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Gaussian quadrature is a quadrature rule that chooses both sample locations and coefficients to achieve unusually high polynomial exactness. For a finite interval $[a,b]$ with $a<b$, assume the integration weight $w$ is integrable and positive almost everywhere. DefineLet $p_n$ be the monic orthogonal polynomial of polynomial degree $n$ for the inner product $⟨u,v{⟩}_w={\int }_a^{b}u(x)v(x)w(x)dx$: it satisfies $⟨p_n,q{⟩}_w=0$ for every polynomial $q$ of polynomial degree below $n$. The polynomial roots of $p_n$ are distinct and inside $(a,b)$; take them as the nodes $x_i$. Integrate the Lagrange cardinal polynomials ${\ell }_i$ to obtain $A_i=I_w({\ell }_i)$. These coefficients are positive, andThe integration weight $w(x)$ and the discrete coefficients $A_i$ play different roles: $w$ specifies the integral being approximated, while $A_i$ tell us how much each sample contributes.^[ref]

A proof outline explains the gain over simply integrating an arbitrary polynomial interpolation. Use polynomial division to write $f=p_{n}q+r$, with $\mathrm{deg}q\le n-1$ and $\mathrm{deg}r\le n-1$. Orthogonality makes $I_w(p_{n}q)=0$, and sampling at the polynomial roots of $p_n$ makes $Q_n(p_{n}q)=0$. The Lagrange cardinal polynomials recover $r$ exactly, so both expressions agree. Also $A_i=Q_n({\ell }_i^2)=I_w({\ell }_i^2)>0$, since ${\ell }_i^2$ has polynomial degree $2n-2$. This is the mechanism behind positive weights of Gaussian quadrature.

For a concrete two-node example, set $w=1$ on $[-1,1]$. The Legendre polynomial $P_2(x)=(3x^2-1)/2$ has polynomial roots $x=\pm 1/\sqrt{3}$. Its monic polynomial version $p_2=x^2-1/3$ has zero integral against both $1$ and $x$, as required. Symmetry and exactness for the constant polynomial give $A_1=A_2=1$. Thus the Gauss–Legendre instance of Gaussian quadrature isDirect evaluation on $1,x,x^2,x^3$ gives $2,0,2/3,0$, exactly their integrals. For the polynomial $f(x)=1+x+x^2+x^3$, two samples therefore give $8/3$ exactly. But for $x^4$ the rule gives $2/9$, whereas the integral is $2/5$: the error $I_1-Q_2$ is $8/45$.

The ceiling $2n-1$ is sharp for $n$ distinct real nodes under the positive-weight assumptions. Their squared nodal polynomial ${\prod }_i(x-x_i{)}^2$ has polynomial degree $2n$, is zero at every sample, and has a strictly positive weighted integral. Thus no such quadrature rule integrates every polynomial of polynomial degree $2n$ exactly. Nor does Gaussian quadrature integrate every nonpolynomial function exactly: the same two-node example gives $Q_2(|x|)=2/\sqrt{3}$, while $I_1(|x|)=1$.

Gaussian quadrature is useful when evaluations are costly and a modest polynomial captures the integrand well. More precisely, for a continuous function $f$ and any polynomial $r$ of polynomial degree at most $2n-1$, positivity and constant exactness giveThe uniform norm on $[a,b]$ measures the approximation between samples. This bound connects quadrature rule accuracy to uniform approximation, rather than promising an error from sample count alone. Corners, sharp unresolved features and roundoff error require their own assessment; the absolute value function example already shows why polynomial exactness is a qualified guarantee.

An orthogonal polynomial family chooses polynomials that do not overlap under a specified inner product. For real polynomials on an interval $(a,b)$, use the weighted inner productAssume $a<b$, the weight $w$ is measurable and positive almost everywhere, and its absolute moments ${\int }_a^b|x{|}^{k}w(x)dx$ are finite for every integer $k\ge 0$. These conditions make the integrals finite and give every nonzero polynomial positive squared norm. A sequence $p_0,p_1,\dots$ is an orthogonal polynomial family when $p_n$ has real coefficients and polynomial degree exactly $n$, and $⟨p_n,p_k{⟩}_w=0$ for $n\ne k$.^[ref]

Here $h_n=\Vert p_n{\Vert }_w^2>0$ and ${\delta }_{nk}$ is the Kronecker delta. Orthogonality does not require $h_n=1$: an orthonormal set is obtained only after replacing $p_n$ by $p_n/\sqrt{h_n}$. Multiplying each member by any nonzero real constant preserves orthogonality, but changes its norm and the corresponding expansion coefficient. The monic orthogonal polynomial convention instead fixes each leading coefficient to one.

