In the quadrature methods studied so far, the nodes were first chosen
and then we determine the weights via some function interpolation
process. The resulting methods have certain degree of precision in terms
of what polynomials they can exactly integrate.
Let us consider the task of computing
I ( f ) = ∫ a b w ( x ) f ( x ) d x I(f) = \int_a^b w(x) f(x) \ud x I ( f ) = ∫ a b w ( x ) f ( x ) d x where w ( x ) w(x) w ( x )
I n ( f ) = ∑ j = 1 n w j f ( x j ) I_n(f) = \sum_{j=1}^n w_j f(x_j) I n ( f ) = j = 1 ∑ n w j f ( x j ) Here w j w_j w j w ( x ) w(x) w ( x )
The evaluation of f f f
8.8.1 Unit weight function ¶ Let us take the interval to be [ − 1 , + 1 ] [-1,+1] [ − 1 , + 1 ] w ( x ) = 1 w(x)=1 w ( x ) = 1
∫ − 1 1 f ( x ) d x ≈ I n ( f ) = ∑ j = 1 n w j f ( x j ) \int_{-1}^1 f(x) \ud x \approx I_n(f) = \sum_{j=1}^n w_j f(x_j) ∫ − 1 1 f ( x ) d x ≈ I n ( f ) = j = 1 ∑ n w j f ( x j ) The
error
E n ( f ) = ∫ − 1 1 f ( x ) − ∑ j = 1 n w j f ( x j ) E_n(f) = \int_{-1}^1 f(x) - \sum_{j=1}^n w_j f(x_j) E n ( f ) = ∫ − 1 1 f ( x ) − j = 1 ∑ n w j f ( x j ) Note thet
E n : C [ − 1 , 1 ] → R E_n : \cts[-1,1] \to \re E n : C [ − 1 , 1 ] → R
E n ( a 0 + a 1 x + … a m x m ) = a 0 E n ( 1 ) + a 1 E n ( x ) + … + a m E n ( x m ) E_n(a_0 + a_1 x + \ldots a_m x^m) = a_0 E_n(1) + a_1 E_n(x) + \ldots + a_m E_n(x^m) E n ( a 0 + a 1 x + … a m x m ) = a 0 E n ( 1 ) + a 1 E n ( x ) + … + a m E n ( x m ) Hence
E n ( p ) = 0 , ∀ p ∈ P m E_n(p) = 0, \qquad \forall p \in \poly_m E n ( p ) = 0 , ∀ p ∈ P m if and only if
E n ( x i ) = 0 , 0 ≤ i ≤ m E_n(x^i) = 0, \qquad 0 \le i \le m E n ( x i ) = 0 , 0 ≤ i ≤ m 8.8.1.1 Case n = 1 n=1 n = 1 ¶ The quadrature formula is
I 1 ( f ) = w 1 f ( x 1 ) I_1(f) = w_1 f(x_1) I 1 ( f ) = w 1 f ( x 1 ) and we have to find w 1 w_1 w 1 x 1 x_1 x 1
E 1 ( 1 ) = 0 ⟹ ∫ − 1 1 d x − w 1 = 0 E 1 ( x ) = 0 ⟹ ∫ − 1 1 x d x − w 1 x 1 = 0 \begin{align}
E_1(1) &= 0 \quad\implies\quad \int_{-1}^1 \ud x - w_1 = 0 \\
E_1(x) &= 0 \quad\implies\quad \int_{-1}^1 x \ud x - w_1 x_1 = 0
\end{align} E 1 ( 1 ) E 1 ( x ) = 0 ⟹ ∫ − 1 1 d x − w 1 = 0 = 0 ⟹ ∫ − 1 1 x d x − w 1 x 1 = 0 which yields
w 1 = 2 , x 1 = 0 w_1 = 2, \qquad x_1 = 0 w 1 = 2 , x 1 = 0 so that
∫ − 1 1 f ( x ) d x ≈ 2 f ( 0 ) \int_{-1}^1 f(x) \ud x \approx 2 f(0) ∫ − 1 1 f ( x ) d x ≈ 2 f ( 0 ) This is just the midpoint-rule. It uses only one point but is exact for linear polynomials. Note that trapezoidal method uses two points but is also exact only for linear polynomials. In terms of the number of nodes used, the mid-point rule is more optimal since it needs less function evaluation to give same degree of precision.
