Timeline for How to minimize $(x-a)^2+(y-b)^2$ subject to $ \sqrt{a}+\sqrt{b}=\sqrt{2}$?
Current License: CC BY-SA 4.0
13 events
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| May 3, 2020 at 11:20 | comment | added | Asaf Shachar | @gandalf61 Thanks. I guess that such a numerical solution can only be computed after we provide some specific choice of initial data $x,y$, right? It is not a way to produce an "approximate function" in $x,y$... | |
| May 1, 2020 at 22:54 | answer | added | Wolfgang Bangerth | timeline score: 1 | |
| May 1, 2020 at 19:01 | answer | added | Federico Poloni | timeline score: 1 | |
| May 1, 2020 at 18:06 | comment | added | nicoguaro♦ | @FedericoPoloni, would you mind summarizing your comments as an answer? | |
| May 1, 2020 at 16:06 | comment | added | gandalf61 | @AsafShachar A numerical solution to an equation is an approximate solution that is calculated to whatever degree of accuracy you require, usually using an iterative method such as Newton-Raphson | |
| May 1, 2020 at 15:52 | comment | added | Asaf Shachar | @FedericoPoloni Hi, thanks that made the program do it! Can you explain why it is beneficial from a computational point of view to reformulate the problem to be with smooth constraints? That is very interesting for me. (I guess most software programs 'don't know' how to make such a substitution by themselves?). I presume this is a well-studied subject which I know nothing about. Where to start looking? What are the "key-words"? | |
| May 1, 2020 at 15:42 | comment | added | Asaf Shachar | @gandalf61 Yes, I am now quite sure that an exact analytic expression would be quite horrible to work with. What exactly do you mean by a numerical solution? (I think I would be happy with approximate simpler expressions). | |
| May 1, 2020 at 15:02 | comment | added | gandalf61 | @AsafShachar The answer in Mathematics Stack Exchange that you linked to reduces the problem to solving a cubic. An analytic solution to a general cubic equation is possible, although the result will be a complicated expression. On the other hand, if you are satisfied with a numerical solution, then that should be straightforward. | |
| May 1, 2020 at 14:25 | comment | added | origimbo | As aside, symbolic math packages (e.g. sympy, since you mentioned Python, but Mathematica will do it) can handle solving cubic equations very readily. | |
| May 1, 2020 at 13:58 | comment | added | Asaf Shachar | I am interested in the function for this specific problem. It seems that trying to solve this analytically (by hand) gets you to a cubic equation with rather messy formulas for the roots. (and it doesn't matter if you express $b$ in terms of $a$, or use Lagrang'es multipliers). | |
| May 1, 2020 at 13:42 | comment | added | origimbo | Are you actually looking for a generic approach, or just to come up with a function for this specific problem? In the latter case, couldn't you just substitute a=c^2, then cancel the b and proceed analytically? | |
| May 1, 2020 at 13:35 | review | First posts | |||
| May 1, 2020 at 14:57 | |||||
| May 1, 2020 at 13:31 | history | asked | Asaf Shachar | CC BY-SA 4.0 |