UNISA NUMERICAL METHODS (COS 233-8) Solutions to October 2000 Final Exams
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1 UNISA NUMERICAL METHODS COS -8 Solutios to October Fil Ems Questio Re: Tble 4. pge 47 o Gerld/Wetley [ d editio] Write dow te itertio scemes or i ii iii te bisectio metod; te metod o lse positio regul lsi; te sect metod. b Te tree tbles below coti results obtied by mes o te metods i bove. Te problem solved is idig te root Ect vlue:.47 o si e wit strtig vlues o d. Use te itertio scemes o to determie wic metod ws used to geerte te results i ec tble respectively. Sow oly te clcultios ecessry to distiguis te metods. Metod A Itertio i Metod B Itertio i Metod C Itertio i c Epli te diereces i te respective covergece speeds o metods A, B d C. d For ec metod, stte coditio uder wic te metod ils.
2 Solutio i Te bisectio metod begis wit two vlues d b tt brcket root o te equtio. Tese two vlues re determied by veriyig tt b <. Te metod te successively divides te itervl i l d replces oe edpoit wit te midpoit so tt te root is brcketed gi. Te error ter itertios is give by b. Te root is oud weever tis error term is less t give vlue ε or te irst time. ii Te lse positio metod ssumes tt te uctio is lier i te viciity o root o te equtio. Strtig wit two vlues tt brcket te root, it mkes sure tt te root remis betwee te successive pirs o boudries. Te et iterte, ulike te bisectio metod, is tke t te itersectio o lie betwee te pir o -vlues d te -is rter t t te midpoit. Te itertio ormul is It is te cecked weter d re o opposite sigs. I tey re, te is replced by else is replced by. Te metod stops util te two boudries o te itervl dier by less t give vlue ε. iii Te sect metod uses te sme priciple s te regul lsi metod ecept tt te two cose strtig vlues do ot ecessrily brcket te root. Te irst vlue is cose er te root d te secod vlue c be o te sme side o te root s te irst but it is possible tt it is o te opposite side. A strigt lie is drw troug te two poits d its -itercept is tke s te et iterte. To esure covergece, te secod strtig vlue sould be erer to te root t te irst vlue. Te ew vlue sould ormlly be closer to te root so tt te oldest vlue is immeditely replced by te most recet vlue beore clcultig te et iterte. Te itertio ormul is te sme s tt o regul lsi but te replcemet is utomtic, tt is, tere is o ceck or brcketig. b Give si e wit strtig vlues o d ote tt is closer to te root t is. It is obvious tt metod C is te itervl bisectio metod rom te cose vlues o. Te irst dierece betwee te vlues o metods A d B occurred t te tird iterte. Tis implies tt te coice o te ew vlue ws ot te sme. For te sect metod, te tird iterte sould be weres, or te regul lsi metod, te tird iterte sould be ,.777 te root beig brcketed i te itervl,.777 sice.4799 d.777 ve te sme sig.
3 c Comprtively, te speeds re s ollows: Itervl bisectio: very poor Regul lsi : itermedite Sect : best Te itervl bisectio metod is muc slower sice it wstes lot o time i bisectig. It ppes quite ote tt erlier iterte is muc better t lter oe. Te regul lsi metod is sligtly ster sice te et iterte is ot tke t te midpoit but rter t te itersectio o lie d te -is. Furtermore, most o te time covergece occurs oly rom oe side o te root d tt prticulrly slows dow te process. Te sect metod is te stest sice its lgoritm is less complicted t tt o regul lsi. No ceck s to be perormed i order to coirm tt te root is brcketed i te itervl. d All tree metods my il i te uctio is ot cotiuous i te cose itervl. Te bisectio metod my eve give lse root. Te sect metod my lso il i te uctio is r rom lier er te root its successive itertes will ly o to poits tt re very r rom te root.
