two margin comparison and one conclusion? I read following notes, and couldn't get it. any idea or hint would highly appreciated. 
a SVM classifier using a second order polynomial kernel. The first polynomial kernel maps each input data x to $ \phi_1(x)=[x, x^2]$ . The second polynomial kernel maps each
input data x to $ \phi_2(x) = [2x, 2x^2]$ 
my question is how we reach the following conclusion? 

In general, is the margin we would attain using $ \phi_2(x)$ is
  greater in comparison to the margin resulting from using $ \phi_1(x)$.

 A: Since the value of $\phi_2$ is twice that of $\phi_1$, all the distances in $\phi_2$-space are twice as big as the distances in $\phi_1$-space. This means that the margin (which is roughly the "thickness" of the separating hyperplane that the SVM learns) is twice as big also. We can show this with an even simpler pair of kernel functions, $\phi_1(x, y) = (x, y)$ and $\phi_2(x) = (2x, 2y)$--the principle is exactly the same as with the pair of kernels you suggested.
If you have a dataset with positive points at $(0,0), (0,1)$ and negative points at $(1,0), (1,1)$, then using $\phi_1$ you'll learn the following SVM:

On the other hand, $\phi_2$ multiplies each coordinate by 2 relative to $\phi_1$, so you learn the following hyperplane instead:

As you can see, because all distances are inflated by a factor of 2, the margin is greater as well.
Appendix: R code for plots
do.plot <- function(D, main, sub) {
    plot(NA, xlim=c(-0.5, 2.5), ylim=c(-0.5, 2.5), xlab='x', ylab='y', main=main, sub=sub)
    points(c(0, 0), c(0, D), pch='+')
    points(c(D, D), c(0, D), pch=4)
    abline(v=0, lty=2)
    abline(v=D, lty=2)
    abline(v=D/2)
}
do.plot(1, 'phi1', 'margin=1')
do.plot(2, 'phi2', 'margin=2')

