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  #1  
Old 07-12-2017, 05:41 PM
tikenn tikenn is offline
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Default Problem 3.6(a)

I think this is similar to the answer provided for problem 3.5, but I am having a difficult time understanding why, in Problem 3.4, the error is continuous and differentiable at the point y_n\textbf{w}^T\textbf{x}_n=1. I have the three cases looking like this so far:

y_n\textbf{w}^T\textbf{x}_n>1 in which I believe E(\mathbf{x}_n) = 0
y_n\textbf{w}^T\textbf{x}_n<1 in which I believe E(\mathbf{x}_n) = (1 - y_n\textbf{w}^T\textbf{x}_n)^2
y_n\textbf{w}^T\textbf{x}_n=1 in which I believe E(\mathbf{x}_n) = 0

These cases would make y_n\textbf{w}^T\textbf{x}_n=1 non-differentiable because the derivatives from the left and right are different. Am I evaluating the error from the book wrong --> e_n(\textbf{w})=(\max(0,1-y_n\textbf{w}^T\textbf{x}_n))^2?
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Old 07-15-2017, 02:27 PM
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htlin htlin is offline
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Default Re: Problem 3.6(a)

Are the derivatives on the two sides really different? :-)
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Old 07-16-2017, 12:36 AM
tikenn tikenn is offline
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Default Re: Problem 3.6(a)

I apologize, I realize I made an error in labeling the problem that my question refers to. My original question actually refers to Problem 3.4a.

Quote:
Originally Posted by htlin View Post
Are the derivatives on the two sides really different? :-)
Anyway, thank you for the hint! My confusion was with how problem 3.5 and problem 3.4 were so different, but I forgot to evaluate the gradients all the way through for both. Thanks!
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Old 08-24-2017, 05:05 AM
subbupd subbupd is offline
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Default Re: Problem 3.6(a)

Correct - evaluating the gradients would work!
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