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

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

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. 
Re: Q4) h(x) = ax
I thought it would simply be (y1/x1 + y2/x2)/2 to find an a that minimizes the mean square error on two points, no?

Re: Q4) h(x) = ax
So, in this procedure we:
Pick two points; Find the best slope for those two points, the one that minimizes the squared error for those two points; Do this N times and average all the s Rather than: Pick two points; Calculate the squared error for those two points as a function of ; Do this N times, then find the that minimizes the sum of all of the squared errors, as we do with linear regression Are we doing the first thing here or the second thing? Either way there's a simple analytic solution, but I'm not sure which procedure we're doing. 
Re: Q4) h(x) = ax
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Re: Q4) h(x) = ax

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