedly not all of them yield integer results.
[ And bear in mind, incidentally, that you will have royally screwed up your symbolic theorem prover --- see examples from Mathematica and Maple in my note. ]
WFL
On 7/23/17, James Propp <jamespropp@gmail.com> wrote:
Maybe n-choose-k should actually be two-valued (like sqrt(z)), with a branch-point at n=k=0?
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