我一直试图用16个方程(线性)和解决命令来解决16个变量。
下面我粘贴我的代码
syms x11 x12 x13 x14 x21 x22 x23 x24 x31 x32 x33 x34 x41 x42 x43 x44;
%defining symbolic variables
x=[x11 x12 x13 x14; x21 x22 x23 x24; x31 x32 x33 x34; x41 x42 x43 x44]; % putting all those variables in matrix form
N4=[-1 3 -3 1;3 -6 3 0;-3 3 0 0;1 0 0 0]; % 4*4 control matrix
digits(4);% setting precision to 4 digits instead of default 32
syms sx;% defining a variable sx
for i=1:8
for j=1:8
u=(i-1)/7;
w=(i-1)/7;
sx(i,j)= [u^3 u^2 u 1]*N4*x*N4'[w^3;w^2;w;1]; % this gives me 8*8 matrix
end
end
for a=1:8
for b=1:8
kx(a,b)=b; % the x coordinates of 64 points of the image
end
end
ex=0 % defining variable ex which is the error function.
for i=1:8
for j=1:8
ex= ex+ (sx(i,j)-kx(i,j))^2;
end
end
diff_x11=vpa(diff(ex,x11)); % I want to find values of x11,x12...
diff_x12=vpa(diff(ex,x12)); % for which the error ex is minimum
diff_x13=vpa(diff(ex,x13)); % so I differentiate ex with all 16
diff_x14=vpa(diff(ex,x14)); % variables and get 16 equations
diff_x21=vpa(diff(ex,x21));% and then try to solve them
diff_x22=vpa(diff(ex,x22));
diff_x23=vpa(diff(ex,x23));
diff_x24=vpa(diff(ex,x24));
diff_x31=vpa(diff(ex,x31));
diff_x32=vpa(diff(ex,x32));
diff_x33=vpa(diff(ex,x33));
diff_x34=vpa(diff(ex,x34));
diff_x41=vpa(diff(ex,x41));
diff_x42=vpa(diff(ex,x42));
diff_x43=vpa(diff(ex,x43));
diff_x44=vpa(diff(ex,x44));
qx=solve(diff_x11,diff_x12,diff_x13,diff_x14,diff_x21,diff_x22,diff_x23,diff_x24,diff_x31,diff_x32,diff_x33,diff_x34,diff_x41,diff_x42,diff_x43,diff_x44,x11,x12,x13,x14,x21,x22,x23,x24,x31,x32,x33,x34,x41,x42,x43,x44);
我使用了如上所示的solve命令。最初我只得到变量x11和x44的2个值。那个时候精度是默认的32位数。当我降低精度并开始使用代码中显示的'vpa'命令时,我得到了另外两个值x22和x33。最后,我只得到16个中的4个变量的解,其中4个是带有可变元素的4 * 4矩阵的对角元素
我需要得到所有16个。任何帮助将不胜感激。谢谢Ankit
答案 0 :(得分:0)
如果你正在解决线性方程,你为什么要这么复杂地做呢?
例如,如果要解决线性方程组的问题
Ax+By=C
Dx+Ey=F
Gx+Hy=I
即。你想知道最好(在最小二乘意义上)适合三个方程的数值[x; y],你写的
%# A,B, etc are numbers, or arbitrarily complex functions
%# that evaluate to a numerical value at the time the equations are solved.
matrixA = [A,B;D,E;G,H];
matrixB = [C;F;I];
xyVector = matrixA \ matrixB;