Question

Consider the solution of the nonlinear system of equations f1(x,y)=2x-4cos(y)=0 f2(x,y)=4xsin(y)=0 We want to start with (x0,90)=(0,7/6) and apply Newton's method to compute (X7,71). Whic one of the following(s) is/are true: (1) The solutions are (0, 1/2) and (-2, 1) 2 (II) The Jacobian evaluated at (0,1/2) is (II) (X1,81) = (0,5 +13) Select one: O a. (I) and (11) b. (II) and (111) c. (l) and (III) O d. None of these O e.(I), (II) and (III)

Answer 1

and filx, y) = ax - Ucesy=0 ģ folx,y) = Ux Sm Cy)=0 Po 2x- uCasy =0 = x= 2losy 4x&mily) = 0 = (a @sy) Smly) = 0 8 (687) Cemy) = 0 Casy =o or smyco y = I g y = T = X= Cos Cop -) X X = 20 (1) => x=0 [Cos I) and (@get) are solutions afi Jy a ubring joriy usmy af afa 48 y usinly) uxlasy if and y. The Po = 4x Cosy ofa OX 4- 2 4 80 At (09) JE Jacobian at 60,9 = (] Xo = (0,5) ► J () I (0,1)=( false) - con las $(99)= ( 13.06url] Xi = Xo - (J6ca) f(x3) AF By Newton Method

[%]-[1][-28] - [14]- [] > [19.106] (19yo) = (0,5+) 6e9 option is correct (I, I 911 are true) 80