高斯积分法 Gaussian Quadrature

In Newton-Cotes formula, we choose \(n+1\) evenly spaced points on the interval to do Lagrange interpolation and deduce a formula from the integral of that polynomial. To improve the precision, we will choose \(n\) special points in the interval; with the same interpolation–integration approach, we can achieve a degree of exactness of \(2n-1\).
(Recall: When an algorithm works exactly right for all polynomials of degree \(\leq m\) but not for degree \(m+1\), we say that it has a degree of exactness of \(m\). )

One of the typical solutions is to choose the \(n\) roots of the Legendre polynomial \(P_n(x)\). The Legendre polynomials form a complete orthogonal system on \([-1,1]\) with weight function \(w(x)=1\) (where \(P_n(x)\) is a polynomial of degree \(n\)), i.e.

\[\int_{-1}^{1}{P_m(x)P_n(x)\,\mathrm{d}x}=0\quad \text{if}\, m\neq n. \]

With the standardization condition \(P_n(1)=1\) for all \(n\), all the polynomials can be uniquely determined. In fact, by Rodrigues' Formula,

\[P_n(x)=\frac{1}{2^{n}n!}\frac{\mathrm{d}^{n}}{\mathrm{d}x^n}{(x^2-1)^n}. \]

Let \(x_1,x_2,\cdots,x_n\) be the roots of \(P_n(x)\). We squeeze our target interval \([a,b]\) to \([-1,1]\) and do Lagrange interpolation with points \((x_i,g(x_i))\), where \(g(x)\) is the transformed \(f(x)\). Gaussian Quadrature is then

\[\int_{a}^{b}{f(x)\,\mathrm{d}x}\approx \frac{b-a}{2}\sum_{i=1}^{n}{w_{i}f(\frac{b-a}{2}x_i+\frac{a+b}{2})} \]

where the weights \(w_i\)​ can be found similarly to the Newton–Cotes \(\alpha_i​\) values, or via the closed form

\[w_i=\frac{2}{(1-x_i^2)[P_n'(x_i)]^2}. \]

We have known the procedure of Gaussian quadrature. But why are Legendre polynomials so useful that our exactness is doubled? It all comes from the orthogonality. Say we have a polynomial \(T(x)\) whose degree is no more than \(2n-1\). Divide \(T(x)\) by \(P_n(x)\)and we have \(T(x)=Q(x)P_n(x)+R(x)\), where \(deg\,Q\leq n-1,\, deg\,R\leq n-1\). We write \(Q(x)\) as a linear combination of lower-order Legendre polynomials: \(Q(x)=\sum_{i=0}^{n-1}{k_{i}P_i(x)}\). Then by the definition of Legendre polynomials, \(\int_{-1}^{1}{P_i(x)P_n(x)=0}\), thus \(\int_{-1}^{1}{Q(x)P_n(x)=0}\).
Now we are ready to prove that

\[\int_{-1}^{1}{T(x)\,\mathrm{d}x}=\sum_{i=1}^{n}{w_{i}T(x_i)}. \]

For the left side,

\[\int_{-1}^{1}{T(x)\,\mathrm{d}x}=\int_{-1}^{1}{Q(x)P_n(x)\,\mathrm{d}x}+\int_{-1}^{1}{R(x)\,\mathrm{d}x}=\int_{-1}^{1}{R(x)\,\mathrm{d}x}. \]

For the right side,

\[\forall i=1,2,\cdots,n\quad P_n(x_i)=0 \]

\[\sum_{i=1}^{n}{w_{i}T(x_i)}=\sum_{i=1}^{n}{w_{i}[Q(x_i)P_n(x_i)+R(x_i)]}=\sum_{i=1}^{n}{w_{i}R(x_i)}. \]

By its construction, an n-point Gaussian quadrature rule is exact for any polynomial of degree up to \(n−1\). Since \(deg\,R\leq n-1\), the quadrature gives the exact integral:

\[\int_{-1}^{1}{R(x)\,\mathrm{d}x}=\sum_{i=1}^{n}{w_{i}R(x_i)}. \]

Thus our quadrature works exactly right for \(T(x)\). On the other hand, the error

\[E_n​(f)=\frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^3}f^{(2n)}(\xi) \]

for some \(\xi\in(-1,1)\).
Therefore, the degree of exactness is \(2n-1\). Above is all of the magic of Gaussian quadrature.