Consider a quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ where $n \in \mathbb{N}$, $\lambda_0, \ldots, \lambda_n \in \mathbb{R}$ and $x_0 < x_1 < \cdots < x_n$ are $n+1$ distinct points in $I$. We assume that the coefficients $(\lambda_j)_{0 \leqslant j \leqslant n}$ are chosen as $$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x,$$ where $(L_0, \ldots, L_n)$ is the Lagrange basis associated with the points $(x_0, \ldots, x_n)$. Thus, the formula $I_n(f)$ is of order $m \geqslant n$. By reasoning with the polynomial $\prod_{i=0}^n (X - x_i)$, show that $m \leqslant 2n+1$.
Consider a quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ where $n \in \mathbb{N}$, $\lambda_0, \ldots, \lambda_n \in \mathbb{R}$ and $x_0 < x_1 < \cdots < x_n$ are $n+1$ distinct points in $I$. We assume that the coefficients $(\lambda_j)_{0 \leqslant j \leqslant n}$ are chosen as
$$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x,$$
where $(L_0, \ldots, L_n)$ is the Lagrange basis associated with the points $(x_0, \ldots, x_n)$. Thus, the formula $I_n(f)$ is of order $m \geqslant n$.
By reasoning with the polynomial $\prod_{i=0}^n (X - x_i)$, show that $m \leqslant 2n+1$.