This is my problem: I am looking for a way to optimally define the number of breaking points I need in order to approximate a quadratic function to a linear one. Right now, I have selected previously the number of breaking points ($m$) involved in the linearisation, but I'm dubious that this is causing me numerical stability issues in the whole problem. So in a general form I have
$y=x^{2}$
where
$x\in \mathbb{R} , \quad 0<x<x_{max}, \quad m\in \mathbb{Z}$
I am working with MILP and used separable programming to add non-linear constraints into the problem. The constraint is:
$x_{1}^{2} - x_{2}^{2} \ge K$
with the linearisation approach we have: \begin{equation}\label{eq:sum_convex_separable2} x_1=\sum_{i=0}^{r}m_{1i}\lambda_{1i} \quad \text{and} \quad \phi_1(x_1)\approx \sum_{i=0}^{r}\phi_1(m_{1i})\lambda_{1i} \end{equation} \begin{equation}\label{eq:sum_convex_separable3} x_2=\sum_{i=0}^{r}m_{2i}\lambda_{2i} \quad \text{and} \quad \phi_2(x_2)\approx \sum_{i=0}^{r}\phi_2(m_{2i})\lambda_{2i} \end{equation} The boundaries for $x_1$ and $x_2$ are:
$0\le x_1 \le x_{1_{max}}\qquad 0\le x_2 \le x_{2_{max}}\quad x_1,x_2 \in \mathbb{R} $
where lambdas need to meet the next adjacency conditions for both cases of $x_1$ and $x_2$: \begin{equation}\label{eq:sum_convex_separable_constraints} \lambda_0\le y_1, \quad \lambda_i \le y_i + y_{i+1} \quad \text{for} \quad i=1,\dots r-1, \quad \lambda_r \le y_r \end{equation}
\begin{equation}\label{eq:sum_convex_separable_constraints2} \sum_{i=0}^{r}\lambda_i=1, \quad \sum_{i=1}^{r}y_i=1, \end{equation} and \begin{equation}\label{eq:sum_convex_separable_constraints3} (\lambda,y)\ge 0, \quad y\in \{0,1\} \end{equation}
At this moment, I selected the number of breaking points ($m$) arbitrarily and the space between kept constant, but I am not sure if this can be improved by changing the space between them.
What approach would be recommended here?
Thank you guys so much in advance,
