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MidtermSolution che 132b

MidtermSolution che 132b - ChE132B Midterm Exam Solutions...

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ChE132B Midterm Exam Solutions 1. (a) ( 4 points ). If μ = 0, 𝑝 𝑛+1 = 𝑝 𝑛 There will be no iteration at all. This is not an acceptable algorithm for finding a root. If μ = 1, 𝑝 𝑛+1 = 𝑔 ( 𝑝 𝑛 ) The modified algorithm becomes the fixed point iteration again. (b) ( 18 points ). 𝑝 𝑛+1 = 𝜇𝑔 ( 𝑝 𝑛 ) + (1 − 𝜇 ) 𝑝 𝑛 (1) 𝑝 𝑛 = 𝜇𝑔 ( 𝑝 𝑛−1 ) + (1 − 𝜇 ) 𝑝 𝑛−1 (2) Taylor expand g(p n-1 ) 𝑔 ( 𝑝 𝑛−1 )~ 𝑔 ( 𝑝 𝑛 ) + 𝑔 ( 𝑝 𝑛 )( 𝑝 𝑛−1 − 𝑝 𝑛 ) (3) Plug in expanded g(p n-1 ) back to Eq(2), 𝑝 𝑛 ~ 𝜇 �𝑔 ( 𝑝 𝑛 ) + 𝑔 ( 𝑝 𝑛 )( 𝑝 𝑛−1 − 𝑝 𝑛 ) + (1 − 𝜇 ) 𝑝 𝑛−1 (4) Rearrange 𝑝 𝑛 = 𝜇𝑔 ( 𝑝 𝑛 ) − 𝜇𝑔 ( 𝑝 𝑛 )( 𝑝 𝑛 − 𝑝 𝑛−1 ) + (1 − 𝜇 ) 𝑝 𝑛−1 (5) Subtract Eq(5) from Eq(1) 𝑝 𝑛+1 − 𝑝 𝑛 = 𝜇𝑔 ( 𝑝 𝑛−1 )( 𝑝 𝑛 − 𝑝 𝑛−1 ) + (1 − 𝜇 )( 𝑝 𝑛 − 𝑝 𝑛−1 ) (6) Divide both sides by p n -p n-1 𝑝 𝑛+1 −𝑝 𝑛 𝑝 𝑛 −𝑝 𝑛−1 = 𝜇𝑔 ( 𝑝 𝑛−1 ) + (1 − 𝜇 ) (7) The absolute value of Eq(7) is the error ratio of this method.
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𝐸 𝑛+1 𝐸 𝑛 = �𝜇𝑔 ( 𝑝 𝑛−1 ) + (1 − 𝜇 ) (8) as n goes to infinity, lim 𝑛→∞ 𝐸 𝑛+1 𝐸 𝑛 = �𝜇𝑔 ( 𝑝 ) + (1 − 𝜇 ) (9) Therefore, Order of convergence = 1 Asymptotic error constant = �𝜇𝑔 ( 𝑝 ) + (1 − 𝜇 ) (c) ( 4 points ).
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