Numerical Mathematics - A Collection of Solved Problems

306 NUMERIČKO DIFERENCIRANJE I NUMERIČKA INEGRACIJANadalje, imamoA 1 = A 2 =R 2 (f) = ‖Q 2‖ 24!Dakle, formula (1) ima oblikZ 1−1‖Q 1 ‖ 2Q 1 (x 1 ) Q ′ 2 (x 1) = ‖Q 1‖ 22 x 2 1=π82 · 14= π 4 ,f (4) (ξ) = π768 f(4) (ξ) (−1 < ξ < 1).p1 − x 2 f(x)dx = π „ “f − 1 ” “ 1” «+ f + π4 2 2 768 f(4) (ξ).7.2.20. Odrediti parametre i ostatak u sledećim kvadraturnim formulamaGaussovog tipa:1 ◦ ∫ 1−1f(x) sin 2 πx2 dx = A 1 f(x 1 ) + A 2 f(x 2 ) + R 2 (f),∫ 12 ◦ (1 + x)f(x)dx = A 1 f(x 1 ) + A 2 f(x 2 ) + R 2 (f),−13 ◦ ∫ 1−111 + x 2 f(x)dx = A 1 f(x 1 ) + A 2 f(x 2 ) + R 2 (f),∫ π/24 ◦ f(x) cos xdx = A 1 f(x 1 ) + A 2 f(x 2 ) + R 2 (f),−π/25 ◦ ∫ +∞0e −x√ xf(x)dx = A 1 f(x 1 ) + A 2 f(x 2 ) + R 2 (f).Primenom treće formule približno izračunatiI =∫ 10arctan x1 + x 2 dx.Rešenje. 1 ◦ Ortogonalni polinomi sa težinom p(x) = sin 2 πx„ 2Q 0 (x) = 1, Q 1 (x) = x, Q 2 (x) = x 2 1−3 + 2 «π 2 , s obzirom da jena (−1, 1) su:(1, Q 0 ) =Z 1−1Zsin 2 πx12 dx = 1, `x2 , Q0´= x 2 sin 2 πx−1 2 dx = 1 3 + 2 π 2 .