Timeline for Conditions for when an off-centre ellipsoid fits inside the unit ball
Current License: CC BY-SA 3.0
5 events
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| Aug 25, 2013 at 9:29 | comment | added | Antony | Right, I agree with this. However, does the fact that the equation in 3 dimensions has 6 roots necessarily mean that there is in general no analytic solution to the problem? It seems like there is some redundancy in this method, since it also finds points on the ellipsoid that are closest to the origin. Could there be some more geometric argument for identifying only the most distant points, which would lead to a a lower order equation that can be solved analytically? | |
| Aug 25, 2013 at 3:20 | comment | added | Wlodek Kuperberg | Assuming no two of the semiaxes are equal, in the plane there will be up to four points; in 3D up to six, and so on. If the equation cannot be solved by hand, there are numerical methods to find approximate solutions. | |
| Aug 25, 2013 at 2:59 | comment | added | Antony | Do you mean four such points for the 3D ellipsoid case? A quick go at this seems to give an equation which is 6th order in terms of a Lagrange multiplier. (That is by maximising $x^2+y^2+z^2$ subject to the constraint $(\frac{x-c_1}{t_1})^2+(\frac{y-c_2}{t_2})^2+(\frac{z-c_3}{t_3})^2-1=0$.) | |
| Aug 25, 2013 at 2:43 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
Added the second sentence.
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| Aug 25, 2013 at 2:37 | history | answered | Wlodek Kuperberg | CC BY-SA 3.0 |