Differential Equations Solutions 37

# Differential Equations Solutions 37 - Chapter 9 Solutions...

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Unformatted text preview: Chapter 9 Solutions: Numerical Methods for Unconstrained Optimization CHALLENGE 9.1. f (x) = x4 + x2 (x2 − 1), 1 4x3 1 2x2 − 1 g(x) = 12x2 1 0 , H(x) = 0 2 . Step 1: p=− 12x2 1 0 0 2 4x3 1 2x2 − 1 −1 so x← 48 0 =− 0 2 32 −3 −1 −2/3 +3/2 = 4/3 1/2 ex1 +x2 (1 + x1 ) ex1 +x2 x1 + 2x2 = 8 4.7726 2 −1 + = −32/48 +3/2 , . CHALLENGE 9.2. g(x) = H(x) = ex1 +x2 (2 + x1 ) ex1 +x2 (1 + x1 ) ex1 +x2 (1 + x1 ) ex1 +x2 x1 + 2 ; 12 8 = 8 6 . Now det(H) = 72 − 64 = 8, so p = −H−1 g = 1 8 6 −8 −8 12 −8 −4.7726 −1.2274 0.8411 = . We’d never use this inverse formula on a computer, except possibly for 2x2 matrices. Gauss elimination is generally better: L= 1 2/3 0 1 ,U = 12 0 8 2/3 . Note that pT g = −5.8050 < 0, so the direction is downhill. 47 ...
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