西安建筑科技大学：《高等数学计算方法》课程教学资源（PPT课件讲稿）Chapter 4.6 Gauss-Legendre Integration（Optional）

2.6 Gauss-Legendre integration (Optional)

2.6 Gauss-Legendre Integration (Optional)

f(a)dx a wn.f(1)+w2f(a2)

)=1: ldx=2=1+2 f(a)=. rdx=0=01x1+2x 1 f(a)=r =1x1+2 dx=o +

1+2 11 w2 e 11+2-3 3 11=-022

Theorem 2.8( Gauss-Legendre Two-Point Rule). If f s continuous on en f(x)dm≈G2()=f()+f (2.90 The Gauss-legendre rule G2(f) has degree of precision n=3. If E C[-1,1, the en (2.91 where E2(f)= 135 (2.92)

Example 2. 17. Use the two-point Gauss-Legendre rule to approximate da ln(3)-ln(1)≈1.09861 1x+2 and compare the result with the trapezoidal rule T(, h) with h= 2 and Simpson s rule S(f, h) with h=

GN(=WN.1f( 1+WN2f(N 2)+.+WN,Nf(aN N

C1f(a. c= Ck=l Wnkf(EN k)+ en(f) I abscissas, IN, k Weights, WN k Truncation error, EN() 0.57735026921000000 f(4(c) 0.57735026921.00000000 135 3/±0.7745966920.5556 0.000000000.888888 15.750 4/±0.8611363116 0.3478548451 8) ±0.33998104360.6521451549 472,875 ±0.90617984590.2369268851 5±0.53846931010.4786286705 (10) 1,237,732,650 0.0000000000.56888888 ±0.93246951420.1713244924 6±0.66120938650.3607615730 213(6! (12)313! ±0.23861918610.679139346 ±0.94910791230.1294849662 7/±0.74153118560.2797053915 15 (14)21°( ±0.40584515140.3818300505 (14!)315! 0.0000000004179591837 ±0.96028985650.1012285363 81±079667402231045 f16)(c)217(8!)4 (16)317! ±0.18343464250.3626837834

Theorem 2.9(Gauss-Legendre three-point rule). If f is continuous on[-1, 1, then 5f(-y3/5)+8f(0)+5f(V3/ f(x)dx≈G3(f) (294 The Gauss-Legendre rule G3(f) has degree of precision n= 5. If f E Co-1,1 then 5f(-y3/5)+8f(0)+5(y3/5) f(r)dac +E3(f 2.95 where E3(f) (2.96 15.750

Example 2.18. Show that the three-point Gauss-Legendre rule is exact for 5n4d=2=G3(f)