The optimal hyperplane in SVM is defined as:
$$\mathbf w \cdot \mathbf x+b=0,$$
where $b$ represents threshold. If we have some mapping $\mathbf \phi$ which maps input space to some space $Z$, we can define SVM in the space $Z$, where the optimal hiperplane will be:
$$\mathbf w \cdot \mathbf \phi(\mathbf x)+b=0.$$
However, we can always define mapping $\phi$ so that $\phi_0(\mathbf x)=1$, $\forall \mathbf x$, and then the optimal hiperplane will be defined as $$\mathbf w \cdot \mathbf \phi(\mathbf x)=0.$$
Questions:
Why many papers use $\mathbf w \cdot \mathbf \phi(\mathbf x)+b=0$ when they already have mapping $\phi$ and estimate parameters $\mathbf w$ and theshold $b$ separatelly?
Is there some problem to define SVM as $$\min_{\mathbf w} ||\mathbf w ||^2$$ $$s.t. \ y_n \mathbf w \cdot \mathbf \phi(\mathbf x_n) \geq 1, \forall n$$ and estimate only parameter vector $\mathbf w$, assuming that we define $\phi_0(\mathbf x)=1, \forall\mathbf x$?
If definition of SVM from question 2. is possible, we will have $\mathbf w = \sum_{n} y_n\alpha_n \phi(\mathbf x_n)$ and threshold will be simply $b=w_0$, which we will not treat separately. So we will never use formula like $b=t_n-\mathbf w\cdot \phi(\mathbf x_n)$ to estimate $b$ from some support vector $x_n$. Right?