Gaussian quadrature

In numerical analysis, an n-point Gaussian quadrature rule, named after Carl Friedrich Gauss,^[1] is a quadrature rule constructed to yield an exact result for polynomials of degree 2n − 1 or less by a suitable choice of the nodes x_i and weights w_i for i = 1, ..., n.

The modern formulation using orthogonal polynomials was developed by Carl Gustav Jacobi in 1826.^[2] The most common domain of integration for such a rule is taken as [−1, 1], so the rule is stated as $${\displaystyle \int _{-1}^{1}f(x),円dx\approx \sum _{i=1}^{n}w_{i}f(x_{i}),}$${\displaystyle \int _{-1}^{1}f(x),円dx\approx \sum _{i=1}^{n}w_{i}f(x_{i}),}

which is exact for polynomials of degree 2n − 1 or less. This exact rule is known as the Gauss–Legendre quadrature rule. The quadrature rule will only be an accurate approximation to the integral above if f (x) is well-approximated by a polynomial of degree 2n − 1 or less on [−1, 1].

The Gauss–Legendre quadrature rule is not typically used for integrable functions with endpoint singularities. Instead, if the integrand can be written as

$${\displaystyle f(x)=\left(1-x\right)^{\alpha }\left(1+x\right)^{\beta }g(x),\quad \alpha ,\beta >-1,}$${\displaystyle f(x)=\left(1-x\right)^{\alpha }\left(1+x\right)^{\beta }g(x),\quad \alpha ,\beta >-1,}

where g(x) is well-approximated by a low-degree polynomial, then alternative nodes x_i' and weights w_i' will usually give more accurate quadrature rules. These are known as Gauss–Jacobi quadrature rules, i.e.,

$${\displaystyle \int _{-1}^{1}f(x),円dx=\int _{-1}^{1}\left(1-x\right)^{\alpha }\left(1+x\right)^{\beta }g(x),円dx\approx \sum _{i=1}^{n}w_{i}'g\left(x_{i}'\right).}$${\displaystyle \int _{-1}^{1}f(x),円dx=\int _{-1}^{1}\left(1-x\right)^{\alpha }\left(1+x\right)^{\beta }g(x),円dx\approx \sum _{i=1}^{n}w_{i}'g\left(x_{i}'\right).}

Common weights include ${\textstyle {\frac {1}{\sqrt {1-x^{2}}}}}${\textstyle {\frac {1}{\sqrt {1-x^{2}}}}} (Chebyshev–Gauss) and ${\textstyle {\sqrt {1-x^{2}}}}${\textstyle {\sqrt {1-x^{2}}}}. One may also want to integrate over semi-infinite (Gauss–Laguerre quadrature) and infinite intervals (Gauss–Hermite quadrature).

It can be shown (see Press et al., or Stoer and Bulirsch) that the quadrature nodes x_i are the roots of a polynomial belonging to a class of orthogonal polynomials (the class orthogonal with respect to a weighted inner-product). This is a key observation for computing Gauss quadrature nodes and weights.

For the simplest integration problem stated above, i.e., f(x) is well-approximated by polynomials on ${\displaystyle [-1,1]}${\displaystyle [-1,1]}, the associated orthogonal polynomials are Legendre polynomials, denoted by P_n(x). With the n-th polynomial normalized to give P_n(1) = 1, the i-th Gauss node, x_i, is the i-th root of P_n and the weights are given by the formula^[3] $${\displaystyle w_{i}={\frac {2}{\left(1-x_{i}^{2}\right)\left[P'_{n}(x_{i})\right]^{2}}}.}$${\displaystyle w_{i}={\frac {2}{\left(1-x_{i}^{2}\right)\left[P'_{n}(x_{i})\right]^{2}}}.}

Some low-order quadrature rules are tabulated below (over interval [−1, 1], see the section below for other intervals).

An integral over [a, b] must be changed into an integral over [−1, 1] before applying the Gaussian quadrature rule. This change of interval can be done in the following way: $${\displaystyle \int _{a}^{b}f(x),円dx=\int _{-1}^{1}f\left({\frac {b-a}{2}}\xi +{\frac {a+b}{2}}\right),円{\frac {dx}{d\xi }}d\xi }$${\displaystyle \int _{a}^{b}f(x),円dx=\int _{-1}^{1}f\left({\frac {b-a}{2}}\xi +{\frac {a+b}{2}}\right),円{\frac {dx}{d\xi }}d\xi }

with ${\displaystyle {\frac {dx}{d\xi }}={\frac {b-a}{2}}}${\displaystyle {\frac {dx}{d\xi }}={\frac {b-a}{2}}}

Applying the ${\displaystyle n}${\displaystyle n} point Gaussian quadrature ${\displaystyle (\xi ,w)}${\displaystyle (\xi ,w)} rule then results in the following approximation: $${\displaystyle \int _{a}^{b}f(x),円dx\approx {\frac {b-a}{2}}\sum _{i=1}^{n}w_{i}f\left({\frac {b-a}{2}}\xi _{i}+{\frac {a+b}{2}}\right).}$${\displaystyle \int _{a}^{b}f(x),円dx\approx {\frac {b-a}{2}}\sum _{i=1}^{n}w_{i}f\left({\frac {b-a}{2}}\xi _{i}+{\frac {a+b}{2}}\right).}

Use the two-point Gauss quadrature rule to approximate the distance in meters covered by a rocket from ${\displaystyle t=8\mathrm {s} }${\displaystyle t=8\mathrm {s} } to ${\displaystyle t=30\mathrm {s} ,}${\displaystyle t=30\mathrm {s} ,} as given by $${\displaystyle s=\int _{8}^{30}{\left(2000\ln \left[{\frac {140000}{140000-2100t}}\right]-9.8t\right){dt}}}$${\displaystyle s=\int _{8}^{30}{\left(2000\ln \left[{\frac {140000}{140000-2100t}}\right]-9.8t\right){dt}}}

Change the limits so that one can use the weights and abscissae given in Table 1. Also, find the absolute relative true error. The true value is given as 11061.34 m.

