Suport Vector Machines optimization – why multiply by 1/2? In the primal version of the SVM problem, the term $\|w\|^2$ is divided by 2 for mathematical convenience. I fail to see this so called mathematical convenience caused by this change. If one were to ignore this division by 2, the resultant dual form of the problem would have a coefficient with a value of 1/4 somewhere. By dividing by 2 in that earlier step, one instead has a coefficient of 1/2 in that same position. I fail to understand why this slight change from 1/4 to 1/2 is considered a major source of mathematical convenience. 
 A: It's to make the Langrange multiplier formulation a little less messy, specifically for the partial derivative to $\mathbf{w}$ as was hinted in the comment:
$$
\begin{align}
L_p &= \frac{1}{2}||\mathbf{w}||^2+C\sum_{i=1}^n\xi_i
-\sum_{i=1}^n\alpha_i\Big[y_i\big(\langle\mathbf{w},\varphi(\mathbf{x}_i)\rangle+b\big)-(1-\xi_i)\Big]-\sum_{i=1}^n\mu_i\xi_i, \\
\frac{\partial L_p}{\partial \mathbf{w}}&=\mathbf{w}-\sum_{i=1}^n \alpha_i y_i \varphi(\mathbf{x}_i),
\end{align}
$$
where $\xi$ is the vector of slack variables, $\alpha$ the vector of support weights, $b$ the bias and $\varphi(\cdot)$ the embedding function.
A direct result of this convention and the KKT conditions (which imply $\frac{\partial L_p}{\partial \mathbf{w}}=0$) is the well-known form of the separating hyperplane $\mathbf{w}$ in the feature space spanned by $\varphi(\cdot)$:
$$
\mathbf{w}=\sum_{i=1}^n \alpha_i y_i \varphi(\mathbf{x}_i),
$$
and the decision function $f(\cdot)$:
$$
f(\mathbf{z}) = \sum_{i=1}^n \alpha_i y_i \kappa(\mathbf{x}_i, \mathbf{z}) + b = \sum_{i=1}^n \alpha_i y_i \langle \varphi(\mathbf{x}_i), \varphi(\mathbf{z})\rangle + b,
$$
with $\kappa(\cdot,\cdot)$ the kernel function.
