chapter2 - i book 2005/4/7 22:31 page S-4 #25 i i i Answers...

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± ± “book” — 2005/4/7 — 22:31 — page S-4 — #25 ± ± ± ± Answers to Selected Exercises for Chapter 2 Section 2.1 (page 69) 1. (a) p 3 =0 . 875, ( a 4 ,b 4 )=(0 . 875 , 1 . 000) (b) p 3 . 375, ( a 4 4 . 375 , 0 . 500) (c) p 3 . 625, ( a 4 4 . 500 , 0 . 625) (d) p 3 . 625, ( a 4 4 . 625 , 0 . 750) 3. n p n | p n p | ( b a ) / 2 n 1 3.50000 0.35841 0.50000 2 3.25000 0.10841 0.25000 3 3.12500 0.01659 0.12500 4 3.18750 0.04591 0.06250 5 3.15625 0.01466 0.03125 5. n p n | p n p | ( b a ) / 2 n 1 1.50000 0.29906 0.50000 2 1.25000 0.04906 0.25000 3 1.12500 0.07594 0.12500 4 1.18750 0.01344 0.06250 5 1.21875 0.01781 0.03125 11. with ( a 1 1 )=(2 , 3) and ² =5 × 10 4 , p 11 =2 . 35107. 13. with ( a 1 1 )=(1 . 44 , 1 . 48) and ² × 10 5 , p 10 =1 . 457539. 15. with ( a 1 1 )=(4 , 5) and ² =10 5 , p 17 =4 . 261482. 17. ζ . 4039731 f . 2625461 m . 3533436 ,F m . 5205986. 19. (a) 3.46%, compounded monthly (b) 3.07%, compounded monthly. Section 2.2 (page 78) 1. (a) p 3 . 884507, ( a 4 4 . 884507 , 1 . 000000) (b) p 3 . 462648, ( a 4 4 . 462648 , 1 . 000000) (c) p 3 . 567703, ( a 4 4 . 000000 , 0 . 567703) (d) p 3 . 738945, ( a 4 4 . 738945 , 1 . 000000) 5. Error n p n | p n p | Estimate 1 1.015748 0.154183 2 1.030366 0.139565 3 1.043882 0.126049 0.165874 4 1.056333 0.113598 0.145556 5 1.067761 0.102169 0.127730 S-4
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± ± “book” — 2005/4/7 — 22:31 — page S-5 — #26 ± ± ± ± Selected answers for Section 2.3 S-5 7. Error n p n | p n p | Estimate 1 0.073000 0.045973 2 0.055990 0.028963 3 0.045274 0.018247 0.018247 4 0.038522 0.011495 0.011495 5 0.034269 0.007242 0.007242 9. (a) n p n | p n p | 1 1.717948717949 1.410 × 10 2 2 1.732218859330 1.681 × 10 4 3 1.732050802076 5.493 × 10 9 Convergence appears to be of order 2, which is better than expected for the method of false position. This happens because f ± ( 3) = 0. (b) n p n | p n p | 1 3.734693877551 1.383 × 10 1 2 3.859328133794 1.366 × 10 2 3 3.871720773366 1.263 × 10 3 4 3.872867347773 1.160 × 10 4 5 3.872972695145 1.065 × 10 5 Each new error is roughly one-tenth the previous error, so convergence is linear. Note that f ± ( 15) ± =0. 11. (a) 1 . 507099, 1.288677 (b) 3 . 042682, 0.685220, 3.357462 (c) 0.827181, 1.109712 13. h =1 . 39 meters 15. h =0 . 482 cm Section 2.3 (page 93) 3. GIVEN: iteration function g starting approximation x 0 convergence parameter ² maximum number of iterations Nmax STEP 1: for iter from 1 to Nmax STEP 2: compute x 1 = g ( x 0 ) STEP 3: if | x 1 x 0 | ,OUTPUT x 1 STEP 4: copy the value of x 1 to x 0 end OUTPUT: “maximum number of iterations has been exceeded” 5. (b) We cannot prove uniqueness because there is no k< 1 such that | g ± ( x ) |≤ k for all real numbers x . (c) Since g ± ( p ) ± = 0, convergence will be linear.
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± ± “book” — 2005/4/7 — 22:31 — page S-6 — #27 ± ± ± ± S-6 Answers to Selected Exercises for Chapter 2 (d) Theoretical n p n | p n p | Error Estimate 1 1.0000000000 0.2609148668 2 0.5403023059 0.1987828273 3 0.8575532158 0.1184680826 0.129542854379 4 0.6542897905 0.0847953427 0.079375374087 5 0.7934803587 0.0543952255
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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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chapter2 - i book 2005/4/7 22:31 page S-4 #25 i i i Answers...

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