[math-fun] http://en.wikipedia.org/wiki/Steiner_chain n=4
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5, but for n=6, {k1,...,k6} bounded by k0 and k7, k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh? And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). WFL> I can prove this via plane tetracyclic coordinates. The later conjectures can undoubtedly be decided in the same fashion, for small fixed n . Fred Lunnon Yeah, but in the States you need a prescription for that. Wait, you can do those weirdly unsymmetrical n=5's? How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!)
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >> All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis? For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation? WFL On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
RWG:
Wait, you can do those weirdly unsymmetrical n=5's?
In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials).
Yeah, but in the States you need a prescription for that.
In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ... WFL On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >>
All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis?
For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation?
WFL
On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For n = 5 --- assuming their validity, which currently remains unproved --- RWG's experimental quadratic relations constitute a polynomial basis for constraints on Steiner ring curvatures, with dimension 4. Also there are no linear constraints. [Established using Gröbner bases.] The constraints comprise cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 ; k1*k0 + k2*k0 - k3*k0 - k4*k0 + 2*k1*k3 - k2^2 + k1*k5 + k3*k4 + k3*k5 = k1*k6 + k2*k6 - k3*k6 - k4*k6 + 2*k2*k4 - k3^2 + k2*k1 + k4*k5 + k5*k2 ; there are only 5 (not 10) in each orbit, since symmetry causes reflections to be equivalent to shifts. One relation in each orbit is redundant. WFL On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG:
Wait, you can do those weirdly unsymmetrical n=5's?
In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials).
Yeah, but in the States you need a prescription for that.
In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ...
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >>
All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis?
For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation?
WFL
On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Red-faces all round alert: the claims made in my previous post concerning dimension and basis of the constraint set are unconditionally withdrawn. The only useful fact to emerge from my Gröbner explorations is that RWG's longer relation is equivalent to the more elegant and succinct (4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5) = (4*k3^2 - 6*k2*k4 + 2*k5*k1 - k0*k3 + k6*k3) . However, even this minuscule improvement was called into question when it occurred to me to compute the dimension of the relation set via the rank of its Jacobian matrix [[d f_i / d k_j]] , where f_i and k_j range over the 10 relations and 7 variables resp. Disaster! The Jacobian rank is 5 ; but Steiner rings have dimension 3 modulo isometry, and there are only n+2 = 7 variables. Since 3+5 > 7 , I conclude that the relations cannot possibly hold for all Steiner 5-rings. Fred Lunnon On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 5 --- assuming their validity, which currently remains unproved --- RWG's experimental quadratic relations constitute a polynomial basis for constraints on Steiner ring curvatures, with dimension 4. Also there are no linear constraints. [Established using Gröbner bases.]
The constraints comprise cyclic shifts (fixing k0,k6 ) of
k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 ;
k1*k0 + k2*k0 - k3*k0 - k4*k0 + 2*k1*k3 - k2^2 + k1*k5 + k3*k4 + k3*k5 = k1*k6 + k2*k6 - k3*k6 - k4*k6 + 2*k2*k4 - k3^2 + k2*k1 + k4*k5 + k5*k2 ;
there are only 5 (not 10) in each orbit, since symmetry causes reflections to be equivalent to shifts. One relation in each orbit is redundant.
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG:
Wait, you can do those weirdly unsymmetrical n=5's?
In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials).
Yeah, but in the States you need a prescription for that.
In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ...
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >>
All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis?
For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation?
WFL
On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
> If the four curvatures are k1, k2, k3, k4, and the bounding circles > are > k0 and k5, > then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of > signs to k0 and k5. > --rwg > Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Panic over --- modulo similarity, the dimension of curvature sets equals 2 ; and 2+5 = 7, so rank 5 (homogeneous) constraints is what we should expect! Unfortunately, this is insufficient to prove that the RWG relations are a basis --- although they probably are, and likewise there probably are no linear constraints. WFL On 8/11/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Red-faces all round alert: the claims made in my previous post concerning dimension and basis of the constraint set are unconditionally withdrawn. The only useful fact to emerge from my Gröbner explorations is that RWG's longer relation is equivalent to the more elegant and succinct
(4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5) = (4*k3^2 - 6*k2*k4 + 2*k5*k1 - k0*k3 + k6*k3) .
However, even this minuscule improvement was called into question when it occurred to me to compute the dimension of the relation set via the rank of its Jacobian matrix [[d f_i / d k_j]] , where f_i and k_j range over the 10 relations and 7 variables resp.
