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For two-node Gaussian quadrature with linear weight, the monic node polynomial $q(x)=x^2-sx+p$ must be orthogonal to $1$ and $x$ for weight $x$. Indeed, exactness through degree three makes the quadrature of $q$ and $xq$ vanish at its nodes. The moment equations areso $s=6/5$ and $p=3/10$. Thus the Gaussian quadrature nodes, labelled in increasing order, areIntegrating the linear Lagrange interpolation cardinal functions gives the weightsThey are positive, sum to $1/2$, and reproduce the first moment $1/3$. Any cubic polynomial has remainder of degree at most one after division by $q$; orthogonality annihilates the other part both in the integral and at the nodes. This proves degree-three exactness, not merely the two moment conditions for the weights.

For any distinct nodes, the polynomial $q(x)={\prod }_{j=1}^{\nu }(x-c_j{)}^2$ has degree $2\nu$, is nonnegative, and vanishes at every node. The quadrature rule sum for $q$ is zero, but ${\int }_0^{\pi }wqdx>0$ because the weight is positive and $q$ is positive except at finitely many points. Therefore no such rule is exact for all polynomials of degree at most $2\nu$, irrespective of the chosen weights.

For the sine weight, its first four moments are $2,\pi ,{\pi }^2-4,{\pi }^3-6\pi$. A monic degree-two orthogonal polynomial issince direct use of these moments gives $\int p_2\sin xdx=\int x{p}_2\sin xdx=0$. Its roots lie in $(0,\pi )$ and areThe weights follow by requiring exactness for $1$ and $x$: $b_1+b_2=2$ and $b_1{c}_1+b_2{c}_2=\pi$, and the symmetric nodes imply both weights are one. To prove cubic exactness, divide any cubic as $f=a{p}_2+r$, with $a,r$ of degree at most one. The integral of $a{p}_2$ is zero by orthogonality, and its nodal values vanish; the remaining linear polynomial is integrated exactly. This is the two-node Gaussian quadrature with sine weight; the argument proves its Gaussian quadrature property without assuming a general quadrature theorem.

Apply the Gaussian quadrature rule to the constant polynomial $1$. Exactness gives ${\sum }_{j=1}^n{A}_j=2$.

Use the Lagrange interpolation polynomialIts square has degree $2n-2$, so the exactness already proved givesThe positive weights of Gaussian quadrature follow because ${\ell }_j$ is a nonzero continuous polynomial. Thus these Gaussian quadrature weights are in fact positive, rather than merely nonnegative.

A degree-$n$ Legendre polynomial $P_n$ is orthogonal, for integration on $[-1,1]$, to every polynomial of degree at most $n-1$, and has $n$ distinct roots in $(-1,1)$. Divide $p$ by $P_n$:The Orthogonality of Legendre polynomials gives $\int p=\int r$. At every quadrature node $x_j$, $p(x_j)=r(x_j)$. The assumed exactness for $r$ therefore proves exactness up to degree $2n-1$, the characteristic property of Gaussian quadrature.

To prove convergence of positive quadrature on continuous functions, for a continuous $f$, the Weierstrass approximation theorem supplies a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. For every sufficiently large $n$, the Gaussian quadrature rule is exact for this fixed polynomial. Positivity of its weights and their sum $2$ then giveSince $\epsilon$ is arbitrary, $Q_n(f)\to {\int }_{-1}^{1}f$. Notice that no differentiability of $f$ is needed; uniform polynomial approximation and positivity provide the whole argument.

Let $x_1,\dots ,x_k$ be the distinct interior roots of $p_n$ where its sign changes, and put $q(x)={\prod }_{j=1}^k(x-x_j)$. If $k<n$, orthogonality gives ${\int }_a^{b}w{p}_{n}q=0$. But $p_{n}q$ has constant sign on $(a,b)$, apart from its isolated zeros, because every sign change has been canceled. It is not identically zero, and the positive weight makes its integral nonzero, a contradiction. Thus $k\ge n$. Degree $n$ forces exactly $n$ distinct simple roots, all inside $(a,b)$.

