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-   -   problem 1.2 (http://book.caltech.edu/bookforum/showthread.php?t=4856)

venkatesh-devale 09-16-2018 01:58 PM
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?

htlin 09-16-2018 09:26 PM
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.

venkatesh-devale 09-16-2018 11:15 PM
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?

htlin 09-20-2018 03:42 AM
Re: problem 1.2
 
Quote:
Originally Posted by venkatesh-devale (Post 13132)
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.

venkatesh-devale 09-20-2018 12:36 PM
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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