Lecture-17

Lecture-17 - 1 Lecture-17 Theorems 5.8 and 5.9...

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Unformatted text preview: 1 Lecture-17 Theorems 5.8 and 5.9 & Levenberg-Marquadet = k b = < K , , 2 2 1 || || k k k k f K , , 2 1 k k Since at the restarts 1 cos- = k q < 2 2 || || cos k k k f q We can use Theorem 5.7 to prove global convergence for algorithms, which are periodically started by setting If restarts occur at = 1 } { j j k If the restarts are done after every n iterations, the sequence is infinite || || lim = j k j f Therefore a subsequence of gradients will approach to zero: || || inf lim = k k f Convergence of Algorithms with restarts 2 Theorem 5.8 Suppose that the function is Lipschitz continuously differentiable, , and Algorithm 5.4 is implemented with a line search that satisfies strong Wolfe conditions, with 0< c 2 <1/2 . Then || || inf lim = k k f g || || k f Proof Proof by contradiction: Assume that: large, ly sufficient all for || || g g k f k 1 , || || || || cos || || || || 2 1 K , k p f p f k k k k k = 2200 c q c < 2 2 || || cos k k k f q < = 2 4 1 || || || || k k k p f (D) || || inf lim = k k f We have proved this (Lecture 16) Now 3 Proof ; 1 1 1 k FR k k k p f p + + + +- b 1- +- = k FR k k k p f p b 1 1 2 1 1 1 2 1 1 1 1 ) ( 2 ) ( ) ( ) (--------- + - = + - - = +- +- = k T k FR k k T k FR k k T k k T k k T k FR k T k FR k k T k FR k k T k k T k k FR k k T k FR k k k T k p p p f f f p p p p f p p f f f p p p f p f p p b b b b b b b 2 1 2 2 1 1 2 1 || || 1 | |---- - - k k T k k T k f c c p f c p f k T k k T k FR k f f f f = + + + 1 1 1 b 2 1 2 2 2 2 2 2 1 2 2 1 1 1 2 2 2 2 1 2 2 1 2 2 2 2 1 2 1 2 2 || || ) ( || || 1 1 || || || || ) ( || || 1 2 || || || || ) ( || || 1 2 || || || || ) ( | | 2 || || || ||--------- + - + + - + = + - + + + k FR k k k k FR k k k T k k T k k k FR k k FR k k k FR k k T k FR k k k p f c c p p f f f f f c c f p f c c f p p f f p b b b b b b From (C) Proof k T k k T k p f c p f - + 2 1 | | 1 , 1 1 2 || || 1 1 2 2 2 2 K , k c c f p f c k k T k = 2200-- -- 1 1 2 1 | |--- - k T k k T k p f c p f 1 1 2 || || 1 1 2 2 2 1 1 1 2 c c f p f c k k T k-- ----- 2 1 2 2 1 1 2 1 || || 1 | |---- - - k k T k k T k f c c p f c p f (A) (B) Combining (A) and (B) (C) Wolfs condition Lemma 5.6 4 Proof 2 1 2 2 2 2 2 || || ) ( || || 1 1 || ||- + - + k FR k k k p f c c p b || || || || || || || || || || || || || || 1 || || 1 || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || || ) ( ) ( || || ) ( || || || || ] || || ) ( || || [ ) ( || || || || 1 2 4 3 2 2 2 4 2 4 2 1 2 4 3 2 2 4 2 4 3 2 1 4 2 3 2 2 2 4 2 4 1 4 1 4 2 1 3 4 1 4 2 3 2 2 2 2 1 2 2 1 3 2 2 3 2 2 2 2 1 2 1 3 2 2 3 2 =-...
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Lecture-17 - 1 Lecture-17 Theorems 5.8 and 5.9...

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