4 Questio Re: Sectio 4. pge 48 o Gerld/Wetley [ d editio] b Stte Newto s metod d te use it to id pproimtio to te solutio o i te itervl,. Use.5 s strtig vlue. Do tree itertios. Use eigt deciml ritmetic wit roudig. Tbulte te bsolute diereces betwee successive pproimtios obtied i, d te use tis tble to commet o te covergece speed o Newto s metod s pplied i. Solutio Newto s metod is bsed o lier pproimtio o te uctio but does so by usig tget to te curve. A strtig vlue o is tke d te tget to te curve t tt -vlue is drw. Te -itercept o te tget is te et pproimtio. Te Newto itertio sceme is give by I tis cse, te give equtio is. Lettig, we ve l so tt te itertio ormul is l Te bove epressio my be simpliied to to icrese eiciecy d speed. l l Strtig wit. 5, we obti te ollowig tree itertios: b.5l.5 l l l l Itertio umber, i i E i i i
5 Te error vlues would seem to suggest cse o qudrtic covergece, tkig ito ccout te umber o deciml plces, tt is, E E i i. Perormig ceck: i kei E For istce,.4788 k.4797 k. 744 d.5 k.4788 k.899 Tese two vlues o k re very close to ec oter. Te dierece betwee tem occurred most probbly due to roudig errors. We coclude tt tere is ideed qudrtic covergece.
6 Questio Re: Emple. pge 9 o Gerld/Wetley [ d editio] Use te Crout reductio metod to obti LU decompositio o te mtri M b Use te LU decompositio o bove to solve te lier system d M wit M s i d [ ] T d. c Cosider te lgoritm below. You my ssume tt te mtri A is digolly domit, tt A b, d tt d b re -vectors. ALGORITHM QUESTION c red,,...,, b,..., b,,...,, ε repet s. 4 or i to do 5 y i i edor 7 or i to do 8 i bi 9 or j to i do i i ij * j edor or j I to do i i ij * y j 4 edor 5 i i / ii s s y i i 7 edor 8 util s < ε 9 prit Approimte solutio is,..., prit Vlue o s is s i Idetiy te metod tt is pseudo-coded bove. ii Wy is it possible to replce te y j i lie wit iii Wt is te purpose o te umber s? Solutio j? ij Usig te Crout reductio tecique, we ve l m l 7 m l m lu l 4 m m lu 4 l l m l u l u
7 u l m m m lu u u l 7 l Tereore, te LU decompositio o M is give by M 7 4 b Give M d, tt is,, Usig orwrd substitutio wit L, we ve so tt d 4, d [ 4] d d [ 4 ]. 7 Te, usig bckwrd substitutio wit U, we obti wic illy yields,, 4. 4 c i Tis is te pseudo code or te Jcobi metod. ii Tis is becuse te vlues o te i s d te iii s is te dierece betwee successive itertios. y i s re te sme t tt stge.
8 Questio 4 Re: Equtio. pge o Gerld/Wetley [ d editio] Complete te divided dierece tble below. You eed oly write dow te vlues o α, β d γ, NOT te wole tble. First divided dierece Secod divided dierece Tird divided dierece Fourt divided dierece α β.85 γ b c Write dow te iterpoltig polyomil o igest possible degree, suitble or pproimtig., bsed o te give divided dierece tble. Now use te polyomil i b to pproimte.. Solutio α β γ b Te iterpoltig polyomil o te igest possible degree, 4, is P c P
9 Questio 5 Re: Formule pge 45 o Gerld/Wetley [ d editio] Is te uctio C, give below, cubic splie wit respect to te kots,,, d? Give mtemticl justiictio or your swer. C,,, < > Solutio Simpliictio o te bove uctio gives,, C,,, < < < > sice, rom deiitio,,, < It is obvious tt we ve cubic polyomils i ll te itervls. Substitutio o te vlues o t te kots yield cotiuous curve: At, At, At, At, At, Te grdiets o te cubic curves t te kots re lso equl: At, At, At, At, At,
