Homework 3 Solutions

Homework 3 Solutions - 6.3 {a} Fixed peim. The equatiml can...

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Unformatted text preview: 6.3 {a} Fixed peim. The equatiml can be solved in two ways. The way that COIWEIgE‘E is firm = RILEY,- — 2.5 The resulthlg iterations are J: I; E; D 5 1 3.391155 4311-423 2 2.9332T4- 15.51% 3 2139246 5.16% 111- 2.151231?! 1 31% 5 2.?25955 BETH: Er 2121359 [1.19% ? 2.?2011'4- [1.06% E- 2.T'1EIE1E [1.02% (h) Newrml—thhson El 5 —1 3.5 -E-.2 1 3.353559 -2.T1 D44 —4 .EEIDT32 49.09% 2 2.301332 43.30 EDIE —3.8 [1266 19329-5 3 2.?211'03 410 [1 E44 6.54222 2.95% 4 5 2.?19341 —3.1E-DE 3.638133 13.05% 2.?19341 JAE-13 3.63863 EDGE-"u 6.5 [a] The Newton—Raphson method can he set up as — 2 + fixj — 4x} + 8.51? 1'r—1 :1" — 1 6—31} +1.51; Using an initial guess of 4.2, this fonnnla jumps around and evenutallv converges on the root at U.4T45?2 after 9 iterations: F xi fig} Fix} ea 8 4.2 —18.318 —1.14 1 484912 —182.132 38.8839? 188.81% 2 —2.8?392 82.4399 38.82898 83.39% 3 —1.13?83 —14.?3? 1?.84118 128.2?‘18 4 —D.2T2?4 8.94412 3.29343? 31 31.8888 8 8.28283 8.94341 4.43EEIUT1 234.4?‘18 8 8.418384 8.18212 2.938943 51.1?‘18 ? 8.4?88?4 8.81821 2.56?58T 1 1 .2388 8 8.4?4882 -8.2E—88 2.841334 [1.34% 9 U.4?45T2 —1.4E—89 2.841249 8.88% The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.2 corresponds to a small negative slope of the function. Hence, the first iteration shoots to a negative value that is far flow. a root. 2|} 1|} I} -1l:r -2fl (b) Using an initial guess of4.43, this fonmda jmnps around even more and EVEfltuallj’ converges on the root at H.4T45T2 after 25 iterations: 35 an: .1. an 5 4.43 —15.4554 —5.55255 1 —3535.13 —3.1E+15 23355553 155.11% 2 —2525.2 —5.1E+55 15355532 55.55% 3 —1?45.25 —2.?E+55 4553?53 55.55% 4 —1155.25 —5E+55 2545132 55.11% 5 —??5.554 —2.4 E+55 5553535 55.1?% 5 515.423 —?E+5? 4541 ?5.4 55.25% ? 543.35? -2.1E+5? 1?5535.5 55.35% 5 —225.545 —5135351 ?5535.54 55.55% 5 —151.152 -1515555 35455.?? 55.5?% 15 —55.552? —535555 15??2.5? 51 .31 % 11 —55.?252 -155542 ?51 1.? 51.55% 12 —42.5552 —4?2?5 311?.??4 53.55% 13 —2?.?535 -13554.1 135?.55? 54.55% 14 —1?.?551 —413?.11 51?.5453 55.55% 15 41.555 -1215.55 2?5.5?55 55.52% 15 —5.555?1 —35?.541 123.??55 5?.14% 1? —3.554?5 -153.552 55.534? ?5.2?% 15 —1 .5355 —25.55 55 25.?5352 155.52% 15 -5.55521 -5.1555 12.21515 15?.25% 25 —5.51555 —2.15551 5144551 3?12.?44% 21 5.325251 -5.45441 3.55555? 155.545% 22 5.453525 —5.55525 2.553545 25.253% 23 5.4?4 —5.55145 2.545515 4.425% 24 5.4?45?2 —1.1E—55 2.541252 5.121% 25 5.4?45?2 —5.5E—1 3 2.541245 5.555% The reason for the behavior is depicted in the following plot. As can be oeen, the gneos of x = 4.43 corresponds to an even smaller near—zero negative slope. Hence. the first iteration shoots to a large positive value that is far fi'om a root. 25 15 5 6.15 {a} A graph of the function indicates that the lowest real root is approximately 414: SID 4 (b) The stopping criterion corresponding to 3 significant figures can be determined with Eq. 3.? is as = (11.5 x103‘3)% = 0.05% Using initial guesses of 15.1 = —l and x: = —C|.6, the secant method can be iterated to this level as summarized in the following table: 1‘ x,_1 113.1 x x ea [1 —1 29.4 41.5 T5984 1 41.6 T5984 41.461155 112541499 30.26391: 2 41.415059 1.125415 41.41 563 [1.1 52122311 9.?61‘13 3 41.41 9153 [1.1551224 41.41541r [1.0039661 13 1 1111213 4 41.41 54? [111113955 41.41 535 9.539 E—UE [1.026% 9.3 [a] The system is first expressed as an augmented matrix: 10 2 —1 2? —3 —6 2 —61.5 1 l 5 —2l.5 Forward elimination: (121 is eliminated by multiplying 1'0“! 1 b3: —3.-"lfl and subtracting the result from row 2. (131 is eliminated by multiple row 1 by l.-"l CI and subtracting the result from row 3. ID 2 —l 2? III —5.4 1.? —53.4 [I [1.3 5.1 —24.2 (132 is eliminated by multiplying row 2 b3: U.8.-"[—5.4} and subtracting the result from row 3. ID 2 —1 2? G —5.4 1.? —:’+3.4 fl 0 5.351852 —32.1].ll Back substitutierii: I _—32_1111__E 3' _ 5.351352 _ (birfi‘heck: I: =LW=3 lfl(fl_i)+2(3)—{—5)=2? —3-. —3(U-5J-fi(3)+2(—5)=—613 — — — _7 I1=W=flj a_5+s+5(45)=—21.5 9.9 {a} The system is first expressed as an augmented matrix: 4 l —1 —2 5 l 2 4 6 l l 6 Forward elimiiiatinn: First. we pivot by switching rows 1 and 3: 6 l l 6 5 l 2 4 4 l —1 —2 Multiply row 1 5.-"6 and subtract fi'om row 2 to eliminate r331. Multiply row 1 by 4.-"6 and subtract fie-m raw 3 t6 eliminate H31. 6 1 l 6 I] 6.16667" 1.1666? —1 I] 0.33333 —l.6667'r —6 Pivot: I] (1.1666? —l.l667" —l 6 1 l 6 I] 6.33333 1.6663 —6 Mllltlpl‘j' row 2 by U. 166613633333 = 13.5 and subtract fi'om row 3 to eliminate (133. Back substitution: 2 2 Check: _ —5—{—l.666?)l _ _13 33— :133333 _ MPH—3:4 _ _ _ 5(3)—13+2(1}=4 I126 1 { 13:3 6 6(3)—13+1=fi ...
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Homework 3 Solutions - 6.3 {a} Fixed peim. The equatiml can...

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