Solution

First, changing the limits of integration from ${\displaystyle \left[8,30\right]}${\displaystyle \left[8,30\right]} to ${\displaystyle \left[-1,1\right]}${\displaystyle \left[-1,1\right]} gives

$${\displaystyle {\begin{aligned}\int _{8}^{30}{f(t)dt}&={\frac {30-8}{2}}\int _{-1}^{1}{f\left({\frac {30-8}{2}}x+{\frac {30+8}{2}}\right){dx}}\\&=11\int _{-1}^{1}{f\left(11x+19\right){dx}}\end{aligned}}}$${\displaystyle {\begin{aligned}\int _{8}^{30}{f(t)dt}&={\frac {30-8}{2}}\int _{-1}^{1}{f\left({\frac {30-8}{2}}x+{\frac {30+8}{2}}\right){dx}}\\&=11\int _{-1}^{1}{f\left(11x+19\right){dx}}\end{aligned}}}

Next, get the weighting factors and function argument values from Table 1 for the two-point rule,

• ${\displaystyle c_{1}=1.000000000}${\displaystyle c_{1}=1.000000000}
• ${\displaystyle x_{1}=-0.577350269}${\displaystyle x_{1}=-0.577350269}
• ${\displaystyle c_{2}=1.000000000}${\displaystyle c_{2}=1.000000000}
• ${\displaystyle x_{2}=0.577350269}${\displaystyle x_{2}=0.577350269}

Now we can use the Gauss quadrature formula $${\displaystyle {\begin{aligned}11\int _{-1}^{1}{f\left(11x+19\right){dx}}&\approx 11\left[c_{1}f\left(11x_{1}+19\right)+c_{2}f\left(11x_{2}+19\right)\right]\\&=11\left[f\left(11(-0.5773503)+19\right)+f\left(11(0.5773503)+19\right)\right]\\&=11\left[f(12.64915)+f(25.35085)\right]\\&=11\left[(296.8317)+(708.4811)\right]\\&=11058.44\end{aligned}}}$${\displaystyle {\begin{aligned}11\int _{-1}^{1}{f\left(11x+19\right){dx}}&\approx 11\left[c_{1}f\left(11x_{1}+19\right)+c_{2}f\left(11x_{2}+19\right)\right]\\&=11\left[f\left(11(-0.5773503)+19\right)+f\left(11(0.5773503)+19\right)\right]\\&=11\left[f(12.64915)+f(25.35085)\right]\\&=11\left[(296.8317)+(708.4811)\right]\\&=11058.44\end{aligned}}} since $${\displaystyle {\begin{aligned}f(12.64915)&=2000\ln \left[{\frac {140000}{140000-2100(12.64915)}}\right]-9.8(12.64915)\\&=296.8317\end{aligned}}}$${\displaystyle {\begin{aligned}f(12.64915)&=2000\ln \left[{\frac {140000}{140000-2100(12.64915)}}\right]-9.8(12.64915)\\&=296.8317\end{aligned}}} $${\displaystyle {\begin{aligned}f(25.35085)&=2000\ln \left[{\frac {140000}{140000-2100(25.35085)}}\right]-9.8(25.35085)\\&=708.4811\end{aligned}}}$${\displaystyle {\begin{aligned}f(25.35085)&=2000\ln \left[{\frac {140000}{140000-2100(25.35085)}}\right]-9.8(25.35085)\\&=708.4811\end{aligned}}}

Given that the true value is 11061.34 m, the absolute relative true error, ${\displaystyle \left|\varepsilon _{t}\right|}${\displaystyle \left|\varepsilon _{t}\right|} is $${\displaystyle \left|\varepsilon _{t}\right|=\left|{\frac {11061.34-11058.44}{11061.34}}\right|\times 100\%=0.0262\%}$${\displaystyle \left|\varepsilon _{t}\right|=\left|{\frac {11061.34-11058.44}{11061.34}}\right|\times 100\%=0.0262\%}

The integration problem can be expressed in a slightly more general way by introducing a positive weight function ω into the integrand, and allowing an interval other than [−1, 1]. That is, the problem is to calculate $${\displaystyle \int _{a}^{b}\omega (x),円f(x),円dx}$${\displaystyle \int _{a}^{b}\omega (x),円f(x),円dx} for some choices of a, b, and ω. For a = −1, b = 1, and ω(x) = 1, the problem is the same as that considered above. Other choices lead to other integration rules. Some of these are tabulated below. Equation numbers are given for Abramowitz and Stegun (A & S).

Let p_n be a nontrivial polynomial of degree n such that $${\displaystyle \int _{a}^{b}\omega (x),円x^{k}p_{n}(x),円dx=0,\quad {\text{for all }}k=0,1,\ldots ,n-1.}$${\displaystyle \int _{a}^{b}\omega (x),円x^{k}p_{n}(x),円dx=0,\quad {\text{for all }}k=0,1,\ldots ,n-1.}

Note that this will be true for all the orthogonal polynomials above, because each p_n is constructed to be orthogonal to the other polynomials p_j for j<n, and x^k is in the span of that set.

