is it possible to use "for loop" for matrix?

v=zeros(1,4);
X1=zeros(1,6);
X2=zeros(1,6);
for k=1:6
    for j=1:n
        X1(1,k)=(j^k)+X1(1,k);
        X2(1,k)=(j^k)+X2(1,k);
    end
end
g1=zeros(1,4);
g2=zeros(1,4);
for i=1:40
    for j=1:4
        g1(1,j)=sum((i.^(j-1)).*V(i,1));
        g2(1,j)=sum((i.^(j-1)).*I(i,1));
    end
end

x1=[40 X1(1,1) X1(1,2) X1(1,3); X1(1,1) X1(1,2) X1(1,3) X1(1,4); X1(1,2) X1(1,3) X1(1,4) X1(1,5); X1(1,3) X1(1,4) X1(1,5) X1(1,6)];
x2=[40 X2(1,1) X2(1,2) X2(1,3); X2(1,1) X2(1,2) X2(1,3) X2(1,4); X2(1,2) X2(1,3) X2(1,4) X2(1,5); X2(1,3) X2(1,4) X2(1,5) X2(1,6)];
k1=inv(x1);
k2=inv(x2);
a1=g1*k1;
a2=g2*k2;
V3=sqrt(((a1(1,2)^2)+(4*(a1(1,3)^2))));
i3=sqrt(((a2(1,2)^2)+(4*(a2(1,3)^2))));
tettav=atan((-2*a1(1,3))/a1(1,2));
tv=rad2deg(tettav);
tettai=atan((-2*a2(1,3))/a2(1,2));
ti=rad2deg(tettai);
Z=V3/i3;
tz=tv-ti;

how can write x1 and x2 with for loop instead of matrix?is it possible?

or is there anway so i could make this code shorter?

and is there any code soi could estimate matlab Measurement time?

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/toeplitz.html perhaps

[Walter Roberson]

You can use tic() tock() to estimate execution time

Answer 1

[Jan]

X1 = zeros(1,6);
X2 = zeros(1,6);
for k=1:6
    for j=1:n
        X1(1,k)=(j^k)+X1(1,k);
        X2(1,k)=(j^k)+X2(1,k);
    end
end

x1 = [40      X1(1,1) X1(1,2) X1(1,3); ...
      X1(1,1) X1(1,2) X1(1,3) X1(1,4); ...
      X1(1,2) X1(1,3) X1(1,4) X1(1,5); ...
      X1(1,3) X1(1,4) X1(1,5) X1(1,6)];

x2 = [40      X2(1,1) X2(1,2) X2(1,3);
      X2(1,1) X2(1,2) X2(1,3) X2(1,4); ...
      X2(1,2) X2(1,3) X2(1,4) X2(1,5); ...
      X2(1,3) X2(1,4) X2(1,5) X2(1,6)];

% Without loops:
X1 = sum((1:6).' .^ (1:6), 1);
X2 = X1;

% More compact:
x1 = [40, X1(1:3); X1(1:4); X1(2:5); X1(3:6)];
x2 = [40, X2(1:3); X2(1:4); X2(2:5); X2(3:6)];

But what is the idea of creating x1 and x2 with loops? Simpler code is easier to debug.

[arian hoseini]

X1 = sum((1:6).' .^ (1:6), 1);
X2 = X1;

about this part....it will change my answer...?!

[Jan]

Maybe. This might depend on what "n" is in for j=1:n . I guessed boldly that n=6. Now with flexibel n:

X1 = zeros(1,6);
X2 = zeros(1,6);
n  = 8;
for k = 1:6
    for j = 1:n
        X1(1,k)=(j^k)+X1(1,k);
        X2(1,k)=(j^k)+X2(1,k);
    end
end

X1_ = sum((1:n).' .^ (1:6), 1);
X2_ = X1;

isequal(X1, X1_)

ans = logical
   1

isequal(X2, X2_)

ans = logical
   1

[arian hoseini]

thank u sir

Answer 2

[Voss]

n = 2;
X1=zeros(1,6);
for k=1:6
    for j=1:n
        X1(1,k)=(j^k)+X1(1,k);
    end
end
X1

X1 = 1×6

3 5 9 17 33 65

% the way you have it now:
x1=[40 X1(1,1) X1(1,2) X1(1,3); X1(1,1) X1(1,2) X1(1,3) X1(1,4); X1(1,2) X1(1,3) X1(1,4) X1(1,5); X1(1,3) X1(1,4) X1(1,5) X1(1,6)]

x1 = 4×4

40 3 5 9 3 5 9 17 5 9 17 33 9 17 33 65

% the same, but written across multiple lines of code:
x1 = [ ...
    40 X1(1,1) X1(1,2) X1(1,3); ...
    X1(1,1) X1(1,2) X1(1,3) X1(1,4); ...
    X1(1,2) X1(1,3) X1(1,4) X1(1,5); ...
    X1(1,3) X1(1,4) X1(1,5) X1(1,6)]

x1 = 4×4

40 3 5 9 3 5 9 17 5 9 17 33 9 17 33 65

% another way to do the same - indexing 
% sets of elements at once instead of 
% each element of X1 individually:
x1 = [ ...
    40 X1(1,1:3); ...
    X1(1,1:4); ...
    X1(1,2:5); ...
    X1(1,3:6)]

x1 = 4×4

40 3 5 9 3 5 9 17 5 9 17 33 9 17 33 65

% another way to do the same - since X1 is
% a row vector, you can omit the first (row)
% index, i.e., X1(1,ii) is X1(ii):
x1 = [ ...
    40 X1(1:3); ...
    X1(1:4); ...
    X1(2:5); ...
    X1(3:6)]

x1 = 4×4

40 3 5 9 3 5 9 17 5 9 17 33 9 17 33 65

% another way to do the same - do all 
% of the indexing into X1 at once:
x1 = 40*ones(4);
idx = (0:3)+(0:3).';
good_idx = idx > 0;
x1(good_idx) = X1(idx(good_idx))

x1 = 4×4

40 3 5 9 3 5 9 17 5 9 17 33 9 17 33 65

[arian hoseini]

best answer ever

[Voss]

Thanks!