Hello, RBF kernel: \(\displaystyle k(x_i , x_j ) = φ(x_i )^Tφ(x_j ) = e^{-\frac{||x_i-x_j||^2}{2\sigma^2}}\) and the mapping function φ is unknown but I know that it maps some x into an infinite-dimensional space, I have to prove that subspace of φ would map to \(\displaystyle \mathbb{R}^∞\) with \(\displaystyle \sigma = \sqrt{2}\)
\(\displaystyle k(x_i , x_j ) = e^{-\frac{||x_i-x_j||^2}{2\sigma^2}} = e^{-\frac{||x_i-x_j||^2}{4}} \)
\(\displaystyle = e^{-\frac{(x_i-x_j)^T(x_i-x_j)}{4}}\)
\(\displaystyle = e^{-\frac{<x_i-x_j, x_i-x_j>}{4}} \)
\(\displaystyle = e^{-\frac{<x_i,x_i−x_j> − <x_j,x_i−x_j>}{4}} \)
\(\displaystyle = e^{-\frac{<x_i,x_i> − <x_i,x_j> − <x_j,x_i>+<x_j,x_j>}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2 + ||x_j||^2 − 2 <x_i,x_j>}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2 + ||x_j||^2}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2}{4}} \cdot e^{-\frac{||x_j||^2}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
\(\displaystyle = e^{-\frac{(x_i^Tx_i)}{4}} \cdot e^{-\frac{(x_j^Tx_j)}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
is it correct and what is the next step? I have to apply \(\displaystyle e^x = \sum\limits_{i=0}^{\infty}\frac{x^i}{i!} \) and \(\displaystyle k_d(x_i,x_j) = (x_i^Tx_j)^d\) somewhen
I guess for first term:
\(\displaystyle = e^{-\frac{(x_i^Tx_i)}{4}} = \sum\limits_{i=0}^{\infty}\frac{{-\frac{(x_i^Tx_i)}{4}}^i}{i!} = \) and this is not correct I think
\(\displaystyle k(x_i , x_j ) = e^{-\frac{||x_i-x_j||^2}{2\sigma^2}} = e^{-\frac{||x_i-x_j||^2}{4}} \)
\(\displaystyle = e^{-\frac{(x_i-x_j)^T(x_i-x_j)}{4}}\)
\(\displaystyle = e^{-\frac{<x_i-x_j, x_i-x_j>}{4}} \)
\(\displaystyle = e^{-\frac{<x_i,x_i−x_j> − <x_j,x_i−x_j>}{4}} \)
\(\displaystyle = e^{-\frac{<x_i,x_i> − <x_i,x_j> − <x_j,x_i>+<x_j,x_j>}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2 + ||x_j||^2 − 2 <x_i,x_j>}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2 + ||x_j||^2}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
\(\displaystyle = e^{-\frac{||x_i||^2}{4}} \cdot e^{-\frac{||x_j||^2}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
\(\displaystyle = e^{-\frac{(x_i^Tx_i)}{4}} \cdot e^{-\frac{(x_j^Tx_j)}{4}} \cdot e^{-\frac{−2(x_i^Tx_j)}{4}} \)
is it correct and what is the next step? I have to apply \(\displaystyle e^x = \sum\limits_{i=0}^{\infty}\frac{x^i}{i!} \) and \(\displaystyle k_d(x_i,x_j) = (x_i^Tx_j)^d\) somewhen
I guess for first term:
\(\displaystyle = e^{-\frac{(x_i^Tx_i)}{4}} = \sum\limits_{i=0}^{\infty}\frac{{-\frac{(x_i^Tx_i)}{4}}^i}{i!} = \) and this is not correct I think
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