10 Questio Re: Equtio. pge 9 o Gerld/Wetley [ d editio] Cosider te eperimetl dt i te ollowig tble: Y Te -vlues re ssumed to be ree o error. b Fid te lest squres polyomil o degree two or tese dt. Write dow epressio or te totl error. Solutio Te summrised iormtio rom te tble bove is y y y 8. 9 Let te lest-squres cubic iterpoltig polyomil be we ve te mtri equtio y. Usig te lest-squres criterio, 4 y y y wic gives I ugmeted orm, te mtri becomes
11 Usig Gussi elimitio wit prtil pivotig, we ve R.88 R R.88 R R R.8 R 4.7R R.8 R R.98 R.47R Bck substitutio gives.5979,. 59 d. 789 so tt te lest squres pproimtio polyomil o degree is give by y..5. i te coeiciets re writte to two deciml plces. b Te totl error is te sum o te squres o te residuls tt is, te qutity to be miimised i te metod o lest squres. Tereore, te totl error is give by Y i i.789 i i
12 Questio 7 Re: Appedi A pges d 4 o Gerld/Wetley [ d editio] Give uctio tt is suicietly dieretible, use Tylor series epsios to derive two dieret pproimtio ormuls or te irst derivtive o. I ec cse give lso te order o te error term. Solutio I give uctio is suicietly dieretible over give itervl, tt is, dieretible t lest times, te, it c be pproimted by te Tylor s series!...!! Remider Eq. weever te epsio is bout poit. By dieretitig te etire equtio, we ve!...! Remider Eq. Te error i te Eq., usully derived i elemetry clculus i te orm o te itegrl dt t t! ], [,! ξ ξ is, s c be see, o order so tt te epsio o will ve error o order. Aoter useul wy o represetig te sme epsio is by sowig ow te series beves t distce rom te ied poit. By lettig i Eq. bove, we ve...!...!! Remider Eq. Tis time, dieretitio leds to te epressio...!...! Remider Eq. 4 Te error term i Eq. is evidetly ], [,! ξ ξ d is o order. Tereore, te error i te epsio o will be o order.
13 Questio 8 Re: Problem 5 o pge 44 o Gerld/Wetley [ d editio] problem o ssigmet 4 Yer Sow tt oe etrpoltio o te trpezoidl rule is ideticl to Simpso s Rule wit comprble vlue or. Te etrpoltio ormul is Better estimte More ccurte More ccurte Less ccurte, were is te epoet o i te error term O ssocited wit te two vlues estimtes tt re used i te etrpoltio. Remider: more ccurte. less ccurte b Fid te lest umber M d te pproprite step size so tt te globl error or te composite trpezoidl rule, give by b ξ 5 wit ξ [, b] is less t or te pproimtio o l d by te composite trpezoidl rule or M subitervls or pels. Solutio b Let I d wit d b. For te ske o coveiece, deote i by i. Furtermore, or te rest o te proo, use strips o widt so tt we ve ordites t,, 4,...,.Tereore, b I d 4 4 [... ] [... ] I we divide ec o te bove strips i two, tt is, ito strips o widt, we will certily obti more ccurte vlue or te itegrl. I tis cse, b d [... ]. I Usig Romberg itegrtio, wit rtio o : te strips ve bee lved, we ve Improved vlue more ccurte vlue more ccurte vlue less ccurte vlue.
14 Hece, improved vlue or I is [ ] [ ] [... ] wic simpliies to [ ] 4. It is observed tt, but or te widt o ec strip, te bove ormul is te sme s Simpso s Rule. I we use strip o widt isted o, we do obti Simpso s Rule. 4 b Te globl error i te composite trpezium rule is b E M ξ or some ξ suc tt < ξ < b were M is te umber o subitervls pels d is te widt o pel. Te itegrl to be pproimted is d l. Tus, give l, l d Wit d b, we ve tt m so tt. E M ξ d E M. Berig i mid tt M, we tereore ve 5 5, tt is,. d M. 5 Te pproprite size would be d te lest umber o pels is 9 sice M 9..
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