NATIONAL OPEN UNIVERSITY OF NIGERIA

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1 NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH COURSE TITLE: NUMERICAL ANALYSIS

2 Course Code MTH Course Title NUMERICAL ANALYSIS Course Developer Dr. Ajibol S. O. Ntiol Ope Uiversity of Nigeri Lgos Cotet Editor Dr. ABIOLA. Bkole Ntiol Ope Uiversity of Nigeri Lgos Course Coorditor Dr. Ajibol S. O. Ntiol Ope Uiversity of Nigeri Lgos Progrmme Leder Dr. ABIOLA.Bkole Ntiol Ope Uiversity of Nigeri Lgos NATIONAL OPEN UNIVERSITY OF NIGERIA ii

3 COURSE GUIDE MTH Ntiol Ope Uiversity of Nigeri Hedqurters 4/6 Ahmdu Bello Wy Victori Isld Lgos Abuj Office 5, Dr Es Slm Street Off Amiu Ko Crescet Wuse II, Abuj Nigeri. e-mil: URL: Ntiol Ope Uiversity of Nigeri 6 First Prited 8 ISBN: All Rights Reserved Prited by: For Ntiol Ope Uiversity of Nigeri iii

4 CONTENT PAGE Module Iterpoltio Uit Iterpoltio (Lgrge s Form).. Uit Newto s Form of the Iterpoltig Polyomil 5 Uit Iterpoltio t Eqully Spced Poits Module Solutio of Lier Algebric Equtios 55 Uit Direct Method 55 Uit Iverse of A Squre Mtrix 9 Uit Itertive Methods Uit 4 Eige-Vlues d Eige-Vectors 5 Module Solutio of No-Lier Equtios i oe Vrible. 59 Uit Review of Clculus 59 Uit Itertio Methods for Loctig Root. 89 Uit Chord Methods for Fidig Root... 8 Uit 4 Approximte Root of Polyomil Equtio... 5 iv

5 COURSE GUIDE MTH COURSE GUIDE MTH NUMERICAL ANALYSIS Course Developer Dr. Ajibol S. O. Ntiol Ope Uiversity of Nigeri Lgos Cotet Editor Dr. ABIOLA. Bkole Ntiol Ope Uiversity of Nigeri Lgos Course Coorditor Dr. Ajibol S. O. Ntiol Ope Uiversity of Nigeri Lgos Progrmme Leder Dr. ABIOLA. Bkole Ntiol Ope Uiversity of Nigeri Lgos NATIONAL OPEN UNIVERSITY OF NIGERIA v

6 Ntiol Ope Uiversity of Nigeri Hedqurters 4/6 Ahmdu Bello Wy Victori Isld Lgos Abuj Office 5, Dr Es Slm Street Off Amiu Ko Crescet Wuse II, Abuj Nigeri. e-mil: URL: Ntiol Ope Uiversity of Nigeri 8 First Prited 8 ISBN: All Rights Reserved Prited by.. For Ntiol Ope Uiversity of Nigeri vi

7 COURSE GUIDE MTH CONTENTS PAGE Itroductio. The Course Course Aims & Objectives.. Workig through the course Course mterils.. Study Uits.. Textbooks Assessmet... 5 Tutor-Mrked Assigmets. 5 Ed of Course Exmitio. 5 Summry. 5 vii

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9 MTH MODULE Itroductio MTH : Discussio of Lgrge s form for; The techique of determiig pproximte vlue of f(x) for o-tbulr vlue of x which lies i the iterl [, b] is clled iterpoltio. The process of determiig the vlue of f(x) for vlue of x lyig outside the itervl [, b] is clled extrpoltio. The Lgrge s form of the iterpoltig polyomil derived bove hs sme drw bcks compred to Newto s form of iterpoltig polyomil. Before derivig Newto s geerl form of iterpoltig polyomil. We itroduce the cocept of divided differece d the tbulr represettio of divided differeces. Numericl solutio of systems of lier lgebric equtios ply promiet role i boudry vlue problems, for ordiry d prtil differetil equtios, sttisticl ifluece, optimiztio theory, lest squre fittigs of dt etc. Numericl methods for solvig lier lgebric system my be divided ito two types, direct d itertive. To uderstd the umericl methods for solvig lier system of equtios, it is ecessry to hve some kowledge of the properties of mtrices. The prerequisite to the course shll be lier Algebr courses. The Course As -credit uit course, study uits grouped ito modules of uits i module, 4 uits i module d 4 uits i module. This course guide gives brief summry of the totl cotets cotied i the course mteril. The fudmetl theorem of lgebr d its useful clories, iverse iterpoltio d errors. Newto s form of the iterpoltig polyomil fetures divided differeces d iterpoltig polyomil error types. Likewise iterpoltig t eqully spced poits, here we tlked bout differeces. For eqully spced odes, we shll del with three types of differeces, mely forwrd, bckwrd d cetrl d discuss their represettio i the form of tble. Also discussed her re some direct d itertive methods for fidig the solutio of system of lier lgebric equtios. Lstly, we discussed three fudmetl theorems, mely; itermedite vlue theorem, Rolle s theorem d Lgrge s me vlue theorem. All these theorems give properties of cotiues fuctios defied o 59

