# Lecture-16 - Lecture-16 Lemma 5.6 & Theorem 5.7 Lemma 5.6...

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1 Lecture-16 Lemma 5.6 Suppose that the Algorithm 5.4 is implemented with a step length such that it satisfies strong wolf conditions with 0< c 2 <1/2. Then the method generates the descent directions p k that satisfies the following inequalities: 1 0 , 1 1 2 || || 1 1 2 2 2 2 K , k c c f p f c k k T k = 2200 - - - - 2 1 0 , 1 1 1 2 2 2 < < - < - - < - c c 2 1 0 , 0 1 1 2 1 2 2 2 < < < - - < - c c c (B)

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Induction, k= 0: 1 0 , 1 1 2 || || 1 1 2 2 2 2 K , k c c f p f c k k T k = 2200 - - - - 1 || || || || 2 0 0 0 2 0 0 0 - = - = f f f f p f T T So by using (B), it is easy to see the both inequalities are satisfied. Assume holds for k. ; 1 1 1 k k k k p f p + + + + -∇ = b Algorithm (5.4) ; 1 1 1 1 1 1 k T k k k T k k T k p f f f p f + + + + + + + -∇ = b 2 1 1 1 2 1 1 1 || || 1 || || + + + + + + + - = k k T k k k k T k f p f f p f b 2 1 0 , 1 1 1 2 2 2 < < - < - - < - c c 2 1 0 , 0 1 1 2 1 2 2 2 < < < - - < - c c c (B) Proof 2 1 1 1 1 2 1 1 1 || || 1 || || + + + + + + + + - = k k T k k T k k T k k k T k f p f f f f f f p f 2 1 1 1 2 1 1 1 || || 1 || || + + + + + + + - = k k T k k k k T k f p f f p f b ; 1 1 1 k T k k T k FR k f f f f + + + b 2 1 2 1 1 1 || || 1 || || k k T K k k T k f p f f p f + - = + + + + | | | | 2 1 k T k k T k p f c p f + pk f c pk f pk f c T k T k T k - + 2
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## This note was uploaded on 06/12/2011 for the course COT 6505 taught by Professor Shah during the Spring '07 term at University of Central Florida.

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Lecture-16 - Lecture-16 Lemma 5.6 & Theorem 5.7 Lemma 5.6...

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