最速下降法与牛顿法结合求无约束最优值（Unconstrained optimization by combining steepest descent method with Newton method）.doc

最速下降法与牛顿法结合求无约束最优值（Unconstrained optimization by combining steepest descent method with Newton method） The steepest descent method combines Newtons method with an unconstrained optimal value,.Txt, to value life -- God keeps you alive, and certainly has his plans. Lei Feng does good without seeking recognition, but everything diary. Ha ha, I have achieved the steepest descent French Newton combination, and can also animate the iterative process of solving its optimal value. Have been implemented on the program. (matlab) when running, the steepest descent accuracy should not be too small, and the Newton accuracy at the back can be taken as an arbitrary value. Tic CLC; clear; SYMS X1 x2 G=[]; G=input (` enter, want x1^2, x2^2, x1*x2, x1, X2, constant coefficients, such as [1,2,3,4,5,6] coefficient vector = = ); A=G (1,1); b=G (1,2); c=G (1,3); d=G (1,4); e=G (1,5); g=G (1,6); F=a*x1^2+b*x2^2+c*x1*x2+d*x1+e*x2+g; % draw the original image Figure; X11=-100:0.5:100; X22=x11; [x11, x22]=meshgrid (X11, X22); F11=a.*x11.^2+b*x22.^2+c*x11.*x22+d.*x11+e.*x22+g; Surf (F11), grid, on, hold, on; % draw the original image Df1=diff (F, x1); df2=diff (F, x2);% for function first order DF=[df1; df2]; Df11=diff (DF1, x1); df12=diff (DF1, X2); Df21=diff (df2, x1); df22=diff (df2, x2);%; here, the two derivative of the function is obtained DEE=[df11, DF12, DF21, df22]; X=input (\ enter the initial value of X is x=[x1, x2], x=: ); X=x; E=input (enter the precision of the steepest descent method you want (usually 3~5) E=: ; %, here are some calculations of the relevant initial values T=[]; D=T; T ((1) =subs (DF, [x1, x2],, [x (1), X (2))]; TH=subs (DEE, [x1, x2], [x (1), X (2)); %, here are some calculations of the relevant initial values Disp (because you enter the initial value, here will start the steepest descent search: ); For k=1:100000 D ((1) =-T (: 1);%d (k) is x (k+1), =x (k), +A (k), *d (k) A (1) = (T ((1)*T) (/: 1)) / (T ((:) 1)*TH*T ((: 1)); TH=subs (DEE, [x1, x2], [x (1, K),