10 MTH NUMERICAL ANALYSIS closed itervl [, b]. Although the theorems re ot proved but their utility ws illustrted with exmples. Course Aim & Objectives O the completio of this course, you re expected to: fid the Lgrge s form of iterpoltig polyomil complete the pproximte vlue of f t o-tbulr poit. Complete the error omitted i iterpoltio, if the fuctio is kow t o-tbulr poit of iterest. Fid upper boud i the mgitude of the error. Write forwrd, bckwrd d cetrl differeces i terms of fuctio vlues from tble of either differece d locte differece of give order t give poit. Obti the iterpoltig polyomil of f(x) for give dt by pplyig y oe of the iterpoltig formule. Obti the solutio of systems of lier lgebric equtios by usig the direct methods such s Crmer s rule, Guss elimitio method Lu decompositio method. Workig through the Course This course ivolves tht you would be required to sped lot of time to red. The cotet of this mteril is very dese d require you spedig gret time to study it. This ccouts for the gret effort put ito its developmet i the ttempt to mke it very redble d comprehesible. Nevertheless, the effort required of you is still tremedous. I would dvice tht you vil yourself the opportuity of ttedig the tutoril sessios where you would hve the opportuity of comprig kowledge with your peers. The Course Mteril You will be provided with the followig mterils: Course Guide Study Uits I dditio, the course comes with list of recommeded textbooks, which through re ot compulsory for you to cquire or ideed red, re ecessry s supplemets to the course mteril. 6

11 MTH MODULE Study Uits The followig re the study uits cotied i this course. The uits re rrged ito idetifible but redble modules. Module Uit Iterpoltio (Lgrge s Form) This uit tkes oe through the defiitio of iterpoltio, iverse iterpoltio d error. Uit Newto s Form of the Iterpoltig Polyomil This uit is sub-divided ito divided differece Newto s Geerl Form of iterpoltig polyomil, d the error of the iterpoltig polyomil. Divided differece d derivtive of the fuctios d further results o iterpoltios error. Uit Iterpoltio t Eqully Spced Poits This uit tkes bout the three types of differeces i.e. forwrd, bckwrd d cetrl differeces. Differece formule which ecompsses: Newto s Forwrd Differece formul d Newto s Bckwrd-Differece formul. Module Uit Solutio of Lier Algebric Equtios. Direct Method This uit etils the prelimiries, Crmer s rule, direct methods for specil mtrices. Guss elimitio methods d LU decompositio method. Uit Iverse of A Squre Mtrix This uit is sub-divided ito method of djoits, the Guss-Jord reductio method d LU decompositio method. Uit Itertive Methods This uit cosists of the geerl itertive methods. The Jccobi s itertio methods d the Guss-Seidel itertio method. 6

12 MTH NUMERICAL ANALYSIS Uit 4 Eige-Vlues d Eige-Vectors. This uit focused o the Eige vlue problem. The power method d the iverse power method. Module Uit Review of Clculus Here, the three fudmetl theorems, Tylor's theorem, error (roud off d tructio errors) re discussed. Uit Itertio Methods for Loctig Root. This uit discussed: The iitil pproximtio to root (tbultio d grphicl methods). Bisectio method d fixed poit itertio method. Uit Chord Methods for Fidig Root This etils Repuler-Flsi method, Newto Rphso method d covergece criterio. Uit 4 Approximte Root of Polyomil Equtio. It c be sub-divided ito some results o roots of polyomil equtio. Birge-Viet method d Greffe s Root squrig method. Textbooks More recet editios of these books re recommeded for further redig. Egieerig Mthemtics P. D. S. Verm. Geerlized fuctios i mthemticl physics by V. S. Vidimirov. Mthemticl methods for sciece studets by G. Stepheso. Geerlized fuctios by R. F. Hoskis. Egieerig mthemtics by K. A. Strod. Egieerig Mthemtics by Kreyszic. 6

13 MTH MODULE Assessmet There re two compoets of ssessmet for this course. The Tutor Mrked Assigmet (TMAS) d the ed of the course exmitio. Tutor Mrked Assigmets (TMAs) The (TMAS) is the cotiuous ssessmet compoet of your course. It ccouts for % of the totl score. You will be give 4 (TMAS) to swer. Three of these must be swered before you re llowed to sit for the ed of course exmitio. The (TMAS) would be give to you by your fcilittor d retured fter you hve doe the ssigmet. Ed Of Course Exmitio This exmitio cocludes the ssessmet for the course. It costitutes 7% of the whole course. You will be iformed of the time for the exmitio. It my or my ot coicide with the uiversity semester exmitio. Summry I summry, we hve see how to desire the Lgrge s form of iterpoltig polyomil for give dt. It hs bee show tht the iterpoltig polyomil for give dt is uique. We hve derived the geerl error formul d its use hs bee illustrted to judge the ccurcy of our clcultios. For system of equtios Ax = b i ukow, where A is osigulr mtrix, the methods of fidig the solutio vector x my be brodly clssified ito two types. i) Direct methods d ii) Itertio methods. For lrger systems, direct methods becomes more efficiet if the coefficiet mtrix A is i oe of the forms D (digol), L (lower trigulr) or U (upper trigulr). We further discussed the followig methods for fidig pproximte roots of polyomil questios: (Birge-Viet d Greffe s root squrig methods). 6

