Hw4 q4

[silvrous]

What is the range of a? [-1,1] of [-INF, +INF] ?

[kkkkk]

From my testing, a is within +/- pi.
This follows from the max/min gradient of sin(pi.x) = pi . cos(pi . x)
It is ok to use a larger range, just that the program will run slower.

[silvrous]

Thanks, but now I think I've hit another snag. Is the bias in class surely 0.21? My calculation shows it as exactly 0.31...

[Tyler]

I can't help with that but I have a more basic question. To get ax on two points, do we take the "a" based on the average of the two points

(y1+y2)/(x1+x2)
-or-

calculate "a" on each point and take the average?
1/2 (y1/x1 + y2/x2)

[silvrous]

The two give very different results, but shouldn't they be equivalent in grading?

[yaser]

Originally Posted by Tyler

I can't help with that but I have a more basic question. To get ax on two points, do we take the "a" based on the average of the two points

(y1+y2)/(x1+x2)
-or-

calculate "a" on each point and take the average?
1/2 (y1/x1 + y2/x2)

Choose the value of that minimizes the average mean-squared error on the two points.

[jmknapp]

I don't think (y1+y2)/(x1+x2) is valid, i.e., not the best choice for the line a*x. Maybe 1/2*(y1/x1 + y2/x2) is close but is it correct?

Seems like the way to go is to get a formula for the distance function (squared distance) and then minimize it. The answer from that process differs from 1/2*(y1/x1 + y2/x2).

EDIT: Prof. Mostafa posted while I was checking that result.

[kurts]

Is the bias in class surely 0.21?

In class, the hypothesis that resulted in a bias of 0.21 was of the form ax+b, where the line was free to be offset from the origin. In this problem, b is restricted to be 0, so one would expect the bias to be different.

[kkkkk]

Originally Posted by kurts

In class, the hypothesis that resulted in a bias of 0.21 was of the form ax+b, where the line was free to be offset from the origin. In this problem, b is restricted to be 0, so one would expect the bias to be different.

In Lecture 8 slide 15, for y=ax+b, g_bar passes through the origin even though it is not forced to. This is reasonable due to the 'symmetry' of the curve.