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=
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