If we pick the n nodes x_i to be the zeros of p_n, then there exist n weights w_i which make the Gaussian quadrature computed integral exact for all polynomials h(x) of degree 2n − 1 or less. Furthermore, all these nodes x_i will lie in the open interval (a, b).^[4]

To prove the first part of this claim, let h(x) be any polynomial of degree 2n − 1 or less. Divide it by the orthogonal polynomial p_n to get $${\displaystyle h(x)=p_{n}(x),円q(x)+r(x).}$${\displaystyle h(x)=p_{n}(x),円q(x)+r(x).} where q(x) is the quotient, of degree n − 1 or less (because the sum of its degree and that of the divisor p_n must equal that of the dividend), and r(x) is the remainder, also of degree n − 1 or less (because the degree of the remainder is always less than that of the divisor). Since p_n is by assumption orthogonal to all monomials of degree less than n, it must be orthogonal to the quotient q(x). Therefore $${\displaystyle \int _{a}^{b}\omega (x),円h(x),円dx=\int _{a}^{b}\omega (x),円{\big (},円p_{n}(x)q(x)+r(x),円{\big )},円dx=\int _{a}^{b}\omega (x),円r(x),円dx.}$${\displaystyle \int _{a}^{b}\omega (x),円h(x),円dx=\int _{a}^{b}\omega (x),円{\big (},円p_{n}(x)q(x)+r(x),円{\big )},円dx=\int _{a}^{b}\omega (x),円r(x),円dx.}

Since the remainder r(x) is of degree n − 1 or less, we can interpolate it exactly using n interpolation points with Lagrange polynomials l_i(x), where $${\displaystyle l_{i}(x)=\prod _{j\neq i}{\frac {x-x_{j}}{x_{i}-x_{j}}}.}$${\displaystyle l_{i}(x)=\prod _{j\neq i}{\frac {x-x_{j}}{x_{i}-x_{j}}}.}

We have $${\displaystyle r(x)=\sum _{i=1}^{n}l_{i}(x),円r(x_{i}).}$${\displaystyle r(x)=\sum _{i=1}^{n}l_{i}(x),円r(x_{i}).}

Then its integral will equal $${\displaystyle \int _{a}^{b}\omega (x),円r(x),円dx=\int _{a}^{b}\omega (x),円\sum _{i=1}^{n}l_{i}(x),円r(x_{i}),円dx=\sum _{i=1}^{n},円r(x_{i}),円\int _{a}^{b}\omega (x),円l_{i}(x),円dx=\sum _{i=1}^{n},円r(x_{i}),円w_{i},}$${\displaystyle \int _{a}^{b}\omega (x),円r(x),円dx=\int _{a}^{b}\omega (x),円\sum _{i=1}^{n}l_{i}(x),円r(x_{i}),円dx=\sum _{i=1}^{n},円r(x_{i}),円\int _{a}^{b}\omega (x),円l_{i}(x),円dx=\sum _{i=1}^{n},円r(x_{i}),円w_{i},}

where w_i, the weight associated with the node x_i, is defined to equal the weighted integral of l_i(x) (see below for other formulas for the weights). But all the x_i are roots of p_n, so the division formula above tells us that $${\displaystyle h(x_{i})=p_{n}(x_{i}),円q(x_{i})+r(x_{i})=r(x_{i}),}$${\displaystyle h(x_{i})=p_{n}(x_{i}),円q(x_{i})+r(x_{i})=r(x_{i}),} for all i. Thus we finally have $${\displaystyle \int _{a}^{b}\omega (x),円h(x),円dx=\int _{a}^{b}\omega (x),円r(x),円dx=\sum _{i=1}^{n}w_{i},円r(x_{i})=\sum _{i=1}^{n}w_{i},円h(x_{i}).}$${\displaystyle \int _{a}^{b}\omega (x),円h(x),円dx=\int _{a}^{b}\omega (x),円r(x),円dx=\sum _{i=1}^{n}w_{i},円r(x_{i})=\sum _{i=1}^{n}w_{i},円h(x_{i}).}

This proves that for any polynomial h(x) of degree 2n − 1 or less, its integral is given exactly by the Gaussian quadrature sum.

To prove the second part of the claim, consider the factored form of the polynomial p_n. Any complex conjugate roots will yield a quadratic factor that is either strictly positive or strictly negative over the entire real line. Any factors for roots outside the interval from a to b will not change sign over that interval. Finally, for factors corresponding to roots x_i inside the interval from a to b that are of odd multiplicity, multiply p_n by one more factor to make a new polynomial $${\displaystyle p_{n}(x),円\prod _{i}(x-x_{i}).}$${\displaystyle p_{n}(x),円\prod _{i}(x-x_{i}).}

This polynomial cannot change sign over the interval from a to b because all its roots there are now of even multiplicity. So the integral $${\displaystyle \int _{a}^{b}p_{n}(x),円\left(\prod _{i}(x-x_{i})\right),円\omega (x),円dx\neq 0,}$${\displaystyle \int _{a}^{b}p_{n}(x),円\left(\prod _{i}(x-x_{i})\right),円\omega (x),円dx\neq 0,} since the weight function ω(x) is always non-negative. But p_n is orthogonal to all polynomials of degree n − 1 or less, so the degree of the product $${\displaystyle \prod _{i}(x-x_{i})}$${\displaystyle \prod _{i}(x-x_{i})} must be at least n. Therefore p_n has n distinct roots, all real, in the interval from a to b.

