Numerical Mathematics - A Collection of Solved Problems

NUMERIČKA INTEGRACIJA 321tj.(4) A k =Kako su redomπS n−1 (x k )S ′ n(x k ) .S ′ n(x) = −(n + 1)√ 1 − x 2 · cos[(n + 1) arccos x] + x sin[(n + 1)arccos x](1 − x 2 ) √ 1 − x 2 ,1S n−1 (x k ) = q sin[n arccos x k ] =1 − x 2 kS ′ n(x k ) ===zaključujemo da je1sin kπn + 11sin kπn + 1· sin(n + 1 − 1)kπn + 1“sin kπ cos1sin kπn + 1=· sin nkπn + 11sin kπ sinn + 1“kπ −kπ ”n + 1kπ kπ”− sinn + 1 n + 1 cos kπ = (−1) k+1 ,−(n + 1) sin kπn + 1 cos kπ + cos kπsin kπn + 1= (−1)k+1 (n + 1),sin 3 kπsin 2 kπn + 1n + 1S n−1 (x k )S n(x ′ k ) = (−1)2k+2 (n + 1)sin 2 kπn + 1= n + 1sin 2kπn + 1.Najzad, zamenom u (4), dobijamoA k =π kπn + 1 sin2 n + 1 ,tako da tražena Gaussova kvadraturna formula (3) postajeZ 1−1p1 − x 2 g(x) dx = πn + 1nXsin 2k=1kπ“n + 1 g coskπ”+ R n (g).n + 1Kako je2x 2 k − 1 = cos 2kπ , k = 1, . . . , n,n + 1