How add another loop in program

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hasan s 2021-3-18

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编辑： Jan 2021-3-22

采纳的回答： Jan

value123.m

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First of all, thanks to all the experts who help others...I appreciate your time, And best wishes to everyone.

I have program ( take column1 column2 until column100 , to get 100 a and 100 b and 100 c)and I need to get the following...which print it as follows ,and I dont know where I add it in the same program since it depend on the output of it (that get 100 values of a ,b,c ).where A,B, C are the intial values.

c2=0;
for w=1:100
    c2=c2+1;
value1=value1+(((a(w))-A)^2)\100;  %a(w) is 100 value of a
value2=value2+(((b(w))-B)^2)\100;   %b(w) is 100 value of b
value3=value3+(((c(w))-C)^2)\100;  %c(w) is 100 value of c
end
value1;
value2;
value3;

and, (if possible) to add to the program" if else "for the error (or "while " of "if" as you show correct in my program)

if any Prof. can help me thanks alot.

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hasan s 2021-3-18

编辑：hasan s 2021-3-18

I donot know how I get the values of a,b,c that obtained inside every loop .

and put it as 100 values in the output as a 100 column to save it all ,, what I change in the program..please

Jan 2021-3-19

I do not exactly understand, what you want to change.

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采纳的回答

Jan 2021-3-19

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The only problems I see is that value1/2/3 is not initialized and that the division is /, not \ .

c2 is not used, so omit it.

value1 = 0;
value2 = 0;
value3 = 0;
for w = 1:100
   value1 = value1 + (a(w) - A)^2 / 100;
   value2 = value2 + (b(w) - B)^2 / 100;
   value3 = value3 + (c(w) - C)^2 / 100;
end
value1
value2
value3

Or vectorized as typical Matlab code:

value1 = sum((a - A)^2) / 100;
value2 = sum((b - B)^2) / 100;
value3 = sum((c - C)^2) / 100;

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Jan 2021-3-19

编辑：Jan 2021-3-20

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In your attached m-File, I assume this line contains a typo:

sum5=sum7+log((Q(i,j))/A);
%       ^  should this be a 5?

c1 and s are not used, so simply omit them.

A simplified version of your code:

Q = unifrnd(0,1,10,100); % matrix 10*100
A = 1; 
B = 2;
C = 3;
for j = 1:100
    sum1 = sum(exp(A*(1-exp((Q(:, j) / A).^C))));       
    sum2 = sum(exp(C*A*(1-exp((Q(:,j) / A).^C))) .* ...
               exp((Q(i,j) / A).^C));
    sum3 = sum(exp(B*A*(1-exp((Q(:,j) / A).^C))) .* ...
               (Q(i,j) / A).^C .* log(Q(i,j) / A).^2);
    sum4 = sum(exp(2*A*B*(1-exp((Q(:,j) / A).^C))));
    sum5 = sum(log(Q(:,j) / A));
    x = real([-2*A*sum1+2*sum2;C*sum3+A*sum4;-2*A*sum5]);
    y1 = -sum1;
    y2 = -sum2;
    y3 = -B*sum3;
    y4 = -sum2;
    y5 = -sum4;
    y6 = -B*A*sum5;
    y7 = -B*sum3;
    y8 = -B*A*sum5;
    y9 = C*sum1;
    d  = real([y1 y2 y3; y4 y5 y6; y7 y8 y9]);
    z0 = [A; B; C];
    p = d \ x;
    z = z0-p;
    a = z(1,1);   % Maybe all you want is: a(j) = ...
    b = z(2,1);   %                        b(j) = ...
    c = z(3,1);
    err=abs(z-z0);
% if any (err>[0.00001;0.00005;0.00002]) then stop  ( I don not know how err choose to continue untill 100 column , sinec I donot want to stop untill get take 100 j and in exam the questions are different)
%if not continue with another j
end  %to loop j

I do not undestand the part with err: You want to stop what? An iteration until err matchs specific limits would not be useful here, because the core of the loop operates on constant inputs, so a next iteration would give you the same result.