8.8.1.2 Case n = 2 n=2 n = 2 ¶ The quadrature formula is
I 2 ( f ) = w 1 f ( x 1 ) + w 2 f ( x 2 ) I_2(f) = w_1 f(x_1) + w_2 f(x_2) I 2 ( f ) = w 1 f ( x 1 ) + w 2 f ( x 2 ) and we have to determine four quantities w 1 , w 2 , x 1 , x 2 w_1, w_2, x_1, x_2 w 1 , w 2 , x 1 , x 2
E 2 ( x i ) = ∫ − 1 1 x i d x − [ w 1 x 1 i + w 2 x 2 i ] = 0 , i = 0 , 1 , 2 , 3 E_2(x^i) = \int_{-1}^1 x^i \ud x - [w_1 x_1^i + w_2 x_2^i] = 0, \qquad i=0,1,2,3 E 2 ( x i ) = ∫ − 1 1 x i d x − [ w 1 x 1 i + w 2 x 2 i ] = 0 , i = 0 , 1 , 2 , 3 which gives us four equations
w 1 + w 2 = 2 w 1 x 1 + w 2 x 2 = 0 w 1 x 1 2 + w 2 x 2 2 = 2 / 3 w 1 x 1 3 + w 2 x 2 3 = 0 \begin{aligned}
w_1 + w_2 &= 2 \\
w_1 x_1 + w_2 x_2 &= 0 \\
w_1 x_1^2 + w_2 x_2^2 &= 2/3 \\
w_1 x_1^3 + w_2 x_2^3 &= 0
\end{aligned} w 1 + w 2 w 1 x 1 + w 2 x 2 w 1 x 1 2 + w 2 x 2 2 w 1 x 1 3 + w 2 x 2 3 = 2 = 0 = 2/3 = 0 The unique solution is
w 1 = w 2 = 1 , x 1 = − 1 3 , x 2 = 1 3 w_1 = w_2 = 1, \qquad x_1 = -\frac{1}{\sqrt{3}}, \qquad x_2 = \frac{1}{\sqrt{3}} w 1 = w 2 = 1 , x 1 = − 3 1 , x 2 = 3 1 This rule uses 2 points but has degree of precision 3. Simpson rule also
has degree of precision 3 but uses 3 points.
8.8.1.3 Case of general n n n ¶ There are 2 n 2n 2 n { w j , x j } \{w_j, x_j\} { w j , x j } j = 1 , 2 , … , n j=1,2,\ldots,n j = 1 , 2 , … , n 2 n − 1 2n-1 2 n − 1
E n ( x i ) = 0 , i = 0 , 1 , 2 , … , 2 n − 1 E_n(x^i) = 0, \qquad i=0,1,2,\ldots,2n-1 E n ( x i ) = 0 , i = 0 , 1 , 2 , … , 2 n − 1 which gives 2 n 2n 2 n
∑ j = 1 n w j x j i = { 0 i = 1 , 3 , … , 2 n − 1 2 i + 1 i = 0 , 2 , … , 2 n − 2 \sum_{j=1}^n w_j x_j^i = \begin{cases}
0 & i=1,3,\ldots,2n-1 \\
\frac{2}{i+1} & i=0,2,\ldots,2n-2
\end{cases} j = 1 ∑ n w j x j i = { 0 i + 1 2 i = 1 , 3 , … , 2 n − 1 i = 0 , 2 , … , 2 n − 2 We have system of non-linear equations and it is not
possible to solve them analytically. We will take a digression to study
some properties of orthogonal polynomials which will help us to derive
these Gauss quadrature rules.
