Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: solving a system of multi-variable polynomial equations in mupad Date: Thu, 18 Oct 2012 02:54:24 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 57 Message-ID: <k5nr10$9mi$1@newscl01ah.mathworks.com> References: <k4ib2c$eel$1@newscl01ah.mathworks.com> <k4l3c0$2ed$1@newscl01ah.mathworks.com> <k4n7o4$1hf$1@newscl01ah.mathworks.com> <k4nprj$9rg$1@newscl01ah.mathworks.com> <k4vg5t$nss$1@newscl01ah.mathworks.com> <k5023o$mvg$1@newscl01ah.mathworks.com> <k527jg$8nq$1@newscl01ah.mathworks.com> <k52cml$phg$1@newscl01ah.mathworks.com> <k54q5c$2pt$1@newscl01ah.mathworks.com> <k55461$5p9$1@newscl01ah.mathworks.com> <k57f0b$4gd$1@newscl01ah.mathworks.com> <k57q0s$9qm$1@newscl01ah.mathworks.com> <k5hrmo$jtt$1@newscl01ah.mathworks.com> <k5i3g7$ie5$1@newscl01ah.mathworks.com> <k5knsd$l1u$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1350528864 9938 172.30.248.48 (18 Oct 2012 02:54:24 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Thu, 18 Oct 2012 02:54:24 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:780873 "Daniel" wrote in message <k5knsd$l1u$1@newscl01ah.mathworks.com>... > ......... > I hope I have got this right by now when I say that each of the two degenerate conics can either be the point ellipse, the x-axis or y-axis (if the conic is a degenerate parabola), or two crossed lines through the origin (if the conic is a degenerate hyperbola). So the scenarios you outline correspond to the following situations; > > 1) all six x-y points are located at the origin - both conics are the point ellipse > 2) all six x-y points lie along one line through the origin - both conics are the same degenerate parabola > 3) 3 x-y points lie along one line through the origin whilst the other three lie on another line through the origin - both conics are the x-axis and y-axis respectively, or are the same hyperbola. > ........ - - - - - - - - - - I think you misinterpreted some of the statements I made, Daniel. For instance, the "parabolic" condition B^2=A*C>0 does not imply the xy points have to be either the x-axis or y-axis; it could be any straight line through the origin, depending on the values of A, B, and C. Also, having all points in a solution lie along a single straight line through the origin does not imply that the parabolic condition must hold; it might be one of the lines of a "hyperbolic" (B^2>A*C) set. It is easiest to first analyze all the solutions to the six homogeneous equations A*x1^2+2*B*x1*y1+C*y1^2 = 0 A*x2^2+2*B*x2*y2+C*y2^2 = 0 A*x3^2+2*B*x3*y3+C*y3^2 = 0 A*x2*x3+B*(x2*y3+x3*y2)+C*y2*y3 = 0 A*x3*x1+B*(x3*y1+x1*y3)+C*y3*y1 = 0 A*x1*x2+B*(x1*y2+x2*y1)+C*y1*y2 = 0 This can be divided into four different cases. 1) B^2 < A*C (elliptical). The only possible solution in this case is for all xy points to lie at the origin: x1=y1=x2=y2=x3=y3=0. 2) B^2 = A*C > 0 (parabolic). The only possible solutions in this case must have all three xy points lying along some particular straight line through the xy origin. For given A, B, and C of this type, the line can be determined by a = 1/2*atan2(-2*B/(C+A),(C-A)/(C+A)) where 'a' is the angle between the line and the x-axis. For a given set of three xy points along a straight line through the origin with angle 'a', the corresponding A, B, C values for this case will be any satisfying A = K*sin(a)^2 B = -K*sin(a)*cos(a) C = K*cos(a)^2 for some non-zero K. 3) B^2 > A*C (hyperbolic). The only possible solution here is for the three points to all lie along one particular straight line through the origin or else all along some particular second, distinct line also through the origin. You have already seen how the two angles 'a' and 'b' of these two lines can be found from A, B, and C of this case. It was the method using two atan2's contained in my Oct. 7 article. If angles 'a' and 'b' of the two lines are given, then A, B, and C can be any numbers satisfying A = K*sin(a)*sin(b) B = -K/2*sin(a+b) C = K*cos(a)*cos(b) for some non-zero K. 4) A = B = C = 0. In this case any set whatever of three xy points will obviously satisfy the six equations. For your 24 equations which involve two sets of three xy points and two sets of ABC values in all four possible combinations, first of all, there will be no solutions at all unless s-t_h=0 and t-s_h=0, making the equations homogeneous. If this is true, then all solutions can be found in terms of appropriate combinations of solutions from the above set of six equations. If both ABC sets are hyperbolic and if both sets of associated lines are the same, then the two sets of A,B,C values must be proportional to one another and the xy solution is to either have all six points on one of the lines or to have the first three on one line and the second three on the other line. If instead, only one of the lines is shared between the two ABC sets, then the xy solution is to have all six points along this common line. If no lines are shared between the two ABC sets, then all six xy points must be at the origin. If A1,B1,C1 are hyperbolic (B1^2>A1*C1) and A2,B2,C2 are parabolic (B2^2=A2*C2>0), and if one of the lines of the (A1,B1,C1) set is the same as the unique line of the (A2,B2,C2) set, then the xy solutions consist of all possible sets of six points along this common line. If there is not this match of lines, then the only solution is with all six points at the origin. If both ABC sets are parabolic and if these both correspond to the same line, then the ABC values must be proportional to one another and the xy solution is any six points along this common line. If their lines are not the same, then having the six points at the origin is the only solution. If either set or both sets are elliptical, then only the xy origin for all six points will be a solution. Roger Stafford