NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: PHY 314 COURSE TITLE: NUMERICAL COMPUTATIONS

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1 NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: PHY 4 COURSE TITLE: NUMERICAL COMPUTATIONS

2 PHY 4: Numericl Computtios Course Code PHY 4 Course Title Numericl Computtios Course Developer Dr. A. B. Adeloe DEPARTMENT OF PHYSICS UNIVERSITY UNIVERITY OF LAGOS Progrmme Leder Dr. Ajibol S. O. Ntiol Ope Uiversit of Nigeri Lgos COURSE GUIDE

3 NATIONAL OPEN UNIVERSITY OF NIGERIA Cotets Itroductio Te Course Course Aims Course Objectives Workig troug te Course Course Mteril Stud Uits Tetbooks Assessmet Tutor Mrked Assigmet Ed of Course Emitio Summr

4 Itroductio Numericl Alsis is importt prt of Psics d Egieerig. Tis is becuse most of te problems ecoutered i rel life do ot led temselves to solutio i closed form. I oter words, we ve to mke do wit pproimte solutios. It is cler, terefore, tt ou eed to be coverst wit te vrious metods of pproimte solutio of problems, s well s te loss of iformtio ieret i replcig te ect solutio wit pproimte oe. It is lso quite cler tt te fstest w of doig umericl computtio is troug te computer. It is impertive, te, tt ou uderstd oe or more of te vilble progrmmig lguges. I tis course, te progrmmig lguge of iterest is C++. It is quite cler from te foregoig tt umericl lsis is iterestig course, d we would epect ou to ppl ourself full to te course, s lot of our future work i te field of psics would wrrt soud kowledge of umericl lsis. THE COURSE PHY 9 ( Credit Uits Tis -uit course itroduces ou to umericl lsis. Uit discusses te vrious tpes of errors d ow te migt be miimised. Uit is o curve-fittig. You would eed to deduce some psicl prmeters from give set of redigs obtied perps i lbortor. Vrious ws of lierisig give formuls is give, preprtor to drwig lie of best fit from wic te psicl qutit is deduced. Uit is ll bout lier sstems of simulteous equtios. You sll ler ow to dle lrge set of lier equtios b writig tem i te form of mtrices. Suc problems will te be solved wit te metods pplicble to mtrices. You would lso ler ow to rrive t solutios troug itertive metods. 4

5 Uit 4 discusses differet metods of fidig te roots of lgebric d trscedetl equtios. I Uit 5, ou will come cross fiite differeces. You will be itroduced to vrious kids of differeces, d ow to detect te error i differece tbles. Numericl itegrtio is te object of Uit 6. I tis Uit, ou sll ler ow to itegrte fuctio witi give set of limits (defiite itegrls. Uit 7, te cocludig prt of te teor prt of te course discusses te umericl solutio of iitil vlue problems of ordir differetil equtios. Te C++ Progrmmig spect of te course is itroductio to progrm-writig i oe of te most verstile progrmmig lguges. We wis ou success. COURSE AIMS Te im of tis course is to tec ou bout te mecics of te tomic d subtomic prticles. COURSE OBJECTIVES After studig tis course, ou sould be ble to Uderstd te vrious tpes of errors d ow to miimise tem. Lierise give epressio i order to brig out psicl costt from te resultt reltiosip. Fit curve to give set of dt. Solve sstem of lier equtios. Fid te roots of give lgebric or trscedetl equtio. Obti te defiite itegrl of give fuctio of sigle vrible. Work wit fiite differece scemes. Solve first order iitil vlue problems of ordir differetil equtio. Solve iger order iitil vlue problems of ordir differetil equtios. Write C++ progrms for solvig te umericl problems. WORKING THROUGH THE COURSE Numericl metods provide powerful w of solvig lmost problem i psics, provided it s bee properl formulted. It is our belief tt te studet would be motivted eoug to put i good effort i uderstdig te teoreticl prt of tis course d be willig to ler to write progrms i C++ lguge. THE COURSE MATERIAL You will be provided wit te followig mterils: 5

6 Course Guide Stud Mteril cotiig stud uits At te ed of te course, ou will fid list of recommeded tetbooks wic re ecessr s supplemets to te course mteril. However, ote tt it is ot compulsor for ou to cquire or ideed red tem. STUDY UNITS for Numericl Alsis Te followig stud uits re cotied i tis course: Uit : Approimtios d Errors i Numericl Computtios Uit : Approimtios d Errors i Numericl Computtios Uit : Lier Sstems of Equtios Uit 4: Roots of Algebric d Trscedetl Equtios Uit 5: Fiite Differeces d Iterpoltio Uit 6: Numericl Itegrtio Uit 7: Iitil Vlue Problems of Ordir Differetil Equtios TEXTBOOKS Some referece books, wic ou m fid useful, re give below:. Numericl Metods i Egieerig d Sciece Grewl, B. S.. Itroductor Metods of Numericl Alsis Sstr, S. S.. A friedl Itroductio to Numericl Alsis Brdie, B. Assessmet Tere re two compoets of ssessmet for tis course. Te Tutor Mrked Assigmet (TMA, d te ed of course emitio. Tutor Mrked Assigmet Te TMA is te cotiuous ssessmet compoet of our course. It ccouts for % of te totl score. You will be give 4 TMA s to swer. Tree of tese must be swered before ou re llowed to sit for te ed of course emitio. Te TMA s would be give to ou b our fcilittor d retured fter te ve bee grded. Ed of Course Emitio Tis emitio cocludes te ssessmet for te course. It costitutes 7% of te wole course. You will be iformed of te time for te emitio. It m or m ot coicide wit te uiversit semester emitio. Summr Tis course is desiged to l foudtio for ou for furter studies i Numericl Alsis. At te ed of tis course, ou will be ble to swer te followig tpes of questios: Wt is te eed for umericl lsis i Psics? Wt re te tpes of error tt c be ecoutered i umericl work? 6

7 Wt te ws of obtiig te lie tt best fits set of lbortor dt? Wt re te vrious ws of umericll solvig sstem of lier equtios? Wt re te ws i wic we c umericll fid te roots of equtio? How do I itegrte fuctio tt does ot led itself to lticl solutio? How do I solve first order ordir differetil equtio? How do I tckle iger order iitil vlue problem of ordir differetil equtio? Wt re te merits d demerits of some of te metods of umericl lsis? We wis ou success. 7

