[math-fun] Atrocious Mma 8&9, WolframAlpha bug:
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.
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Groan: earlier should have read << ... n_C_m = 0 for m < 0 . But if you define them (less naïvely) via n_C_m = 1/( (n+1)*Beta[m+1, n-m+1] ) ... >> On 6/25/13, James Propp <jamespropp@gmail.com> wrote:
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
Not sure what Jim had in mind here; maybe connected with the fact that --- choosing the nuclear option --- for m > 0, n_C_{-m} = (-1)^n times degree-(m-1) polynomial in n . For example the line m = -2 looks at a casual glance like a misprinted ..., -3, 2, -1, 0, 1, -2, 3, ... instead of Bill's (correct) diagram. If there were a winding point near the origin, it would imply that the Beta function had a corresponding essential singularity --- as far as I know, it has only poles --- Bill? Fred Lunnon
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
_______________________________________________ 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
participants (3)
-
Bill Gosper -
Fred lunnon -
James Propp