Determine the degree of precision of the quadrature formulas No integral interval is specified in the definition of the quadrature formulas. All the polynomials of degree isn can be given by the linear combination of the set of functions{xx The degree of precision of the quadrature formulas can be determined by computing the integral of the power function x'where isn over a suitable interval
Determine the degree of precision of the quadrature formulas ◼ No integral interval is specified in the definition of the quadrature formulas. ◼ All the polynomials of degree i≤n can be given by the linear combination of the set of functions {1, x, x 2 , x 3 ,…, x n}. ◼ The degree of precision of the quadrature formulas can be determined by computing the integral of the power function x i where i≤n over a suitable interval
Since awa-会ra-宫a(-各六心 and ma-蓉w蓉会小-宫客wn We want to prove that QIP(]=P()dx it's just to prove thatdforj01,. That means we can prove the precision of O[f]is n by proving that O[f]is exact for all xi,j户0,1，…，n
න 𝑎 𝑏 𝑃𝑖 (𝑥) 𝑑𝑥 = න 𝑎 𝑏 𝑗=0 𝑖 𝑎𝑗𝑥 𝑗𝑑𝑥 = 𝑗=0 𝑖 𝑎𝑗 න 𝑎 𝑏 𝑥 𝑗𝑑𝑥 = 𝑗=0 𝑖 𝑎𝑗 1 𝑗 + 1 𝑥 ቚ 𝑗+1 𝑎 𝑏 Since , and 𝑄 𝑃𝑖 𝑥 = 𝑘=0 𝑀 𝑤𝑘𝑃𝑖 (𝑥𝑘) = 𝑘=0 𝑀 𝑤𝑘 𝑗=0 𝑖 𝑎𝑗𝑥𝑘 𝑗 = 𝑗=0 𝑖 𝑎𝑗 𝑘=0 𝑀 𝑤𝑘 𝑥𝑘 𝑗 = 𝑗=0 𝑖 𝑎𝑗𝑄[𝑥 𝑗 ]. We want to prove that 𝑄 𝑃𝑖 𝑥 = �� 𝑏 𝑃𝑖 (𝑥) 𝑑𝑥, it’s just to prove that 𝑄 𝑥 �� = �� 𝑏 𝑥 𝑗𝑑𝑥 , for j = 0,1, …, i. That means we can prove the precision of Q[ f ] is n by proving that Q[ f ] is exact for all x j , j=0, 1, …, n
Example 5.4.Determine the degree of precision of Simpson's 3/8 rule. It will suffice to apply Simpson's 3/8 rule over the interval [0,3]with the five test functions fx)=1,x,x2,x3 and x4.For the first four functions,Simpson's 3/8 rule is exact. 1dx=3=81+3①)+3)+1) 93 ,xx=2-80+3四+3(2)+3) 3 x2=9=80+3(1+3(④+9) 3 813 x3dx= 4=8(0+3(1)+3(8)+27) The function fx)x4 is the lowest power of x for which the rule is not exact. xax=9≈婴=0+3(1)+316)+8 Therefore,the degree of precision of Simpson's 3/8 rule is n=3
Example 5.4. Determine the degree of precision of Simpson’s 3/8 rule. It will suffice to apply Simpson’s 3/8 rule over the interval [0, 3] with the five test functions f(x)=1, x, x 2 , x 3 and x 4 . For the first four functions, Simpson’s 3/8 rule is exact. 0 3 1 𝑑𝑥 = 3 = 3 8 1 + 3 1 + 3 1 + 1 න 0 3 𝑥 𝑑𝑥 = 9 2 = 3 8 0 + 3 1 + 3 2 + 3 න 0 3 𝑥 2 𝑑𝑥 = 9 = 3 8 0 + 3 1 + 3 4 + 9 න 0 3 𝑥 3 𝑑𝑥 = 81 4 = 3 8 0 + 3 1 + 3 8 + 27 The function f(x)=x 4 is the lowest power of x for which the rule is not exact. 0 3 𝑥 4 𝑑𝑥 = 243 5 ≈ 99 2 = 3 8 0 + 3 1 + 3 16 + 81 Therefore, the degree of precision of Simpson’s 3/8 rule is n=3
Newton-Cotes Precision Corollary 5.1.Assume that fx)is sufficiently differentiable;then E[f]for Newton- Cotes quadrature involves an appropriate higher derivative. The trapezoidal rule has degree of precision n=1.If feC2[a,b],then h-6+0花r传 ■ Simpson's rule has degree of precision n=3.IffeC4[a,b],then -3+4+0-re Simpson's 3/8 rule has degree of precision n=3.IffeC4[a,b],then -6++3近+-站阳 80 Boole's rule has degree of precision n=5.If fe co[a,b],then h=若6+2看+126+26+7 81f6() 945
Newton-Cotes Precision ◼ Corollary 5.1. Assume that f(x) is sufficiently differentiable; then E[f] for NewtonCotes quadrature involves an appropriate higher derivative. ◼ The trapezoidal rule has degree of precision n=1. If f∈C2 [a,b], then ◼ Simpson’s rule has degree of precision n=3. If f∈C4 [a,b], then ◼ Simpson’s 3/8 rule has degree of precision n=3. If f∈C4 [a,b], then ◼ Boole’s rule has degree of precision n=5. If f∈C6 [a,b], then 1 0 3 (2) 0 1 ( ) ( ) ( ) 2 12 x x h h f x dx f f f = + − 2 0 5 (4) 0 1 2 ( ) ( 4 ) ( ) 3 90 x x h h f x dx f f f f = + + − 3 0 5 (4) 0 1 2 3 3 3 ( ) ( 3 3 ) ( ) 8 80 x x h h f x dx f f f f f = + + + − 4 0 7 (6) 0 1 2 3 4 2 8 ( ) (7 32 12 32 7 ) ( ) 45 945 x x h h f x dx f f f f f f = + + + + −
The Step Sizes In Example 5.1 we applied the quadrature rules with h=0.5.If the endpoints of the interval [a,b]are held fixed,the step size must be adjusted for each rule.The step sizes are h=b-a,h=(b-a)/2,h=(b-a)/3, and h=(b-a)/4 for the trapezoidal rule,Simpson's rule,Simpson's 3/8 rule,and Boole's rule,respectively.The next example illustrates this point. Example 5.2.Consider the integration of the function fx)=1+e*sin(4x) over the fixed interval [a,b]=[0,1].Apply the first four closed Newton- Cotes quadrature formulas on it
The Step Sizes ◼ In Example 5.1 we applied the quadrature rules with h=0.5. If the endpoints of the interval [a, b] are held fixed, the step size must be adjusted for each rule. The step sizes are h=b-a, h=(b-a)/2, h=(b-a)/3, and h=(b-a)/4 for the trapezoidal rule, Simpson’s rule, Simpson’s 3/8 rule, and Boole’s rule, respectively. The next example illustrates this point. ◼ Example 5.2. Consider the integration of the function f(x)=1+e -x sin(4x) over the fixed interval [a, b]=[0, 1]. Apply the first four closed NewtonCotes quadrature formulas on it