8 UNIT : Approimtios d Errors i Numericl Computtios. Itroductio. Objectives. Mi Cotet. Accurc of Numbers.. Approimte Numbers.. Sigifict digits (figures.. Roudig off..4 Aritmetic precisio. Accurc of Mesuremet. Errors.. Roudig Errors.. Ieret Errors.. Tructio Errors..4 Absolute Error, Reltive Error d Percetge Error 4. Coclusio 5. Summr 6. Tutor-Mrked Assigmet (TMA 7. Refereces/Furter Redigs. Itroductio Psics is ect sciece. However, it is strictl impossible to cieve ifiite ccurc i prctice. You re quite wre tt our pprtus or istrumet is ot perfect, eiter is our ee or our mesurig bilit. We te see tt errors rise i everd observtios d mesuremets. Te stud of errors is ver importt i ll res of Sciece d Tecolog. Tis is ecessitted b te fct tt errors sould ot swmp our procedure eoug to lter, sigifictl, coclusios tt m be drw from suc observtios or mesuremets. Aprt from te limittios of observtio d mesuremet, tere re some errors ieret i te problem itself. A good emple i Qutum Mecics is give b te Heiseberg Ucertit Priciple, wic mitis tt we cot mesure some pirs of qutities ccurtel simulteousl, for emple, te positio of bod d its mometum. A ttempt to mesure eiter qutit ccurtel gives ifiite error i te oter. Some oter errors rise s result of represetig ifiite series wit tructed oe. We sll tlk little bit more bout tis i wile.. Objectives At te ed of tis uit, ou would be ble to: uderstd te importce of errors i umericl lsis. roud umber to certi umber of sigifict figures kow ow to reduce te errors ivolved i our umericl work. uderstd ritmetic precisio 8

9 . Mi Cotet. Accurc of Numbers.. Approimte Numbers For te ske of umericl computtio, ll umbers c be clssified uder two brod edigs: ect umbers d pproimte umbers. As te me implies, te former comprises umbers tt re full represeted b some digits. Emples iclude te itegers, d rtiol umbers tt c d ve bee completel writte, e.g.,.58. Approimte umbers re tose tt re ot full specified b te digits represetig tem. As emple, we could write te rtiol umber 7 s.. You re quite wre tt te ctul umber is ot ectl.. B tis stge of our stud, ou must ve worked wit te rtiol umbers. Tese re umbers wic c be writte s frctio of two itegers. Altoug certi rtiol umbers re ect umbers, ou ve lso come cross lot of rtiol umbers tt cot be writte s ect umbers s i te emple bove. Te irrtiol umbers re eve more troublesome. A emple of irrtiol umber is : suc umbers cot be writte s te rtio of two itegers. Tere re two fmilies of umbers tt re uedig: te oes tt repet certi sequeces, d te oes tt do ot. For istce, d Te order of preferece i delig wit umbers i umericl computtios is: turl umbers, rtiol umbers tt ve fiite strig of digits, rtiol umbers tt ve uedig strigs of digits d irrtiol umbers... Sigifict digits (figures We s umber is of r sigifict digits (figures if r digits re used to epress it. As emple,.6,.48 d 86 ll ve four sigifict figures. You would otice tt ec of tem could be writte (wit o loss of iformtio, were is of 4 (r = 4 i tis cse digits, ot strtig or edig wit zero d is iteger, positive or egtive. Te followig rules will be of ssistce to ou. Mke sure te become prt of ou.. Te leftmost o-zero digit is te most sigifict digit, e.g., i.4, is te most sigifict digit.. I te cse were tere is o deciml poit, te rigtmost o-zero digit is te lest sigifict, e.g., 456, is te lest sigifict figure.. If tere is deciml poit, te rigtmost digit is te lest sigifict, eve if it is zero, e.g., i 5.4, te lst is te lest sigifict. Te umber is ot 5.4 or All digits betwee te lest sigifict d te most sigifict (iclusive re sigifict, e.g., i te emple uder rule, re sigifict. I te emple i rule, re sigifict. Tke oter emple:.4 s oe sigifict figure, wile s 8 sigifict figures. It sould be obvious to ou w te ve bee clssified tis w. 9

10 Tere is eceptio, owever: We zero is obtied b roudig, for emple, 9.5 is rouded to sigifict figures. Tis becomes, te lst zero beig sigifict i tis cse. You c compre tis wit rule bove... Roudig off Te irrtiol umbers re perfect emple of umbers wit uedig digits. Eve i te cse of rtiol umbers tere c still be uedig umber of digits d i some oter cses we m decide to reduce te umber of digits b wic umber is represeted. Tis process is clled roudig off. Rules for Roudig off umber to sigifict figures ( Discrd ll digits to te rigt of te t digit (b If te discrded prt of te umber is (I less t lf uit i te t plce leve te t digit ucged (II (III greter t lf uit i te t plce, icrese te t digit b uit ectl equl to lf uit i te t plce, leve te digit ucged if it is eve; icrese b uit if oterwise. Emples: Roud te followig umbers to 5 sigifict figures: (i (ii (iii.459 Solutio: To 5 sigifict figures, te umbers re: (i.48 (rules ( d (b(iii t digit ucged s it is eve (ii 6.4 (rules ( d (b(i s te discrded prt of te umber is less t lf uit i te t plce. (iii.44 (rules ( d (b(iii t digit icresed b uit s it is odd Note: A umber rouded off to sigifict figures is sid to be correct to sigifict plces...4 Aritmetic precisio As we ve sid before, it migt be ecessr to roud off our umbers to mke tem useful for umericl computtio, moreso s it would require ifiite computer memor to store uedig umber. Te precisio of umber is idictio of te umber of digits tt ve bee used to epress it. I scietific computig, it is te umber of sigifict digits or umbers, wile i ficil sstems, it is te umber of deciml plces. You re quite wre tt most currecies i te world re quoted to two deciml plces. I our ow cse, ritmetic precisio (ofte referred to simpl s precisio is te specified umber of sigifict figures or digits to wic te umber of iterest is to be rouded.

11 .4 Errors We sid erlier, tt we sll be revisitig te differet tpes of errors. Tese re:.4. Roudig Errors Tese re errors icurred b tructig sequece of digits represetig umber, s we sw i te cse of represetig te rtiol umber 7 b., isted of..., wic is uedig umber. Aprt from beig uble to write tis umber i ect form b d, our istrumets of clcultio, be it te clcultor or te computer, c ol dle fiite strig of digits. Roudig errors c be reduced if we cge te clcultio procedure i suc w s to void te subtrctio of erl equl umbers or divisio b smll umber. It c lso be reduced b retiig t lest oe more sigifict figure t ec step t te oe give i te dt, d te roudig off t te lst step..4. Ieret Errors As te me implies, tese re errors tt re ieret i te sttemet of te problem itself. Tis could be due to te limittios of te mes of clcultio, for istce, te clcultor or te computer. Tis error could be reduced b usig iger precisio of clcultio..4. Tructio Errors If we tructe Tlor s series, wic sould be ifiite series, te some error is icurred. Tis is te error ssocited wit tructig sequece or b termitig itertive process. Tis kid of error lso results we, for istce, we crr out umericl differetitio or itegrtio, becuse we re replcig ifiitesiml process wit fiite oe. I eiter cse, we would ve required tt te elemetl vlue of te idepedet vrible ted to zero i order to get te ect vlue...4 Absolute Error, Reltive Error d Percetge Error Te bsolute error i mesuremet is te bsolute differece betwee te mesured vlue d te ctul vlue of te qutit. Tus, we c write Absolute error = ctul vlue mesured vlue Te rtio of te bsolute error to te ctul vlue is te reltive error. We c terefore write te reltive error s ctul vlue mesured vlue Absolute error = ctul vlue Te reltive error tke to percetge is te percetge error. Percetge error c terefore be writte s

