Numerical Mathematics - A Collection of Solved Problems

314 NUMERIČKO DIFERENCIRANJE I NUMERIČKA INEGRACIJAom funkcijom p(x) = p x(2 − x). Konstruišimo ovaj niz polinoma. Imamo redomk = 0 : Q 0 (x) = 1;k = 1 : (Q 0 , Q 0 ) = C 0 = π 2 , (x, Q 0) = C 1 = π 2 ,Q 1 (x) = x − (x, Q 0)(Q 0 , Q 0 ) Q 0(x) = x − 1;k = 2 : (x 2 , Q 0 ) = C 2 = 5π 8 , (x2 , Q 1 ) = C 3 − C 2 = π 4 ,(Q 1 , Q 1 ) = C 2 − 2C 1 + C 0 = π 8 ,k = 3 :Q 2 (x) = x 2 − (x, Q 0)(Q 0 , Q 0 ) Q 0(x) − (x2 , Q 1 )(Q 1 , Q 1 ) Q 1(x) = x 2 − 2x + 3 4 ;” “x 3 , Q 0 = C 3 = 7π 8 , (x3 , Q 1 ) = C 4 − C 3 = 7π16 ,(x 3 , Q 2 ) = C 5 − 2C 4 + 3 4 C 3 = 3π32 ,(Q 2 , Q 2 ) = C 4 − 4C 3 + 11 2 C 2 − 3C 1 + 9 16 C 0 = π 32 ,Q 3 (x) = x 3 − (x3 , Q 0 )(Q 0 , Q 0 ) Q 0(x) − (x3 , Q 1 )(Q 1 , Q 1 ) Q 1(x) − (x3 , Q 2 )(Q 2 , Q 2 ) Q 2(x)= x 3 − 3x 2 + 5 2 x − 1 2 .S obzirom da je“Q 3 (x) = (x − 1) x 2 − 2x + 1 ”,2jednostavno odred¯ujemo čvorovex 1 = 1 −Težinski koeficijenti su tada0A 1 = A 3 =A 2 =√22 , x 2 = 1 , x 3 = 1 +√22 .‖Q 2 ‖ 2Q 2 (x 1 ) Q ′ 3 (x 1) = π/32(1/4) · 1 = π 8 ,‖Q 2 ‖ 2Q 2 (x 2 )Q ′ 3 (x 2) = π/32(−1/2)(−1/4) = π 4 .Dakle, kvadraturna formula ima oblikZ 2„ „ √ «„p π 2x(2−x)f(x)dx = f 1− + 2f(1) + f 1+8 2√22««+ R 3 (f).