8.8.2 Unit weight: solution for general case ¶ We can find the quadrature rule without having to solve a matrix problem. Let f : [ − 1 , 1 ] → R f : [-1,1] \to \re f : [ − 1 , 1 ] → R n ≥ 1 n \ge 1 n ≥ 1 { x j , 1 ≤ j ≤ n } \{ x_j, 1 \le j \le n \} { x j , 1 ≤ j ≤ n } P n P_n P n [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
f n ( x ) = ∑ j = 1 n f ( x j ) ℓ j ( x ) f_n(x) = \sum_{j=1}^n f(x_j) \ell_j(x) f n ( x ) = j = 1 ∑ n f ( x j ) ℓ j ( x ) where ℓ j \ell_j ℓ j n − 1 n-1 n − 1
f ( x j ) − f n ( x j ) = 0 , 1 ≤ j ≤ n f(x_j) - f_n(x_j) = 0, \qquad 1 \le j \le n f ( x j ) − f n ( x j ) = 0 , 1 ≤ j ≤ n we can write
f ( x ) = f n ( x ) + P n ( x ) R ( x ) f(x) = f_n(x) + P_n(x) R(x) f ( x ) = f n ( x ) + P n ( x ) R ( x ) for some function R R R
I ( f ) = ∫ − 1 1 f ( x ) d x = ∫ − 1 1 f n ( x ) d x + ∫ − 1 1 P n ( x ) R ( x ) d x = ∑ j = 1 n w j f ( x j ) + ∫ − 1 1 P n ( x ) R ( x ) d x = I n ( f ) + ∫ − 1 1 P n ( x ) R ( x ) d x \begin{align*}
I(f) &= \int_{-1}^1 f(x) \ud x \\
&= \int_{-1}^1 f_n(x) \ud x + \int_{-1}^1 P_n(x) R(x) \ud x \\
&= \sum_{j=1}^n w_j f(x_j) + \int_{-1}^1 P_n(x) R(x) \ud x \\
&= I_n(f) + \int_{-1}^1 P_n(x) R(x) \ud x \\
\end{align*} I ( f ) = ∫ − 1 1 f ( x ) d x = ∫ − 1 1 f n ( x ) d x + ∫ − 1 1 P n ( x ) R ( x ) d x = j = 1 ∑ n w j f ( x j ) + ∫ − 1 1 P n ( x ) R ( x ) d x = I n ( f ) + ∫ − 1 1 P n ( x ) R ( x ) d x where
I n ( f ) = ∑ j = 1 n w j f ( x j ) , w j = ∫ − 1 1 ℓ j ( x ) d x I_n(f) = \sum_{j=1}^n w_j f(x_j), \qquad w_j = \int_{-1}^1 \ell_j(x) \ud x I n ( f ) = j = 1 ∑ n w j f ( x j ) , w j = ∫ − 1 1 ℓ j ( x ) d x If f ∈ P 2 n − 1 f \in \poly_{2n-1} f ∈ P 2 n − 1 R ∈ P n − 1 R \in \poly_{n-1} R ∈ P n − 1
R = a 0 P 0 + a 1 P 1 + … + a n − 1 P n − 1 ⇓ ∫ − 1 1 P n ( x ) R ( x ) d x = 0 , ∀ R ∈ P n − 1 \begin{gather}
R = a_0 P_0 + a_1 P_1 + \ldots + a_{n-1}P_{n-1} \\
\Downarrow \\
\int_{-1}^1 P_n(x) R(x) \ud x = 0, \qquad \forall R \in \poly_{n-1}
\end{gather} R = a 0 P 0 + a 1 P 1 + … + a n − 1 P n − 1 ⇓ ∫ − 1 1 P n ( x ) R ( x ) d x = 0 , ∀ R ∈ P n − 1 Thus
I n ( f ) = I ( f ) , ∀ f ∈ P 2 n − 1 I_n(f) = I(f), \qquad \forall f \in \poly_{2n-1} I n ( f ) = I ( f ) , ∀ f ∈ P 2 n − 1 the quadrature formula has degree of precision atleast 2 n − 1 2n-1 2 n − 1