Numerical Mathematics - A Collection of Solved Problems

330 NUMERIČKO DIFERENCIRANJE I NUMERIČKA INEGRACIJAgde jeL 1 (x) = 1 − x2f(−1) + 1 + x f(1)2Lagrangeov interpolacioni polinom, a H 1 (x) = ax + b Hermiteov polinom prvogstepena. Dakle,H 3 (x) = 1 − x2f(−1) + 1 + x2f(1) + (x + 1)(x − 1)(ax + b),H ′ 3(x) = − 1 2 f(−1) + 1 2 f(1) + 2x(ax + b) + (x2 − 1)a.Zamenom x sa −1, odnosno 1, imamo8>< H 3(−1) ′ = − 1(2)2 f(−1) + 1 2 f(1) + 2(a − b) = f ′ (−1),>: H 3(1) ′ = − 1 2 f(−1) + 1 2 f(1) + 2(a + b) = f ′ (1).Rešavanjem sistema (2) po a i b dobijamoDakle,a = 1 4 [f(−1) − f(1) + f ′ (−1) + f ′ (1)], b = − 1 4 [f ′ (−1) − f ′ (1)].H 3 (x) = x3 − 3x + 24+ x3 − x 2 − x + 14f(−1) + −x3 + 3x + 24Integracijom poslednje jednakosti nalazimof(1)f ′ (−1) + x3 + x 2 − x − 14f ′ (1).(3)Z 1−1H 3 (x)dx = f(−1)4+ f ′ (−1)4Z 1−1Z 1−1(x 3 − 3x + 2) dx + f(1)4(x 3 − x 2 − x + 1) dx + f ′ (1)4Z 1−1Z 1−1(−x 3 + 3x + 2) dx(x 3 + x 2 − x − 1) dx= f(−1) + f(1) + 1 3 f ′ (−1) − 1 3 f ′ (1).Sada, pod uslovom da f ∈ C 4 [−1, 1], imamo (videti [2, str. 54])(4) f(x) = H 3 (x) + r(f, x),gde sur(f, x) = f(4) (η)4!Ω(x) (−1 < η < 1) i Ω(x) = (x − 1) 2 (x + 1) 2 = (x 2 − 1) 2 .