Re: [math-fun] Atrocious Mma 8&9, WolframAlpha bug:
JimP>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 For simplicity I was avoiding Z^2 and cowering in R^2, but I can't for the life of me imagine a branch point at (0 choose 0). Can you expound this sign change? 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 Sorry to bifurcate on your carpet (or a least leave crumbs), but I *can* have my cake and eat it too. (n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] The surface is clearly symmetrical under y -> x-y. The asymmetry comes from the order of the limitry. (x choose y) "goes vertical" at negative integer x, integer y, and can take on any value depending on the direction of approach. This is why it's so hard to intelligibly plot. One way to think of the negative x realm is adjacent, diagonal strips of curly-edged lasagna, uncooked. Except the curls have infinite amplitude at the edges. Now, slide alternate strips half a ruffle, and "glue" each upward-going vertical edge to the downward- going vertical edge of the adjacent strip. Now that's a surface. --rwg Noninteger arguments are totally piece of cake. 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. _______________________________________________
I understand now what Bill is driving at. The conclusion seems to be that the Gamma (or Beta) function definition is essentially useless for defining n_C_m for integers m, n < 0 . It looks like fans of the radiation symbol option need to come up with some convincing alternative justification for taking (*) as an axiom, to extend to the region m < 0, n < 0 . At any rate when (n,m) = (-1,-1) , there are indeed situations where the value 1 gives a consistent combinatorial result; I'm not sure though whether there may be others where 0 is preferable. Fred Lunnon On 6/26/13, Bill Gosper <billgosper@gmail.com> wrote:
JimP>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
For simplicity I was avoiding Z^2 and cowering in R^2, but I can't for the life of me imagine a branch point at (0 choose 0). Can you expound this sign change?
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
Sorry to bifurcate on your carpet (or a least leave crumbs), but I *can* have my cake and eat it too.
(n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n]
The surface is clearly symmetrical under y -> x-y. The asymmetry comes from the order of the limitry. (x choose y) "goes vertical" at negative integer x, integer y, and can take on any value depending on the direction of approach. This is why it's so hard to intelligibly plot. One way to think of the negative x realm is adjacent, diagonal strips of curly-edged lasagna, uncooked. Except the curls have infinite amplitude at the edges. Now, slide alternate strips half a ruffle, and "glue" each upward-going vertical edge to the downward- going vertical edge of the adjacent strip. Now that's a surface. --rwg Noninteger arguments are totally piece of cake.
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
Bill Gosper asks "Can you expound this sign change?" I can't remember the full story, but here's a summary of part of it (which I learned about from Ira Gessel): Assume that we can define C(m,n) for all integers m and n so that C(m,n)=(-1)^n C(-m+n-1,n) and C(m,n)=C(m,m-n) for all m and n. Then applying these two transformations alternately three times we get C(m,n) = -C(m,n): C(m,n) = (-1)^n C(-m+n-1,n) = (-1)^n C(-m+n-1,-m-1) = (-1)^(m+n+1) C(-n-1,-m-1)= (-1)^(m+n+1) C(-n-1,-n+m) = -C(m,-n+m) = -C(m,n). Jim Propp On Tue, Jun 25, 2013 at 10:42 PM, Bill Gosper <billgosper@gmail.com> wrote:
JimP>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
For simplicity I was avoiding Z^2 and cowering in R^2, but I can't for the life of me imagine a branch point at (0 choose 0). Can you expound this sign change?
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
Sorry to bifurcate on your carpet (or a least leave crumbs), but I *can* have my cake and eat it too.
(n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n]
The surface is clearly symmetrical under y -> x-y. The asymmetry comes from the order of the limitry. (x choose y) "goes vertical" at negative integer x, integer y, and can take on any value depending on the direction of approach. This is why it's so hard to intelligibly plot. One way to think of the negative x realm is adjacent, diagonal strips of curly-edged lasagna, uncooked. Except the curls have infinite amplitude at the edges. Now, slide alternate strips half a ruffle, and "glue" each upward-going vertical edge to the downward- going vertical edge of the adjacent strip. Now that's a surface. --rwg Noninteger arguments are totally piece of cake.
