Prove: Gaussian RBF kernel would map into infinity space

[Sauraj]

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

 

[Sauraj]

Taking the Taylor expansion of the third factor,
$\displaystyle e^{-\frac{−2(x_i^Tx_j)}{4}} = e^{\frac{(x_i^Tx_j)}{2}} = \sum\limits_{i=0}^{\infty}\frac{ \frac{(x_i^Tx_j)^i}{2}}{i!} = \sum\limits_{i=0}^{\infty}\frac{(x_i^Tx_j)^i}{2i!} = 2 + \frac{(x_i^Tx_j)}{2} + \frac{(x_i^Tx_j)^2}{4} + ...$ now we see that RBF kernel is formed by taking an infinite sum over polynomial kernels. Is it correct?

 

[Sauraj]

ok does not matter, the solution is here http://pages.cs.wisc.edu/~matthewb/pages/notes/pdf/svms/RBFKernel.pdf