If you make a trip through the six sectors of Z^2 that surround n=k=0, extrapolating analytically (in the discrete sense of the term), you'll find that the binomial coefficient function changes sign. So there's a sense in which (n choose k) is best thought of as living not in Z^2 but a double-cover thereof, with a branch point at (0 choose 0). I have been puzzled by this for a long time. Jim Propp On Tue, Jun 25, 2013 at 3:59 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
There does seem to be a genuine ambiguity involved in extending binomial coefficients to negative arguments: if they are to be Taylor coefficients in the expansion of (1+x)^n , then plainly it follows that n_C_k = 0 for k < 0 . But if you define them (less naïvely) via n_C_k = 1/( (n+1)*Beta[m+1, n-m+1] ) it is evident from (say) the plot at http://mathworld.wolfram.com/BetaFunction.html that the "radiation warning" and bilateral symmetry are indeed correct.
Then it's not the symmetry which breaks down: Binomial[n, m] - Binomial[n, n-m] = 0 , for all integer n,m but the recursion: Binomial[n, m] - Binomial[n-1, m] - Binomial[n-1, m-1] = 0 _except_ at m = n = 0, where it equals -1 .
I haven't investigated the situation for non-integer arguments. But below Bill --- along with the CAS's --- seems to be trying to have his cake and eat it [which might help explain why he's having trouble plotting it. Mind the carpet!] Either both triangle and surface are symmetric, or neither is; and either you define 'em via Taylor series, or via Gamma functions --- but not both at once.
Fred Lunnon
On 6/25/13, Bill Gosper <billgosper@gmail.com> wrote:
In[509]:= Table[Binomial[-1, k], {k, -3, 3}]
Out[509]= {1, -1, 1, 1, -1, 1, -1}
But In[550]:= Table[Binomial[m1, k], {k, -3, 3}]
Out[550]= {0, 0, 0, 1, m1, 1/2 (-1 + m1) m1, 1/6 (-2 + m1) (-1 + m1) m1}
In[551]:= % /. m1 -> -1
Out[551]= {0, 0, 0, 1, -1, 1, -1}
Macsyma, since 40 years, (c1) makelist(binom(-1,k),k,-3,3) (d1) [0, 0, 0, 1, - 1, 1, - 1]
I expect as much from Maple.
Mma Doc: "As a bivariate function, Binomial is not continuous in both variables at negative integers:
In[1]:= Binomial[-3, -5]
Out[1]= 0
In[2]:= Binomial[-3.0000001, -5.0000001]
Out[2]= 6. "
But in inconvenient fact, In[552]:= Binomial[-3, -5]
Out[552]= 6
I don't see how this eluded their test suite unless the latter is corrupted.
Mma Doc: "In general, (n m) is defined by \[CapitalGamma] (n+1)/(\[CapitalGamma](m+1)\[CapitalGamma](n-m+1)) or suitable limits of this."
For the record, these limits are Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] This is how the bilateral symmetry of Pascal's triangle breaks down, even though the surface x!/(y!*(x-y)!) (appropriately skewed) is bilaterally symmetric. (I have repeatedly failed to make a convincing plot of this.)
For anyone still doubting the symmetry breakdown for negative integer n: The recurrence that generates each row of Pascal's Triangle from the previous would place a "2" at the apex (0,0), if Out[509] were correct! --rwg Here is an almost correct technique for printing number triangles, of which Julian must repeatedly remind me:
pt[n_Integer, from_Integer: 0] := TableForm[Table[If[EvenQ[i + j], "", Binomial[-1 + i, -Floor[n/4] + (1 + i + j)/2] /. 0 -> ""], {i, from, n}, {j, -Ceiling[3*n/4], n + 1}]]
10 -4 1 1 -4
6 -3 1 1 -3 6
-4 3 -2 1 1 -2 3
-1 1 -1 1 1 -1 1 -1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
(Assuming GMail hasn't murdered it,) the upper left wedge is the spurious subject of this message. Pascal's Triangle is really Pascal's biwedge. NOT Pascal's Radiation Warning. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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