{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 3 Solutions

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online