12 ctul vlue mesured vlue Percetge error = ctul vlue Emples 4. Coclusio I tis Uit ou lert tt errors occur i mesuremet, becuse te imperfect observer mkes use of imperfect mesurig istrumets. Some errors re ievitble s te re prt of te problem uder ivestigtio. Moreover, te istrumets of clcultio, suc s te computer, c ol dle fiite umber of digits, s te memor is fiite. You lso lert to write certi umber i specified umber of deciml poits. You got to kow ow to roud umber to umber of sigifict figures. Some ws of reducig some of tese errors were lso discussed. 5. Summr I tis Uit, ou lert te followig: Errors re itegrl prt of life. How to roud umber to specific umber of sigifict figures. Te differet tpes of error d ow some of tem m be reduced. 6. Tutor-Mrked Assigmet. Roud te followig to te umber of sigifict figures idicted. (.48 4 sigifict figures (b sigifict figures (c.4589 sigifict figures. A studet mesured te legt of strig of ctul legt 7.5 cm s 7.4 cm. Clculte te bsolute error d te percetge error. 7. Refereces/Furter Redigs

13 Solutios to Tutor Mrked Assigmet. Roud te followig to te umber of sigifict figures idicted. (.48 4 sigifict figures =. (b sigifict figures = 95.5 (c.4589 sigifict figures =.458. A studet mesured te legt of strig of ctul legt 7.5 cm s 7.4 cm. Clculte te bsolute error d te percetge error. Absolute error is Te percetge error is

14 UNIT : Approimtios d Errors i Numericl Computtios. Itroductio. Objectives 8. Mi Cotet. Lier Grp. Lieristio. Curve Fittig.. Metod of Lest Squres.. Metod of group verges 9. Coclusio. Summr. Tutor-Mrked Assigmet (TMA. Refereces/Furter Redigs. Itroductio I most eperimets s psicist, ou would be epected to plot some grps. Tis cpter eplis i detils, ow ou c iterpret te equtio goverig prticulr peomeo, plot te pproprite grp wit te dt obtied, to illustrte te ieret psicl fetures, d deduce te vlues of some psicl qutities. Te process of fittig curve to set of dt is clled curve-fittig. We sll ow tke look t te possible cses tt could rise i curve-fittig.. Objectives At te ed of tis uit, ou sould be ble to: Lierise give equtio i order to plot lier grp from wic some psicl costts c be determied. Derive te equtio for lest squres lier fit. Derive te equtio for te metod of movig verges. Fit lier grp to set of dt.. Mi Cotet. Lier Grp Te lw goverig te psicl peomeo uder ivestigtio could be lier, of te form m c. It follows tt grp could be plotted of te poits ( i, i, i =,,, were is te umber of observtios (or sets of dt. We could obti te lie of best fit vi of umber of metods: More o tis.. Lieristio A olier reltiosip c be lierised d te resultig grp lsed to brig out te reltiosip betwee vribles. We sll cosider few emples: Cse : e. (i We could tke te logritm of bot sides to bse e: l l( e l l e l, 4

15 sice l e. Tus, plot of l gist gives lier grp wit slope uit d -itercept of l. (ii We could lso ve plotted gist origi, wit slope equl to. e. Te result is lier grp troug te l Cse : T g We c write tis epressio i tree differet ws: l (i l T l( l l( (l l l g. g Rerrgig, we obti, l T l l l( l g writig tis i te form m c, we see tt plot of l T gist l l gives slope of.5 d l T itercept of l( l g. Oce te itercept is red of te grp, ou c te clculte te vlue of g. (ii T g l A plot of T versus l gives lier grp troug te origi (s te itercept is zero. Te slope of te grp is, from wic te vlue of g c be recovered. g (iii Squrig bot sides, 4 T l g A plot of T versus l gives lier grp troug te origi. Te slope of te grp is 4, d te vlue of g c be obtied ppropritel. g t Cse : N N e Te studet c sow tt plot of, d l N itercept is l N. l N versus t will give lier grp wit slope Wt oter fuctios of N d t could ou plot i order to get d N? 5

16 Cse 4: f u v We rerrge te equtio: v f u A plot of. f v (-is versus u (-is gives slope of d verticl itercept of Emple A studet obtied te followig redig wit mirror i te lbortor. u 4 5 v Lierise te reltiosip. Plot te grp of v versus u d drw te lie v f u of best fit. Hece, fid te focl legt of te mirror. All distces re i cm. Solutio u v /u /v Te grp is plotted i Fig... 6

17 . -. /u (/cm /v (/cm Fig..: Lier grp of te fuctio v f u Te slope is. 5 d te itercept. 4.4, or f f = 5 cm..4. From v f, we see tt te itercept is u. Curve Fittig Wt we did i Sectio., geerll, ws to plot te vlues of depedet vrible gist te correspodig vlues of te idepedet vrible. Wit tis doe, we got te lie of best fit. Te ltter could ve bee obtied b ee judgmet. Tere re some oter ws of deducig te reltiosip betwee te vribles. We sll first cosider te oes bsed o lier reltiosip, or te oes tt c be someow reduced to suc reltiosips... Metod of Lest Squres Suppose i, i,, re te poits of te idepedet vrible were te depedet vrible vig respective vlues i, i,, is mesured. Cosider te grp below, were we ve ssumed lier grp of equtio m c. Te t ec poit i, i,,, m c. i i Te lest squre metod etils miimizig te sum of te squres of te differece betwee te mesured vlue d te oe predicted b te ssumed equtio. 7

18 ( m c m c Fig..: Illustrtio of te error i represetig set of dt wit te lie of best fit S i mi c i (. We ve tke te squre of te differece becuse tkig te sum loe migt give te impressio tt tere is o error if te sum of positive differeces is blced b te sum of egtive differeces, just s i te cse of te relevce of te vrice of set of dt. Now, S is fuctio of m d c, tt is, S S( m, c. Tis is becuse we seek lie of best fit, wic will be determied b pproprite slope d suitble itercept. I cse, i d i re ot vribles i tis cse, vig bee obtied i te lbortor, for istce. You ve bee tugt t oe poit or oter tt for fuctio of sigle vrible df f (, te etrem re te poits were. However, for fuctio of more t d oe vrible, prtil derivtives re te relevt qutities. Tus, sice S S( m, c, te coditio for etrem is S m S d c. S m [ ( m c](. i i i i 8

