48/50 Linear System Solver
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02-17-2014, 07:19 AM
Post: #8
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RE: 48/50 Linear System Solver
I don't consider that the calculator is giving a "meaningless result", although it is certainly the case that the offered result does not satisfy the equation with zero or negligible residuals.
To start with a simple illustration, consider the two-dimensional case with the equations x - y = 1 and x - y = -1 . If you put this system into the linear solver you get the "solution" of x = 0, y = 0 . Obviously this does not solve the equations in the usual sense, but it, and any other point midway between the two lines defined by the equations, will give smaller residuals than other points. Among all these midway points, the calculator chooses the one with the smallest distance from the origin. In the three dimensional case differing pathologies are possible. In Tim's case it seems that the three planes defined by the three equations intersect in three parallel distinct lines. Since these never meet in a single point, there is no exact solution. In order to have something a bit easier to think about, let us consider an analogous case where the three equations are y + z = 1, y = 0, z = 0 . Here we have three planes, and their three lines of intersection are parallel. If we put this system into the linear solver, we get the "solution" x = 0, y = 0.3333..., z = 0.3333... . Again, this is not a solution in the usual sense, but it, and any other point with the same y and z values will give smaller residuals than all other points. Again, the calculator offers the one closest to the origin. I agree that it would be nice if the calculator gave warning for this kind of a situation. I don't know whether there might be cases where this type of "solution" might be of practical value, but it is not meaningless. |
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