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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