19 S [ i c i ( mi c](.4 From equtio., i i m i i i i d from equtio.4, i m i i i i It follows from te fct tt.6 give, respectivel, c i =.5 c =.6 i i i d similr epressios, tt equtios.5 d m c.7 m c.8 Multiplig equtio.8 b gives m c.9 Fill, from equtios.7 d.9, m. d from equtio.8, c m. Emple A studet obtied te followig dt i te lbortor. B mkig use of te metod of lest squres, fid te reltiosip betwee d t. Tus, for te followig set of redigs: t Te tble c be eteded to give t =95 t = =6 =.4 t =4 t =68 t =95 =459 t t m.657 t t c mt

20 Hece, te reltiosip betwee d t is,.657t Metod of group verges As te me implies, set of dt is divided ito two groups, ec of wic is ssumed to ve zero sum of residuls. Tus, give te equtio m c.4 we would like to fit set of observtios s close s possible. Te error i te mesured vlue of te vrible d te vlue predicted b te equtio is (s we ve see i Fig. : ( m c.5 i i i Te fitted lie requires two ukow qutities: m d c. Tus, two equtios re eeded. We would cieve tese two equtios b dividig te dt ito two, oe of size l d te oter of size -l, were is te totl umber of observtios. Te ssumptio tt te sum of errors for ec group is zero, requires tt d l i l From equtio.6, l [ ( m c] =.6 i i [ ( m c] =.7 i m i i i l i lc d equtio.7 ields il i i m i ( l c il te ltter equtio beig true sice l is te umber of observtios tt fll ito tt group..8.9 Dividig troug b l d l, respectivel, equtio.8 gives l l i m i c l l i i. d from equtio.9, i m i l il l il c. Tus, m c m c.

21 Subtrctig, ( m. m.4 d c m.5 Emple Let us solve te emple i Sectio.. usig te metod of group verges. t We sll divide te dt ito two groups, suc s: d t t Te tbles c be eteded to give, for Tble : d for Tble 4: t 5 9 =6 t = 8 =8 = t 6 =59 t = =79 =9.5 d m t = t c mt = (.6769 = Tus, te equtio of best fit is,.6769t

22 4. Coclusio I tis Uit, ou lert ow to lierise epressio i order to obti some relevt iformtio we writte s lier equtio. You lso derived te equtios for two differet metods of drwig te lie of best fit. I dditio, ou pplied tese formuls to set of dt d ws ble to write te equtio of best fit i ec cse. 5. Summr I tis Uit, ou lert: How to lierise olier epressio i order to deduce some desired prmeters. How to drw te lie of best fit wit te metod of lest squres. How to drw te lie of best fit wit te metod of group verges. 6. Tutor Mrked Assigmet (TMA. Te curret flowig i prticulr R-C circuit is tbulted gist te cge i te time t t, suc tt t time t t, te curret is. A. Usig te lestsqures metod, fid te slope d te itercept of te lier fuctio reltig te curret i to te time t. Hece, determie te time-costt of te circuit. t i Solve te problem i TMA wit te metod of group verges b dividig ito two groups of tree dt sets ec. t..4 i..6. d t.6.8 i A studet performig te simple pedulum eperimet obtied te followig tble, were t is te time for 5 oscilltios. l (cm t (s Fid te ccelertio due to grvit t te loctio of te eperimet, usig (i te metod of lest squres, d (ii te metod of group verges. 7. Refereces/Furter Redigs

23 Solutios to Tutor Mrked Assigmet. Te curret flowig i prticulr R-C circuit is tbulted gist te cge i te time t t, suc tt t time t t, te curret is. A. Usig te lestsqures metod, fid te slope d te itercept of te lier fuctio reltig te curret i to te time t. Hece, determie te time-costt of te circuit. Tkig logs: t i i t / RC ie. gist t gives slope d itercept log i. RC t / RC t log i log i log( e log i. A plot of log i RC t I tsqure log l tlogl Sum Averge Slope -.44 Itercept (.5.96 m c log l mt.844 m, or RC. 568 = time costt of te circuit. RC m. Solve te problem i TMA wit te metod of group verges b dividig ito two groups of tree dt sets ec. t..4 i..6. d t.6.8 i..9.7

24 m Group t i log i Group t i log i ( c m.7964 ( A studet performig te simple pedulum eperimet obtied te followig tble, were t is te time for 5 oscilltios. l (cm t (s Fid te ccelertio due to grvit t te loctio of te eperimet, usig (iii te metod of lest squres, d (iv te metod of group verges. Metod of lest squres (tkig logs log T log log l : A plot of log T gist log l gives slope.5 d g itercept c = log, from wic te vlue of g is g log ( c l t log l log T (log l*(log l (log l*(log T

25 Sum Averge Metod of lest squres (tkig squres 4 T l. A plot of g from wic g 4 : m slope.499 itercept.857 pi 6.84 log pi.7986 log pi -iter.5679 (log piiter.59log g.84g T gist l gives lie troug te origi wit slope m = L t L Tsqure lsqure Tsqure l Sum Averge slope g.4 itercept , g Metod of group verges (tkig logs Group L t log l log T

26 Sum Averge Group L t log l log T Sum Averge slope itercept.8999 g.8 Metod of group verges (tkig squres Group L t l Tsqure Sum Averge Group L t l Tsqure Sum Averge slope.8775 itercept.5 g.8 6

27 UNIT : Lier Sstems of Equtios. Itroductio. Objectives. Mi Cotet. Sstem of Lier Equtios. Gussi Elimitio. Guss-Jord Elimitio.4 LU Decompositio.5 Jcobi Itertio.6 Guss-Seidl Itertio 4. Coclusio 5. Summr 6. Tutor Mrked Assigmet 7. Refereces/Furter Redig. Itroductio Perps i ll res of Psics, ou would come cross sstem of lier equtios. For emple, ou migt wt to kow wt proportios of two or more vribles ou would eed to cieve some specific vlues of desired composite product. Tis kid of problem could led to set of lier equtios. Tis uit will equip ou wit te ecessr tools to solve sstem of lier equtios. You sll come cross direct metods s well s itertive ws of solvig suc problems.. Objectives You sould be ble to do te followig fter studig tis Uit: Write sstem of lier equtios i ugmeted mtri form Solve sstem of lier equtios.. Sstem of Lier Equtios It is ecessr for us to set te stge b gettig to kow ow to write te geerl set of simulteous lier equtios. Let us cosider lier sstem of equtios b... b b. Tis c be writte i te form 7

