$\mathrm{LagrangeBasis}\left(k,\mathrm{nodes},x\right)=w_k\prod _{j\ne k}\left(x-{\mathrm{nodes}}_j\right)$ defines the $k$th Lagrange polynomial of degree $n$ which is either $1$ or $0$ on the given nodes. By convention, the nodes are indexed from $0$, so $\mathrm{nodes}=\left[x_0,x_1,...,x_n\right]$ , and the barycentric weights $w_k$ are defined as $w_k=\prod _{j\ne k}\left(\frac{1}{x_k-x_j}\right)$.

At present, this can only be evaluated in Maple by prior use of the object-oriented representation obtained by P:=convert(p,MatrixPolynomialObject,x) and subsequent call to P:-Value(<x-value>) , which uses the numerically stable barycentric form to evaluate the polynomial p.

$\mathrm{nodes}≔\left[-1\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]$

$\mathrm{nodes}≔\left[\mathrm{-1}\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]$

$p≔3\mathrm{LagrangeBasis}\left(0\&comma;\mathrm{nodes}\&comma;x\right)+5\mathrm{LagrangeBasis}\left(2\&comma;\mathrm{nodes}\&comma;x\right)+7\mathrm{LagrangeBasis}\left(3\&comma;\mathrm{nodes}\&comma;x\right)$

$p≔3\mathrm{LagrangeBasis}\left(0\&comma;\left[\mathrm{-1}\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]\&comma;x\right)+5\mathrm{LagrangeBasis}\left(2\&comma;\left[\mathrm{-1}\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]\&comma;x\right)+7\mathrm{LagrangeBasis}\left(3\&comma;\left[\mathrm{-1}\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]\&comma;x\right)$

$P≔\mathrm{convert}\left(p\&comma;\mathrm{MatrixPolynomialObject}\&comma;x\right)$

$P≔\mathrm{Record}\left(\mathrm{Value}={\mathrm{Default}}_{\mathrm{value}}\&comma;\mathrm{Variable}=x\&comma;\mathrm{Degree}=3\&comma;\mathrm{Coefficient}=\mathrm{coe}\&comma;\mathrm{Dimension}=\left[1\&comma;1\right]\&comma;\mathrm{Basis}=\mathrm{LagrangeBasis}\&comma;\mathrm{BasisParameters}=\left[\left[\mathrm{-1}\&comma;-\frac{1}{3}\&comma;\frac{1}{3}\&comma;1\right]\right]\&comma;\mathrm{IsMonic}=\mathrm{mon}\&comma;\mathrm{OutputOptions}=\left[\mathrm{shape}=\left[\right]\&comma;\mathrm{storage}=\mathrm{rectangular}\&comma;\mathrm{order}=\mathrm{Fortran\_order}\&comma;\mathrm{fill}=0\&comma;\mathrm{attributes}=\left[\right]\right]\right)$

$P:-\mathrm{Degree}\left(\right)$

$3$

$\mathrm{seq}\left(P:-\mathrm{Value}\left(\mathrm{nodes}\left[k\right]\right)\left[1,1\right]\&comma;k=1..\mathrm{nops}\left(\mathrm{nodes}\right)\right)$

$3,0,5,7$

$P:-\mathrm{Value}\left(0.3\right)$

$\left[\begin{array}{c}4.729812500\end{array}\right]$

$\mathrm{factor}\left(P:-\mathrm{Value}\left(t\right)\left[1,1\right]\right)$

$-\frac{\left(33t^2-26t-35\right)\left(3t+1\right)}{16}$