Can my code be written like this?

i want to write this formula as code in matlab,but i am not very sure that is my code right.

The range of j and k are both 1~K

for k= 1:K
     for j=1:K
        if k==j
            a=mu(k).*( abs(h_(k)'*f_(k)^2 ) / hat_r(k);
        else
           b=mu(k).*( abs(h_(k)'*f_(j)^2 );
        end      
    end
    total=a+b;
end

[Walter Roberson]

No, your are overwriting b for every iteration of for j except the k==j case, and you are not totalling those b cases.

[yang-En Hsiao]

can you show me your modification? thx a lot

[Walter Roberson]

It is not obvious to me what K_k is

[Jan]

Neither the elementwise .* does not matter, when you work with scalars. Using conj looks less confusing than using the transposition.

In abs(h_(k)'*f_(k)^2) you are squaring f only. Use this instead:

abs(conj(h_(k)) * f_(k))^2

[yang-En Hsiao]

K_k means that the range of j is the same as k,that is 1 ~K.but j can't not be equal to k

Answer 1

[Jan]

There is no "hat_r(k)" in your formula. A trailing parenthesis is missing also.

total = 0;
hc    = conj(h_);
for k = 1:K
    c1   = ;
    sumj = 0;
    for j = 1:K
        if j ~= k   % [EDITED]
          sumj = sumj + abs(hc(k) * f_(j)) ^ 2;
        end
    end
    total = total + mu(k) * (abs(hc(k) * f_(k)) ^ 2 - sumj) / r_hat(k);
end

[yang-En Hsiao]

it should be bar_r(k).there is a bar_r(k) ,1.,in the picture.

and the 2 means that the range of j is the same as k,that is 1 ~K.but j can't not be equal to k

[Jan]

If "it" should be "bar_r(k)", than simply insert this in my code. You do not have to wait until I do it. By the way, the character is a greek "gamma", not an "r". But names do not matter.

There is still a missing closing parenthesis. Please post a valid formula. A smaller zoom-level would be nice.