SOLVE 4 NONLINEAR EQUATIONS (HELP PLEASE)
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From: Walter Roberson on 29 Apr 2010 19:42 Roger Stafford wrote: > Walter Roberson <roberson(a)hushmail.com> wrote in message > <hrd0jl$bsc$1(a)canopus.cc.umanitoba.ca>... >> ...... >> Xc = (1/2) * a1 + (1/2) * a2, >> ..... > > There is something decidedly fishy about those answers, Walter. In > each of the answers 2, 3, and 4, they gave Xc as the average between a1 > and a2. However, it would be easy to place the points (a1,b1) and > (a2,b2) on an ellipse so that their x-coordinates are well off to one > side of the ellipse's center and not equal to its average. What gives? Good question. When I tried running the original forms through my symbolic package, it was taking an indefinite time. I normalized the forms and although the solution wasn't exactly the fastest, it didn't take unduely long. But it could be that I miswrote the normalization. If I have time, I'll try again later... but I have to finish my taxes by tomorrow and tomorrow is also the day my license for the symbolic package expires.
From: GAURAV on 29 Apr 2010 19:44 Hi Walter and Roger, Hey guys! The solution of four equations you gave works MUCH MUCH BETTER than the earlier ellipse fit thing I was using. Thanks a lot!!! REALLY REALLY APPRECIATE.. Still it is not the best fit obtained but I gotta a hell lotta improvement. Also, if something else you guys think would work for me to form a surface, let me know PLEASE!!! Thank You, Regards, Gaurav.
From: Roger Stafford on 29 Apr 2010 20:28 "Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <hrd4m4$2h1$1(a)fred.mathworks.com>... > A note on Roger's solution: I would prefer normalize the eqt by fixing F = 1, for example. This will replace the calculation of svd by solving a 4x4 linear system. > > Bruno Hi Bruno. Normalizing by setting F to 0 doesn't always work. You can very easily be dealing with an ellipse that runs right through the origin, and the right value for F is zero in such a case. Even if it runs near the origin, forcing F to be 1 will encounter awkward accuracy problems. Using 'svd' has the virtue that the sum of the squares of the five coefficients is constrained to be unity which will easily allow F to be zero or some appropriately small value relative to the others. To me it seems the more "natural" method. Roger Stafford
From: GAURAV on 29 Apr 2010 20:28 Hey Walter, Can you tell me how you solved these four equations? Thanks, Gaurav
From: Roger Stafford on 29 Apr 2010 20:59 "Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hrd86k$fje$1(a)fred.mathworks.com>... > "Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <hrd4m4$2h1$1(a)fred.mathworks.com>... > > A note on Roger's solution: I would prefer normalize the eqt by fixing F = 1, for example. This will replace the calculation of svd by solving a 4x4 linear system. > > > > Bruno > > Hi Bruno. Normalizing by setting F to 0 doesn't always work. You can very easily be dealing with an ellipse that runs right through the origin, and the right value for F is zero in such a case. Even if it runs near the origin, forcing F to be 1 will encounter awkward accuracy problems. > > Using 'svd' has the virtue that the sum of the squares of the five coefficients is constrained to be unity which will easily allow F to be zero or some appropriately small value relative to the others. To me it seems the more "natural" method. > > Roger Stafford Bruno, to make it seem a little less complicated, we could just as well have applied the 'null' function to the 4 x 5 M matrix I defined. We are simply looking for a non-zero vector which is orthogonal to its four rows in R^5, which would therefore satisfy the equation. There is certain to be one. Any one or more of the coefficients might naturally be zero without disrupting the calculations. Roger Stafford
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