## 2 Methods: Sparse Regularization and Identification  
When addressing parametric problems, the use of regressions is quite normal [8]. We consider, for the sake of simplicity, a simple model \(y(\mu)\), with \(y\) being the quantity of interest that depends on parameter \(\mu\). If three data are assumed to be known, \(\left\{(\mu_{1},y_{1}),(\mu_{2},y_{2})\text{ and }(\mu_{3},y_{3})\right\}\), one is tempted to approximate function \(y(\mu)\) by using a polynomial basis according to the following. \[y(\mu)=a_{0}+a_{1}\mu+a_{2}\mu^{2}. \tag{1}\]  
If we assume that these data are free of noise, it is possible to calculate the three approximation coefficients, \(a_{0}\), \(a_{1}\) and \(a_{2}\), from relation (1) by solving the following system. \[\left( \begin{array}{ccc}1&\mu_{1}&\mu_{1}^{2}\\ 1&\mu_{2}&\mu_{2}^{2}\\ 1&\mu_{3}&\mu_{3}^{2}\end{array} \right)\left( \begin{array}{c}a_{0}\\ a_{1}\\ a_{2}\end{array} \right)=\left( \begin{array}{c}y_{1}\\ y_{2}\\ y_{3}\end{array} \right). \tag{2}\]  
In the case of having M data, the approximation generalizes to the following:  
\[y(\mu)=\sum_{k=0}^{\mathsf{M}-1}a_{k}\mu^{k}, \tag{3}\]  
and the procedure for computing the coefficients \(a_{k}\), \(k=0,\ldots,\mathsf{M}-1\) follows a similar procedure. Thus, the approximation richness follows the data's availability. However, we would like to increase the richness without necessarily increasing the amount of data. Imagine, for a while, that the problem solution reads \(y(\mu)=\mu^{3}\). The question is as follows: Can we determine this solution with only three data? Obviously, by using approximation basis \(\{1,\mu,\mu^{2}\}\), it is not possible. A possibility consists of taking a richer basis, e.g., \(\{1,\mu,\mu^{2},\mu^{3},\mu^{4}\}\), that, as it can be noticed, contains the solution; however, if only three data are available, the resulting algebraic system  
\[\left( \begin{array}{ccccc}1&\mu_{1}&\mu_{1}^{2}&\mu_{1}^{3}&\mu_{1}^{4}\\ 1&\mu_{2}&\mu_{2}^{2}&\mu_{2}^{2}&\mu_{2}^{4}\\ 1&\mu_{3}&\mu_{3}^{3}&\mu_{3}^{3}&\mu_{3}^{4}\end{array} \right)\left( \begin{array} []{c}a_{0}\\ a_{1}\\ a_{2}\\ a_{3}\\ a_{4}\end{array} \right)=\left( \begin{array}{c}y_{1}\\ y_{2}\\ y_{3}\end{array} \right), \tag{4}\]  
is underdetermined (there are infinite solutions). Thus, depending on the strategy followed for solving it (backslash, pseudo-inverses, optimization formulation with different norm choices, etc. ), different solutions will be obtained, all them representing perfectly the three data, but they will differ when considering any other value \(\mu\
eq\{\mu_{1},\mu_{2},\mu_{3}\}\). If an algorithm can provide the right solution, in our case \(a_{3}=1\) and \(a_{0}=a_{1}=a_{2}=a_{4}=0\), one can consider an extremely rich approximation basis, enabling the approximation of the searched solution by selecting very few terms in it. The amount of data will scale with the number of terms in the rich basis needed for approximating the searched solution instead of scaling with the approximation's basis size. Thus, when addressing our problem \(y(\mu)\), we consider a very rich approximation basis composed of M linearly independent functions \(\mathcal{G}_{i}(\mu)\), \(i=1,\ldots,\text{M}\) and D data \(y_{k}\), \(k=1,\ldots,\text{D}\), with \(\text{D}\ll\text{M}\), and the resulting algebraic system reads:  
\[\left( \begin{array}{ccccc}\mathcal{G}_{1}(\mu_{1})&\mathcal{G}_{2}(\mu_{1}) &\cdots&\mathcal{G}_{\text{M}}(\mu_{1})\\ \mathcal{G}_{1}(\mu_{2})&\mathcal{G}_{2}(\mu_{2})&\cdots&\mathcal{G}_{\text{ M}}(\mu_{2})\\ \vdots&\vdots&\ddots&\vdots\\ \mathcal{G}_{1}(\mu_{\text{D}})&\mathcal{G}_{2}(\mu_{\text{D}})&\cdots& \mathcal{G}_{\text{M}}(\mu_{\text{D}})\end{array} \right)\left( \begin{array} []{c}a_{0}\\ a_{1}\\ \vdots\\ a_{\text{M}-1}\end{array} \right)=\left( \begin{array}{c}y_{1}\\ y_{2}\\ \vdots\\ y_{\text{D}}\end{array}