Question

求高手！用牛顿迭代法解三元三次方程的MATLAB程序！

求高手！用牛顿迭代法解下面的三元三次方程，用MATLAB实现，急用，非常感谢！
(1-c)*(1-y)*(1-m)*10.61+y(1-c)*(1-m)*30.56+c*(1-y)*(1-m)*60.67+m*(1-c)*(1-y)*0.95+y*m*(1-c)*28.45+y*c*(1-m) *8.19+c*m*(1-y) *2.51+c* y*m* 76.69-3.21=0
(1-c)*(1-y)*(1-m)*15.21+y*(1-c)*(1-m)*14.58+c*(1-y)*(1-m)*65.2+m*(1-c)*(1-y)*0.96+y*m*(1-c)*14.23+y*c*(1-m) *17.41+c*m*(1-y) *1.62+m*y *c*80.32=0
(1-c)*(1-y)*(1-m)*44.77+y*(1-c)*(1-m)*9.9+c*(1-y)*(1-m)*3.97+m*(1-c)*(1-y)*0.78+y*m*(1-c)*0.8+y*c*(1-m) *3.3+c*m*(1-y) *7.95+ m*c*y*66.79=0
c和m和y分别都是未知数，解的范围都是从0-1。
小妹财富不多，求高手解答。
非常感谢 jimtien的解答，我的方程漏掉了一些数据，在这里我把完整的写出来，此方程中X、Y、Z是已知数据，我的意思是通过这个方程，将一组X、Y、Z数据，转换成c、m、y数据。能给出一个能加载 三维数据的 最终程序么？
(1-c)*(1-y)*(1-m)*10.61+y(1-c)*(1-m)*30.56+c*(1-y)*(1-m)*60.67+m*(1-c)*(1-y)*0.95+y*m*(1-c)*28.45+y*c*(1-m) *8.19+c*m*(1-y) *2.51+c* y*m* 76.69-X=0
(1-c)*(1-y)*(1-m)*15.21+y*(1-c)*(1-m)*14.58+c*(1-y)*(1-m)*65.2+m*(1-c)*(1-y)*0.96+y*m*(1-c)*14.23+y*c*(1-m) *17.41+c*m*(1-y) *1.62+m*y *c*80.32-Y=0
(1-c)*(1-y)*(1-m)*44.77+y*(1-c)*(1-m)*9.9+c*(1-y)*(1-m)*3.97+m*(1-c)*(1-y)*0.78+y*m*(1-c)*0.8+y*c*(1-m) *3.3+c*m*(1-y) *7.95+ m*c*y*66.79-Z=0

补充：X Y Z的参考数据
X Y Z

38.98 38.38 32.29
7.07 9.73 10.97
0.67 0.85 0.79
69.85 68.99 56.01
1.11 1.7 1.6
52.05 54.56 49.08
3.21 2.15 3.76
44.6 46.98 41.77
1.21 1.56 1.46
63.46 63.94 51
3 4.05 4.41
13.01 13.6 11.03

Answer 1

用fsolve可解出来：
先构造函数：
function output=solveproblem(X)
c=X(1);
m=X(2);
y=X(3);
output(1)=(1-c)*(1-y)*(1-m)*10.61+y*(1-c)*(1-m)*30.56+c*(1-y)*(1-m)*60.67+m*(1-c)*(1-y)*0.95+y*m*(1-c)*28.45+y*c*(1-m) *8.19+c*m*(1-y) *2.51+c* y*m* 76.69-3.21
output(2)=(1-c)*(1-y)*(1-m)*15.21+y*(1-c)*(1-m)*14.58+c*(1-y)*(1-m)*65.2+m*(1-c)*(1-y)*0.96+y*m*(1-c)*14.23+y*c*(1-m) *17.41+c*m*(1-y) *1.62+m*y *c*80.32
output(3)=(1-c)*(1-y)*(1-m)*44.77+y*(1-c)*(1-m)*9.9+c*(1-y)*(1-m)*3.97+m*(1-c)*(1-y)*0.78+y*m*(1-c)*0.8+y*c*(1-m) *3.3+c*m*(1-y) *7.95+ m*c*y*66.79

然后求解：
x0=[1 1 1]; %初值
ans1 = ones(20,3);
options=optimset('fsolve');
for k=1:20
[x,fval,exitflag,output]=fsolve(@solveproblem,x0,options);
x0=x;
ans1(k,:)=x;
end

ans1 =
-0.2892 0.9143 0.1893
带入三个方程验证，分别等于0.0012，4.8363e-004，-0.0016

追问

能看下我的问题补充，非常感谢~

追答

function output=solveproblem(X1,Y1)
c=X1(1);
m=X1(2);
y=X1(3);

X=Y1(1);
Y=Y1(2);
Z=Y1(3);

output(1)=(1-c)*(1-y)*(1-m)*10.61 + y*(1-c)*(1-m)*30.56 + c*(1-y)*(1-m)*60.67 + m*(1-c)*(1-y)*0.95 + y*m*(1-c)*28.45+y*c*(1-m) *8.19 + c*m*(1-y)*2.51 + c*y*m*76.69 - X
output(2)=(1-c)*(1-y)*(1-m)*15.21 + y*(1-c)*(1-m)*14.58 + c*(1-y)*(1-m)*65.2 + m*(1-c)*(1-y)*0.96 + y*m*(1-c)*14.23+y*c*(1-m) *17.41+ c*m*(1-y)*1.62 + m*y*c*80.32 - Y
output(3)=(1-c)*(1-y)*(1-m)*44.77 + y*(1-c)*(1-m)*9.9 + c*(1-y)*(1-m)*3.97 + m*(1-c)*(1-y)*0.78 + y*m*(1-c)*0.8+y*c*(1-m) *3.3 + c*m*(1-y)*7.95 + m*c*y*66.79 - Z
-----------------------------------------
XYZ=[38.98 38.38 32.29;
7.07 9.73 10.97;
0.67 0.85 0.79;
69.85 68.99 56.01;
1.11 1.7 1.6;
52.05 54.56 49.08;
3.21 2.15 3.76;
44.6 46.98 41.77;
1.21 1.56 1.46;
63.46 63.94 51;
3 4.05 4.41;
13.01 13.6 11.03];

[m,n]=size(XYZ);
x0=[1 1 1]; %初始值
ans1 = [zeros(m,3) XYZ]; %每行1~3列保存结果，4~6列保存X,Y,Z值
options = optimset('fsolve');

for k=1:m
[x,fval,exitflag,output]=fsolve(@solveproblem,x0,options,XYZ(k,:));
ans1(k,1:3)=x;
end

Answer 2

你这能不能把它给简化一下？晕

追问

其实简化一下也是这样，现在这个式子看着很复杂，其实不是很复杂的。最多都是不超过三个，求导的时候应该也比较好求