求高手指导这个程序错在哪里?结果为什么总是1?(这是matlab 程序)clear all;clc;x(1)=1.3;x(2)=1.6;x(3)=1.9;f(x(1))=0.6200860;f(x(2))=0.4554022;f(x(3))=0.2818186;df(x(1))=-0.5220232;df(x(2))=-0.5698959;df(x(3))=-0.5811571;for i=1
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![求高手指导这个程序错在哪里?结果为什么总是1?(这是matlab 程序)clear all;clc;x(1)=1.3;x(2)=1.6;x(3)=1.9;f(x(1))=0.6200860;f(x(2))=0.4554022;f(x(3))=0.2818186;df(x(1))=-0.5220232;df(x(2))=-0.5698959;df(x(3))=-0.5811571;for i=1](/uploads/image/z/12078675-27-5.jpg?t=%E6%B1%82%E9%AB%98%E6%89%8B%E6%8C%87%E5%AF%BC%E8%BF%99%E4%B8%AA%E7%A8%8B%E5%BA%8F%E9%94%99%E5%9C%A8%E5%93%AA%E9%87%8C%3F%E7%BB%93%E6%9E%9C%E4%B8%BA%E4%BB%80%E4%B9%88%E6%80%BB%E6%98%AF1%3F%EF%BC%88%E8%BF%99%E6%98%AFmatlab+%E7%A8%8B%E5%BA%8F%EF%BC%89clear+all%3Bclc%3Bx%281%29%3D1.3%3Bx%282%29%3D1.6%3Bx%283%29%3D1.9%3Bf%28x%281%29%29%3D0.6200860%3Bf%28x%282%29%29%3D0.4554022%3Bf%28x%283%29%29%3D0.2818186%3Bdf%28x%281%29%29%3D-0.5220232%3Bdf%28x%282%29%29%3D-0.5698959%3Bdf%28x%283%29%29%3D-0.5811571%3Bfor+i%3D1)
求高手指导这个程序错在哪里?结果为什么总是1?(这是matlab 程序)clear all;clc;x(1)=1.3;x(2)=1.6;x(3)=1.9;f(x(1))=0.6200860;f(x(2))=0.4554022;f(x(3))=0.2818186;df(x(1))=-0.5220232;df(x(2))=-0.5698959;df(x(3))=-0.5811571;for i=1
求高手指导这个程序错在哪里?结果为什么总是1?(这是matlab 程序)
clear all;
clc;
x(1)=1.3;
x(2)=1.6;
x(3)=1.9;
f(x(1))=0.6200860;
f(x(2))=0.4554022;
f(x(3))=0.2818186;
df(x(1))=-0.5220232;
df(x(2))=-0.5698959;
df(x(3))=-0.5811571;
for i=1:3
Z(2*i-1)=x(i);
Z(2*i)=x(i);
Q(2*i-1,1)=f(x(i));
Q(2*i,1)=f(x(i));
Q(2*i,2)=df(x(i));
if(i~=1)
Q(2*i-1,2)=( Q(2*i-1,1)-Q(2*i-2,1) )/( Z(2*i-1)-Z(2*i-2) );
end
end
for i=3:6
for j=3:i
Q(i,j)=( Q(i,j-1)-Q(i-1,j-1) )/( Z(i)-Z(i-j+1) );
end
end
Q
求高手指导这个程序错在哪里?结果为什么总是1?(这是matlab 程序)clear all;clc;x(1)=1.3;x(2)=1.6;x(3)=1.9;f(x(1))=0.6200860;f(x(2))=0.4554022;f(x(3))=0.2818186;df(x(1))=-0.5220232;df(x(2))=-0.5698959;df(x(3))=-0.5811571;for i=1
给你改好了.matlab的语法问题.不是f(x(1))是fx(1). 同里应该是dfx(1)
clear all;
clc;
x(1)=1.3;
x(2)=1.6;
x(3)=1.9;
fx(1)=0.6200860;
fx(2)=0.4554022;
fx(3)=0.2818186;
dfx(1)=-0.5220232;
dfx(2)=-0.5698959;
dfx(3)=-0.5811571;
for i=1:3
Z(2*i-1)=x(i);
Z(2*i)=x(i);
Q(2*i-1,1)=fx(i);
Q(2*i,1)=fx(i);
Q(2*i,2)=dfx(i);
if(i~=1)
Q(2*i-1,2)=( Q(2*i-1,1)-Q(2*i-2,1) )/( Z(2*i-1)-Z(2*i-2) );
end
end
for i=3:6
for j=3:i
Q(i,j)=( Q(i,j-1)-Q(i-1,j-1) )/( Z(i)-Z(i-j+1) );
end
end
Q