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TAMU PETE 301 - Numerical Methods for Engineers Ch. 4 Solutions

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1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 4 4.1 (a) For this case xi = 0 and h = x. Thus, 2)0(")0(')0()(2xfxffxf 1)0(")0(')0(0 efff 21)(2xxxf (b) 231()26ii i ixx x xihhfx e e h e e      for xi = 0.2, xi+1 = 1 and h = 0.8. True value = e–1 = 0.367879. zero order: 818731.0)1(2.0ef %55.122%100367879.0818731.0367879.0t first order: 163746.0)8.0(818731.0818731.0)1( f %49.55%100367879.0163746.0367879.0t second order: 42574.028.0818731.0)8.0(818731.0818731.0)1(2f%73.15%100367879.042574.0367879.0t third order: 355875.068.0818731.028.0818731.0)8.0(818731.0818731.0)1(32f %26.3%100367879.0355875.0367879.0t 4.2 Use the stopping criterion: s = 0.5102–2% = 0.5% True value: cos(/3) = 0.5 zero order: 13cos  %100%1000.510.5t first order: 451689.023/13cos2 %66.9t %4.121%1000.45168910.451689a second order:2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 501796.0243/451689.03cos4 %359.0t %986.9%100501796.0451689.0501796.0a third order: 499965.07203/501796.03cos6 %00709.0t %366.0%100499965.0501796.0499965.0a Since the approximate error is below 0.5%, the computation can be terminated. 4.3 Use the stopping criterion: s = 0.5102–2% = 0.5% True value: sin(/3) = 0.866025… zero order: 047198.133sin  %92.20%1000.866025047198.10.866025t first order: 855801.063/047198.13sin3 %18.1t %36.22%100855801.0047198.1855801.0a second order: 866295.01203/855801.03sin5 %031.0t %211.1%100866295.0855801.0866295.0a third order: 866021.050403/866295.03sin7 %000477.0t %0316.0%100866021.0866295.0866021.0a Since the approximate error is below 0.5%, the computation can be terminated. 4.4 (a) 01232(0) 1 2(1) 1 2(2) 1 2(3) 1357(1) (1) (1) (1)arctan 2(0) 1 2(1) 1 2(2) 1 2(3) 1 357xx x x xxxxx   (b) Use the stopping criterion: s = 0.5102–2% = 0.5%3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. True value: arctan(/6) = 0.482347907 zero order: arctan 0.52359966 0.482348 0.523599100% 8.55%0.482348t first order: 47575.004785.0523599.03)6/(66arctan3 1. 3 7 %t 0.47575 0.523599100% 10.06%0.47575a second order: 35/6 /6arctan 0.47575 0.007871 0.4836266 3 5     0.26%t 0.48362 0.47575100% 1.63%0.48362a third order: 357/6 /6 /6arctan 0.48362 0.00154 0.48207966357      0.06%t 0.482079 0.48362100% 0.32%0.482079a fourth order: 3579/6 /6 /6 /6arctan 0.482079 0.000329 0.48240866 3 5 7 9       0.012%t 0.482408 0.482079100% 0.07%0.482408a Since the approximate error is below 0.5%, the computation can be terminated. 4.5 True value: f(3) = 554. zero order: 62)1()3(  ff %191.111%10055462)(554t first order: 78)2(7062)13)(1('62)3( ff %921.85t second order: 3544213878)13(2)1("78)3(2ff %101.36t third order: 55486150354)13(6)1(354)3(3)3(ff %0t Thus, the third-order result is perfect because the original function is a third-order polynomial. 4.6 True value: f(2.5) = ln(2.5) = 0.916291... zero order:4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 0)1()5.2(  ff %100%1000.91629100.916291t first order: 5.1)5.1(10)15.2)(1(')1()5.2( fff %704.63%1000.9162915.10.916291t second order: 375.05.1215.1)15.2(2)1("5.1)5.2(22ff %074.59%1000.916291375.00.916291t third order: 5.15.162375.0)15.2(6)1(375.0)5.2(33)3(ff %704.63%1000.9162915.10.916291t fourth order: 234375.05.12465.1)15.2(24)1(5.1)5.2(44)4(ff %421.74%1000.916291234375.00.916291t Thus, the process seems to be diverging suggesting that a smaller step would be required for convergence. 4.7 True value: 2837)2(12)2(75)2('71275)('22fxxxf function values: xi–1 = 1.8 f(xi–1) = 50.96 xi = 2


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