Cost Function, Linear Regression, trying to avoid hard coding theta. Octave.

You can use vectorize of operations in Octave/Matlab. Iterate over entire vector - it is really bad idea, if your programm language let you vectorize operations. R, Octave, Matlab, Python (numpy) allow this operation. For example, you can get scalar production, if theta = (t0, t1, t2, t3) and X = (x0, x1, x2, x3) in the next way: theta * X' = (t0, t1, t2, t3) * (x0, x1, x2, x3)' = t0*x0 + t1*x1 + t2*x2 + t3*x3 Result will be scalar.

For example, you can vectorize h in your code in the next way:

H = (theta'*X')';
S = sum((H - y) .^ 2);
J = S / (2*m);

Above answer is perfect but you can also do

H = (X*theta);
S = sum((H - y) .^ 2);
J = S / (2*m);

Rather than computing

(theta' * X')'

and then taking the transpose you can directly calculate

(X * theta)

It works perfectly.

The below line return the required 32.07 cost value while we run computeCost once using θ initialized to zeros:

J = (1/(2*m)) * (sum(((X * theta) - y).^2));

and is similar to the original formulas that is given below.

It can be also done in a line- m- # training sets

J=(1/(2*m)) * ((((X * theta) - y).^2)'* ones(m,1));