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
As I see it, there's only one natural way to extend the binomial symbol n_C_k to almost all reals (and also complexes), and that is via the gamma function 𐅃(z), since for positive integers z we have from 𐅃(z) = (z-1)! that (*) n_C_k = 𐅃(n+1) / (𐅃(k+1) 𐅃(n-k+1)) This definition would imply that 0_C_0 is equal to 1, and also that for integer n and k then for either a) n >= 0 AND (k <= -1 OR k >= n+1) we have n_C_k = 0; OR b) n <= -1 AND (k <= -1 AND k >= n+1), --or more simply-- n+1 <= k <= -1 we have n_C_k = 0. (Since, the two simple poles in the denominator force the quotient to equal 0, despite the simple pole in the numerator.) Since the gamma function never takes the value 0 and has poles only at the nonpositive integers, a) and b) should cover all cases where the definition (*) of n_C_k = 0. --Dan On 2013-06-26, at 8:53 AM, James Propp wrote:
Bill Gosper asks "Can you expound this sign change?"
I can't remember the full story, but here's a summary of part of it (which I learned about from Ira Gessel):
Assume that we can define C(m,n) for all integers m and n so that C(m,n)=(-1)^n C(-m+n-1,n) and C(m,n)=C(m,m-n) for all m and n. Then applying these two transformations alternately three times we get C(m,n) = -C(m,n):
C(m,n) = (-1)^n C(-m+n-1,n) = (-1)^n C(-m+n-1,-m-1) = (-1)^(m+n+1) C(-n-1,-m-1)= (-1)^(m+n+1) C(-n-1,-n+m) = -C(m,-n+m) = -C(m,n).
Jim Propp
On Tue, Jun 25, 2013 at 10:42 PM, Bill Gosper <billgosper@gmail.com> wrote:
JimP>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
For simplicity I was avoiding Z^2 and cowering in R^2, but I can't for the life of me imagine a branch point at (0 choose 0). Can you expound this sign change?
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
Sorry to bifurcate on your carpet (or a least leave crumbs), but I *can* have my cake and eat it too.
(n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n]
The surface is clearly symmetrical under y -> x-y. The asymmetry comes from the order of the limitry. (x choose y) "goes vertical" at negative integer x, integer y, and can take on any value depending on the direction of approach. This is why it's so hard to intelligibly plot. One way to think of the negative x realm is adjacent, diagonal strips of curly-edged lasagna, uncooked. Except the curls have infinite amplitude at the edges. Now, slide alternate strips half a ruffle, and "glue" each upward-going vertical edge to the downward- going vertical edge of the adjacent strip. Now that's a surface. --rwg Noninteger arguments are totally piece of cake.
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Big mistake. Never mind. Red-facedly, Dan On 2013-06-26, at 12:38 PM, Dan Asimov wrote:
As I see it, there's only one natural way to extend the binomial symbol n_C_k to almost all reals (and also complexes), and that is via the gamma function 𐅃(z), since for positive integers z we have from 𐅃(z) = (z-1)! that
(*) n_C_k = 𐅃(n+1) / (𐅃(k+1) 𐅃(n-k+1))
This definition would imply that 0_C_0 is equal to 1, and also that for integer n and k then for either
a) n >= 0 AND (k <= -1 OR k >= n+1)
we have n_C_k = 0;
OR
b) n <= -1 AND (k <= -1 AND k >= n+1),
--or more simply--
n+1 <= k <= -1
we have n_C_k = 0.
(Since, the two simple poles in the denominator force the quotient to equal 0, despite the simple pole in the numerator.)
Since the gamma function never takes the value 0 and has poles only at the nonpositive integers, a) and b) should cover all cases where the definition (*) of n_C_k = 0.
participants (4)
-
Bill Gosper -
Dan Asimov -
Fred lunnon -
James Propp