 Solution

The dual space $V^{\ast }$ consists of all real linear functionals on $V$. If $v_1,\dots ,v_d$ is a basis, its dual basis ${\epsilon }^1,\dots ,{\epsilon }^d$ is characterized by ${\epsilon }^i(v_j)={\delta }_{ij}$; explicitly ${\epsilon }^i$ extracts the $i$th coordinate. The natural map $J:V\to V^{\ast \ast }$ is $J(v)(\varphi )=\varphi (v)$. It is linear and injective, because the coordinate functionals separate nonzero vectors, and both spaces have dimension $d$. Hence it is an isomorphism. Its definition uses no choice of basis, which is the meaning of “naturally” here.

Evaluation $e_x(p)=p(x)$ is linear, so belongs to $V_n^{\ast }$. For distinct nodes $x_i$, define the Lagrange interpolation polynomialsThey lie in $V_n$ and obey $e_{x_j}({\ell }_i)={\delta }_{ij}$. Applying a putative relation $\sum a_i{e}_{x_i}=0$ to ${\ell }_j$ gives $a_j=0$, so the $n+1$ evaluations are independent and form a basis of $V_n^{\ast }$. The polynomials ${\ell }_i$ give their dual basis in $V_n$, using the canonical identification with $V_n^{\ast \ast }$.

Every $p\in V_n$ has the interpolation expansion $p(t)={\sum }_{i}p(x_i){\ell }_i(t)$: the difference has $n+1$ distinct roots but degree at most $n$. Integrate to get the numerical integration weights ${\lambda }_i={\int }_{-1}^1{\ell }_i(t)dt$. For the specified symmetric five nodes, symmetry gives weights $(a,b,c,b,a)$. Exactness on $1,t^2,t^4$ givesOdd moments vanish automatically. Solving yields Boole's rule