The weights can be expressed as

where ${\displaystyle a_{k}}${\displaystyle a_{k}} is the coefficient of ${\displaystyle x^{k}}${\displaystyle x^{k}} in ${\displaystyle p_{k}(x)}${\displaystyle p_{k}(x)}. To prove this, note that using Lagrange interpolation one can express r(x) in terms of ${\displaystyle r(x_{i})}${\displaystyle r(x_{i})} as $${\displaystyle r(x)=\sum _{i=1}^{n}r(x_{i})\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}}$${\displaystyle r(x)=\sum _{i=1}^{n}r(x_{i})\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}} because r(x) has degree less than n and is thus fixed by the values it attains at n different points. Multiplying both sides by ω(x) and integrating from a to b yields $${\displaystyle \int _{a}^{b}\omega (x)r(x)dx=\sum _{i=1}^{n}r(x_{i})\int _{a}^{b}\omega (x)\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}dx}$${\displaystyle \int _{a}^{b}\omega (x)r(x)dx=\sum _{i=1}^{n}r(x_{i})\int _{a}^{b}\omega (x)\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}dx}

The weights w_i are thus given by $${\displaystyle w_{i}=\int _{a}^{b}\omega (x)\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}dx}$${\displaystyle w_{i}=\int _{a}^{b}\omega (x)\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}dx}

This integral expression for ${\displaystyle w_{i}}${\displaystyle w_{i}} can be expressed in terms of the orthogonal polynomials ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)} and ${\displaystyle p_{n-1}(x)}${\displaystyle p_{n-1}(x)} as follows.

We can write $${\displaystyle \prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}\left(x-x_{j}\right)={\frac {\prod _{1\leq j\leq n}\left(x-x_{j}\right)}{x-x_{i}}}={\frac {p_{n}(x)}{a_{n}\left(x-x_{i}\right)}}}$${\displaystyle \prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}\left(x-x_{j}\right)={\frac {\prod _{1\leq j\leq n}\left(x-x_{j}\right)}{x-x_{i}}}={\frac {p_{n}(x)}{a_{n}\left(x-x_{i}\right)}}}

where ${\displaystyle a_{n}}${\displaystyle a_{n}} is the coefficient of ${\displaystyle x^{n}}${\displaystyle x^{n}} in ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)}. Taking the limit of x to ${\displaystyle x_{i}}${\displaystyle x_{i}} yields using L'Hôpital's rule $${\displaystyle \prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}\left(x_{i}-x_{j}\right)={\frac {p'_{n}(x_{i})}{a_{n}}}}$${\displaystyle \prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}\left(x_{i}-x_{j}\right)={\frac {p'_{n}(x_{i})}{a_{n}}}}

We can thus write the integral expression for the weights as

In the integrand, writing $${\displaystyle {\frac {1}{x-x_{i}}}={\frac {1-\left({\frac {x}{x_{i}}}\right)^{k}}{x-x_{i}}}+\left({\frac {x}{x_{i}}}\right)^{k}{\frac {1}{x-x_{i}}}}$${\displaystyle {\frac {1}{x-x_{i}}}={\frac {1-\left({\frac {x}{x_{i}}}\right)^{k}}{x-x_{i}}}+\left({\frac {x}{x_{i}}}\right)^{k}{\frac {1}{x-x_{i}}}}

yields $${\displaystyle \int _{a}^{b}\omega (x){\frac {x^{k}p_{n}(x)}{x-x_{i}}}dx=x_{i}^{k}\int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx}$${\displaystyle \int _{a}^{b}\omega (x){\frac {x^{k}p_{n}(x)}{x-x_{i}}}dx=x_{i}^{k}\int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx}

provided ${\displaystyle k\leq n}${\displaystyle k\leq n}, because $${\displaystyle {\frac {1-\left({\frac {x}{x_{i}}}\right)^{k}}{x-x_{i}}}}$${\displaystyle {\frac {1-\left({\frac {x}{x_{i}}}\right)^{k}}{x-x_{i}}}} is a polynomial of degree k − 1 which is then orthogonal to ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)}. So, if q(x) is a polynomial of at most nth degree we have $${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {1}{q(x_{i})}}\int _{a}^{b}\omega (x){\frac {q(x)p_{n}(x)}{x-x_{i}}}dx}$${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {1}{q(x_{i})}}\int _{a}^{b}\omega (x){\frac {q(x)p_{n}(x)}{x-x_{i}}}dx}

We can evaluate the integral on the right hand side for ${\displaystyle q(x)=p_{n-1}(x)}${\displaystyle q(x)=p_{n-1}(x)} as follows. Because ${\displaystyle {\frac {p_{n}(x)}{x-x_{i}}}}${\displaystyle {\frac {p_{n}(x)}{x-x_{i}}}} is a polynomial of degree n − 1, we have $${\displaystyle {\frac {p_{n}(x)}{x-x_{i}}}=a_{n}x^{n-1}+s(x)}$${\displaystyle {\frac {p_{n}(x)}{x-x_{i}}}=a_{n}x^{n-1}+s(x)} where s(x) is a polynomial of degree ${\displaystyle n-2}${\displaystyle n-2}. Since s(x) is orthogonal to ${\displaystyle p_{n-1}(x)}${\displaystyle p_{n-1}(x)} we have $${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {a_{n}}{p_{n-1}(x_{i})}}\int _{a}^{b}\omega (x)p_{n-1}(x)x^{n-1}dx}$${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {a_{n}}{p_{n-1}(x_{i})}}\int _{a}^{b}\omega (x)p_{n-1}(x)x^{n-1}dx}

We can then write $${\displaystyle x^{n-1}=\left(x^{n-1}-{\frac {p_{n-1}(x)}{a_{n-1}}}\right)+{\frac {p_{n-1}(x)}{a_{n-1}}}}$${\displaystyle x^{n-1}=\left(x^{n-1}-{\frac {p_{n-1}(x)}{a_{n-1}}}\right)+{\frac {p_{n-1}(x)}{a_{n-1}}}}

