1、k=0;while k100 lanmed=-(g0)*s0/(s0*A*s0); x=x0+lanmed*s0; g=g(x); k=k+1; if norm(g) x=zeros(100,1); di1ti(x)After 1 iterations,obtain the optimal solution. The optimal solution is -0.250000.The optimal .ans =0.005*ones(100,1).第二题1.最速下降法。function zy_x=di2titidu(x)%该函数用来解大作业第二题。 yimuxulong=1e-5; k=0;
2、s0=-g0;while k=0 if norm(g0)=0&i(f(x0)+c*lanmed*g0*s0) lanmed=lanmed/2; i=i+1; x0=x; g0=g(x);zy_x=x;zyj=f(x);fprintf(after %d iterations,obtain the optimal solution.nnThe optimal solution is %f.nn The optimal .n,k,zyj);x1=1 0 0 0*x;x2=0 1 0 0*x;x3=0 0 1 0*x;x4=0 0 0 1*x;f=(x1-1)2+(x3-1)2+100*(x2-x12
3、)2+100*(x4-x32)2;g=2*(x1-1)-400*x1*(x2-x12);200*(x2-x12);2*(x3-1)-400*x3*(x4-x32);200*(x4-x32); x=-1.2 1 -1.2 1; di2titidu(x)after 5945 iterations,obtain the optimal solution.The optimal solution is 0.000000. The optimal ans = 1.00001.00002.牛顿法function zy_x=di2tinewton(x) h0=h(x0);s0=-inv(h0)*g0;k g0=g(x0,M);s0=-inv(H(x0,M)*g0; while k %牛顿法; g0=g(x,M); s0=-inv(H(x0,M)*g0; if max(abs(h(x),g1(x),g2(x),g3(x)0.5 M=M*c; m=m+1;,m,zyj);function F=F(x,M) x1=1 0*x;x2=0 1*x;F=4*x1-x22-12+M*(h2+g12+g22+g32);function g=g(x,M) x