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At equally replicated quantitative treatment levels, evaluations of orthogonal polynomials generate pairwise orthogonal treatment contrasts, using the replication-weighted inner product. Five equally spaced levels give four contrast directions representing degrees one through four, besides the constant direction. This decomposes treatment variation without assuming the response has low polynomial degree.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

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.

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 $p_0=1$.

Multiply the Gegenbauer differential equation byIt becomes the Sturm-Liouville formFor two polynomial solutions of degrees $m\ne n$, multiply their equations crosswise, subtract, and integrate over $(-1,1)$. The boundary term vanishes because $\alpha >0$ and $(1-x^2{)}^{\alpha +1/2}\to 0$. Since the eigenvalues $n(n+2\alpha )$ are distinct,Thus $a=-1$, $b=1$, and the displayed function is the weight.

Every interior root is simple: a solution and its derivative cannot both vanish at an ordinary point of a second-order linear differential equation unless the solution is identically zero. Let $x_1,\dots ,x_k$ be all interior roots and put $q(x)={\prod }_i(x-x_i)$. If $k<n$, then $q$ has degree below $n$, so orthogonality givesBut $C_n^{\alpha }/q$ has no zero and constant sign on $(-1,1)$, whence the integrand $q(x{)}^2(C_n^{\alpha }(x)/q(x))w(x)$ has one strict sign except at finitely many points. Its integral cannot vanish, a contradiction. Therefore $k=n$: all $n$ roots are real, simple, and lie in $(-1,1)$.

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.

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.

For integration against $\sin x$ on $[0,\pi ]$, the monic quadratic orthogonal polynomial is $x^2-\pi x+2$. Its two roots are the Gaussian quadrature nodes $c_{1,2}=\pi /2\mp \sqrt{{\pi }^2/4-2}$, with weights one. The rule integrates every cubic exactly: divide it by the orthogonal quadratic and integrate the degree-one remainder.

Evaluating arbitrary monic polynomials of successive degrees $0,\dots ,m-1$ gives the same determinant as evaluating the monomials of those degrees. The basis-change matrix is triangular with diagonal entries one, so its determinant is one. This permits an orthogonal polynomial basis to replace the monomials in eigenvalue densities.