The term in the brackets is a polynomial of degree ${\displaystyle n-2}${\displaystyle n-2}, which is therefore orthogonal to ${\displaystyle p_{n-1}(x)}${\displaystyle p_{n-1}(x)}. The integral can thus be written as $${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {a_{n}}{a_{n-1}p_{n-1}(x_{i})}}\int _{a}^{b}\omega (x)p_{n-1}(x)^{2}dx}$${\displaystyle \int _{a}^{b}\omega (x){\frac {p_{n}(x)}{x-x_{i}}}dx={\frac {a_{n}}{a_{n-1}p_{n-1}(x_{i})}}\int _{a}^{b}\omega (x)p_{n-1}(x)^{2}dx}

According to equation (2 ), the weights are obtained by dividing this by ${\displaystyle p'_{n}(x_{i})}${\displaystyle p'_{n}(x_{i})} and that yields the expression in equation (1 ).

${\displaystyle w_{i}}${\displaystyle w_{i}} can also be expressed in terms of the orthogonal polynomials ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)} and now ${\displaystyle p_{n+1}(x)}${\displaystyle p_{n+1}(x)}. In the 3-term recurrence relation ${\displaystyle p_{n+1}(x_{i})=(a)p_{n}(x_{i})+(b)p_{n-1}(x_{i})}${\displaystyle p_{n+1}(x_{i})=(a)p_{n}(x_{i})+(b)p_{n-1}(x_{i})} the term with ${\displaystyle p_{n}(x_{i})}${\displaystyle p_{n}(x_{i})} vanishes, so ${\displaystyle p_{n-1}(x_{i})}${\displaystyle p_{n-1}(x_{i})} in Eq. (1) can be replaced by ${\textstyle {\frac {1}{b}}p_{n+1}\left(x_{i}\right)}${\textstyle {\frac {1}{b}}p_{n+1}\left(x_{i}\right)}.

Consider the following polynomial of degree ${\displaystyle 2n-2}${\displaystyle 2n-2} $${\displaystyle f(x)=\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {\left(x-x_{j}\right)^{2}}{\left(x_{i}-x_{j}\right)^{2}}}}$${\displaystyle f(x)=\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {\left(x-x_{j}\right)^{2}}{\left(x_{i}-x_{j}\right)^{2}}}} where, as above, the x_j are the roots of the polynomial ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)}. Clearly ${\displaystyle f(x_{j})=\delta _{ij}}${\displaystyle f(x_{j})=\delta _{ij}}. Since the degree of ${\displaystyle f(x)}${\displaystyle f(x)} is less than ${\displaystyle 2n-1}${\displaystyle 2n-1}, the Gaussian quadrature formula involving the weights and nodes obtained from ${\displaystyle p_{n}(x)}${\displaystyle p_{n}(x)} applies. Since ${\displaystyle f(x_{j})=0}${\displaystyle f(x_{j})=0} for j not equal to i, we have $${\displaystyle \int _{a}^{b}\omega (x)f(x)dx=\sum _{j=1}^{n}w_{j}f(x_{j})=\sum _{j=1}^{n}\delta _{ij}w_{j}=w_{i}>0.}$${\displaystyle \int _{a}^{b}\omega (x)f(x)dx=\sum _{j=1}^{n}w_{j}f(x_{j})=\sum _{j=1}^{n}\delta _{ij}w_{j}=w_{i}>0.}

Since both ${\displaystyle \omega (x)}${\displaystyle \omega (x)} and ${\displaystyle f(x)}${\displaystyle f(x)} are non-negative functions, it follows that ${\displaystyle w_{i}>0}${\displaystyle w_{i}>0}.

There are many algorithms for computing the nodes x_i and weights w_i of Gaussian quadrature rules. The most popular are the Golub-Welsch algorithm requiring O(n²) operations, Newton's method for solving ${\displaystyle p_{n}(x)=0}${\displaystyle p_{n}(x)=0} using the three-term recurrence for evaluation requiring O(n²) operations, and asymptotic formulas for large n requiring O(n) operations.

Orthogonal polynomials ${\displaystyle p_{r}}${\displaystyle p_{r}} with ${\displaystyle (p_{r},p_{s})=0}${\displaystyle (p_{r},p_{s})=0} for ${\displaystyle r\neq s}${\displaystyle r\neq s} for a scalar product ${\displaystyle (\cdot ,\cdot )}${\displaystyle (\cdot ,\cdot )}, degree ${\displaystyle (p_{r})=r}${\displaystyle (p_{r})=r} and leading coefficient one (i.e. monic orthogonal polynomials) satisfy the recurrence relation $${\displaystyle p_{r+1}(x)=(x-a_{r,r})p_{r}(x)-a_{r,r-1}p_{r-1}(x)\cdots -a_{r,0}p_{0}(x)}$${\displaystyle p_{r+1}(x)=(x-a_{r,r})p_{r}(x)-a_{r,r-1}p_{r-1}(x)\cdots -a_{r,0}p_{0}(x)}

and scalar product defined $${\displaystyle (f(x),g(x))=\int _{a}^{b}\omega (x)f(x)g(x)dx}$${\displaystyle (f(x),g(x))=\int _{a}^{b}\omega (x)f(x)g(x)dx}

