HW6Solution

# HW6Solution - a maximum 1[5pts Fit the data set given in...

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Homework 6 Solution February 28, 2011 1. Consider the following function: f ( x, y ) = - ( x - 1) 2 - ( y - 2) 2 + xy (a) [2pt] Find the gradient vector and the Hessian matrix. (b) [2pt] Perform one iteration of the Newton’s method with the initial point ~ x 0 = ( x 0 , y 0 ) T = (0 , 0) T ; check the function values and the gradient vector at ~ x 0 and ~ x 1 and discuss the results. (c) [1pt] Determine whether the optimum found in (b) is the maximum or the minimum. Q1: (a) gradient vector ∂f ∂x = - 2 x + 2 + y, ∂f ∂y = - 2 y + 4 + x f = [ - 2 x + 2 + y, - 2 y + 4 + x ] T Hessian matrix 2 f ∂x 2 = - 2 , 2 f ∂y 2 = - 2 , 2 f ∂x∂y = 2 f ∂y∂x = 1 H - 1 = - 2 1 1 - 2 , f ( x 0 ) = [2 , 4] T (b) Newton’s Method Given initial point ~x 0 = ( x 0 , y 0 ) T = (0 , 0) T ~ x 1 = x 1 y 1 = x 0 y 0 - H - 1 f H - 1 = - 0 . 6667 - 0 . 3333 - 0 . 3333 - 0 . 6667 ~ x 1 = 0 0 - - 0 . 6667 - 0 . 3333 - 0 . 3333 - 0 . 6667 2 4 = 2 . 6667 3 . 3333 1

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(c) figure out f ( x 0 ) and f ( x 1 ) f ( x 0 ) = - 5 , f ( x 1 ) = 4 . 3333 The result indicates that the iteration is reaching to a maximum On the other hand, if we calculate the eigenvalues of H, we could obtain λ 1 = - 1 , λ 2 = - 3 which means H is negative definite. So the optimum is therefore
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Unformatted text preview: a maximum 1. [5pts] Fit the data set given in Table 1 to the following function, y = a + a 1 x 2 + a 2 sin ( y ). Find a , a 1 and a 2 . x y f ( x, y ) 1.8 1 0.5 2.3 2 1 0.5 3 1.5-4 4 2-11 Q2: For y = a + a 1 x 2 + a 2 sin ( y ), let z = 1 , z 1 = x 2 , z 2 = sin ( y ),then Z = 1 1 1 . 4794 1 4 . 8415 1 9 . 9975 1 16 0 . 9093 , Y = 1 . 8 2 . 3 . 5-4-11 Z T Z = 5 30 3 . 2277 30 354 27 . 3715 3 . 2277 27 . 3715 2 . 7597 Z T Y = [-10 . 4000 ,-207 . 7000 ,-12 . 4688] T Z T Z~a = Z T Y, ~a = ( Z T Z )-1 ( Z T Y ) = [1 . 8144 ,-. 9740 , 3 . 0202] T 2...
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