28 b b b.. Gussi Elimitio A recll of te solutio of sstem of two equtios will elp i itroducig te Gussi Elimitio metod. For istce, let (, be solutio set (,. Te te followig equtios re i order...4 You migt wt to verif tt tese equtios re cosistet wit te give solutio set. We could multipl equtio. b - d dd to equtio.. Tis ields Equivletl,. Substitutig tis vlue of i eiter equtio. or.4 gives. Te ugmeted mtri represetig our sstem of two equtios is B Gussi elimitio, we seek to mke ever etr below te mi digol zero. Tis we cieve b reducig to zero, mkig use of te first row. ( ii' ( i ( ii Tus, Substitutig tis i te first row gives (.8 from wic we obti. 8

29 Te process of reducig ever elemet below te mi digol to zero (row ecelo form is clled Gussi Elimitio. Tt of substitutig obtied vlues to clculte oter vribles is clled Bck Substitutio. You c see tt tere is otig ew bout Gussi elimitio. It is process ou ve bee crrig out ll log, but wic ou ever clled tis me. Te sme process c be crried over to te cse of sstem of tree equtios. Let (,, be solutio set. Te, te equtios below re vlid: z 5 z 5 4z.9 Te ugmeted mtri is Tis ields (b Gussi elimitio ( ii' ( i ( ii ( iii ( i ( / ( iii 5 7 / 5 5 / 5 5 ( iii'' ( ii' (5 / 7( iii ' Upo bck substitutio, z or z z ; z ; z Trditioll, i Mtemtics, it is usul to use idices suc s,, etc. isted of,, z. Do ou ve ide w tis is so? It is becuse if we st wit te lpbets, we sll soo ru out of smbols. Ber i mid tt ot ll te lpbets c be emploed s vribles; s emple,, b, c re commol used s costts. I dditio, it mkes it es to ssocite te coefficiets,, etc. wit,, etc. respectivel. More importtl i umericl work, it mkes progrmmig esier. For istce for our sstem of tree equtios, we could use te more geerl ottio: 9

30 4 4 4 ( ii ' ( i ( / ( ii ( iii ( i ( / ( iii ' ' ' ' ' ' ' ' ( iii'' ( ii' ( / ( iii' ' ' ' ''. We would like to soud ote of wrig ere. How do ou set ' equl to zero? From te epressio., '. I order to void vig to del wit frctios wic could led to roudig errors, it is better to put tis i te form: ( ii' ( i ( ii. A better w of writig equtio. is, ( ii ' ( i ( ii ( iii ( i ( iii ' ' ' ' ' ' ' ' ( iii'' ( ' ii ( iii' '' ' ''.. Guss-Jord Elimitio Tis etils elimitig i dditio to te etries below te mjor digol, te etries bove it, so tt te mi mtri is digol mtri. I tt cse, te solutio to te sstem is give b dividig te elemet i te ugmeted prt of te mtri b te digol elemet for tt row. I oter words, te ed product of Guss-Jord elimitio looks like '' '' '''' 4 '' 4 ''' '''' 4. from wic it follows tt ''/ ''' '''/ ''''. 4 ''/ ' 4 4 ' Emple

31 We sll solve problem usig te Guss-Jord elimitio. Luckil, we ve lred completed te Gussi elimitio prt of tis metod. We cotiue from were we stopped. 5 5 ( i' ( iii ( i 5 ( ii' ( iii ( ii 5 4 It follows tt, ( i'' ( ii' ( i'.4 or ; or ; d z or z.4 LU Decompositio Suppose we could write te mtri l = l l l l u l u u u u u.5 Tis implies tt lu, lu, lu.6 lu, lu lu, lu lu.7 lu, lu lu, lu lu lu.8 Witout loss of geerlit, we could set te digol elemets of te L mtri equl to. Te, u u u = l u u.9 l l u Multiplig out te rigt side of equtio.9, u u = lu lu u lu lu lu l u l u u l u u u.9b From te equlit of mtrices, tis requires tt, u.

32 u. u. lu l / u /. lu l / u /.4 lu u, or u lu u u l u lu u, or u lu u u.5 u.6 lu u u u u l.7 u l u l.8 l u l u u u u lu lu.9 You c see tt we ve determied ll te ie elemets of te two mtrices i terms of te elemets of te origil mtri. Oce we ve obtied L d U, te we c write te origil equtio s LU. were d re colum vectors. We sll write w U Te, L w. Emple Solve te followig sstem of equtios usig te metod of LU decompositio..

33 z 5 z 5 4z. Te correspodig mtri is 4 u =.4 u =.5 u.6 l / = /.7 l / = /.8 u ( 5/.9 u ( 5/ l ( 7 / 5 5/.4 u u lu lu 4 (/ ( ( 7 / 5(5/.4 Tus, / / 5/ 5/ = 7 / As ou c see, we got te decompositio rigt, s te multiplictio of te L d U gives te origil mtri. Te origil equtio is equivlet to LU Lw.4 L w implies / / w 5 w 5 7 / 5 w.44

34 4 Solvig, w w w or 5 (5 5 5 w w w w w, or ( w w w.47 w U implies: 5/ 5 5/ 5 /.48 B bck substitutio, ( ( Te solutio set is terefore,,, z Jcobi Itertio Give te sstem of equtios d z c b.55 d z c b.56 d z c b.57 Solvig for, d z, gives z c b d.58 ] [ z c d b.59 ] [ b d c z.6 It is es to see tt provided te digol elemets re lrge reltive to te oter coefficiets, te sequece of itertio would coverge.

35 For iitil vlues, d z, te sceme would be s sow below: d b cz.6 [ d c z ] b.6 z [ d b ] c.6 We c ow write, for = d bove, d b c z.64 [ d c z ] b.65 z [ d b ] c.66 Te sequece of itertio cotiues util tere is covergece, i te sese tt, d z z re less t te prescribed tolerce. Emple We sll solve te followig sstem of equtios usig te Jcobi itertio metod. 5 z 8 z 59 z 9 Equivletl, 8 z 59 z 9 5,, z Let us ssume tt te iitil guess of solutio is (,,. Te, te first set of vlues for te itertio is: z / 9 /

36 59 z ( Tble. sows te rest of te computtio. Tble.: Tble for Jcobi itertio z Guss-Seidl Itertio You would recll tt i ec of te Jcobi itertios, we clculted te vlue of te vribles usig te old vribles. Te Guss-Seidl itertio is modifictio of tis metod, i wic te vlue of obtied i prticulr itertio d te old vlue of z is put ito te formul for to obti ew vlue for. Te ew vlues of d re substituted ito te equtio for z. Tus, give te sstem of equtios b cz d.77 b cz d.78 b c z.79 d wit te iitil coditio,, z, d b cz.8 [ d c z ].8 b z [ d b ].8 c 6

