Thread: Q4) h(x) = ax
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Old 02-03-2013, 04:20 PM
sanbt sanbt is offline
Join Date: Jan 2013
Posts: 35
Default Re: Q4) h(x) = ax

Originally Posted by Axonymous View Post
I wanted confirmation of the result that I got using the method we are supposed to implement. So I derived the slope of the "best" line, shown to us in slide 11 of lecture 8. (Which also applies in our case because it goes through the origin.) I did this by minimizing the area in yellow on that slide. (You can actually see that slope is close to 1 from the slide.)

I was surprised that the answer I got for question 4 is so different from this "perfect" approximation line that was found by minimizing the integral. It stands to reason that it should vary a little, but there is quite a difference between the two values.
So the slope of the best line is just the slope of the line passing through each 2 points you picked each time. (implies that Ein= 0) But then you need 2D integral to average over that expression, over [-1, 1] x [-1,1]. The result should be close to the simulation.
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