Disaster! The Jacobian rank is 5 ; but Steiner rings have dimension 3 modulo isometry, and there are only n+2 = 7 variables. Since 3+5 > 7 , I conclude that the relations cannot possibly hold for all Steiner 5-rings.
Fred Lunnon
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 5 --- assuming their validity, which currently remains unproved --- RWG's experimental quadratic relations constitute a polynomial basis for constraints on Steiner ring curvatures, with dimension 4. Also there are no linear constraints. [Established using Gröbner bases.]
The constraints comprise cyclic shifts (fixing k0,k6 ) of
k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 ;
k1*k0 + k2*k0 - k3*k0 - k4*k0 + 2*k1*k3 - k2^2 + k1*k5 + k3*k4 + k3*k5 = k1*k6 + k2*k6 - k3*k6 - k4*k6 + 2*k2*k4 - k3^2 + k2*k1 + k4*k5 + k5*k2 ;
there are only 5 (not 10) in each orbit, since symmetry causes reflections to be equivalent to shifts. One relation in each orbit is redundant.
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG:
Wait, you can do those weirdly unsymmetrical n=5's?
In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials).
Yeah, but in the States you need a prescription for that.
In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ...
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >>
All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis?
For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation?
WFL
On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
> On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> > wrote: > >> If the four curvatures are k1, k2, k3, k4, and the bounding circles >> are >> k0 and k5, >> then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment >> of >> signs to k0 and k5. >> --rwg >> > Found nothing for n=5, >
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7, > k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* > k1+k3+k5=k2+k4+k6. (Four relations.) > --rwg >
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I can prove this via plane tetracyclic coordinates. The later conjectures can undoubtedly be decided in the same fashion, for small fixed n . Fred Lunnon On 8/8/14, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Theorem: The sum of the curvatures of opposite circles in a Steiner ring of n circles with n even is constant. Proof: Consider the canonical n-ring C having concentric inner and outer circles, centred at the origin, with the first circle of the ring centred on the x-axis. Denote C_j the j-th circle of the ring, with j taken modulo n ; Ci , Co the inner and outer circles tangential to all C_j . Easily, C_j has centre ( cos(2pi j/n), sin(2pi j/n) ) , radius s = sin(pi/n) ; C_(j+n/2) has centre ( -cos(2pi j/n), -sin(2pi j/n) ) , radius s ; Ci has centre ( 0, 0 ) , radius 1-s ; Co has centre ( 0, 0 ) , radius -1-s (note negative sign). Any n-ring D equals C transformed by some Moebius inversion X , and conversely. We represent oriented circles and inversions in the plane by Lie-sphere (contact geometry) coordinate vectors with 5 homogeneous components: then X = [x, y, u, v, 0] ; C_j = [(1-t^2)/(1+t^2), 2*t/(1+t^2), s^2/2, 1-s^2/2, s] ; C_(j+n/2) = [-(1-t^2)/(1+t^2), -2*t/(1+t^2), s^2/2, 1-s^2/2, s] ; D_j = -X * C_j * X ; D_(j+n/2) = -X * C_(j+n/2) * X ; for some x,y,u,v , and appropriate t ; and `*' denotes product in the Clifford algebra Cl(3, 2) . The signed curvature of a circle D is given by (D[3] + D[4])/D[5] ; hence X increases the sum of opposite curvatures by factor ( (D_j[3] + D_j[4])/D_j[5] + (D_(j+n/2)[3] + D_(j+n/2)[4])/D_(j+n/2)[5] ) / ( (C_j[3] + C_j[4])/C_j[5] + (C_(j+n/2)[3] + C_(j+n/2)[4])/C_(j+n/2)[5] ) which reduces to -(s^2 u^2 + 2 s^2 u v + s^2 v^2 - u^2 - 2 u v - v^2 - x^2 - y^2) / (u^2 - v^2 + x^2 + y^2) . This is independent of t , and hence of j . The sum is plainly constant for C , increases by a constant factor for D , therefore remains constant wrt j . QED A similar computation for inner and outer curvatures shows that the same increase applies to their sum, which equals the first sum times 1/s^2 - 1 . The result generalises immediately to m-fold rings with n replaced by n/m . In fact, n plays no essential part in the main computation, which relies only C ring circles having centres lying on the unit circle. Their parametrisation by t is a convenient ruse avoiding complications in the CAS program. Fred Lunnon On 8/9/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
On 8/8/14, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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