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hw5_solution

hw5_solution - Numerical Analysis in Engineering ME 140A...

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Numerical Analysis in Engineering ME 140A, Fall 2007 Homework #5 Solution November 14, 2007 By: Mohamad Nasr-Azadani: [email protected] 1. For the given value of ρ = 3 . 2, the solution of logistic difference equation will converge to a steady 2-cycle behavior. See “p1.m” Matlab code for the implementation of this 0 5 10 15 20 25 30 35 40 45 50 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 u 0 = 0.2 n u(n) 0 5 10 15 20 25 30 35 40 45 50 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 u 0 = 0.4 n u(n) 0 5 10 15 20 25 30 35 40 45 50 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 n u(n) u 0 = 0.6 0 5 10 15 20 25 30 35 40 45 50 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 u 0 = 0.8 n u(n) Figure 1: (Problem 1) “ u n ” vs. “ n ” for different initial conditions ( ρ = 3 . 2). problem. Note that this code also uses “u n function.m” function. 1
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2. a) See “p2parta.m” for the implementation of this part. The obtained results for 0.01% accuracy were as: ρ = 2 . 90 , MaxN = 74 ρ = 2 . 95 , MaxN = 147 ρ = 2 . 99 , MaxN = 682 Note that the relative error was defined as = u n +1 - u n u n +1 × 100% . (1) 0 10 20 30 40 50 60 70 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 rho = 2.90 n u(n) 0 20 40 60 80 100 120 140 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 rho = 2.95 n u(n) 0 100 200 300 400 500 600 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 rho = 2.99 n u(n) Figure 2: (Problem 2-a) “ u n ” vs. “ n ” for different values of ρ .
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