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, 11 Oct 2012 21:51:07 +0000 (UTC) Organization: Univ College London Lines: 47 Message-ID: <k57f0b$4gd$1@newscl01ah.mathworks.com> References: <k4ib2c$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> 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 1349992267 4621 172.30.248.48 (11 Oct 2012 21:51:07 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Thu, 11 Oct 2012 21:51:07 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 2788389 Xref: news.mathworks.com comp.soft-sys.matlab:780482 "Roger Stafford" wrote in message <k55461$5p9$1@newscl01ah.mathworks.com>... > "Daniel" wrote in message <k54q5c$2pt$1@newscl01ah.mathworks.com>... > > ........ > > If we had AC<B^2 for our choice of A,B, and C, then we would have a hyperbola. I have not checked that the above all of the above holds in this scenario - you assure me it does, and so I assume it will hold (I will check in due course). ...... > - - - - - - - - - - > What you have written looks substantially correct to me. > > I add here here the reasoning that in the "hyperbolic" case you must have F = 0 to have solutions to your six basic equations. (It is obvious in the "parabolic" case.) > > As has already been established, after the appropriate rotation the six equations would have this form: > > X1^2/a^2 - Y1^2/b^2 = 1 > X2^2/a^2 - Y2^2/b^2 = 1 > X3^2/a^2 - Y3^2/b^2 = 1 > X2*X3/a^2 - Y2*Y3/b^2 = 0 > X3*X1/a^2 - Y3*Y1/b^2 = 0 > X1*X2/a^2 - Y1*Y2/b^2 = 0 > > where a and b are positive and where the X's and Y's refer to the rotated coordinates. However, there can be no solution to such a set of equations as shown by the following impossibility: > > (X1^2/a^2)*(X2^2/a^2)*(X3^2/a^2) = > (Y1^2/b^2+1)*(Y2^2/b^2+1)*(Y3^2/b^2+1) > > (Y1^2/b^2)*(Y2^2/b^2)*(Y3^2/b^2) = > (Y2*Y3/b^2)*(Y3*Y1/b^2)*(Y1*Y2/b^2) = > (X2*X3/a^2)*(X3*X1/a^2)*(X1*X2/a^2) = > (X1^2/a^2)*(X2^2/a^2)*(X3^2/a^2) . > > The first quantity would have to be greater than the last and yet is also identical to it! > > Roger Stafford Roger thanks very much for confirming my summary is OK, and thanks for showing me the contradiction in the hyperbolic case. A very quick general question: what is your strategy for finding these contradictions? For example do you say try to solve them (manually or with Matlab), then if you can't solve them do you make the assumption there is no solution, and proceed to try and find a contradiciton which shows there cannot be solutions? Or do you do the reverse, that is do you try and find a contradicition, and then if you cannot find one do you then move on to trying to find a solution? If so is this a general strategy you employ for all problems? Specifically, I am just moving on now to the code you have written in message 6. I am not sure if I have forgotten a lot of trig or simply have never learnt it well enough to forget it, but are the two equations in parentheses in the equation K*(x*sin(a)-y*cos(a))*(x*sin(b)-y*cos(b)) = A*x^2+2*B*x*y+C*y^2 equivalent to the two lines y=x*tan(a) and y=x*tan(b) respectively, where tan(a), and tan(b) are the slopes of the lines? These two lines are then the crossed lines of the degenerate hyperbola, and so you are initially trying to find two lines that are defined by the right-hand side, by changing the angles, and hence the gradients of the lines? As ever I appreciate your help on this, and I will at some point get through everything you have said!