37 d b c z.8 [ d c z ] b.84 z [ d b ] c.85 As i te cse of te Jcobi itertio, te sequece of itertio cotiues util tere is covergece, i te sese tt, d z z re less t te prescribed tolerce. Emple We sll solve te followig sstem of equtios usig te Guss-Seidl itertio metod. Assume (,, is te iitil guess of solutio. 5 z 8 z 59 z z z.8 9 z z You c verif te remiig clcultios o Tble.. Tble.: Tble for Guss-Seidl itertio z

38 Observtio: As epected, te Guss-Seidl itertio coverged fster t te Jcobi itertio. 4. Coclusio I tis Uit, ou lert vrious metods for solvig sstem of lier lgebric or trscedetl equtios usig vrious metods: some were direct, wile te oters were itertive i ture. You lso got to kow te merits d te demerits of direct d itertive metods. You lso foud out tt it is importt, i elemetr row opertios, to void vig to del wit frctios, so s to keep roudig errors miiml. 5. Summr You lert te followig i tis Uit: How to write mtri i te form meble for progrmmig. How to umericll solve set of lier equtios. Tt te Guss-Seidl itertio coverges fster t te Jcobi itertio. I umericl work, for te ske of voidig roudig errors, it is better to reti frctios for s log s possible. Itertio is dvisble ol if te mi digol elemets re lrge compred wit te oter etries of te equivlet mtri. 6. Tutor Mrked Assigmet. Solve te sstem of lier equtios z, z 4, 9 6 z 7 usig te metod of (i Gussi elimitio (ii Guss-Jord elimitio (iii LU decompositio (iv Jcobi itertio (v Guss-Seidl itertio. Solve te sstem of lier equtios z, z 4, 9 6 z 4 usig te metod of (i Gussi elimitio (ii Guss-Jord elimitio (iii LU decompositio (iv Jcobi itertio (v Guss-Seidl itertio 8

39 7. Refereces/Furter Redig 9

40 Solutios to Tutor Mrked Assigmet. Solve te sstem of lier equtios z, z 4, 9 6 z 7 usig te metod of (i Gussi elimitio Iitil ugmeted mtri First roud of Gussi elimitio Secod roud of Gussi elimitio (ii Guss-Jord elimitio Lst mtri for Gussi elimitio First roud of Jord elimitio Secod roud of Jord elimitio (iii LU decompositio z z z 7. Te correspodig mtri is 4

41 9 6 u =.4 u =.5 u.6 l / = / =.7 l / = 9 / = 9.8 u ((.9 u (( 9 l 6 (.4 u u lu lu (9( ( ( 5.4 Tus, 9 = W got te decompositio rigt, s te multiplictio of te L d U gives te origil mtri. Te origil equtio is equivlet to L w implies w w 4 9 w 7 LU Lw,.44 Solvig, w.45 w w 4 or w 4 w 4 (.46 w w w 7, or w w 9w 7 ( 9( U w implies: 7 4

42 B bck substitutio, 7 / 5 = ( 7 / / ( 8/ 5 ( 7 / Te solutio set is terefore,.54 Notice tt, were ecessr, we reverted to frctios to void icurrig roudig errors.. Solve te sstem of equtios 5 z 6, z 5, 4 5z usig: (i Jcobi itertio (ii Guss-Seidl itertio Assume strtig set of vlues z d tolerce of 6 i i 5, i i, z i z i 5. (i (ii Jcobi itertio Guss-Seidl itertio

43 Observtio: Te Guss-Seidl itertio sceme coverged fster t te Jcobi itertio, s ws epected.. Solve te sstem of lier equtios z, z 4, 9 6 z 4 usig te metod of (iv Gussi elimitio Iitil ugmeted mtri First roud of Gussi elimitio Secod roud of Gussi elimitio Aswers - z (v Guss-Jord elimitio Lst mtri for Gussi elimitio First roud of Jord elimitio Secod roud of Jord elimitio (vi LU decompositio 4

44 z z z 7. Te correspodig mtri is 9 6 u =.4 u =.5 u.6 l / = / =.7 l / = 9 / = 9.8 u ((.9 u (( 9 l 6 (.4 u u lu lu (9( ( ( 5.4 Tus, 9 = W got te decompositio rigt, s te multiplictio of te L d U gives te origil mtri. Te origil equtio is equivlet to L w implies w w 4 9 w 7 Solvig, LU Lw,.44 44

45 w.45 w w 4 or w 4 w 4 (.46 w w w 7, or w w 9w 7 ( 9( U w implies: B bck substitutio, 7 / 5 = ( 7 / / ( 8/ 5 ( 7 / Te solutio set is terefore,.54 Notice tt, were ecessr, we reverted to frctios to void icurrig roudig errors. 4. Solve te sstem of equtios 5 z 6, z 5, 4 5z usig: (i Jcobi itertio (ii Guss-Seidl itertio Assume strtig set of vlues z d (i Jcobi itertio tolerce of 7 z i z i. z i i, 7 i i, 45

46 (ii Guss-Seidl itertio of 6 z i z i 5. 6 i i 5, 6 i i 5, z

47 Uit 4: Roots of Algebric d Trscedetl Equtios. Itroductio. Objectives. Mi Cotet. Itroductio. Bisectio Metod.. Merits of te Bisectio Metod.. Demerits of te Bisectio Metod. Newto-Rpso Metod.. Merits of te Newto-Rpso Metod.4 Regul-flsi metod.5 Sect Metod 4. Coclusio 5. Summr 6. Tutor Mrked Assigmet 7. Refereces/Furter Redig.. Demerits of te Newto-Rpso Metod. Itroductio I Psics, s well s i m oter scietific fields, tere is lws te eed to fid te root of equtio. You ve o doubt bee tcklig suc problems from ig scool ds. However, up till ow, ou ve bee ble to dle simple cses tt clcultor could be emploed to do. I tis Uit, ou sll ler ow to dle te more complicted cses of roots of lgebric d trscedetl equtios.. Objectives B te time ou re troug wit tis Uit, ou sould be ble to: Fid te root of equtio or equivletl te zero of fuctio. You would lso be ble to compre te vrious metods of obtiig te zero of fuctio.. Mi Cotet. Itroductio You re probbl quite fmilir wit te cocept of te fuctio of cotiuous vrible f (, cotiuous over certi itervl of te idepedet vrible. If we equte f ( to zero, we obti te equtio f (. You migt eve see te process s tt of equtig two differet fuctios f ( d f (, were te ltter is ideticll zero. 47