for ${\displaystyle r=0,1,\ldots ,n-1}${\displaystyle r=0,1,\ldots ,n-1} where n is the maximal degree which can be taken to be infinity, and where ${\textstyle a_{r,s}={\frac {\left(xp_{r},p_{s}\right)}{\left(p_{s},p_{s}\right)}}}${\textstyle a_{r,s}={\frac {\left(xp_{r},p_{s}\right)}{\left(p_{s},p_{s}\right)}}}. First of all, the polynomials defined by the recurrence relation starting with ${\displaystyle p_{0}(x)=1}${\displaystyle p_{0}(x)=1} have leading coefficient one and correct degree. Given the starting point by ${\displaystyle p_{0}}${\displaystyle p_{0}}, the orthogonality of ${\displaystyle p_{r}}${\displaystyle p_{r}} can be shown by induction. For ${\displaystyle r=s=0}${\displaystyle r=s=0} one has $${\displaystyle (p_{1},p_{0})=(x-a_{0,0})(p_{0},p_{0})=(xp_{0},p_{0})-a_{0,0}(p_{0},p_{0})=(xp_{0},p_{0})-(xp_{0},p_{0})=0.}$${\displaystyle (p_{1},p_{0})=(x-a_{0,0})(p_{0},p_{0})=(xp_{0},p_{0})-a_{0,0}(p_{0},p_{0})=(xp_{0},p_{0})-(xp_{0},p_{0})=0.}

Now if ${\displaystyle p_{0},p_{1},\ldots ,p_{r}}${\displaystyle p_{0},p_{1},\ldots ,p_{r}} are orthogonal, then also ${\displaystyle p_{r+1}}${\displaystyle p_{r+1}}, because in $${\displaystyle (p_{r+1},p_{s})=(xp_{r},p_{s})-a_{r,r}(p_{r},p_{s})-a_{r,r-1}(p_{r-1},p_{s})\cdots -a_{r,0}(p_{0},p_{s})}$${\displaystyle (p_{r+1},p_{s})=(xp_{r},p_{s})-a_{r,r}(p_{r},p_{s})-a_{r,r-1}(p_{r-1},p_{s})\cdots -a_{r,0}(p_{0},p_{s})} all scalar products vanish except for the first one and the one where ${\displaystyle p_{s}}${\displaystyle p_{s}} meets the same orthogonal polynomial. Therefore, $${\displaystyle (p_{r+1},p_{s})=(xp_{r},p_{s})-a_{r,s}(p_{s},p_{s})=(xp_{r},p_{s})-(xp_{r},p_{s})=0.}$${\displaystyle (p_{r+1},p_{s})=(xp_{r},p_{s})-a_{r,s}(p_{s},p_{s})=(xp_{r},p_{s})-(xp_{r},p_{s})=0.}

However, if the scalar product satisfies ${\displaystyle (xf,g)=(f,xg)}${\displaystyle (xf,g)=(f,xg)} (which is the case for Gaussian quadrature), the recurrence relation reduces to a three-term recurrence relation: For ${\displaystyle s<r-1,xp_{s}}${\displaystyle s<r-1,xp_{s}} is a polynomial of degree less than or equal to r − 1. On the other hand, ${\displaystyle p_{r}}${\displaystyle p_{r}} is orthogonal to every polynomial of degree less than or equal to r − 1. Therefore, one has ${\displaystyle (xp_{r},p_{s})=(p_{r},xp_{s})=0}${\displaystyle (xp_{r},p_{s})=(p_{r},xp_{s})=0} and ${\displaystyle a_{r,s}=0}${\displaystyle a_{r,s}=0} for s < r − 1. The recurrence relation then simplifies to $${\displaystyle p_{r+1}(x)=(x-a_{r,r})p_{r}(x)-a_{r,r-1}p_{r-1}(x)}$${\displaystyle p_{r+1}(x)=(x-a_{r,r})p_{r}(x)-a_{r,r-1}p_{r-1}(x)}

or $${\displaystyle p_{r+1}(x)=(x-a_{r})p_{r}(x)-b_{r}p_{r-1}(x)}$${\displaystyle p_{r+1}(x)=(x-a_{r})p_{r}(x)-b_{r}p_{r-1}(x)}

(with the convention ${\displaystyle p_{-1}(x)\equiv 0}${\displaystyle p_{-1}(x)\equiv 0}) where $${\displaystyle a_{r}:={\frac {(xp_{r},p_{r})}{(p_{r},p_{r})}},\qquad b_{r}:={\frac {(xp_{r},p_{r-1})}{(p_{r-1},p_{r-1})}}={\frac {(p_{r},p_{r})}{(p_{r-1},p_{r-1})}}}$${\displaystyle a_{r}:={\frac {(xp_{r},p_{r})}{(p_{r},p_{r})}},\qquad b_{r}:={\frac {(xp_{r},p_{r-1})}{(p_{r-1},p_{r-1})}}={\frac {(p_{r},p_{r})}{(p_{r-1},p_{r-1})}}}

(the last because of ${\displaystyle (xp_{r},p_{r-1})=(p_{r},xp_{r-1})=(p_{r},p_{r})}${\displaystyle (xp_{r},p_{r-1})=(p_{r},xp_{r-1})=(p_{r},p_{r})}, since ${\displaystyle xp_{r-1}}${\displaystyle xp_{r-1}} differs from ${\displaystyle p_{r}}${\displaystyle p_{r}} by a degree less than r).

