《数值分析》课程PPT教学课件（英文版）Chapter 02 Numerical Integration 2.6 Gauss-Legendre Integration（Optional）

2.6 GauSs-Legendre Integration(Optional)

2.6 Gauss-Legendre Integration (Optional)

f(ad c N wn.f()+w2f(a2)

f(x)=1 ldx=2=1+2 f(x)=x:/xd=0=0101+ f()=2/ 心心 =a=l12i+2x d=0= +

1+2 101+22=

Theorem 2.8( Gauss-Legendre Two-Point Rule). If f is continuous on -1,1,the )d≈G2(f)=f()+f( 2.90) The Gauss-legendre rule G2(f) has degree of precision n=3. If fE C-1,1] the en f(x)dx≈G2)=f()+f(-5)+E2(f) (2.91) where E2(f) 135 (2.92)

Example 2. 17. Use the two-point Gauss-Legendre rule to approximate dx +2 =ln(3)-ln(1)≈1.0961 and compare the result with the trapezoidal rule T(f, h) with h= 2 and Simpson s rule S(, h )with h=

GN(=WNIf(CN. 1)+WN 2f( N, 2 )+...+WN N f(EN, N)

门1(x)d=∑=10Nk(xk)+Ex(f) abscissas, IN, k Weights, WN, k Truncation error, EN(f) 0.5773502692 000000000 f(4(c) 0.57735026921.0000000 135 3/±0.7745966020.55555 f(6)(c) 0.000000000.88888 15.750 4/±0.861363116 0.3478548451 f(8)(c) ±0.33998104360.6521451549 3,472,875 ±0.90617984590.2369268851 5±0.53846931010.4786286705 f(10)(c) 37,732,650 0.00000000056888888 ±0.93246951420.1713244924 6±0.66120938650.3607615730 f(12)(c)213(6)4 (12)313! ±0.23861918610.4679139346 ±0.94910791230.1294849662 7/±0.74153118560.2797053915 f(14)215(8)4 ±0.40584515140.3818300505 (14)315! 0.0000000.479591837 ±0.060289856501012285363 810796674021045 f(16)(c)217(8!)4 ±0.52553240990.3137066459 (16)317! ±0.18343464250.3626837834

Theorem 2.9(Gauss-Legendre three-point rule). If f is continuous on -1, 1, then 5f(-V3/5)+8f0)+5f(y35 f(x)dm≈G3(f) 2.94 The Gauss-Legendre rule G3(f)has degree of precision n=5. Iff E C6[-1, 11 en 5f(-y3/5)+8f0)+5f(3/5) f()d. +E3(f) (2.95) where E3(f) 2.96

Example 2. 18. Show that the three-point Gauss-Legendre rule is exact for 5rx=2=G3(f)