Linear Separator in Higher Dimension - Theory Example Explanation We have the following dataset:
$$ \begin{bmatrix}
x_1 & x_2 & y\\ 
+1 & +1 & +1\\ 
-1 & +1 & +1\\ 
0 & -1 & +1\\ 
0 & 0 & -1
\end{bmatrix} $$
I was asked to find a mapping $\varphi$ to 3-dimenssion such that in the new dimension this data will become linear separable, and find this linear separator.
Here is the answer, I'd love if someone can explain me how to get it, and give me some intuition, because I didn't understand it:
$$\varphi(x)=(x_1~,~~x_2~,~~x_1^2+x_2^2)$$
By choosing the weights $(w_1,w_2,w_3,b)=(0,0,1,-0.5)$ resulting the predictor:
$$h(x_1,x_2)=sign(<w,~\varphi(x_1,x_2)>)=sign(<(0,0,1,−0.5), (_1 , _2 , _1^2+x_2^2, 1) >) = sign(x_1^2+x_2^2-0.5)$$
Thanks!

It's just mathematical intuition. The table says that if any of $x_i$ has absolute value $1$, the result is $1$, otherwise $0$. You can fit any other mapping to satisfy this condition. For example,
$|x_1|+|x_2|$ would be another answer, so as $x_1^4+x_2^4$.
As for the weights, since we have found a mapping that defines the label, other features have no importance. Thus, their weights can be $0$, and weight corresponding to the mapping can be $1$.
The bias term is adjusted such that if the features are $0$, $\varphi(x)+b$ is negative (such that sign function can give $-1$). This could be $-0.6$ as well, but typically, the middle point is chosen, due to robustness reasons.
