I enquired << Indeed, is there another credible limiting strategy which yields the nuclear option instead? >> Taking the limit along the line x = y , instead of in the order y then x , has an interesting effect: instead of just the lower and right-hand of the three wedges, we get just the lower and left-hand! But no fixed limit direction generates all three wedges. Fred Lunnon On 6/27/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
Why should not similar reasoning apply to w/z at (0, 0) ?
WFL
On 6/27/13, Dan Asimov <dasimov@earthlink.net> wrote:
[Note: Fred posted his comment about two (real) variables before I finished writing this. Dinner intervened.]
P.S. It's more sensible to refer to meromorphic functions of *two* variables, that is, f: C^2 -> S^2.
These are all ratios f(z,w) = g(z,w) / h(z,w) of entire functions g,h: C^2 -> C.
For binomial coefficients, I prefer the more symmetrical definition of
(*) D(u,v) := u! / ((u/2 - v)! (u/2 + v)!)
implying that D(u,v) = D(u,-v).
Hence D(u,v) = C(u, u/2 - v) = C(u, u/2 + v), for C( , ) as below.
Conversely, C(z,w) = D(z, z/2 - w) = D(z, w - z/2).
But in any case, to express C(z,w) as the quotient of two entire functions, we need z = -1 and w = either -1 or 0.
Recall that 1/Γ(z) is entire, and so
(*) C(z,w) = (1/Γ(w+1))*(1/Γ((z-w+1)) / 1/Γ(z+1)
is such a quotient (of entire functions of two variables).
But
Then we should have C(-1,-1) = D(-1, 1/2) = D(-1, -1/2), which again gives
C(-1,-1) = limit as (z,w) -> (-1,-1) of (1/Γ(w+1))*(1/Γ((z-w+1)) / 1/Γ(z+1)
and (-1)! / ((-1)! 0!), and no matter how you take the limit, 1 is the limit.
--Dan
On 2013-06-26, at 7:40 PM, Dan Asimov wrote:
I confess I don't see why limits in a given direction (rather than some other) should make any difference.
The Gamma function is a meromorphic function on C, or in other words a holomorphic function from C -> S^2, the Riemann sphere.
All meromorphic functions on a connected Riemann surface (like C) form a field: They can be uniquely added, subtracted, multiplied, and (excluding the constant function 0) divided.
So defining z_C_w = C(z,w) = (z)! / (w! (z-w)!) makes it a well-defined meromorphic function on C (where of course we define z! to be Γ(z+1)). With this definition, it's clearly always true that C(z,w) = C(z,z-w).
(This is the same as RWG's definition, no?)
For instance, (-1)_C_(-1) would then be defined as C(-1,-1) = (-1)! / ((-1)! 0!) = 1 / 0! = 1.
But there are certainly cases where C(z,w) has a pole for integer arguments, which is as it should be:
For any positive integer z and nonpositive integer w (or nonpositive integer z-w), this definition gives the value of oo for C(z,w).
--Dan
On 2013-06-26, at 6:41 PM, Fred lunnon wrote:
Bill Gosper enquired << Knuth Vol 1 must clarify this. What does Maple say?
Anybody still know how where to find an old-fashioned library with books? I used to know where there were several ...
Maple espouses the nuclear option, at any rate when sticking to specific numerical values. In common with the other playes under discussion, once it has to cope with variable sets, knickers may become seriously twisted --- as I recently had occasion to mention in these hallowed pages.
Magma in contrast prefers biwedge.
<< I'd be interested to see a situation where (-1 Choose -1) := 1 avoids absurdity, barring reckless use of C(m,n)=C(m,m-n)).
I think you may have shot my fox here --- looking more carefully, I now am obliged to admit that my expressions can be recast using symmetry in the undisputed region, so that the value required for consistency becomes {-1}_C_0 = 1 instead.
<< Huh? That limit I keep repeating seems to do it all. >>
Including consistency with the Taylor-series model, agreed. My gripe with it is simply on the grounds of arbitrariness: why take limits in this particular direction, rather than (say) reversing them? Indeed, is there another credible limiting strategy which yields the nuclear option instead?
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