Hi, motterlet and Steer Thanks a lot your helps.
Unfortunately 'lsqrsolve' did't give the smallest initial point for my function. The local minimum of f^2+a*x^2 is attained at x s.t. f(x)=-a*x. So if a is small such x is a neighborhood of x s.t. f(x)=0. But it is not necessarily of x which is the smallest one. Probably my function is not well-behaved as like cos(x) so that it fail. Now I get an awkward method. I find the first x(i) s.t. f(x(i))>0 and f(x(i-1))<0 where x(i)=x0+i*dx and x0 is some constant which is smaller than smallest solution of f(x)=0. Then I modify f(x) to new function fnew(x) as f(x)=f(x(i-1))+[f(x(i))-f(x(i-1))]/[x(i)-x(i-1)]*[x-x(i)] for x>=x(i) and fnew(x)=f(x) for x<=x(i). Using fsolv(x(i-1),fnew) gives the smallest solution of f(x)=0. With your helps I could get a practical solution. Thanks again. Best regards. -- View this message in context: http://mailinglists.scilab.org/fsolver-tp4033340p4033350.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. _______________________________________________ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users