1 CHAPTER 22 22 1 Although it s not required the analytical solution can be evaluated simply as I 3 0 3 xe2 x dx 0 25e 2 x 2 x 1 504 53599 0 The tableau depicting the implementation of Romberg integration to s 0 5 is iteration t a 1 2 4 8 1 259 8216 1815 42957072 952 90724344 630 87848089 537 22588842 2 31 8835 43 2082 665 39980101 523 53556004 506 00835760 3 1 8912 1 8397 514 07794398 504 83987744 4 0 0312 0 0290545 504 69324146 22 2 The integral can be evaluated analytically as I 2 1 2 1 x dx x 2 x 2 2 x 2 dx 1 2 x3 1 23 1 13 1 I 2x 2 2 2 1 4 8333 x 1 3 2 3 1 3 The tableau depicting the implementation of Romberg integration to s 0 5 is iteration t a 1 2 4 1 6 0345 5 12500000 4 90972222 4 85274376 2 0 0958 1 4833 4 83796296 4 83375094 3 0 0028 0 0058 4 83347014 2 7 9715 1 97282684 1 94377297 3 0 0997 1 94183605 Thus the result is 4 83347014 22 3 1 n 1 2 4 a 1 34376994 1 81556261 1 91172038 22 4 Change of variable 2 1 2 1 xd 1 5 0 5 xd 2 2 2 1 dx dxd 0 5dxd 2 x I 1 1 2 1 1 5 0 5 xd 0 5dxd 1 5 0 5 xd 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 2 Therefore the transformed function is 1 f xd 0 5 1 5 0 5 xd 1 5 0 5 xd 2 Two point formula 1 1 I f f 2 074414 2 755961 4 830375 3 3 4 833333 4 830375 t 100 0 0612 4 833333 Three point formula I 0 5555556 f 0 7745967 0 8888889 f 0 0 5555556 f 0 7745967 0 5555556 2 022895 0 8888889 2 347222 0 5555556 2 921322 1 123831 2 08642 1 622957 4 833207 4 833333 4 833207 t 100 0 0026 4 833207 Four point formula I 0 3478548 f 0 861136312 0 6521452 f 0 339981044 0 6521452 f 0 339981044 0 3478548 f 0 861136312 0 3478548 2 009026 0 6521452 2 16712 0 6521452 2 573718 0 3478548 2 997699 0 698849 1 413277 1 678438 1 042764 4 833328 4 833333 4 833328 100 0 00010 t 4 833333 22 5 Change of variable 3 0 3 0 xd 1 5 1 5 xd 2 2 3 0 dx dxd 1 5dxd 2 x I 1 1 1 5 1 5 xd e3 3 xd 1 5dxd Therefore the transformed function is f xd 2 25 2 25 xd e3 3 xd Two point formula 1 I f 3 1 f 3 379298 402 9157 406 295 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 3 t 504 53599 406 295 100 19 47 504 53599 Three point formula I 0 5555556 f 0 7745967 0 8888889 f 0 0 5555556 f 0 7745967 0 5555556 0 99728 0 8888889 45 19246 0 5555556 819 1714 0 554044 40 17107 455 0952 495 8204 504 53599 495 8204 t 100 1 727 504 53599 Four point formula I 0 3478548 f 0 861136312 0 6521452 f 0 339981044 0 6521452 f 0 339981044 0 3478548 f 0 861136312 0 3478548 0 473908 0 6521452 10 75639 0 6521452 167 9269 0 3478548 1113 793 0 164851 7 014731 109 5127 387 4382 504 1305 504 53599 504 1305 t 100 0 0804 504 53599 22 6 Change of variable 2 0 2 0 xd 1 xd 2 2 2 0 dx dxd dxd 2 1 e1 xd sin 1 x d I dxd 1 1 1 xd 2 x Therefore the transformed function is f xd e1 xd sin 1 xd 1 1 xd 2 Five point formula I 0 236927f 0 90618 0 478629f 0 53847 0 568889f 0 0 478629f 0 53847 0 236927f 0 90618 0 236927 0 102 0 478629 0 582434 0 568889 1 143678 0 478629 1 382589 0 236927 1 370992 0 024166442 0 278769811 0 650625504 0 661746769 0 324824846 1 940133372 1 940130022 1 940133372 t 100 0 000173 1 940130022 22 7 Here is the Romberg tableau for this problem 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 4 1 2 5 5616 288 56033084 289 37640049 a n 1 2 4 224 36568786 272 51167009 285 16021789 3 0 0188 289 43080513 Therefore the estimate is 289 430805 22 8 Change of variable 3 3 3 3 xd 3xd 2 2 3 3 dx dxd 3dxd 2 1 3 I dxd 1 1 9 x 2 d x Therefore the transformed function is f xd 3 1 9 xd2 Two point formula 1 1 I f f 0 75 0 75 1 5 3 3 2 49809 1 5 100 39 95 t 2 49809 The remaining formulas can be implemented with the results summarized in this table n 2 3 4 5 6 Integral 1 5 3 1875 2 189781 2 671698 2 411356 t 39 95 27 60 12 34 6 95 3 47 Thus the results are converging but at a very slow rate Insight into this behavior can be gained by looking at the function and its derivatives f xd f xd 54 xd 1 9 xd2 2 54 1 9 xd2 2 1944 xd2 1 9 xd2 1 9 xd2 4 We can plot the second derivative as 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 …
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