48 Figure 4. sows te grp of Fig. 4. f (. Te -is c be see s te fuctio f (. Equtig te two fuctios gives f f (. Te resultig ( equtio,, s two solutios d (te two solutios re idicted i Figure 4.. Let us slide f ( dow to f (, te lower orizotl lie. Te equtio becomes. Tis is perps oe of te commoest qudrtic equtios ou ever cme cross. Te solutios re:. d.. You c ceck tis out o Fig. s well. Siftig f ( lower to. 5 would esure tt te resultig equtio s o rel solutios s te curve would ot itersect te lie. Te equtios we ve delt wit so fr ve bee suc tt could esil be solved usig lticl metods. It sould be obvious to ou tt suc equtios sould form smll subset of muc lrger fmil of equtios, te solutios of most of wic do ot redil led temselves to lticl metods, especill s te power of te polomil beig equted to zero becomes lrge. Equtig polomil to zero gives lgebric equtio. A trscedetl fuctio is fuctio tt trsceds te orml lws of lgebr s it cot be epressed s sequece of te lgebric opertios of dditio/subtrctio, multiplictio/divisio, emple beig te squre root of oter fuctio. Oter emples iclude logritmic, trigoometric, epoetil fuctios d teir iverses. If equtio ivolves te trscedetl epressios, suc s epoetils, trigoometric, logritmic fuctios, te equtio is sid to be trscedetl equtio. We sll ssume tt te fuctio wose roots we desire, f (, is fuctio of, wose zeros (or te roots of te resultig equtio lie o te rel is. Tt is, te roots 48

49 of te equtio f ( re rel umbers. Tere re umber of metods of fidig te roots. We sll ow tke some of tese.. Bisectio Metod Fig. As te me implies, we obti te poits d, suc tt f ( f (, meig tt te vlue of f s opposite sigs t te two poits, wic poits to te fct tt root eists betwee d. We pproimte tis root b te verge of te two, i.e., ( /. Let tis be. Te we evlute f (. is te combied wit or, depedig o te oe t wic te sig of te fuctio is opposite tt of f. Tis gives 4. Tis process is repeted util f ( ttis te prescribed tolerce. We ve illustrted tis i Fig for te root of te equtio, give tt te root lies betwee =. d =.4. Te, ( / =.8. f (, so we combie it wit to rrive t ( / 4, d so o. Te covergece of te Bisectio metod is slow d sted.. Merits of te Bisectio Metod. As ou c see, te root bisectio metod lws coverges. Tis is becuse ou would get closer d closer to te root s te distce betwee te two poits of iterest is lved ec time.. You c lso keep tb o te error. If te root lies betwee te poits d b, tere will be sequece: ( 49

50 b ( b ( b... ( b. But ou would 4 b recll tt b b d. Tus, b. O te oter d, we ote tt te first itertio poit is t lest s close to te root s lf te itervl b b b, i.e.,. Similrl, for te t itertio. b b b But b. Hece, ( b. We b b coclude tt, d tis gives us ide of te mimum error i our estimte of te root... Demerits of te Bisectio Metod. Te covergece is geerll slow.. You migt ctull be pprocig sigulrit, for emple, wile delig wit fuctios tt re ot cotiuous betwee te two iitil poits. A clssicl emple is te fuctio f (, egtive for d positive for. As ou strt out wit te bisectio metod wit poit o te rigt of d oter o te left of, ou re uder te impressio tt tere sould be root ibetwee. If te fuctio is cotiuous betwee te iitil guesses, tis problem is elimited.. Te bisectio metod will ot work if te fuctio is tgetil to te -is t te desired root. For emple, f ( is tgetil to te -is t te poit wic is te root of te equtio. Te fuctio is positive o eiter side of, so ou would ot eve tr to get it i te first plce, s te bisectio metod imposes te coditio tt te sigs o eiter side be differet. 4. If oe of te iitil poits is close to te root, ou would eed m itertios to rrive t te root. 5. It does ot work for repeted roots. If tere re multiple roots witi te itervl give, te sceme rrows dow o ol oe of te roots. 6. It does ot work for repeted roots. Emple: Fid zero of te fuctio f ( betwee te poits.4 5 d.7, usig te bisectio metod. Tke te tolerce to be j j. Solutio f (.4.9 f (

51 f ( f ( You c cofirm tt Tble 4. is ideed true. Tble 4.: Tble for Bisectio metod f ( E-.55.E E E E E E E E E E-5. Newto-Rpso Metod Cosider Tlor Series f ''( ( f ( f ( f '( 4.! To first order pproimtio, we c eglect secod order d iger order terms. I tt cse, if f (, te we tructe equtio 4., levig ol te first two terms o te rigt. Te, f ( f ( f '( 4. or f (, 4. f '( so tt wit iitil guess of, we obti better pproimtio, i.e., f ( 4.4 f '( 5

52 It is quite cler tt te fuctio f ( must be differetible for ou to be ble ppl te Newto-Rpso metod. More geerll, f ( i i i i 4.5 f '( i Wit iitil guess of, we c te get sequece,,, wic we epect to coverge to te root of te equtio. We c rerrge equtio 4.5 to obti, f ( i f '( i 4.6 i i meig tt Newto-Rpso metod is equivlet to tkig te slope of te fuctio f ( t te i t itertive poit, d te et pproimtio is te poit were te slope itersects te is. See te Fig 4.: f ( f ( f ( i i i Fig. 4.: Grp sowig te grdiet reltiosip of Newto-Rpso metod.. Merits of te Newto-Rpso Metod. Te Newto-Rpso metod s fst rte of covergece.. It c idetif repeted roots, sice it does ot eplicitl look for cges i te sig of f (.. It c fid comple roots of polomils if ou strted wit comple iitil guess. 5

53 .. Demerits of te Newto-Rpso Metod. It requires tt we compute bot f ( d f '(, wic mkes te sceme tig.. Some fuctios migt ot be so es to differetite. I tt cse, it migt be f ( f ( useful to tke pproimte differetil,.. It is quite sesitive to iitil coditio d m diverge for te wrog coice of iitil poit. 4. It will ot work if f '(. Also, if te differetil is sufficietl close to zero, te sequece m diverges w from te root, or coverge ver slowl. 5. If te derivtive cges sigs t test poit, te sequece m oscillte roud poit tt m ot eve be te root. 6. It cot detect repeted roots. Emple: Fid te zeros of te fuctio f ( usig te Newto- 5 Rpso metod, strtig wit =.4. Tke te tolerce to be j j. Solutio f ( f '( f ( f '( f '( 6 f '( f ( 6 4(.4 (.4 6(.4 6( ,. 4.55,. 77.5,. 5.5, Regul-flsi metod A regul-flsi or metod of flse positio ssumes test vlue for te solutio of te equtio. 5

