Gaussian Quadrature

In M245 we used the method of undetermined coefficients to find the quadrature points (x_i)_i=1ⁿ and weights (H_i)_i=1ⁿ in the Gaussian formula $\sum _{i=1}^nH_if(x_i)$ for the evaluation of $\int _a^bf(x)w(x)dx$. This amounts to setting the error

$$E(f)=\int _a^bf(x)w(x)dx-\sum _{i=1}^nH_if(x_i)$$

to zero for $f(x)=x^p,~p=0,1,\ldots ,2n-1$ to obtain the equations  

$$\int _a^bx^pw(x)dx=\sum _{i=1}^nH_ix_i^p,\quad p=0,1,\ldots ,2n-1.$$ (4)

Since there are n quadrature points, we assume they are the zeros of

$$\Pi (x)=\prod _{i=1}^n(x-x_i)=\sum _{k=0}^na_kx^k,$$

with $a_n\equiv 1$.Now, for $p=0,1,\ldots ,n-1$, we form linear combinations of equations ():

$$\begin{array} {lcl}\sum _{k=0}^na_k\int _a^bx^{p+k}w(x)dx &... ... \Rightarrow~~~~~\int _a^b\Pi (x)x^pw(x)dx & = & 0,\end{array}$$

since $\sum _{k=0}^na_kx_i^k=\Pi (x_i)=0$ for $i=1,2,\ldots ,n$. This is a set of n linear equations, which we can solve for the coefficients of $\Pi$, whose zeros provide the quadrature points.

Theorem 3926

If distinct points $x_1,x_2,\ldots x_n$ can be chosen so that

$$\int _a^b\Pi (x)x^sw(x)dx=0,\quad s=0,1,\ldots ,n-1,$$

where $\Pi (x)=\Pi _{i=1}^n(x-x_i)$ and $w(x)\ge 0$, and if $H_1,H_2,\ldots ,H_n$ are chosen so that

$$E(f)=\int _a^bf(x)w(x)dx-\sum _{s=1}^nH_sf(x_s)$$

vanishes when f is any polynomial of degree n-1, then E(f) vanishes when f is any polynomial of degree 2n-1.

Find a 2-point Gaussian formula for $\int _0^1f(x)\log xdx$.

Let the required formula be H₁f(x₁)+H₂f(x₂) and let $\Pi (x)=(x-x_1)(x-x_2)=x^2-ax+b$. Then

$$\begin{array} {lcl}0 & = & \int _0^1(x^2-ax+b)\log xdx=\frac... ...ax+b)\log xdx=\frac{a}{9}-\frac{b}{4}-\frac{1}{16}.\end{array}$$

This is the same example we solved in M245, and the rest of the solution is identical. We note that the number of points is n=2, so the resulting formula is exact for polynomials of degree 2n-1=3.

We finally establish that the quadrature points always exist, and moreover that they are real, distinct and in (a,b).

Theorem 3940

If $w(x)\geq 0$ then there exists a unique monic polynomial $\Pi$ of degree n such that

$$\int _a^bx^s\Pi (x)w(x)dx=0,\quad s=0,1,\ldots ,n-1$$

and the zeros of $\Pi$ are real, distinct and strictly in (a,b).

[] Proof.Existence and Uniqueness

Let $\Pi (x)=\sum _{s=0}^{n-1}a_sx^s+x^n$. Then for $k=0,1,\ldots ,n-1$

$$\begin{array} {lcl}0 & = & \int _a^bx^k\Pi (x)w(x)dx\\ & = &... ...sum _{s=0}^{n-1}a_sx^sw(x)dx+\int _a^bx^kx^nw(x)dx.\end{array}$$

We can write these as a set of linear equations for $a_0,a_1,\ldots ,a_{n-1}$

$$\sum _{s=0}^{n-1}a_s\int _a^bx^{k+s}w(x)dx=-\int _a^bx^{k+n}w(x)dx,\quad k=0,1,\ldots ,n-1.$$

We show that they have a unique solution by showing that the corresponding homogeneous set

$$\sum _{s=0}^{n-1}a_s\int _a^bx^{k+s}w(x)dx=0,\quad k=0,1,\ldots ,n-1$$

has only the trivial solution $a_s=0,~s=0,1,\ldots ,n-1$. Writing them as

$$\int _a^bx^k\sum _{s=0}^{n-1}a_sx^sw(x)dx=0,\quad k=0,1,\ldots ,n-1,$$

we multiply the kth of these by a_k for $k=0,1,\ldots ,n-1$and add to obtain

$$\begin{array} {lcl}0 & = & \sum _{k=0}^{n-1}a_k\int _a^bx^k\... ..._a^b\left(\sum _{k=0}^{n-1}a_kx^k\right) ^2w(x)dx, \end{array}$$

which implies that $\sum _{k=0}^{n-1}a_kx^k=0$ for all $x\in (a,b)$, and hence that a_k=0 for $k=0,1,\ldots ,n-1$.