Problem 3.6(a)

tikenn · #1 07-12-2017, 06:41 PM

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 --> $e_n(\textbf{w})=(\max(0,1-y_n\textbf{w}^T\textbf{x}_n))^2$ ?

htlin · #2 07-15-2017, 03:27 PM

Are the derivatives on the two sides really different? :-)

tikenn · #3 07-16-2017, 01:36 AM

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

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!

subbupd · #4 08-24-2017, 06:05 AM

Correct - evaluating the gradients would work!

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