The three-term recurrence relation can be written in matrix form ${\displaystyle J{\tilde {P}}=x{\tilde {P}}-p_{n}(x)\mathbf {e} _{n}}${\displaystyle J{\tilde {P}}=x{\tilde {P}}-p_{n}(x)\mathbf {e} _{n}} where ${\displaystyle {\tilde {P}}={\begin{bmatrix}p_{0}(x)&p_{1}(x)&\cdots &p_{n-1}(x)\end{bmatrix}}^{\mathsf {T}}}${\displaystyle {\tilde {P}}={\begin{bmatrix}p_{0}(x)&p_{1}(x)&\cdots &p_{n-1}(x)\end{bmatrix}}^{\mathsf {T}}}, ${\displaystyle \mathbf {e} _{n}}${\displaystyle \mathbf {e} _{n}} is the ${\displaystyle n}${\displaystyle n}th standard basis vector, i.e., ${\displaystyle \mathbf {e} _{n}={\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}^{\mathsf {T}}}${\displaystyle \mathbf {e} _{n}={\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}^{\mathsf {T}}}, and J is the following tridiagonal matrix, called the Jacobi matrix: $${\displaystyle \mathbf {J} ={\begin{bmatrix}a_{0}&1&0&\cdots &0\\b_{1}&a_{1}&1&\ddots &\vdots \0円&b_{2}&\ddots &\ddots &0\\\vdots &\ddots &\ddots &a_{n-2}&1\0円&\cdots &0&b_{n-1}&a_{n-1}\end{bmatrix}}.}$${\displaystyle \mathbf {J} ={\begin{bmatrix}a_{0}&1&0&\cdots &0\\b_{1}&a_{1}&1&\ddots &\vdots \0円&b_{2}&\ddots &\ddots &0\\\vdots &\ddots &\ddots &a_{n-2}&1\0円&\cdots &0&b_{n-1}&a_{n-1}\end{bmatrix}}.}

The zeros ${\displaystyle x_{j}}${\displaystyle x_{j}} of the polynomials up to degree n, which are used as nodes for the Gaussian quadrature can be found by computing the eigenvalues of this matrix. This procedure is known as Golub–Welsch algorithm.

For computing the weights and nodes, it is preferable to consider the symmetric tridiagonal matrix ${\displaystyle {\mathcal {J}}}${\displaystyle {\mathcal {J}}} with elements $${\displaystyle {\begin{aligned}{\mathcal {J}}_{k,i}=J_{k,i}&=a_{k-1}&k&=1,2,\ldots ,n\\[2.1ex]{\mathcal {J}}_{k-1,i}={\mathcal {J}}_{k,k-1}={\sqrt {J_{k,k-1}J_{k-1,k}}}&={\sqrt {b_{k-1}}}&k&={\hphantom {1,,円}}2,\ldots ,n.\end{aligned}}}$${\displaystyle {\begin{aligned}{\mathcal {J}}_{k,i}=J_{k,i}&=a_{k-1}&k&=1,2,\ldots ,n\\[2.1ex]{\mathcal {J}}_{k-1,i}={\mathcal {J}}_{k,k-1}={\sqrt {J_{k,k-1}J_{k-1,k}}}&={\sqrt {b_{k-1}}}&k&={\hphantom {1,,円}}2,\ldots ,n.\end{aligned}}}

That is,

$${\displaystyle {\mathcal {J}}={\begin{bmatrix}a_{0}&{\sqrt {b_{1}}}&0&\cdots &0\\{\sqrt {b_{1}}}&a_{1}&{\sqrt {b_{2}}}&\ddots &\vdots \0円&{\sqrt {b_{2}}}&\ddots &\ddots &0\\\vdots &\ddots &\ddots &a_{n-2}&{\sqrt {b_{n-1}}}\0円&\cdots &0&{\sqrt {b_{n-1}}}&a_{n-1}\end{bmatrix}}.}$${\displaystyle {\mathcal {J}}={\begin{bmatrix}a_{0}&{\sqrt {b_{1}}}&0&\cdots &0\\{\sqrt {b_{1}}}&a_{1}&{\sqrt {b_{2}}}&\ddots &\vdots \0円&{\sqrt {b_{2}}}&\ddots &\ddots &0\\\vdots &\ddots &\ddots &a_{n-2}&{\sqrt {b_{n-1}}}\0円&\cdots &0&{\sqrt {b_{n-1}}}&a_{n-1}\end{bmatrix}}.}

J and ${\displaystyle {\mathcal {J}}}${\displaystyle {\mathcal {J}}} are similar matrices and therefore have the same eigenvalues (the nodes). The weights can be computed from the corresponding eigenvectors: If ${\displaystyle \phi ^{(j)}}${\displaystyle \phi ^{(j)}} is a normalized eigenvector (i.e., an eigenvector with euclidean norm equal to one) associated with the eigenvalue x_j, the corresponding weight can be computed from the first component of this eigenvector, namely: $${\displaystyle w_{j}=\mu _{0}\left(\phi _{1}^{(j)}\right)^{2}}$${\displaystyle w_{j}=\mu _{0}\left(\phi _{1}^{(j)}\right)^{2}}

where ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}} is the integral of the weight function $${\displaystyle \mu _{0}=\int _{a}^{b}\omega (x)dx.}$${\displaystyle \mu _{0}=\int _{a}^{b}\omega (x)dx.}

See, for instance, (Gil, Segura & Temme 2007) for further details.

