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Aug 20, 2019 at 16:51 history edited babkr CC BY-SA 4.0
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Aug 20, 2019 at 15:20 history edited babkr CC BY-SA 4.0
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Aug 20, 2019 at 9:56 comment added Fabian Werner Note: We are not using this particular setup. For example, for the RBF kernel this transformation $\Phi$ does not map into a finite dimensional space but into the infinite dimensional space $l^2$. What you are referring to is then $k(x,x') = \langle \Phi(x), \Phi(x')\rangle$, i.e. a kernel is almost a dot product (up to this weird and often complicated transformation $\Phi$).
Aug 20, 2019 at 9:45 review First posts
Aug 20, 2019 at 10:37
Aug 20, 2019 at 9:43 history answered babkr CC BY-SA 4.0