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#1
10-08-2013, 06:43 PM
 luwei0917 Junior Member Join Date: Aug 2013 Posts: 3
Problem 3.6

for part (a). why it is instead of 0? Thanks.
#2
10-10-2013, 08:15 AM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 597
Re: Problem 3.6

The point is to show that weights giving >0 implies the existence of weights giving >=1.
Quote:
 Originally Posted by luwei0917 for part (a). why it is instead of 0? Thanks.
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#3
10-09-2014, 09:09 AM
 squidsforbreakfast Junior Member Join Date: Oct 2014 Posts: 1
Re: Problem 3.6

I think I have an answer for parts b and c, but I'm not really sure if what I'm doing is correct. Can I check it with you?
For b), the thing we're trying to optimize is w, so then w would be z. So then we want to multiply the weights of the data (A) with w (z) to ensure that they are less than the signs of the data (y). I'm pretty sure that's right, however it feels a little hand-wavy to me, so even if it is right, I'm worried I may be missing some of the underlying logic of why it is the right answer.
For c), then, we want to minimize error, so then I think (c^T)z is the summation of the errors, b=y(w^Tx) and Az = 1-e (I couldn't figure out how to write xi in here).
But ten we have z being related to the summation of the errors in one equation, but related to just a single error instance in the other, so I think that what I have can't be right, even if it makes sense to me.
Can you offer a hint on what I'm doing wrong here?

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