Slides polynomial kernel #109
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ludwigbothmann
requested changes
Mar 16, 2023
| \begin{equation*} | ||
| \begin{aligned} | ||
| \left ( \xv^T \xtil + b \right)^2 &= \left ( (x_1 \ x_2) \cdot \mat{\xt_1 \\ \xt_2} + b \right) \cdot \left ( (x_1 \ x_2) \cdot \mat{\xt_1 \\ \xt_2} + b \right) \\[10pt] | ||
| &= (x_1 \xt_1 + x_1 \xt_2 + x_2 \xt_1 + x_2 \xt_2 + b) \cdot \\ |
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should be
| &= (x_1 \xt_1 + x_1 \xt_2 + x_2 \xt_1 + x_2 \xt_2 + b) \cdot \\ | ||
| & \ \ \ \ \ \ (x_1 \xt_1 + x_1 \xt_2 + x_2 \xt_1 + x_2 \xt_2 + b) \\[10pt] | ||
| &= x_1^2 \xt_1^2 + x_2^2 \xt_2^2 + 2 x_1 \xt_1 x_2 \xt_2 + 2 b x_1 \xt_1 + 2 b x_2 \xt_2 + b^2 \\[10pt] | ||
| &= ( x_1^2 , x_2^2 , \sqrt{2} x_1 x_2 , \sqrt{2b} x_1 , \sqrt{2b} x_2 , b )^T \cdot \\ |
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this vector is not transposed - but the one in the next line is :)
| \begin{aligned} | ||
| \left ( \xv^T \xtil + b \right)^2 &= \left ( (x_1 \ x_2) \cdot \mat{\xt_1 \\ \xt_2} + b \right) \cdot \left ( (x_1 \ x_2) \cdot \mat{\xt_1 \\ \xt_2} + b \right) \\[10pt] | ||
| &= (x_1 \xt_1 + x_1 \xt_2 + x_2 \xt_1 + x_2 \xt_2 + b) \cdot \\ | ||
| & \ \ \ \ \ \ (x_1 \xt_1 + x_1 \xt_2 + x_2 \xt_1 + x_2 \xt_2 + b) \\[10pt] |
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- same as line above => remove the two summands in the middle
- why not use proper alignment here instead of
\\\\\\? :)
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Adding new slide for the non-homogeneus poly kernel. Also solving the noted typo
Let me know if it fits what you and Bernd expected for the slide or if you want to change anything.