For distinct nodes, let ${\ell }_i(x)={\prod }_{j\ne i}(x-c_j)/(c_i-c_j)$. Lagrange interpolation polynomial gives $f={\sum }_{i}f(c_i){\ell }_i$ for every polynomial of degree at most $n-1$. Integrating this identity proves exactness with $b_i={\int }_a^{b}w{\ell }_i$.

For the Gaussian quadrature nodes, divide any $f\in {\mathbb{P}}_{2n-1}$ as $f=q{p}_n+r$, with $q,r\in {\mathbb{P}}_{n-1}$. The integral of $q{p}_n$ vanishes by orthogonality, and its nodal values vanish because $p_n(c_i)=0$. The remaining polynomial $r$ is integrated exactly by the preceding interpolation result. HenceIts weights are positive: exactness applied to ${\ell }_i^2$ gives $b_i={\int }_a^{b}w{\ell }_i^2>0$.

Define the error functional $L(f)={\int }_a^{b}w(x)f(x)dx-{\sum }_i{b}_{i}f(c_i)$. It annihilates polynomials of degree at most $2n-1$. For $f\in C^{2n}[a,b]$, the Peano kernel theorem giveswhere a formal Peano kernel with the normalization made explicit isNo evaluation of $K$ is needed. The supplied inner product presupposes an integrable positive weight on the finite interval, so the integral and point-evaluation error functional is bounded on continuous functions.

For fixed interpolation nodes $c_0,\dots ,c_{s-1}$, interpolate any $p\in {\mathbb{P}}_{s-1}[x]$ asIntegrating shows that the unique weights making the quadrature rule exact through degree $s-1$ are

Now let the nodes be the zeros of the degree-$s$ orthogonal polynomial $q_s$. Given any $p\in {\mathbb{P}}_{2s-1}[x]$, polynomial division givesOrthogonality gives ${\int }_a^b{q}_{s}h=0$, while $q_s(c_k)=0$ makes the quadrature sum for $q_{s}h$ vanish. The rule is already exact on $r$, so it is exact on $p$. This is the defining precision property of Gaussian quadrature.

The Legendre polynomial $p_n$ is orthogonal on $[-1,1]$ to every polynomial of degree below $n$. Suppose it changes sign at only $r<n$ distinct interior points $x_1,\dots ,x_r$, and putchoosing its overall sign so that $p_{n}q\ge 0$ on $[-1,1]$. This product is positive except at finitely many points, socontradicting orthogonality because $\mathrm{deg}q=r<n$. Thus $p_n$ has at least $n$ sign-changing roots in $(-1,1)$; its degree is $n$, so these are all its roots and they are distinct.

For arbitrary distinct nodes $x_1,\dots ,x_n$, letbe the Lagrange basis polynomials. Every polynomial $P$ of degree below $n$ satisfies $P(t)={\sum }_{i}P(x_i)L_i(t)$, so the required weights exist and areThey are unique because applying any such formula to $P=L_i$ recovers $A_i$.

Now assume exactness for every $P$ of degree below $2n$ and define $Q(t)={\prod }_i(t-x_i)$. For every polynomial $R$ of degree below $n$, the product $QR$ has degree below $2n$ and vanishes at every node. ThereforeThere is a unique monic polynomial of degree $n$ orthogonal to all lower-degree polynomials, so $Q$ is the monic normalization of $p_n$ and the $x_i$ are precisely the roots of $p_n$. Taking $P=1$ givesFinally $L_i^2$ has degree $2n-2$, so exactness givesThese are the positive weights of Gaussian quadrature.

Let ${\ell }_j$ be the Lagrange interpolation polynomial for the roots ${\omega }_j$, and defineFor $\mathrm{deg}Q\le 2n-1$, polynomial division gives $Q=S{P}_n+R$ with $\mathrm{deg}S,\mathrm{deg}R<n$. Orthogonality kills the first term, while interpolation gives $R={\sum }_{j}R({\omega }_j){\ell }_j$. ThereforeTaking $Q={\ell }_j$ proves uniqueness. Taking $Q={\ell }_j^2$ also showsThis is Gaussian quadrature.