Re: [math-fun] Atrocious Mma 8&9, WolframAlpha bug:
DanA> 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 Why settle for almost all? Using z! := Gamma[z+1], for the 3rd time, _nC_m := (n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] covers *all* the cases. Picture: Overlay on the symmetrical surface x!/(y!*(x-y)!) a symmetrical grid of arrows pointing (unsymmetrically) leftward at each gridpoint, indicating that if we always approach from the *right*, we always get the *right* answer. JPropp> 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 OK, that just says we can't define C(m,n) for all integers m and n so that C(m,n)=C(m,m-n). Good! (I'm surprised Ira, a CAS guy, would believe C(m,n)=C(m,m-n) for all m and n. Knuth Vol 1 must clarify this.) What does Maple say? Fred> 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 . rwg> Huh? That limit I keep repeating seems to do it all. Fred> 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 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)). The radiation symbol is more toxic than radiation. This is (n choose m): 1 1 -5 1 -4 1 -3 6 1 -2 3 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 This biwedge extension is typical of many number triangles, including Stirling and Eulerian. --rwg
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? Fred Lunnon On 6/26/13, Bill Gosper <billgosper@gmail.com> wrote:
DanA>
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
Why settle for almost all? Using z! := Gamma[z+1], for the 3rd time,
_nC_m := (n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n]
covers *all* the cases.
Picture: Overlay on the symmetrical surface x!/(y!*(x-y)!) a symmetrical grid of arrows pointing (unsymmetrically) leftward at each gridpoint, indicating that if we always approach from the *right*, we always get the *right* answer.
JPropp>
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
OK, that just says we can't define C(m,n) for all integers m and n so that C(m,n)=C(m,m-n). Good! (I'm surprised Ira, a CAS guy, would believe C(m,n)=C(m,m-n) for all m and n. Knuth Vol 1 must clarify this.) What does Maple say?
Fred> 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 .
rwg> Huh? That limit I keep repeating seems to do it all.
Fred> 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
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)).
The radiation symbol is more toxic than radiation. This is (n choose m):
1
1 -5
1 -4
1 -3 6
1 -2 3
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
This biwedge extension is typical of many number triangles, including Stirling and Eulerian. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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?
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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By sheer coincidence I just stumbled upon http://arxiv.org/abs/1105.3689 Can anyone comment on validity of the statements given? Best, jj * Fred lunnon <fred.lunnon@gmail.com> [Jun 27. 2013 07:15]:
[...]
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
[...]
Seconds out! WFL On 6/27/13, Joerg Arndt <arndt@jjj.de> wrote:
By sheer coincidence I just stumbled upon http://arxiv.org/abs/1105.3689 Can anyone comment on validity of the statements given?
Best, jj
* Fred lunnon <fred.lunnon@gmail.com> [Jun 27. 2013 07:15]:
[...]
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
[...]
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[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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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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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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participants (4)
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Bill Gosper -
Dan Asimov -
Fred lunnon -
Joerg Arndt