%阻尼Newton算法(k=2,mul_count=44,sum_count=28,最少迭代)
syms x1 x2;
f=50*(x2-x1^2)^2+(1-x1)^2;
v=[x1,x2];
df=jacobian(f,v);
df=df.';
G=jacobian(df,v);
epson=1e-12;x0=[0,0]';g1=subs(df,{x1,x2},{x0(1,1),x0(2,1)});G1=subs(G,{x1,x2},{x0(1,1),x0(2,1)});k=0;mul_count=0;sum_count=0;
mul_count=mul_count+12;sum_count=sum_count+6;
while(norm(g1)>epson)
    p=-G1\g1;
    x0=x0+p;
    g1=subs(df,{x1,x2},{x0(1,1),x0(2,1)});
    G1=subs(G,{x1,x2},{x0(1,1),x0(2,1)});
    k=k+1;
    mul_count=mul_count+16;sum_count=sum_count+11;
end;
k
x0
mul_count
sum_count