Rank: Advanced Member Groups: Registered, Advanced Member Joined: 10/11/2010(UTC) Posts: 1,494 Was thanked: 1274 time(s) in 745 post(s)
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Quote:Thank you for considering this . If you include the sum of squares to your example above, then it is equivalent as finding the roots of this three equations sum(((el(pp,i)-f(el(xx,i),a1,a2,a3))*diff((f(el(xx,i),a1,a2,a3)),a1)),i,1,n)≡0sum(((el(pp,i)-f(el(xx,i),a1,a2,a3))*diff((f(el(xx,i),a1,a2,a3)),a2)),i,1,n)≡0sum(((el(pp,i)-f(el(xx,i),a1,a2,a3))*diff((f(el(xx,i),a1,a2,a3)),a3)),i,1,n)≡0It is a well known and maybe redundant to say, sorry - the first derivatives over three unknowns (a1,a2,a3) should be zero Now I knew where it came from. I also fixed a bug with double index ii. I've used it twice. File Attachment(s): uni attached the following image(s): |
Russia ☭ forever Viacheslav N. Mezentsev |