
#1




Q4) h(x) = ax
This question is similar to that in the lectures i.e.,
in the lecture H1 equals h(x) = ax + b Is this question different to the lecture in the respect we shouldn't add "b" (i.e., X0 the bias/intercept) when applying? Or should I treat the same? My confusion is because in many papers etc a bias/intercept is assumed even if not specified i.e., h(x) = ax could be considered the same as h(x) = ax + b 
#2




Re: Q4) h(x) = ax
Quote:
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#3




Re: Q4) h(x) = ax
Thanks for comfirmation, much appreciated

#4




Re: Q4) h(x) = ax
Is there a best way to minimize the meansquared error? I am doing gradient descent with a very low learning rate (0.00001) and my solution is diverging! not converging. Is it not feasible to do gradient descent with two points when approximating a sine?
Thanks 
#5




Re: Q4) h(x) = ax
Never mind, I got my solution to converge, though I do not trust my answer. Oh well.

#6




Re: Q4) h(x) = ax
Quote:
(since linear regression is basically analytical formula for minimizing mean square error). Also, you can confirm if your g_bar from simulation makes sense by calculate it directly. (calculate expectation of the hypothesis from each (x1,x2) over [1,1] x [1,1] ). This involves two integrals but you can plug in the expression to wolfram or mathematica. 
#7




Re: Q4) h(x) = ax
Quote:
and note that X has just a single column (no column of 1's).
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