54 You would recll tt wit te root-bisectio metod, we kew tt root eisted betwee d if te fuctio ws smoot d f ( f (. Let us gi coose tese two poits s i te cse of root-bisectio. Te, for rbitrr d te correspodig, f ( f ( f ( gives te equtio of te cord joiig te poits, f ( d, f (. ( ( 4.9 Settig, tt is, were te cord crosses te -is, f ( 4. f ( f ( Te, we evlute f. Just s i te cse of root-bisectio, if te sig is opposite tt ( of f (, te root lies i-betwee d. Te, we replce b i equtio 4.. I just te sme w, if te root lies betwee d, we replce b. We sll repet tis procedure util we re s close to te root s desired. Emple Fid te root of te equtio f ( betwee =.4 d.7 b te regul-flsi metod. f (.4.9, f ( A solutio lies betwee =.4 d.7. Let. 4 d. 7. Te,.7.4 f (.4 (.9 f ( f (.756 (.9 = f ( Te root lies betwee.4676 d.7. Let d

55 4 = f ( f ( f ( ( (.8898 Tble 4. gives te remiig itertios. Tble 4.: Tble for Regul-flsi metod f ( Sect Metod I te cse of te sect metod, it is ot ecessr tt te root lie betwee te two iitil poits. As suc, te coditio f ( f ( is ot eeded. Followig te sme lsis wit te cse of te regul-flsi metod, f ( f ( f ( 4. Settig gives f ( 4. f ( f ( Tus, vig foud, we c obti s, f (, =,, 4. f ( f ( B ispectio, if f ( f (, te sequece does ot coverge, becuse te formul fils to work for. Te regul-flsi sceme does ot ve tis problem s te ssocited sequece lws coverges. Emple Fid te root of te equtio f ( betwee =.4 d.7 b te regul-flsi metod..4,.7 f (.4.9, f (

56 A solutio lies betwee =.4 d.7. Let. 4 d. 7. Te,.7.4 f (.4 (.9 f ( f (.756 (.9 = f ( = f ( f ( f ( ( You c cotiue wit tis sceme. Tble 4. sows te oter vlues obtied from te opertio. Tble 4.: Tble for Sect Metod f ( Coclusio I tis Uit, ou lert to fid te zeros of lgebric or trscedetl fuctio. We eplored umber of metods, d outlied teir merits d demerits. We were lso ble to estimte te mimum error i te bisectio metod. 5. Summr I tis Uit, ou lert: to fid te zeros of lgebric or trscedetl fuctio usig severl metods. te merits d demerits of te metods. to te mimum error tt c be icurred i usig te bisectio metod. 6. Tutor Mrked Assigmet. Fid te upper boud of te error ou re likel to icur i usig te bisectio metod i fidig te root of equtio if te two strtig poits re.4 d.5 d ou eeded 8 steps to cieve te required tolerce.. Fid root of te equtio. 5 usig te followig metods (tolerce..: (i Root bisectio [strtig poits.9 d. (tolerce f (.]. (ii Newto-Rpso strtig poit. (iii Regul-flsi [strtig poits.9 d.]. (iv Sect [strtig poits.9 d.]. 56

57 . Fid root of te equtio si usig (i Te bisectio metod, give tt te root is betwee.5 d, wit tolerce f (.. (ii Newto-Rpso metod, wit te strtig poit.5, wit tolerce 6 f (. (iii Regul-flsi [strtig poits.5 d.]. (iv Sect [strtig poits.5 d.]. 7. Refereces/Furter Redig 57

58 Solutios to Tutor Mrked Assigmet. Fid te upper boud of te error ou re likel to icur i usig te bisectio metod i fidig te root of equtio if te two strtig poits re.4 d.5 d ou eeded 8 steps to cieve te required tolerce Fid root of te equtio. 5 usig te followig metods (tolerce..: (v Root bisectio [strtig poits.9 d. (tolerce f (.]. Itertio No. i f ( i E E E E E E E-4 (vi Newto-Rpso strtig poit. Itertio No. i f ( i e-6 (vii Regul-flsi [strtig poits.9 d.]. Itertio No. i f ( i

59 (viii Sect [strtig poits.9 d.]. Itertio No. i f ( i E E E-5. Fid root of te equtio si usig (v Te bisectio metod, give tt te root is betwee.5 d, wit tolerce f (.. Itertio No. i f ( i (vi Newto-Rpso metod, wit te strtig poit.5, wit tolerce 6 f (. Itertio No. i f ( i (vii Regul-flsi [strtig poits.5 d.]. Itertio No. i f ( i

60 (viii Sect [strtig poits.5 d.]. Itertio No. i f ( i

61 UNIT 5: FINITE DIFFERENCES AND INTERPOLATION. Itroductio. Objectives. Mi Cotet. Fiite Differeces.. Forwrd Differeces.. Error i Fiite Differece Tble. Iterpoltio.. Newto forwrd iterpoltio formul.. Newto s Bckwrd Iterpoltio Formul 4. Coclusio 5. Summr 6. Tutor Mrked Assigmet 7. Refereces/Furter Redig. Itroductio Give te fuctio f ( we c evlute te vlues of f t differet, tereb represetig cotiuous fuctio wit set of discrete dt. O te oter d, it could be tt we ve set of dt d we would like to see if te could ve bee got from polomil or if ideed we could represet te poits b polomil. Fiite differeces would elp us i tis regrd. Wit te id of fiite differeces, we sll te derive Newto s forwrd d Newto s bckwrd iterpoltio formuls.. Objectives At te ed of tis Uit, ou would be ble to: Deduce polomil from its differece tble. Derive Newto s forwrd d Newto s bckwrd iterpoltio formuls. Fit polomil to give set of dt Iterpolte d etrpolte wit Newto s forwrd differece Iterpolte d etrpolte wit Newto s bckwrd differece. Mi Cotet. Fiite Differeces We proceed b defiig te fiite differece i. First Forwrd differece: f i f i f i 5. ii. First Bckwrd differece: f i f i f i 5. iii. First Cetrl differece: f i f i i / 5. Te tble for forwrd differece would look like Tble 5.. Wt do ou otice bout tis tble? You c see tt d te differeces relted to it pper o te first lie sltig dow to te rigt. 6

62 Tble 5.: Forwrd differece Tble You c see tt differeces wit similr subscripts form lie sltig dowwrd to te rigt from te top. Tble 5. is te bckwrd differece tble. 4 4 Tble 5.: Bckwrd differece tble C ou spot wt mkes tis tble uique? Differeces wit similr subscripts form lie sltig upwrd to te rigt from te bottom. Note tt for forwrd differece,, or geerll, d for bckwrd differece, 5.4 6