chapter4 - i book 2005/4/7 22:31 page S-24 #45 i i i...

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± ± “book” — 2005/4/7 — 22:31 — page S-24 — #45 ± ± ± ± Answers to Selected Exercises for Chapter 4 Section 4.1 (page 277) 1. j x ( j ) T λ ( j ) 1 £ 1 . 000000 1 . 000000 0 . 333333 ¤ 3.000000 2 £ 0 . 333333 1 . 000000 1 . 000000 ¤ 0.333333 3 £ 0 . 333333 1 . 000000 0 . 555556 ¤ 3.000000 4 £ 0 . 066667 1 . 000000 1 . 000000 ¤ 1.666667 5 £ 0 . 090909 1 . 000000 0 . 878788 ¤ 1.933333 3. j x ( j ) T λ ( j ) 1 £ 0 . 000000 0 . 780869 0 . 624695 ¤ 0.600000 2 £ 0 . 442395 0 . 663593 0 . 603266 ¤ 1.390244 3 £ 0 . 563740 0 . 570098 0 . 597649 ¤ 1.875374 4 £ 0 . 594159 0 . 534393 0 . 601165 ¤ 1.933988 5 £ 0 . 602420 0 . 521955 0 . 603865 ¤ 1.939643 5. j x ( j ) T λ ( j ) 1 £ 0 . 428571 0 . 285714 0 . 857143 ¤ 9.000000 2 £ 0 . 581809 0 . 085143 0 . 808857 ¤ 9.734694 3 £ 0 . 490757 0 . 204716 0 . 846906 ¤ 9.998792 4 £ 0 . 546767 0 . 133027 0 . 826650 ¤ 10.095834 5 £ 0 . 513665 0 . 176210 0 . 839701 ¤ 10.131039 7. j x ( j ) T λ ( j ) 1 £ 0 . 800000 0 . 200000 0 . 200000 0 . 000000 1 . 000000 0 . 200000 ¤ 1.000000 2 £ 1 . 000000 0 . 352941 0 . 294118 0 . 058824 0 . 588235 0 . 529412 ¤ 0.500000 3 £ 1 . 000000 0 . 270270 0 . 135135 0 . 081081 0 . 486486 0 . 621622 ¤ 1.088235 4 £ 1 . 000000 0 . 253165 0 . 113924 0 . 063291 0 . 303797 0 . 696203 ¤ 1.067568 5 £ 1 . 000000 0 . 190476 0 . 071429 0 . 059524 0 . 238095 0 . 726190 ¤ 1.063291 S-24
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± ± “book” — 2005/4/7 — 22:31 — page S-25 — #46 ± ± ± ± Selected answers for Section 4.2 S-25 9. (a) With x (0) = £ 1 . 000000 0 . 800821 0 . 191217 0 . 440118 ¤ T , the eigen- vector sequence converges in 23 iterations while the eigenvalue sequence con- verges in 9 iterations. Had the convergence of the eigenvalue sequence been used as a stopping condition, the inFnity norm of the error in the eigenvector estimate would have been 1 . 9696 × 10 2 . (b) With x (0) = £ 0 . 829897 0 . 415026 0 . 372860 ¤ T , the eigenvector se- quence converges in 25 iterations while the eigenvalue sequence converges in 16 iterations. Had the convergence of the eigenvalue sequence been used as a stopping condition, the Euclidean norm of the error in the eigenvector estimate would have been 3 . 389 × 10 3 . 11. (a) r = £ 0 ² ¤ T (b) ± r ± = ², ± ˆ x v 1 ± = ± ˆ x v 2 ± =1 13. (a) λ =5 , v = £ 0 . 510 . 5 ¤ T (b) λ , v = £ 1 / 3111 / 3 ¤ T 15. (a) The eigenvalue sequence converges to λ . 925950; the rate of conver- gence is, however, erratic. The eigenvector sequence converges to v = £ 0 . 259612 0 . 394409 0 . 574050 1 . 000000 ¤ T . (b) The eigenvalue sequence appears to be converging toward λ = 1, and the eigenvector sequence appears to be converging toward v = £ 1000 ¤ T . The rate of convergence, however, is extremely slow, with the asymptotic rate approaching 1. 17. growth rate: 25.72%; age bracket 0–1: 42.07%; age bracket 1–2: 23.43%; age bracket 2–3: 16.77%; age bracket 3–4: 12.00%; age bracket 4–5: 5.73% 19. growth rate: 5 . 50%; group 1: 20.65%; group 2: 66.97%; group 3: 11.46%; group 4: 0.66%; group 5: 0.04%; group 6: 0.03%; group 7: 0.18% Section 4.2 (page 291) 1. λ =0 . 938341 , v = £ 0 . 607239 0 . 616746 0 . 500885 ¤ T 3. λ . 657175 , v = £ 0 . 217785 0 . 219228 0 . 635271 1 . 000000 ¤ T 5. λ . 459582 , v = £ 0 . 449005 0 . 401246 0 . 830887 1 . 000000 ¤ T 7. λ . 000015 , v = £ 0 . 000005 1 . 000000 0 . 999998 ¤ T 9. λ = 0 . 135614 and λ = 7 . 980299 11. λ = 8 . 186163 and λ =8 . 566777 13. (a) The eigenvalue sequence appears to converge very rapidly, but not to an eigenvalue of the matrix A and with the Fnal value depending upon the initial vector. The eigenvector sequence appears to settle into a 2-cycle, flipping between two di±erent vectors.
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This note was uploaded on 10/13/2009 for the course MATH 471 taught by Professor Anna during the Spring '09 term at University of Michigan-Dearborn.

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chapter4 - i book 2005/4/7 22:31 page S-24 #45 i i i...

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