14 MTH NUMERICAL ANALYSIS MODULE INTERPOLATION Uit Uit Uit Iterpoltio (Lgrge s Form) Newto s Form of the Iterpoltig Polyomil Iterpoltio t Eqully Spced Poits UNIT INTERPOLATION (LAGRANGE S FORM) CONTENTS. Itroductio. Objectives. Mi Cotet. Lgrge s Form. Iverse Iterpoltio. Geerl Error Term 4. Coclusio 5. Summry 6. Tutor Mrked Assigmet 7. Refereces/Further Redigs. INTRODUCTION Let f be rel-vlued fuctio defied o the itervl [, b] d we deote f(x k ) by f k. suppose tht the vlues of the fuctio f(x) re give to be f, f, f,, f whe x = x, x, x,, x respectively where x < x < x < x lyig i the itervl [, b]. The fuctio f(x) my ot be kow to us. The techique of determiig pproximte vlue of f(x) for o-tbulr vlue of x which lies i the itervl [, b] is clled iterpoltio. The process of determiig the vlue of f(x) for vlue of x lyig outside the itervl [, b] is clled extrpoltio. I this uit, we derive polyomil P(x) of degree which grees with the vlues of f(x) t the give ( + ) distict poits, clled odes or bscisss. I other words, we c fid polyomil P(x) such tht P(x j ) = f j, j =,,,,. such polyomil P(x) is clled the iterpoltig polyomil of f(x). I sectio. we prove the existece of iterpoltig polyomil by ctully costructig oe such polyomil hvig the desired property. The uiqueess is proved by ivokig the corollry of the fudmetl theorem of Algebr. I sectio. we derive geerl expressio for error i pproximtig the fuctio by the iterpoltig polyomil t poit d this llows us to clculte boud o the error over itervl. I provig this we mke use of the geerl Rolle s theorem. 64

15 MTH MODULE. OBJECTIVES After redig this uit, you should be ble to: fid the Lgrge s form of iterpoltig polyomil iterpoltig f(x) t + distict odl poits compute the pproximte vlue of f t o-tbulr poit compute the vlue of x (pproximtely) give umber y such tht f( x ) = ( y ) (iverse iterpoltio) compute the error committed i iterpoltio, if the fuctio is kow, t o-tbulr poit iterest fid upper boud i the mgitude of the error.. MAIN CONTENT. Lgrge s Form Let us recll the fudmetl theorem of lgebr d its useful corollries. Theorem If P(x) is polyomil of degree, tht is P(x) = x + - x x +,, rel or complex umbers d, the P(x) hs t lest oe zero, tht is, there exists rel or complex umber ξ such tht p(ξ )=. Lemm If z, z,, z k re distict zeros of the polyomil P(x), the for some polyomil R(x). Corollry P(x) = (x z ) (x z ) (x z k )R(x) If P k (x) d Q k (x) re the two polyomils of degree k which gree t the k + distict poits z, z, z,, z k the P k (x) = Q k (x) ideticlly. You hve come cross Rolle s Theorem i the perquisite course. But we eed geerlized versio of this theorem. (Geerl Error Term). This is stted below. 65

16 MTH NUMERICAL ANALYSIS Theorem (Geerlised Rolle s Theorem). Let f be rel-vlued fuctio defied o [, b] which is times differetible o ], b[. If f vishes t the + distict poits x,, x i [, b], the umber c i ], b[ exists such tht f () (c) =. We ow show the existece of iterpoltig polyomil d lso show tht it is uique. The form of the iterpoltig polyomil tht we re goig to discuss i this sectio is clled the Lgrge s form of the iterpoltig polyomil. We strt with relevt theorem. Theorem : Let x, x, x be + distict poits o the rel lie d let f(x) be rel-vlued fuctio defied o some itervl I = [, b] cotiig these poits. The, there exists exctly oe polyomil P (x) of degree, which iterpoltes f(x) t x, x, tht is, P (x j ) = f(x j ), i =,,,,. Proof: First we discuss the uiqueess of the iterpoltig polyomil, d the exhibit oe explicit costructio of iterpoltig polyomil (Lgrge s Form). Let P (x) d Q (x) be two distict iterpoltig polyomils of degree, which iterpolte f(x) t ( + ) distict poits x, x, x. Let h(x) = P (x) - Q (x). Note tht h(x) is lso polyomil of degree. Also h(x j ) = P (x j ) - Q (x j ) = f(x j ) - f(x j ) =, i =,,,,. Tht is, h(x) hs ( + ) distict zeros. But h(x) is of degree d from the Corollry to Lemm, we hve h(x) º. Tht is P (x) Q (x). This proves the uiqueess of the polyomil. Sice the dt is give t the poits (x, f ), (x, f ),, (x, f ) let the required polyomil be writte s P (x j ) = L (x)f + L (x)f + + L (x)f = i= L i (x)f i () Settig x = x j i (), we get P (x j ) = i= L i (x j )f i () 66

17 MTH MODULE Sice this polyomil fits the dt exctly, we must hve L j (x j ) = d or L j (x j ) =, i j L j (x j ) = ij () The polyomil L i (x) which re of degrees re clled the Lgrge fudmetl polyomils. It is esily verified tht these polyomil re give by L j (x) = (x - x )(x - x )...(x - x i- )(x - x i+ )...(x - x ) (x - x )(x - x )...(x - x )(x - x )...( x - x ) i i i i- i i+ i = i= i= j (x x j ) / i= i= j (x i x j ) (4) Substitutig of (4) i () gives the required Lgrge form of the iterpoltig polyomil. Remrk The Lgrge form (Eq. ()) of iterpoltig polyomil mkes it esy to show the existece of iterpoltig polyomil. But its evlutio t poit x i ivolves lot computtio. A more serious drwbck of the Lgrge form rises i prctice due to the followig: Oe clcultes lier polyomil P (x), qudrtic polyomil P (x) e.t.c., by icresig the umber of iterpoltio poits, util stisfctory pproximtio to f(x) hs bee foud. I such situtio Lgrge form does ot tke y dvtge of the vilbility of P k- (x) i clcultig P k (x). Lter o, we shll see how i this respect, Newto form, discussed i the ext uit, is more useful. Let us cosider some exmple to costruct this form of iterpoltio polyomils. Exmple 67

