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» A1 = 0, A2 = 0. 79057, A3 = -0. 79057 Step4 d) We now need to compute the spectral radius of T;, denoted by p(T;) and given by p(T;) = 1 In our case, we find that p(T;)= 0. 79057

» A1 = 0, A2 = 0. 79057, A3 = -0. 79057 Step4 d) We now need to compute the spectral radius of T;, denoted by p(T;) and given by p(T;) = 1 In our case, we find that p(T;)= 0. 79057

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Step 4 “P(T) will give me a matrix How do we calculate it to find this value ? P(T)=0.79067”
5:57 1 which we can get: + 4X( 16X² – 1 = 36A → A = 0 and 16X2 10 %3D 5. → A = 0 and ) = ±/§ » A1 = 0, X2 = 0. 79057, A3 = -0. 79057 Step4 d) We now need to compute the spectral radius of T;, denoted by p(T;) and given by p(T;) = 1 js In our case, we find that p(T;)= 0. 79057 Given that the formula to compute the optimal w is wopt 1+/1- (p(T)*) Hence, we get Wopt = 1. 24040 I70.61237 1+/1- (0.79057²) Thus, we have found the optimal w = 1.240408 - 1. 204
Transcribed Image Text:5:57 1 which we can get: + 4X( 16X² – 1 = 36A → A = 0 and 16X2 10 %3D 5. → A = 0 and ) = ±/§ » A1 = 0, X2 = 0. 79057, A3 = -0. 79057 Step4 d) We now need to compute the spectral radius of T;, denoted by p(T;) and given by p(T;) = 1 js In our case, we find that p(T;)= 0. 79057 Given that the formula to compute the optimal w is wopt 1+/1- (p(T)*) Hence, we get Wopt = 1. 24040 I70.61237 1+/1- (0.79057²) Thus, we have found the optimal w = 1.240408 - 1. 204
5:57 1 b) 4 3 We have A = 3 4 -1 Thus, we get 0 -1 4 0 3 4 0 0 U = 0 0 -1 D = 0 4 0 0 0 0 0 4 We know that 1 00 T; D-1 + U 1 0 0 0 1 1/4 1/4 because D-1 1/4 Now, we need to find the eigenvalues of T;. Step3 c) For doing that, we need to find the possible value of A by putting the determinant of T; - A. I3 as 0, where I3 is the 3 x 3 identity matrix. Thus, we get 4X -3 0 |
Transcribed Image Text:5:57 1 b) 4 3 We have A = 3 4 -1 Thus, we get 0 -1 4 0 3 4 0 0 U = 0 0 -1 D = 0 4 0 0 0 0 0 4 We know that 1 00 T; D-1 + U 1 0 0 0 1 1/4 1/4 because D-1 1/4 Now, we need to find the eigenvalues of T;. Step3 c) For doing that, we need to find the possible value of A by putting the determinant of T; - A. I3 as 0, where I3 is the 3 x 3 identity matrix. Thus, we get 4X -3 0 |
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