Initially this confused me as well, until I remembered that we're discussing a function of not one complex variable but two real variables. Just to kick this firmly into the long grass, inspect the values of x_C_y along a small circle around x,y = -1,-1 , as computed using Maple's Gamma function: bifurc := proc(x, y) evalf(GAMMA(x+1)/GAMMA(y+1)/GAMMA(x-y+1)) end; n,m := -1,-1; # lattice arguments eps := 0.00001; k := 17; # radius & steps [seq(bifurc(n + eps*cos(2*Pi*i/k), m + eps*sin(2*Pi*i/k)), i = 1..k)]; binomial(n, m); # tri-wedge! [.3874020639, .9116204527, 2.008270726, 10.79171867, -3.514635436, -1.324214008, -.6191737859, -.1869323972, .1869323973, .6191737859, 1.324214009, 3.514635440, -10.79171867, -2.008270725, -.9116204528, -.3874020637, 0.+0.*I] 1 Returning to examples of tri-wedge convention, here is a classic example actually dubbed in some sources the "hockey stick" identity, and given traditionally in the form for 0 <= k <= n {n\choose k} = \sum_{j = 1}^{n-k+1} {n-j\choose k-1} ; when k = 0 , this relies on defining {-1}_C_{-1} = 1 . But under the bi-wedge convention --- which I am reluctantly coming round to finding increasingly compelling --- the hockey stick has to give way to the "broken hockey stick" (bifurcated, even?) {n\choose k} = \sum_{j = 1}^{n-k+1} {n-j\choose n-j-k+1} ; not nearly so memorable, sadly. Fred Lunnon On 6/27/13, Dan Asimov <dasimov@earthlink.net> 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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