18 MTH NUMERICAL ANALYSIS If f() = -, f() = 9, f(4) = d f(6) =, fid the Lgrge s iterpoltio polyomil of f(x). Solutio We hve x =, x =, x = 4, x = 6 d f = -, f = 9, f =, f =. The Lgrge s iterpoltig polyomil P(x) is give by P(x) = L (x)f + L (x)f + L (x)f + L (x)f (5) where (x - x )(x - x )(x - x ) L (x) = (x - x )(x - x )(x - x ) (x - )(x - 4)(x - 6) = ( - )( - 4)( - 6) = (x x + 54x 7) L (x) = = (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) (x - )(x - 4)(x - 6) ( - )( - 4)( - 6) = 6 (x x + 4x 4) L (x) = = (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) (x - )(x - 4)(x - 6) (4 - )(4 - )(4-6) = 6 (x x + 7x 8) L (x) = (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) 68

19 MTH MODULE = (x - )(x - )(x - 4) (6 - )(6 - )(6-4) = (x 8x + 9x ) Substitutig L j (x) d f j j =,,, i Eq. (5), we get P(x) = - [x x + 54x 7] (-) + 6 [x x + 4x 4] (9) - 6 [x x + 7x 8] () + [ x 8x + 9x ] () = [ x x + 54x 7] + [x x + 4x 4] -5 [x x + 7x 8] + 5 [ x 8x = 9x ] which gives o simplifictio P(x) = x x = 5x 6 which is the Lgrge s iterpoltig polyomil of f(x). Exmple Usig Lgrge s iterpoltio formul, fid the vlue of f whe x =.4 from the followig tble. Solutio x f the Lgrge s iterpoltig formul with 4 poits is (x - x )(x - x )(x - x ) P(x)= f + (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) f + (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) (x - x )(x - x )(x - x ) f + (x - x )(x - x )(x - x ) f (6) (x - x )(x - x )(x - x ) Substitutig x =., x =.7, x =.8, x =. d f =., f = 5.479, f = 6.496, f =

20 MTH NUMERICAL ANALYSIS i (6), we get (x -.7)(x -.8)(x -.) P(x) = (. -.7)(. -.8)(. -.) *. + (x -.)(x -.8)(x -.) * (.7 -.)(.7 -.8)(.7 -.) (x -.)(x -.7)(x -.) (.8 -.)(.8 -.7)(.8 -.) (x -.)(x -.7)(x -.8) (. -.)(. -.7)(. -.8) * * 7.89 (7) Puttig x =.4 o both sides of (7), we get f (.4) = P (.4) = (.4 -.7)(.4 -.8)(.4 -.) (-.5)( -.6)( -.8) *. + (.4 -.)(.4 -.8)(.4 -.) (.5)(-.)(.) (.4 -.)(.4 -.7)(.4 -.) (.6)(.)(-.) (.4 -.)(.4 -.7)(.4 -.8) (.8)(.)(.) * * * 7.89 = (-.)( -.4)( -.6) (-.5)( -.6)( -.8) *. + (.)(-.4)( -.6) (.5)(-.)( -.) (.)(-.)( -.6) (.6)(.)(-.) (.)(-.)( -.4) (.8)(.)(.) * * * 7.89 =

21 MTH MODULE = Therefore f(x) = Iverse Iterpoltio I iverse iterpoltio for tble of vlues of x d y = f(x), oe is give umber y d wishes to fid the poit x so tht f( x ) = y, where f(x) is the tbulted fuctio. This problem c lwys be solved if f(x) is (cotiuous/d) strictly icresig or decresig (tht is, the iverse of f exists). This is doe by cosiderig the tble of vlues x i, f(x i ), i =,,, to be tble of vlues y i g(y i ), i =,,,, for the iverse fuctio g(y) = f - (y) = x by tkig y i = f(x i ), g(y i ) = x i, i =,,,,. The we c iterpolte for the ukow vlue g( y ) i this tble. P ( y ) = ( y - y j ) x i i= ( y - y ) i= i= j i j d x = P ( y ). This process is clled iverse iterpoltio. Let us cosider some exmples. Exmple From the followig tble, fid the Lgrge s iterpoltig polyomil which grees with the vlues of x t the give vlues of y. Hece fid the vlue of x whe y =. Solutio x 9 49 y 4 5 Let x = g(y). the Lgrge s iterpoltig polyomil P(y) of g(y) is give by P(y) = (y - )(y - 4)(y - 5) ( - )( - 4)( - 5) * + (y - )(y - 4)(y - 5) ( - )( - 4)( - 5) * 9 + (y - )(y - )(y - 5) (4 - )(4 - )(4-5) * 49 + (y - )(y - )(y - 4) * (5 - )(5 - )(5-4) 7

22 MTH NUMERICAL ANALYSIS = - 4 [y y + 47y 6] [y y + 9y ] - 49 [y 9y + y 5] + 8 [y 8y + 9y ] which, o simplifictio, gives P(y) = y y +. The Lgrge s iterpoltig polyomil of x is give by P(y). There fore, x = P(y) = y y + Therefore, whe y =, x = P() = 5. Exmple 4 Fid the vlue of x whe y = from the followig tble of vlues. Solutio x 4 7 y - 4 The Lgrge s iterpoltio polyomil of x is give by P(y) = (y - )(y - )(y - 4) (- )( - )( - 5) (4) + (y + )(y - )(y - 4) ()( - ) (7) + (y + )(y - )(y - 4) ()()( - ) () + (y + )(y - )(y - ) (5) () () () Therefore P() = () () (- ) - () () (5) (4) + (4) () (- ) () () (7) + (4) () (- ) - () () () + (4) () () (5)() () () = = 8 5 =. 7

