How can I find the iteration error? The norm function is giving me a dimension error.

Question

Ruten9 2015-11-28

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/257992-how-can-i-find-the-iteration-error-the-norm-function-is-giving-me-a-dimension-error

编辑： Stephen23 2015-11-30

%%Part A
clear all;clc;
% 4x1 + 2x2 + x3 = 11
% -x1 + 2x2 = 3
% 2x1 + x2 + 4x3 = 16
a= [4 2 1;-1 2 0;2 1 4];
b=[11 3 16]';
X=a\b;
xpart1=X
%%Part B
clc;
% x1= 2.75 - .5x2 -.25x3
% x2= 1.5 + .5x1
% x3 = 4 -.5x1 -.25x2
B= [0 -.5 -.25; .5 0 0; -.5 -.25 0];
c= [2.75;1.5;4];
x0=[0;1;2];
tol = 1e-10;
X = x0; k = 1;
while 1
   xnew = B*X + c;
   if norm(xnew-X) < tol
      break;
   end
   X = xnew;
   k = k+1;
end
k
xpart2= X
error= abs(xpart1-xpart2)
%%Part C
clc;
X(:,1)=x0; k=1;
while 1
   X(:,k+1) = B*X(:,k) + c;
   if norm(X(:,k+1)-X(:,k)) < tol
      break;
   end
   k = k+1;
end
k
x=X(:,k)
%%Iteration Plot
K=0:28;
bottom=X(1,:);
middle=X(2,:);
top=X(3,:);
figure;
plot(K,bottom,'b-','LineWidth',2)
hold on
plot(K,middle,'g-','LineWidth',2)
hold on
plot(K,top,'r-','LineWidth',2)
axis([0 37 0 4])
xlabel('iteration index,  k')
ylabel('x(k)')
title('iterative solution')
ax=gca;
ax.YTick=[0:4]
legend('x_{1}','x_{2}','x_{3}','Location','se')
%%Iteration Error
E(K)=norm(a*X -b);
% u=0:28
% plot(u,E(k))

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Answer 1

Walter Roberson 2015-11-29

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/257992-how-can-i-find-the-iteration-error-the-norm-function-is-giving-me-a-dimension-error#answer_201487

在 MATLAB Online 中打开

You define

K=0:28;

and you try to do

E(K)=norm(a*X -b);

so you are trying to index E at 0, 1, 2, ... 28. But MATLAB does not allow you to index a variable at index 0. You should just be leaving that (K) out and using

E = norm(a*X - b);

You will still have a problem, though. At that point in your code, a is 3 x 3 and X is 3 x 29 so a*X is going to be 3 x 29 . But your b is 3 x 1 and you cannot subtract a 3 x 1 from a 3 x 29. And if you expand the 3 x 1 to be 3 x 29 then you will be taking norm() of a 3 x 29 matrix which is going to give you a scalar result. You should be considering looping.

Possibly, though, you would consider it acceptable to use

E = sqrt(sum(bsxfun(@minus,a*X,b).^2));