Numerical Mathematics - A Collection of Solved Problems

NUMERIČKA INTEGRACIJA 307Čvorovi kvadrature su−x 1 = x 2 =r13 + 2 π 2 ∼ = 0.732104 ,a koeficijentiA 1 = A 2 = 1 2 .Kako je‖Q 2 ‖ 2 =Z 1−1`x2 − x22´sin2 πx2 dx = 4 45 + 83π 2 − 28π 4 ,ostatak u klasi funkcija C 4 [−1, 1] ima oblikR 2 (f) = ‖Q 2‖ 24! · 1 f(4) (ξ) = 1 „ 16 45 + 23π 2 − 7 «π 4 f (4) (ξ),tj.R 2 (f) ∼ = 2.98 · 10 −3 f (4) (ξ) (−1 < ξ < 1) .2 ◦ Uovomslučaju imamo Q 0 (x) = 1, Q 1 (x) = x − 1 3 , Q 2(x) = x 2 − 2 5 x − 1 5 ,pa je x 1 = 1 √ `1− 6´, x2 = 1 √ `1+ 6´, A1 = 1 √ `9− 6´, A2 = 1 √ `9+ 6´i5599R 2 (f)= 1225 f(4) (ξ) (1− < ξ < 1).Z 13 ◦ x nNeka je C n =−1 1 + x 2 dx. Tada je C 0 = π 2 , C 2 = 1 2 (4 − π), C 4 =16 (3π − 8), C 1 = C 3 = 0, pa suodakle nalazimoQ 0 (x) = 1 , Q 1 (x) = x , Q 2 (x) = x 2 − 1 (4 − π) ,π− x 1 = x 2 =r4π − 1 ∼ = 0.522723 ,A 1 = A 2 = π 4 ∼ = 0.785398 ,R 2 (f) =8π − 2472πPrimenom ove formule na integral I dobijamoI = 1 2Z 1−1f (4) (ξ) (−1 < ξ < 1) .arctan |x|1 + x 2 dx = 1 2 · π4 · 2arctan x ∼ = 0.3783 .