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Hi gAr;
Hi bobbym,
It will get closer and closer because this one happens to converge. Sometimes they do not and you have to pull out a different x and y.
Hi bobbym,
If you look you will see that the two equations have been turned into two recurrences. They are just minus the subscripts. Tell me what you get for x and y.
Hi bobbym,
find_root is one command. Also Subtract from both sides. You get Multiply both sides by You get. Divide through by: 2300 You get: Now you have x on the right by itself. Do the same moves with the second equation to get a y by itself on the right. Do you follow up to here?
It may be having since there are so many packages, I haven't used it yet.
Did you check for a numerical solver in Sage, all packages have them?
Hi bobbym,
Hi gAr;
Okay.
Fermat posed this problem with weights = 1 to Torricelli a long time ago. He solved it in a couple of ways. They were all geometric constructions. I feel the problem can be solved numerically but I can not find any analytical methods.
Is this still in active research or is there any good approximation algorithm?
The problem that I took this one from was so artificial that one of the vertices was the answer. I mean he could solve it using a small inequality. I know solutions like that make students happy and look great in books. The fact is in the real world there might be a couple of hundred points. Often you have to settle on a solution that is good but maybe not the best. 