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: Tue, 16 Oct 2012 22:42:21 +0000 (UTC) Organization: Univ College London Lines: 40 Message-ID: <k5knsd$l1u$1@newscl01ah.mathworks.com> References: <k4ib2c$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$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-01-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1350427341 21566 172.30.248.46 (16 Oct 2012 22:42:21 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 16 Oct 2012 22:42:21 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 2788389 Xref: news.mathworks.com comp.soft-sys.matlab:780799 "Roger Stafford" wrote in message <k5i3g7$ie5$1@newscl01ah.mathworks.com>... > "Daniel" wrote in message <k5hrmo$jtt$1@newscl01ah.mathworks.com>... > > ..... But why can't the K be dropped? ...... > - - - - - - - - - - > The purpose of K lies in deriving an algebraic identity with the expression > > A*x^2+2*B*x*y+C*y^2 > > regardless of whether it is to be equal to zero or not. For that you do need K. Otherwise such equations as > > -2*B = K*sin(a+b) > C-A = K*cos(a+b) > 2*sqrt(B^2-A*C) = K*sin(a-b) > C+A = K*cos(a-b) > > would all be inherently false with K absent and with arbitrary A, B, and C. You will notice however that K does drop out of the picture with the two 'atan2' operations and does not play a role in solving the two line angles. So, yes, K is definitely needed in the theory behind the matlab code but is never explicitly solved for in the code because the lines' angles are independent of its value. > > (It is misleading to say that K scales the points on the lines.) > > Roger Stafford Hi Rodger Thanks again for your helpful response. I see now the use of K - I forgot the important fact that sin(a+b) and cos(a+b) are in the range [-1,1]! So yes I see that -2*B = sin(a+b) will always be false if -2*B is not in the range [-1,1], and the K can be chosen to make the equality work. I think I convinced myself it must be superfluous when it dissappered from the subsequent analysis. I have finally got on to your message number 5. You give some very nice geometrical interpretations to the solution space of the A1-B1-C1, and A2-B2-C2 points given that the six x-y points lie somewhere on the two degenerate conics defined by equations 1/4/6 and 13/16/18 (for the coefficients A1,B1,C1), and equations 7/10/12 and 19/22/24 (for the coefficients A2,B2,C2) - here the equation numbers relate to the 24 equations I give in my original post. 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. Your geometric interpretations of the A-B-C solution space, i.e. the whole of R^3, a plane through the origin, or points in R^3 in the direction of a cross-product, show that, given the x-y points are located as discussed above, then there are an infinite number of (A1,B1,C1), and (A2,B2,C2) points that are solutions? I suppose this says a similar thing as your Matlab code but in reverse - i.e., for the 12 equations using (A1,B1,C1) chosen arbitrarily, we can always find two lines from which to draw an infinite number of x-y points as solutions. If the coefficient differences making up (A2,B2,C2) are chosen so that A1=A2, B1=B2, and C1=C2, then the 12 equations using (A2,B2,C2) define the same conic, and there are an infinite number of solutions. Thanks Dan