23 MTH MODULE Hece, x() = P() =.. Now we re goig to fid the error committed i pproximtig the vlue of the fuctio by P (x).. Geerl Error Term Let E (x) = f(x) P (x) be the error ivolved i pproximtig the fuctio f(x) by iterpoltig polyomil. We derive expressio for E (x) i the followig theorem. This result helps us i estimtig useful boud o the error s explied i exmple. Theorem 4 Let x, x,, x be distict umbers i the itervl [, b] d f hs (cotiuous) derivtives upto order ( + ) i the ope itervl ], b[. if P (x) is the iterpoltig polyomil of degree, which iterpoltes f(x) t the poits x,, x, the for ech x [, b], umber ξ (x) i ], b[ exists such tht E (x) = f(x) P (x) = Proof f ( + ) ( ξ ( x) ) ( x x )( x x )... ( x x ) ( + )! (8) If x x k for y k =,,,,, defie the fuctio g for t i [, b] by g(t) = f(t) P (t) [f(x) P (x)] ( t x j ) ( x x ) j= sice f(t) hs cotiuous derivtives up to order ( + ) d P(t) h derivtives of ll orders, g(t) hs cotiuous derivtives up to ( + ) order. Now, for k =,,,,, we hve g(x k ) = f(x k ) = P (x k ) [f(x) - P (x)] j. ( xk x j ) ( x x ) j= = [f(x) - P (x)]. = Furthermore, g(x) = f(x) - P (x) - [f(x) - P (x)] j. ( x x j ) ( x x ) j= = f(x) - P (x) - [f(x) - P (x)]. = j. 7

24 MTH NUMERICAL ANALYSIS Thus g hs cotiuous derivtives up to order ( + ) d g vishes t the ( + ) distict poits x, x,, x. By the geerlized Rolle s Theorem (Theorem ) there exists ξ (x) i ], b[ for which g (+) ξ (x) =. Differetitig g(t), ( + ) times (with respect to t) d evlutig t ξ (x) i, we get = g (+) ξ (x) = f (+) ξ (x) ( + )! Simplifyig we get (error t x = x ) [ f ( x) P ( x)] ( x xi ) i= E ( x ) = f( x ) P ( x ) = ( x x ) i= i f ( + ) ς ( x) ( + )! (9) The error formul (Eq. (9)) derived bove, is importt theoreticl results becuse Lgrge iterpoltig polyomils re extesively used i derivig importt formule for umericl differetitio d umericl itegrtio. It is to be oted tht ξ( x) ξ = depeds o the ;poit x t which the error estimte is required. This depedece eed ot eve be cotiuous. This error formul is of limited utility sice f (+) (x) is ot kow (whe we re give set of dt t specific odes) d the poit x is hrdly kow. But the formul c be used to obti boud o the error of iterpoltig polyomil. Let us see how, by exmple. Exmple 5 The followig tble gives the vlues of f(x) = e x. If we fit iterpoltig polyomil of degree four to the dt, fid the mgitude of the mximum possible error i the computed vlue of f(x) whe x =.5. Solutio x y From Eq. (9), the mgitude of the error ssocited with the 4th degree polyomil pproximtio is give by E 4 (x) = ( x x )( x x )( x x )( x x )( x x ) 4 f ( 5 ) ( ξ ) 5! 74

25 MTH MODULE = ( x.)( x.)( x.4)( x.5)( x.6) ( 5) ( ξ ) f () 5! Sice f(x) = c x, f (5) (x) = e x. Whe x lies i the itervl [.,.6], Mx f (5) (x) = e.6 = 4.95 () Substitutig () i (), d puttig x =.5, the upper boud o the mgitude of the error = (.5 (-.5) (-.5) (-.5) (-.5) * 4.95 = CONCLUSION Let us tke brief look t wht you hve studied i this uit s the cocludig pth of this uit to the summry. 5. SUMMARY I this uit, we hve see how to derive the Lgrge s form of iterpoltig polyomil for give dt. It hs bee show tht he iterpoltig polyomil for give dt is uique. Moreover the Lgrge form of iterpoltig polyomil c be determied for eqully spced or ueqully spced odes. We hve lso see how the Lgrge s iterpoltio formul c be pplied with y s the idepedet vrible d x s the depedet vrible so tht the vlue of x correspodig to give vlue of y c be clculted pproximtely whe some coditios re stisfied. Filly, we hve derived the geerl error formul d its use hs bee illustrted to judge the ccurcy of our clcultio. The mthemticl formule derived i this uit re listed below for your esy referece. ) Lgrge s Form where P (x) = f ( x ) L ( x) i= i L i (x) = ( ) x x j / ( xi x j ) j= j i i j= j i 75

26 MTH NUMERICAL ANALYSIS ) Iverse Iterpoltio P (y) = y y j xi i= j= yi y j j i ) Iterpoltio Error ( ) ( ) E ( x ) = f( x ) P ( x ) = ( x x ) i= i f ( + ) ς ( x) ( + )! 6. TUTOR-MARKED ASSIGNMENT ) Show tht i) i= L i (x) = ii) i= L i (x) k x i = x k, k where L i (x) re Lgrge fudmetl polyomils ) Let w(x) = ( x ) k = x k. Show tht the iterpoltig polyomil of degree with the odes x, x,, x c be writte s P (x) = w(x) i= f(x k) (x - x )w '(x ) k k ) Fid the Lgrge s iterpoltio polyomil of f(x) from the followig dt. Hece obti f(). x 4 5 f(x) ) Fid the vlue of y whe x = 6 from the followig tble: x 7 8 y

27 MTH MODULE 5) Usig the Lgrge s iterpoltio formul, fid the vlue of y whe x =. x y 4 6 6) For the dt of Exmple 5 with lst oe omitted, i.e., cosiderig oly first four odes, if we fit polyomil of degree, fid estimte of the mgitude of the error i the computed vlue of f(x) whe x =.5. Also fid upper boud i the mgitude of the error. 7) Fid the vlue of x whe y = 4 from the tble give below: x y - 5 8) Usig Lgrge s iterpoltio formul, fid the vlue of f(4) from the followig dt: x y REFERENCES/FURTHER READINGS Egieerig Mthemtics P.D.S. Verm. Geerlized Fuctios i Mthemticl Physics by V.S. Vidimirov. Fudmetls of the Fiite Elemet Method. Hrtley Grdi, Fr. 77

28 MTH NUMERICAL ANALYSIS UNIT NEWTON FORM OF THE INTERPOLATING POLYNOMIAL CONTENTS. Itroductio.. Objectives.. Mi Cotet.. Divided Differeces.. Newto s Geerl Form of Iterpoltig Polyomil.. The Error of the Iterpoltig Polyomil..4 Divided Differece d Derivtive of the Fuctio..5 Further Results o Iterpoltio Error. 4. Coclusio. 5. Summry. 6. Tutor Mrked Assigmet. 7. Refereces/Further Redigs.. INTRODUCTION The Lgrge s form of the iterpoltig polyomil derived i Uit hs some drwbcks compred to Newto form of iterpoltig polyomil tht we re goig to cosider ow. I prctice, oe is ofte ot sure s to how my iterpoltio poits to use. Oe ofte clcultes P (x), P (x), icresig the umber of iterpoltio poits, d hece the degrees of the iterpoltig polyomils till oe gets stisfctory pproximtio P k (x), o dvtge is tke of the fct tht oe hs lredy costructed P k- (x), wheres i Newto form it is ot so. Before derivig Newto s geerl form of iterpoltig polyomil, we itroduce the cocept of divided differece d the tbulr represettio of divided differeces. Also the error of the iterpoltig polyomil i this cse is derived i terms of divided differeces. Usig the two differet expressios for the error term we get reltioship betwee th order divided differece d th order derivtive.. OBJECTIVES After studyig this uit, you should be ble to: obti divided differece i terms of fuctio vlues form tble of divided differeces d fid divided differeces with give set of rgumets from the tble 78

29 MTH MODULE show tht divided differece is idepedet of the order of its rgumets obti the Newto s divided differeces iterpoltig polyomil for give dt fid estimte of f(x) for give o-tbulr vlue of x from tble of vlues of x d y [f(x)] relte the k th order derivtive of f(x) with the k th order divided differece from the expressio for the error term.. MAIN CONTENTS. Divided Differeces Suppose tht we hve determied polyomil P k- (x) of degree k which iterpoltes f(x) t the poits x, x, x k-. I order to mke use of P k- (x) i clcultig P k (x) we cosider the followig problem: Wht fuctio g(x) should be dded to P k- (x) to get P k (x)? Let g(x) = P k (x). - P k- (x). Now, g(x is polyomil of degree k d g(x i ) = P k (x i ) - P k- (x i ) = f(x i ) - f(x i ) = for i =,,, k. Suppose tht P (x) is the Lgrge polyomil of degree t most tht grees with the fuctio f t the distict umbers x, x, x. P (x) c hve the followig represettio, clled Newto form. P (x) = A + A (x x ) + A (x x ) (x x ) + + A (x x ) (x x - ) () for pproprite costt A, A,, A. Evlutig P (x) (Eq. ()) t x we get A = P (x ). Similrly whe f(x) - f(x ) P (x) is evluted t x, we get A =. Let us itroduce the x - x ottio for divided differeces d defie it t this stge: The zeroeth divided differece of the fuctio f, with respect to x i, is deoted by f[x i ] d is simply the evlutio of f t x i, tht is, f[x i ] = f(x i ). the first divided differece of f with respect to x i d x i+ is deoted by f[x i, x i+ ] d defied s f[x i+ ] - f[x i ] f[x i, x i+ ] = x - x i+ i The remiig divided differeces of higher orders re defied iductively s follows. The kth divided differeces reltive to x i, x i+,, x i+k is defied s 79

30 MTH NUMERICAL ANALYSIS f[x i+..., x i+ k ]- f[x i,..., x i+ k- ] f[x i, x i+,, x i+k ] =. x - x i+ k i where the (k )st divided differeces f[x i,.., x i+k ] hve bee determied. This shows tht kth divided differece is the divided differeces of (k )st divided differeces justifyig the me. The divided differece f[x i, x,., x k ] is ivrit uder ll permuttios of the rgumets x i, x,., x k. To show this we proceed givig other expressio for the divided differece. For y iteger k betwee d. let Q k (x) be the sum of the first k + terms i form (), i.e. Q k (x) = A + A (x x ) + + A k (x x ) ( x x k- ).. Sice ech of the remiig terms i Eq. () hs the fctor (x x ) (x x ) (x x k ), Eq. () c be rewritte s P (x) = Q k (x) + (x x ) (x x k ) R(x) for some polyomil R(x). s the term (x x ) (x x ) (x x k ) R(x) vishes t ech of the poits x, x k, we hve f(x i ) = P (x i ) = Q k (x i ), i =,,,, k. Sice Q k (x) is polyomil of degree k, by uiqueess of iterpoltig polyomil Q k (x) = P k (x). This shows tht P (x) c be costructed step by step with the dditio of the ext term i Eq. (), s oe costruct the sequece P (x), P (x) with P k (x) obtied from P k- (x) i the form P k (x) = P k- (x) + A k (x x ) (x x k- ) () Tht is, g(x) is polyomil of degree k hvig (t lest) the k distict zeros x,, x k-. \ P k (x) - P k- (x) = g(x) = A k (x x ) (x x k- ), for some costt A k. this costt A k is clled the kth divided differece of f(x) t the poits x,, x k for resos discussed below d is deoted by f[x, x,, x k ]. this coefficiet depeds oly o the vlues of f(x) t the poit x,, x k. thus Eq. () c be writte s P k (x) =P k- (x) + f[x,, x k ] (x x ) (x x k- ), sice (x x ) (x x ) (x x k- ) = x k + polyomil of degree < k, we c rewrite P k (x) s P k (x) = f[x,, x k ] x k + polyomil of degree < k (4) (s P k- (x) is polyomil of degree < k). 8

31 MTH MODULE But cosiderig the Lgrge form of iterpoltig polyomil we hve P k (x) = f ( x ) k k ( x x j ) ( x x ) i i= j= i j i j = k i= ( x ) k C( xi x j ) j= i j f i k x k + polyomil of degree < k. Therefore, o compriso with Eq. (4) we hve f[x,, x k ] = k f ( x ) i ( x x )... ( x x )... ( x x ).. ( x x ) i= i i i+ i k. (5) This shows tht f[y,, y k ] = f[x,, x k ] if y,, y k is reorderig of the sequece x,, x k. We hve defied the zeroeth divided differece of f(x) t x by f[x ] = f(x ) which is cosistet with Eq. (5). For k =, we hve from Eq. (5) f(x ) f[x, x k ] = x - x f(x ) x - x + f(x ) - f(x ) x - x + f[x ]- f[x ] x - x = This shows tht the first divided differece is relly divided differece of divided differeces. We show below i Theorem tht for k > f[x,..., x k ]- f[x..., x k- ] f[x,, x k ] = x - x k (6) This shows tht the k th divided differece is the divided differece of (k )st divided differeces justifyig the me. If M = (x,, x ) d N deotes y elemets of M d the remiig two elemets re deoted by d b, the (f [x,.., x = 8

32 MTH NUMERICAL ANALYSIS [( - st divided differece o N d - ( - )st divided differece o N d b] - b (7) Theorem : f [x,.., x j ] = f[x,..., x j] - f[x, x..., x j- ] x - x j (8) Proof: Let P i- (x) be the polyomil of degree i which iterpoltes f(x) t x,, x i- d let Q j- (x) be the polyomil of degree j which iterpoltes f(x) t the poits x,, x j. Let us defie P(x) s P(x) = x - x - x x j Q j- (x) + x x j - - x x j P j- (x). This is polyomil of degree j, d P(x i ) = f(x i ) for i =,,, j. By uiqueess of the iterpoltig polyomil we hve P(x) = P j (x). Therefore P j (x) x - x - x x j Q j- (x) + x x j - - x x j P j- (x). Equtig the coefficiet of x j from both sides of Eq. (8), we obti (ledig) coefficiet of x j i P j (x) = - ledig coefficiet of Q x j - x ledig coefficiet of P x - x j j- j- (x) (x) Tht is f [x,..., x j ] = f[x,..., x j] - f[x,..., x j- ]. x - x j We ow illustrte this theorem with the help of few exmples but before tht we give the tble of divided differeces of vrious orders. Tble of divided differeces Suppose we deote, for coveiece, first order divided differece of f(x) with y two rgumets by f[.,.], secod order divided differece with y three rgumets by f[.,.,.] d so o. The the tble of divided differece c be writte s follows 8

33 MTH MODULE Tble x f[.] f[.,.] f[.,.,.] f[.,.,.,.] f[.,.,.,.,.] x f f[x,x ] x f f[x,x x ] f[x,x ] f[x,x x x ] x f f[x,x x ] f[x,x x x x 4 ] f[x,x ] f[x x x x 4 ] x f f[x x x 4 ] f[x,x 4 ] x 4 f 4 Exmple : If f(x) = x, fid the vlue of f[, b, c]. Solutio: f[, b] = f(b) - f() b - = b - b - = b + b + = + b + b Similrly, f[, b] = c + cb + b = b + bc + c f[, b, c] = f[b, c] - f[, b] c - = = = (b + bc + c ) - ( + b + b ) c - (c - ) + b(c - ) c - (c - )(c + + b) (c - ) = + b + c f[, b, c] = + b + c. Exmple : If f(x) =, show tht x f[, b, c, d] = - bcd 8

34 MTH NUMERICAL ANALYSIS Solutio: f[, b] = - b b - = - b b(b - ) = - b Similrly, f[b, c] = -, f[c, d] = - bc cd f[, b, c] = = + - bc b = b bc c - c - c bc = c bc Similrly, f[b, c, d] = bcd however f[, b, c, d] = = c bc = c d bcd d bc = - bcd Cosequetly, f[, b, c, d] = - bcd I ext sectio we shll mke use of the divided differece to derive Newto s geerl form of iterpoltig polyomil. 84

35 MTH MODULE. Newto s Geerl Form of Iterpoltig Polyomil I sectio. we hve show how P (x) c be costructed step by step s oe costruct the sequece P (x), P (x), P (x),..., with P k (x) obtied from P k- (x) with the dditio of the ext term i Eq. (), tht is, P k (x) = P k- (x) + (x x ) (x x )...(x x k- ) f[x,..., x k ] Usig this Eq. () c be rewritte s P (x) = f[x ] + (x x ) f[x,x ] + (x x ) (x x ) f[x,x,x ] (x x ) (x x )... (x x - ) f[x,x,...,x ]. (9) This c be writte compctly s follows: j P (x) = f x,..., xi ] ( x x j ) i= [ () j= This is the Newto s form of iterpoltig polyomil. Exmple : From the followig tble of vlues, fid the Newto s form of iterpoltig polyomil pproximtig (x). Solutio: x f(x) We otice tht the vlues of x re ot eqully spced. We re required to fid polyomil which pproximtes f(x). We form the tble of divided differeces of f(x). Tble x f[.] f[.,.] f[.,.,.] f[.,.,.,.] f[.,.,.,.,.]

36 MTH NUMERICAL ANALYSIS Sice the divided differece up to order 4 re vilble, the Newto s iterpoltig polyomil P 4 (x) is give by P 4 (x) = f(x ) + (x x ) f[x,x ] + (x x ) (x x ) f[x,x,x ] + (x x ) (x x ) (x x ) f[x,x,x,x ] + (x x ) (x x ) (x x ) (x x ) f[x,x,x,x,x 4 ] () where x = -, x =, x =, x = 6 d x 4 = 7. The divided differeces f(x ), f[x,x ], f[x,x,x ], f[x,x,x,x ] d f[x,x,x,x,x 4 ] re those which lie log the digol t f(x ) s show by the dotted lie. Substitutig the vlues of x i d the vlues of the divided differeces i Eq. (), we get P 4 (x) = + (x + ) (-9) + (x + ) x (6) + (x + ) x (x ) (5) + (x + ) x (x ) (x 6) () which o simplifictio gives P 4 (x) = x 4 x + 5x 6 Therefore, f(x) =P 4 (x) = x 4 - x + 5x 6 We ow cosider exmple to show how Newto s iterpoltig polyomil c be used to obti the pproximte vlue of the fuctio f(x) t y o-tbulr poit. Exmple 4: Fid the pproximte vlues of f(x) t x = d x = 5 i Exmple. Solutio: Sice f(x) = P 4 (x), from Exmple, we get f() = P 4 () = = 6 d f(5) = P(5) = = 69 Note : Whe the vlues of f(x) for give vlues of x re required to be foud, it is ot ecessry to fid the iterpoltig polyomil P 4 (x) i its simplified form give bove. We c obti the required vlues by substitutig the vlues of x i Eq. () itself. Thus, P 4 () = + () (-9) + () () (6) + () () (-) (5) + () () (-) (-4) 86

37 MTH MODULE Therefore, P 4 () = = 6. Similrly, P 4 (5) = + (6) (-9) + (6) (5) (6) + (6) (5) () (5) + (6) (5) () (-) () = = 69. The f() = P 4 () = 6 Ad f(5) = P(5) = 69. Exmple 5: Obti the divided differeces iterpoltio polyomil d the Lgrge s iterpoltig polyomil of f(x) from the followig dt d show tht they re sme. Solutio: x 4 f(x) () Divided differeces iterpoltio polyomil: Tble x f[x] f[.,.] f[.,.,.] f[.,.,.,.] P(x) = -4 + x(5) + x(x ) (5) + x(x ) (x ) () = x + x 4 \ P(x) = x + x 4 b) Lgrge s iterpoltio polyomil: P(x) = (x - )(x - )(x - 4) x(x - )(x - 4) (- 4) + (- )( - )( - 4) ()(- )( - ) (6) 87

38 MTH NUMERICAL ANALYSIS + x(x - )(x - 4) ()()( - ) (6) + x(x - )(x - ) (4)()() (64) = 6 (x 9x + 6x 4) + (x 7x + x) - 6 (x 6x + 8x) + 8(x 5x + 6x). O simplifyig, we get P(x) = x + x 4. Thus, we fid tht both polyomils re sme. I Uit we hve derived the geerl error term i.e. error committed i pproximtig f(x) by P (x). I ext sectio we derive other expressio for the error term i term of divided differece.. The Error of the Iterpoltig Polyomil Let P (x) be the Newto form of iterpoltig polyomil of degree which iterpoltes f(x) t x..., x. The iterpoltig error E (x) of P (x) is give by E (x) = f(x) P (x) () Let x be y poit differet from x,..., x. If P (x) is the Newto form of iterpoltig polyomil which iterpoltes f(x) t x,..., x d x, the P + ( x ) = f( x ). The by () we hve P + (x) = P (x) + f[x,..., x, x ] ( x x j ) j= Puttig x = x i the bove, we hve f( x ) = P + ( x ) = P ( x ) + f[x,..., x, x ] ( x x j ) j= i.e. E ( x ) = f( x ) - P ( x ) = f[x,..., x, x ] ( x x j ) j= () This shows tht the error is like the ext term i the Newto form. 88

39 MTH MODULE.4 Divided Differece d Derivtive of the Fuctio Comprig Eq. () with the error formul derived i Uit Eq. (9), we c estblish reltioship betwee divided differece d the derivtives of the fuctio E ( x ) = ( + ) f [ x(x)] ( + )! ( x x j ) j= = f[x, x,..., x, x ] ( x x j ) j= Comprig, we hve f[x, x,..., x + ] = (cosiderig x = x + ) ( + ) f ς ( + )! Further it c be show tht x Î ]mi x i, mx x i [. We stte these results i the followig theorem. Theorem : Let f(x) be rel-vlued fuctio, defied o [, b] d times differetible i ], b[. If x,..., x re + distict poits i [, b], the there exists ς ], b[ such tht f[x,..., x ] = Corollry : ( + ) f ς! If f(x) = x, the f[x,..., x ] =!! =. Corollry : If f(x) = x k, k <, the f[x,..., x k ] = sice th derivtive of x k, k <, is zero. 89

40 MTH NUMERICAL ANALYSIS For exmple, cosider the first divided differece f(x ) - f(x ) f[x,x ] = x - x by Me Vlue Theorem f(x ) = f(x ) + (x x ) f (ς ), x < ς < x, substitutig, we get f[x,x ] = f (ς ), x < ς < x. Exmple 6: If f(x) = x + - x , the fid f[x, x,..., x ] = *!! + =. Let us cosider other exmple. