# UCF COT 6505 - Lecture Notes (9 pages)

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# Lecture Notes

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## Lecture Notes

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Lecture Notes

Pages:
9
School:
University of Central Florida
Course:
Cot 6505 - Computational Methods/Analysis I
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Lecture 16 Lemma 5 6 Theorem 5 7 Lemma 5 6 Suppose that the Algorithm 5 4 is implemented with a step length such that it satisfies strong wolf conditions with 0 c2 1 2 Then the method generates the descent directions pk that satisfies the following inequalities 1 f T p 1 2c 1 k k2 2 k 0 1K 1 c2 f k 1 c2 2 c2 1 1 0 0 c2 1 c2 2 2 1 1 1 0 c2 1 c2 2 B 1 Proof 1 1 f T p 2c 1 k k2 2 k 0 1K 1 c2 f k 1 c2 2 c2 1 1 0 0 c2 1 c2 2 Induction k 0 2 1 1 1 0 c2 1 c2 2 B f 0T p0 f 0T f 0 1 f 0 2 f 0 2 So by using B it is easy to see the both inequalities are satisfied Assume holds for k Algorithm 5 4 pk 1 f k 1 k 1 pk f kT 1 pk 1 f kT 1 f k 1 k 1 f kT 1 pk f kT 1 pk 1 f kT 1 pk 1 k 1 f k 1 2 f k 1 2 f T p 1 2c 1 k k2 2 k 0 1K 1 c2 f k 1 c2 f kT 1 pk 1 f kT 1 pk 1 k 1 f k 1 2 f k 1 2 f pk 1 f f k 1 f pk 1 2 f k 1 f f k f k 1 2 T f k 1 pk 1 f KT 1 pk 1 C f k 1 2 f k 2 T k 1 T k 1 T k f kT 1 pk c2 f kT pk T k 1 FR k 1 f kT 1 f k 1 f kT f k Wolf s strong condition 2 c2 f kT pk f kT 1 pk c2 f kT pk 2 f T p 1 2c 1 k k2 2 k 0 1K 1 c2 f k 1 c2 c2 f kT pk f kT 1 pk c2 f kT pk c2 f kT pk f kT 1 pk c2 f kT pk f k 2 f k 2 f k 2 1 c2 f kT pk f kT 1 pk c2 f kT pk 1 1 f k 2 f k 2 f k 2 f kT 1 pk 1 f KT 1 pk 1 f k 1 2 f k 2 From C T c2 f kT pk f kT 1 pk 1 c f p 1 1 2 k 2k 2 2 f k f k 1 f k 1 f T p 1 2c 1 k k2 2 k 0 1K 1 c2 f k 1 c2 c2 f kT pk f kT 1 pk 1 c2 f kT pk 1 f k 2 f k 1 2

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