## 2 A Review of a New Class of Polynomials
### The Chatterjea Polynomials  
Let us first denote the Chatterjea polynomials (ChPs) of order \(h\) by \(\mathcal{S}_{h}\). In Rodrigues' format, the ChPs are defined by  
\[\mathcal{S}_{h}(x):=(3x)^{-h}\,e^{\frac{3}{2}}D^{h}(x^{3h}e^{-\frac{3}{2}}).\]  
On the other hand, they can be represented by the hypergeometric functions as  
\[\mathcal{S}_{h}(x):={}_{2}F_{0}(-h,1+2h;-;-\frac{1}{3}x).\]  
Some mathematical connections between this class of polynomials and the Laguerre and Bessel polynomials were also proved, as mentioned above. However, we are interested in the explicit form of ChPs. It is as follows  
\[\mathcal{S}_{h}(x)=\sum_{k=0}^{h}\frac{h!\,(2h+k)!}{k!\,(h-k)!\,(2h!)!}\Big{(} \frac{x}{3}\Big{)}^{k},\quad h\in\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}. \tag{3}\]  
According to the former formula, we easily find that \(\mathcal{S}_{0}(x)=1\). The list of the next five ChPs is given for the convenience of reference as  
\[\mathcal{S}_{1}(x) =1+x,\] \[\mathcal{S}_{2}(x) =1+\frac{10}{3}x+\frac{10}{3}x^{2},\] \[\mathcal{S}_{3}(x) =1+7x+\frac{56}{3}x^{2}+\frac{56}{3}x^{3},\] \[\mathcal{S}_{4}(x) =1+12x+60x^{2}+\frac{440}{3}x^{3}+\frac{440}{3}x^{4},\] \[\mathcal{S}_{5}(x) =1+\frac{55}{3}x+\frac{440}{3}x^{2}+\frac{5720}{9}x^{3}+\frac{40 040}{27}x^{4}+\frac{40040}{27}x^{5}.\]  
Clearly, we have \(\mathcal{S}_{h}(0)=1\) for all \(h\in\mathbb{N}\). Additionally, note that the coefficients of all ChPs are positive. The next result indicates that these ChPs are the solutions of a differential equation of the second order given by  
\[x^{2}\,\mathcal{S}_{h}^{\prime\prime}(x)+[3+(h+2)x]\mathcal{S}_{h}^{\prime}(x) =h(2h+1)\mathcal{S}_{h}(x),\quad h\in\mathbb{N}_{0}. \tag{4}\]Finally, we derive a decomposition for the vector of ChPs basis functions that will be utilized below. In this respect, we set  
\[\mathbf{S}_{H}(x):=[\mathcal{S}_{0}(x)\quad\mathcal{S}_{1}(x)\quad\mathcal{S}_{2}(x) \quad\ldots\quad\mathcal{S}_{H}(x)], \tag{5}\]  
where \(H\) is a given integer. In terms of monomial bases, we obtain the next result.  
**Lemma 1**.: _For an integer \(H\), we can reformulate \(\mathbf{S}_{H}(x)\) in (5) as_  
\[\mathbf{S}_{H}(x)=\mathbf{Y}_{H}(x)\,\mathbf{R}_{H}, \tag{6}\]  
_where we have_  
\[\mathbf{Y}_{H}(x)=\begin{bmatrix}1&x&x^{2}&\ldots&x^{H}\end{bmatrix},\]  
_and the structured matrix \(\mathbf{R}_{H}\) has an upper-triangular form. Its entries are extracted from relation (3). The matrix representation of \(\mathbf{R}_{H}\) is written as follows:_  
\[\mathbf{R}_{H}=\begin{bmatrix}\frac{0\,!\,0\,!}{0\,!\,0\,!\,3\,!}&\frac{1\,!2\,!}{0 \,!\,1\,!2\,!3\,!}&\frac{2\,!4\,!}{0\,!\,1\,!2\,!4\,!3\,!}&\ldots&\frac{(H-1)!\,!\,(2H-2)!\,!}{0\,!\,(H-1)!\,!\,(2H-2)!\,!\,3\,!}&\frac{H\,!\,(2H)!\,!}{0\,! \,H\,!(2H)!\,!\,3\,!}\\ 0&\frac{1\,!3\,!}{1\,!\,0\,!2\,!3\,!}&\frac{2\,!5\,!}{1\,!\,1\,!1\,!4\,!3\,!}& \ldots&\frac{(H-1)!\,!\,(2H-1)!\,!}{1\,!\,(H-2)!\,!\,(2H-2)!\,!\,3\,!}&\frac{H \,!\,(2H+1)!}{1\,!\,(H-1)!\,!\,(2H)!\,3\,!}\\ 0&0&\frac{2\,!6\,!}{2\,!\,0\,!4\,!3\,!}&\ldots&\frac{(H-1)!\,(2H)!\,}{2\,!\, (H-3)!\,\,(2H-2)!\,!\,3\,!}&\frac{H\,!\,(2H+2)!}{2\,!\,(H-2)!\,(2H)!\,!\,3\,!} \\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&\ldots&\frac{(H-1)!\,(3H-3)!}{(H-1)!\,!\,0\,!\,(2H-2)!\,3\,!^{H-1}}& \frac{H\,!\,(3H-1)!}{(H-1)!\,!\,1\,!(2H)!\,3\,!^{H-1}}\\ 0&0&0&\ldots&0&\frac{H\,(3H)!}{H!\,\,0\,!\,(2H)!\,3\,!}\\ \end{bmatrix}.\]  
_This matrix is nonsingular due to the fact that the diagonal elements are \(r_{i,i}=(3i)!/((2i)!\,3^{i})\) for \(0\leq i\leq H\)._  
Proof.: The proof is an easy exercise.