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Subject: Re: solving a system of multi-variable polynomial equations in mupad
Date: Tue, 9 Oct 2012 08:02:08 +0000 (UTC)
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"Daniel" wrote in message <k4vg5t$nss$1@newscl01ah.mathworks.com>...
> ....... I am equally interested in understanding when the system has no solutions as when it does have solutions. .......
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  Daniel, you have said, "... I am equally interested in understanding when the system has no solutions as when it does have solutions."  I will give you a summary here of my conclusions on that question.

  To begin with, there can be no solutions unless the two quantities which I have called F are zero.  That is, s-t_h and t-s_h must both be zero.  In fact none of the sets of just six equations can have a solution unless the associated F is zero.  This can be demonstrated using the rotation method which I described earlier which with nonzero F's leads to contradictions in each of the three separate cases B^2-A*C < 0, B^2-A*C = 0, and B^2-A*C > 0.

  Assuming zero F's, I find that seeing the situation from the point of view of six x-y points in two-dimensional space and two A-B-C points in three dimensional space to be useful.  The six points are the two sets, (a1,b1), (c1,d1), (e1,f1) and (a2,b2), (c2,d2), (e2,f2) which I have called (x1,y1), (x2,y2), (x3,y3), and (x4,y4), (x5,y5), and (x6,y6).  The two A-B-C points are

 (A1,B1,C1) = (alpha1-beta1_h,alpha2-beta2_h,alpha3-beta3_h), and
 (A2,B2,C2) = (beta1-alpha1_h,beta2-alpha2_h,beta3-alpha3_h) .

  First, if the six x-y points are all located at the origin, that is, all zero, then any two A-B-C points anywhere in their 3D space will constitute a solution.  That is quite obvious.

  Second, if all six x-y points lie along some one line through the origin with not all points at the origin and if a is the angle between that line and the x-axis, then any two points (A1,B1,C1) and (A2,B2,C2) will constitute a solution to the 24 equations provided that each is orthogonal to the 3D vector (cos(a)^2,2*sin(a)*cos(a),sin(a)^2), that is if

 A1*cos(a)^2 + 2*B1*sin(a)*cos(a) + C1*sin(a)^2 = 0  and
 A2*cos(a)^2 + 2*B2*sin(a)*cos(a) + C2*sin(a)^2 = 0 .

Thus the A-B-C points must lie in a certain plane containing the origin.  Showing that to be true for the equations of the form of

 A*x2*x3 + B*(x2*y3+x3*y2) + C*y2*y3 = 0

is also accomplished using the aforementioned rotation method.

  Finally, if the three points of the first x-y set lie along one line through the origin, not all at the origin, and if the second set of three points lie along a different line through the origin, again not all at the origin, and if a and b are the respective angles between the x-axis and these lines, then the set of A-B-C point pairs that provide solutions are all pairs that are orthogonal to both

 (cos(a)^2,2*sin(a)*cos(a),sin(a)^2) and 
 (cos(b)^2,2*sin(b)*cos(b),sin(b)^2) .

Stated equivalently, it is those A-B-C points which lie along the cross product of these vectors.  However we can remove the common nonzero factor, sin(a-b), from this cross product and arrive at the simpler vector

 (-2*sin(a)*sin(b),sin(a+b),-2*cos(a)*cos(b)) .

Thus any two A-B-C points along this direction will constitute a solution to the 24 equations.  The A-B-C points must therefore lie in a certain one-dimensional line through the origin for such a set of six x-y points.

  All other combinations of points in these respective x-y and A-B-C spaces are non-solutions.

  Again I point out that MuPAD would have had a very difficult time explaining to humans, not to mention arriving at, such a complicated situation.

Roger Stafford