hw06_p02 - A Problem 2 Newtons method(x0 y0 =(1 1 1 f(x y =...

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A. Problem 2. Newton’s method ( x 0 , y 0 ) = (1 , 1) f ( x, y ) = 2 x - y + 1 9 e - x + 1 f x ( x, y ) = 2 - 1 9 e - x f y ( x, y ) = - 1 g ( x, y ) = - x + 2 y + 1 9 e - y - 1 g x ( x, y ) = - 1 g y ( x, y ) = 2 - 1 9 e - y Iteration I = 1 . 95912 - 1 - 1 1 . 95912 δ x δ y = - 2 . 04088 0 . 04088 x 1 y 1 = x 0 y 0 + δ x δ y = 1 1 + 1 . 42318 - 0 . 74731 = - 0 . 42318 0 . 25270 Iteration II = 1 . 83035 - 1 - 1 1 . 91370 δ x δ y
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