Rank: Advanced Member Groups: Registered, Advanced Member Joined: 10/11/2010(UTC) Posts: 963 Was thanked: 878 time(s) in 493 post(s)

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): 
