A possible solution, we compare "diffcode_jacobian" and "numderivative". in this case seem to calculate with the same accuracy
function val = HH(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x))).^2+x.^2-1; endfunction; function z=f(x) z(1,1:2)=diffcode_jacobian(HH,x) z(1,3:4)=numderivative(HH,x) endfunction xt = -1.4:0.01:1.4; function Z=g(X,Y,f) for i=1:size(X,"*") T=[X(i) Y(i)]; Z(i,:)=f(T); end endfunction //And for 280 values calculate fast F=g(xt,xt) -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html _______________________________________________ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users