The error of a Gaussian quadrature rule can be stated as follows.^[5] For an integrand which has 2n continuous derivatives, $${\displaystyle \int _{a}^{b}\omega (x),円f(x),円dx-\sum _{i=1}^{n}w_{i},円f(x_{i})={\frac {f^{(2n)}(\xi )}{(2n)!}},円(p_{n},p_{n})}$${\displaystyle \int _{a}^{b}\omega (x),円f(x),円dx-\sum _{i=1}^{n}w_{i},円f(x_{i})={\frac {f^{(2n)}(\xi )}{(2n)!}},円(p_{n},p_{n})} for some ξ in (a, b), where p_n is the monic (i.e. the leading coefficient is 1) orthogonal polynomial of degree n and where $${\displaystyle (f,g)=\int _{a}^{b}\omega (x)f(x)g(x),円dx.}$${\displaystyle (f,g)=\int _{a}^{b}\omega (x)f(x)g(x),円dx.}

In the important special case of ω(x) = 1, we have the error estimate^[6] $${\displaystyle {\frac {\left(b-a\right)^{2n+1}\left(n!\right)^{4}}{(2n+1)\left[\left(2n\right)!\right]^{3}}}f^{(2n)}(\xi ),\qquad a<\xi <b.}$${\displaystyle {\frac {\left(b-a\right)^{2n+1}\left(n!\right)^{4}}{(2n+1)\left[\left(2n\right)!\right]^{3}}}f^{(2n)}(\xi ),\qquad a<\xi <b.}

Stoer and Bulirsch remark that this error estimate is inconvenient in practice, since it may be difficult to estimate the order 2n derivative, and furthermore the actual error may be much less than a bound established by the derivative. Another approach is to use two Gaussian quadrature rules of different orders, and to estimate the error as the difference between the two results. For this purpose, Gauss–Kronrod quadrature rules can be useful.

If the interval [a, b] is subdivided, the Gauss evaluation points of the new subintervals never coincide with the previous evaluation points (except at zero for odd numbers), and thus the integrand must be evaluated at every point. Gauss–Kronrod rules are extensions of Gauss quadrature rules generated by adding n + 1 points to an n-point rule in such a way that the resulting rule is of order 2n + 1. This allows for computing higher-order estimates while re-using the function values of a lower-order estimate. The difference between a Gauss quadrature rule and its Kronrod extension is often used as an estimate of the approximation error.

In some applications, it is desirable to have quadrature rules that have the high accuracy of Gauss formulas, but that also include the end points of the interval among the evaluation points. Such rules are known as Gauss-Lobatto, or simply Lobatto quadrature,^[7] named after Dutch mathematician Rehuel Lobatto. Because for an n point rule, one can no longer freely choose the locations of all quadrature points (2 of the points are fixed at the end points), one needs to expect that the rule is less accurate than regular Gaussian quadrature. Indeed, an n point Gauss-Lobatto rule is only accurate for polynomials up to degree 2n − 3.^[8]

Lobatto quadrature of function f(x) on interval [−1, 1]: $${\displaystyle \int _{-1}^{1}{f(x),円dx}={\frac {2}{n(n-1)}}[f(1)+f(-1)]+\sum _{i=2}^{n-1}{w_{i}f(x_{i})}+R_{n}.}$${\displaystyle \int _{-1}^{1}{f(x),円dx}={\frac {2}{n(n-1)}}[f(1)+f(-1)]+\sum _{i=2}^{n-1}{w_{i}f(x_{i})}+R_{n}.}

Abscissas: x_i is the ${\displaystyle (i-1)}${\displaystyle (i-1)}st zero of ${\displaystyle P'_{n-1}(x)}${\displaystyle P'_{n-1}(x)}, here ${\displaystyle P_{m}(x)}${\displaystyle P_{m}(x)} denotes the standard Legendre polynomial of m-th degree and the dash denotes the derivative.

Weights: $${\displaystyle w_{i}={\frac {2}{n(n-1)\left[P_{n-1}\left(x_{i}\right)\right]^{2}}},\qquad x_{i}\neq \pm 1.}$${\displaystyle w_{i}={\frac {2}{n(n-1)\left[P_{n-1}\left(x_{i}\right)\right]^{2}}},\qquad x_{i}\neq \pm 1.}

Remainder: $${\displaystyle R_{n}={\frac {-n\left(n-1\right)^{3}2^{2n-1}\left[\left(n-2\right)!\right]^{4}}{(2n-1)\left[\left(2n-2\right)!\right]^{3}}}f^{(2n-2)}(\xi ),\qquad -1<\xi <1.}$${\displaystyle R_{n}={\frac {-n\left(n-1\right)^{3}2^{2n-1}\left[\left(n-2\right)!\right]^{4}}{(2n-1)\left[\left(2n-2\right)!\right]^{3}}}f^{(2n-2)}(\xi ),\qquad -1<\xi <1.}

Some of the weights are:

An adaptive variant of this algorithm with 2 interior nodes^[9] is found in GNU Octave and MATLAB as quadl and integrate.^{[10] [11]}

• "Gauss quadrature formula", Encyclopedia of Mathematics , EMS Press, 2001 [1994]
• ALGLIB contains a collection of algorithms for numerical integration (in C# / C++ / Delphi / Visual Basic / etc.)
• GNU Scientific Library — includes C version of QUADPACK algorithms (see also GNU Scientific Library)
• From Lobatto Quadrature to the Euler constant e
• Gaussian Quadrature Rule of Integration – Notes, PPT, Matlab, Mathematica, Maple, Mathcad at Holistic Numerical Methods Institute
• Weisstein, Eric W. "Legendre-Gauss Quadrature". MathWorld .
• Gaussian Quadrature by Chris Maes and Anton Antonov, Wolfram Demonstrations Project.
• Tabulated weights and abscissae with Mathematica source code, high precision (16 and 256 decimal places) Legendre-Gaussian quadrature weights and abscissas, for n=2 through n=64, with Mathematica source code.
• Mathematica source code distributed under the GNU LGPL for abscissas and weights generation for arbitrary weighting functions W(x), integration domains and precisions.
• Gaussian Quadrature in Boost.Math, for arbitrary precision and approximation order
• Gauss–Kronrod Quadrature in Boost.Math
• Nodes and Weights of Gaussian quadrature Archived 2021年04月14日 at the Wayback Machine

• Numerical integration

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