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  #1  
Old 09-16-2018, 12:58 PM
venkatesh-devale venkatesh-devale is offline
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Post problem 1.2

Hello All,

I am not able to understand proceed on this problem. Please can some one just explain how can we do this?

I converted x2 = a*x1 + b to a*x1 + b - x2 = 0 for the line equation and hence the w0 = b, w1 = a and w2 = -1. Is this even correct?

Can someone please help me on this?
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  #2  
Old 09-16-2018, 08:26 PM
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htlin htlin is offline
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Default Re: problem 1.2

The problems ask you to express a and b by \mathbf{w}, not the other way around. So you should probably repeat what you do but from another angle.
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Old 09-16-2018, 10:15 PM
venkatesh-devale venkatesh-devale is offline
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Default Re: problem 1.2

So as x2 = a*x1 + b classifies the set correctly which means for every x1, x2 on this line a*x1 + b - x2 = 0 hence w0*x0 + w1 * x1 + w2 * x2 = 0 for this line.

Considering these equations can we say that a = w1, b = w0 ? Is this correct now. Further what does the part 2 means what type of picture is expected?
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Old 09-20-2018, 02:42 AM
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Default Re: problem 1.2

Quote:
Originally Posted by venkatesh-devale View Post
So as x2 = a*x1 + b classifies the set correctly which means for every x1, x2 on this line a*x1 + b - x2 = 0 hence w0*x0 + w1 * x1 + w2 * x2 = 0 for this line.

Considering these equations can we say that a = w1, b = w0 ? Is this correct now. Further what does the part 2 means what type of picture is expected?
If w_1 = 4, w_2 = 2, w_0 = 2 then the equation of the corresponding line is x_2 = -2 x_1 - 1. That is, a = -2, b = -1. The problem asks you to describe general procedure for any \mathbf{w} and then draw the corresponding line (along with which side to be positive) on the 2D plane for the specific \mathbf{w}. Hope this helps.
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Old 09-20-2018, 11:36 AM
venkatesh-devale venkatesh-devale is offline
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Default Re: problem 1.2

Thanks I got it, Thank you so much! It's great to have you guys around. There will be a lot of questions from me on this book as I am going through it and have no background.
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