[proofplan] We prove Gaussian quadrature with $\nu$ nodes is exact for polynomials of degree $\leq 2\nu - 1$ by dividing any such polynomial by $p_\nu$ (whose zeros are the nodes) and using orthogonality to eliminate the quotient term. Positivity of weights follows from evaluating the rule on $\ell_j^2$. Optimality follows from evaluating on $\prod(x - c_k)^2$, which vanishes at all nodes but has positive integral. [/proofplan] [step:Prove exactness for $\deg f \leq 2\nu - 1$] Let $f \in P_{2\nu-1}[x]$. Divide $f$ by $p_\nu$: $f(x) = p_\nu(x)\,q(x) + r(x)$ where $q, r \in P_{\nu-1}[x]$. Then \begin{align*} \int_a^b w(x)\,f(x)\,d\mathcal{L}^1(x) &= \int_a^b w(x)\,p_\nu(x)\,q(x)\,d\mathcal{L}^1(x) + \int_a^b w(x)\,r(x)\,d\mathcal{L}^1(x). \end{align*} The first integral vanishes by orthogonality ($q \in P_{\nu-1}[x]$). For the quadrature sum: $f(c_k) = r(c_k)$ since $p_\nu(c_k) = 0$. Since $r \in P_{\nu-1}[x]$, the Lagrange interpolant of $r$ at $c_1, \ldots, c_\nu$ is $r$ itself by [uniqueness](/theorems/473). Therefore $\int_a^b w\,r\,d\mathcal{L}^1 = \sum_{k=1}^\nu b_k r(c_k) = \sum_{k=1}^\nu b_k f(c_k)$. [/step] [step:Prove positivity of weights and optimality] For each $j$, $g_j(x) = \ell_j(x)^2 \in P_{2\nu-2}[x] \subseteq P_{2\nu-1}[x]$. By exactness, $\int_a^b w\,\ell_j^2\,d\mathcal{L}^1 = \sum_{k=1}^\nu b_k\,\ell_j(c_k)^2 = b_j$ (since $\ell_j(c_k) = \delta_{jk}$). The left side is positive ($w > 0$ and $\ell_j \not\equiv 0$), so $b_j > 0$. For optimality: $f(x) = \prod_{k=1}^\nu (x - c_k)^2 \in P_{2\nu}[x]$ satisfies $f(c_k) = 0$ for all $k$, so $\sum b_k f(c_k) = 0$. But $\int_a^b w\,f\,d\mathcal{L}^1 > 0$ (since $w > 0$ and $f \geq 0$ with $f \not\equiv 0$). Therefore no $\nu$-point rule can be exact for all of $P_{2\nu}[x]$. [/step]