The first four Legendre polynomials, named for Adrien-Marie Legendre, use $w=1$ on $[-1,1]$ and the normalization $P_n(1)=1$:Their orthogonality and squared norms areThus their Gram matrix has a positive diagonal and zero entries elsewhere, although their function graphs cross and overlap on the page. Orthogonality measures an integral of products, rather than separation of function graphs.^[ref]

For these four polynomials, a short verification expands each product and uses ${\int }_{-1}^1{x}^{2j+1}dx=0$ and ${\int }_{-1}^1{x}^{2j}dx=2/(2j+1)$. For example, $⟨P_0,P_2⟩=(3(2/3)-2)/2=0$ and $⟨P_1,P_3⟩=(5(2/5)-3(2/3))/2=0$. The same calculation gives the diagonal entries $2,2/3,2/5,2/7$. More generally, apply the Gram-Schmidt process to the monomial basis $1,x,x^2,\dots$: at step $n$, subtract projections onto the earlier polynomials. The result retains its leading term, has degree $n$, and has positive norm, proving existence without a differential equation. Different degrees also give linear independence, so $p_0,\dots ,p_m$ form a basis of the polynomials of degree at most $m$.

This basis turns least-squares solutions for functions into independent scalar calculations, as the decoupled coefficients of orthogonal polynomial projections show. In a spectral method, increasing the degree enlarges the approximation space while keeping earlier projection coefficients fixed. Weighted integrals also connect the same orthogonal polynomials to Gaussian quadrature: its nodes are polynomial roots of a member of the family. The projection problem minimizes an integrated squared error; Gaussian quadrature uses the family to choose evaluation points for an integral.

For real monic orthogonal polynomials $p_j$ with squared weighted norms $h_j$ under a positive weight $w$, the functions $p_j\sqrt{w/h_j}$ form an orthonormal set for the unweighted reference measure. Their finite-rank projection kernel turns the squared weighted Vandermonde determinant into a kernel determinant. Normalization on the full labelled configuration space is $1/m!$.

For the inner product obtained by integrating with weight $|x|$ on $[-1,1]$, odd moments vanish and the moment of $x^{2m}$ is $1/(m+1)$. The first monic orthogonal polynomials are $1$, $x$ and $x^2-1/2$. Positive definiteness follows because the weight is positive almost everywhere and a nonzero polynomial cannot vanish almost everywhere.

Let $t_1,\dots ,t_m$ be the distinct roots in $(0,1)$ at which ${\varphi }_r$ changes sign, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Both ${\varphi }_r$ and $q$ change sign at each such point, while roots of even multiplicity do not change sign. Consequently ${\varphi }_{r}q$ has one constant sign on $(0,1)$ except at its zeros, and is not identically zero. Its integral cannot vanish.

If $m<r$, however, orthogonality of the monic orthogonal polynomial ${\varphi }_r$ to every polynomial of lower degree gives $({\varphi }_r,q)=0$, a contradiction. Hence $m\ge r$. Since $\mathrm{deg}{\varphi }_r=r$, there must be exactly $r$ such roots, each of multiplicity one. They exhaust the degree, leaving no complex, exterior, or endpoint roots. Thus all roots are simple and lie strictly in $(0,1)$, which proves the requested weaker inclusion in $[0,1]$. This is the sign-change proof of the zeros of orthogonal polynomials theorem.

The monic orthogonal polynomials ${\varphi }_0,\dots ,{\varphi }_{r+1}$ form a basis of $P_{r+1}$, ordered by degree. Since $x{\varphi }_r$ is monic of degree $r+1$,For $j\le r-2$, symmetry of multiplication by $x$ gives $(x{\varphi }_r,{\varphi }_j)=({\varphi }_r,x{\varphi }_j)=0$, because $\mathrm{deg}(x{\varphi }_j)<r$. Only $a_r$ and $a_{r-1}$ remain. Taking inner products with ${\varphi }_r$ yields $a_r=({\varphi }_r,x{\varphi }_r)/({\varphi }_r,{\varphi }_r)$.

Similarly,since $x{\varphi }_{r-1}-{\varphi }_r$ has degree at most $r-1$. Rearranging proves the three-term recurrence for monic orthogonal polynomials:

The monic orthogonal polynomial $p_n$ is orthogonal to every polynomial of degree less than $n$, since $p_0,\dots ,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\dots ,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)={\prod }_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_{n}q$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.

If $k<n$, orthogonality instead gives ${\int }_a^b{p}_{n}qwdx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore there are exactly $n$ distinct simple zeros in $(a,b)$. The weighted inner products are understood to exist, as required by the stipulated orthogonality.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

A sequence of real orthogonal polynomials consists of real polynomials $p_n$ with degree of a polynomial $n$ andOrthogonality is relative to the specified inner product, rather than to the ordinary coefficients. A monic polynomial has leading coefficient one, so a monic orthogonal polynomial has the form $p_n(x)=x^n+\text{terms of lower degree}$. In particular, the monic degree-zero polynomial is $p_0=1$.

This is the Gram-Schmidt process applied degree by degree. Inductively, subtracting projections onto $p_0,\dots ,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real monic orthogonal polynomial of degree exactly $n+1$.

Now take $q_{n+1}=x{p}_n$. For $k\le n-2$, symmetry of the inner product givesbecause $x{p}_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the three-term recurrence for monic orthogonal polynomialswhereThe second identity uses $x{p}_{n-1}=p_n$ plus lower-degree terms.

For the symmetric Legendre inner product, parity gives ${\alpha }_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, thenThus the first four monic Legendre polynomials are