Numerical Mathematics - A Collection of Solved Problems

146 NELINEARNE JEDNAČINE I SISTEMIKako je za x = x(0) = ˆ0 0 0 ˜⊤ ,imamopa, na osnovu (3), dobijamoDalje imamo2f 0 = 4 −0.1320.2 5 i W 0 = 4 1 0 030 1 0 5 ,−0.30 0 1µ 0 = (f 0, f 0 )“”(f 0 , f 0 ) = 1 W 0 = W0 ⊤ = I ,2x(1) = x(0) − 1 · I f 0 = 4 −0.130.2 5 .−0.32f 1 = 4 0.133 2321.2 −0.6 0.40.05 5 , W 1 = 4 0.9 1.4 0.3 5 , W 1 W1 ⊤ f 1 = 4 0.274830.2098 5 ,0.05−0.4 0.2 1.60.1632pa jes obzirom naµ 1 =0.13 · 0.2748 + 0.05 · 0.2098 + 0.05 · 0.16320.2748 2 + 0.2098 2 + 0.1632 2 ∼ = 0.3720 ,2x(2) = 4 0.13 2−0.2 5 − 0.3720 4 0.1813 20.002 5 ∼ = 4 0.03273−0.2007 5 ,0.3 0.147 0.24532W1 ⊤ f 1 = 4 0.18130.002 5 .0.147Ako se zadržimo na drugom koraku, približne vrednosti odgovarajućeg rešenjasux ∼ = 0.0327 , y ∼ = −0.2007 , z ∼ = 0.2453 ,dok jef 2 = f (x(2)) ∼ =24 0.032−0.017−0.00835 .