## 3 Higher-Order Equivalent Layer-Wise Approach
The first possible choice consists in assuming them equal to the corresponding Legendre polynomials. In other words, they can be defined as follows  
\[F_{\tau}=\left\{ \begin{array}{ll}\frac{L_{1}-L_{2}}{2}=\frac{1-z_{k}}{2}& \text{for }\tau=0\\ L_{\tau+2}-L_{\tau}&\text{for }\tau=1,2,\ldots,N-1\\ \frac{L_{1}+L_{2}}{2}=\frac{1+z_{k}}{2}&\text{for }\tau=N\end{array} \right. \tag{12}\]  
in which the symbol \(L_{\tau+1}\) is introduced to specify the \(\tau\)-th order Legendre polynomial. They can be computed conveniently in the closed interval \(z_{k}\in[-1,1]\) by using the following recursive formula (as shown in [3])  
\[L_{\tau}(z_{k})=\frac{(2\tau-3)z_{k}L_{\tau-1}(z_{k})-(\tau-2)L_{\tau-2}(z_{k })}{\tau-1} \tag{13}\]  
for \(\tau=3,4,\ldots,N\), assuming \(L_{1}(z_{k})=1\) and \(L_{2}(z_{k})=z_{k}\). For clarity purposes, let us consider a fourth order of kinematic expansion (\(N=4\)). The displacement field (9) becomes  
\[ \begin{split} U_{1}&=F_{0}u_{1}^{(0)}+F_{1}u_{1}^{( 1)}+F_{2}u_{1}^{(2)}+F_{3}u_{1}^{(3)}+F_{4}u_{1}^{(4)}+F_{5}u_{1}^{(5)}\\ U_{2}&=F_{0}u_{2}^{(0)}+F_{1}u_{2}^{(1)}+F_{2}u_{2}^{(2)}+ F_{3}u_{2}^{(3)}+F_{4}u_{2}^{(4)}+F_{5}u_{2}^{(5)}\\ U_{3}&=F_{0}u_{3}^{(0)}+F_{1}u_{3}^{(1)}+F_{2}u_{3}^{(2)}+ F_{3}u_{3}^{(3)}+F_{4}u_{3}^{(4)}+F_{5}u_{3}^{(5)}\end{split}  \tag{14}\]  
if the [PERSON]'s function is embedded in the model. Due to expressions (12) the first five Legendre polynomials are required  
\[ \begin{split} L_{1}&=1\\ L_{2}&=z_{k}\\ L_{3}&=\frac{3z_{k}^{2}-1}{2}\\ L_{4}&=\frac{5z_{k}^{3}-3z_{k}}{2}\\ L_{5}&=\frac{35z_{k}^{4}-30z_{k}^{2}+3}{8}\end{split}  \tag{15}\]  
As a consequence, the thickness functions needed for the kinematic expansion at issue take the following aspect \[F_{0} =\frac{1-z_{k}}{2} \tag{16}\] \[F_{1} =\frac{3}{2}\big{(}z_{k}^{2}-1\big{)}\] \[F_{2} =\frac{5}{2}z_{k}\big{(}z_{k}^{2}-1\big{)}\] \[F_{3} =\frac{7}{8}\big{(}5z_{k}^{4}-6z_{k}^{2}+1\big{)}\] \[F_{4} =\frac{1+z_{k}}{2}\] \[F_{5} =(-1)^{k}z_{k}\]  
It should be pointed out that \(F_{5}\) is evidently the Murakami's function (\(F_{5}=F_{N+1}=Z\)). If the Legendre polynomial are used, the coordinates \(\zeta_{\tau}\) which define the kinematic expansion can be located by evaluating the roots of the Legendre polynomial of highest order. Alternatively, the Lagrange polynomials can be used as thickness functions simply assuming \(F_{\tau}(\zeta)=l_{\tau+1}(\zeta)\), in which \(l_{\tau+1}(\zeta)\) denotes the Lagrange polynomial of order \(N\) defined as follows  
\[l_{\tau+1}(\zeta)=\prod_{m=0,m\
eq\tau}^{N}\frac{(\zeta-\zeta_{m})}{(\zeta_{ \tau}-\zeta_{m})} \tag{17}\]  
for \(\tau=0,1,\ldots,N-1,N\). The coordinates \(\zeta_{m}\), for \(m=0,1,\ldots,N-1,N\), can be chosen taking into account an arbitrary distribution of points in the closed interval \(\zeta_{m}\in[-h/2,h/2]\). In the present paper, a uniform (or equally spaced) grid distribution is considered for the sake of simplicity. If the Lagrange polynomials are employed as thickness functions, one gets  
\[F_{\tau}=\left\{ \begin{array}{ll}l_{1}(\zeta)&\text{for $\tau=0$}\\ l_{\tau+1}(\zeta)&\text{for $\tau=1,2,\ldots,N-1$}\\ l_{N+1}(\zeta)&\text{for $\tau=N$}\end{array} \right. \tag{18}\]  
For the sake of completeness, it should be noted that the displacement field (9) without the [PERSON]'s function is equivalent to the displacement approximation of the sampling surfaces (SaS) formulation presented in the works [104; 105], if the Lagrange polynomials (17)-(18) are employed as thickness functions. As in the previous approach, the \((N+1)\)-th thickness function is still given by the [PERSON]'s function (\(F_{N+1}=Z\)). Using both the Legendre polynomials and the Lagrange ones, the thickness functions have the following properties  
\[F_{\tau}(\zeta=\zeta_{m})=\left\{ \begin{array}{ll}0&\text{for $\tau\
eq m$}\\ 1&\text{for $\tau=m$}\end{array} \right. \tag{19}\]  
with \(\tau,m=0,1,2,\ldots,N\).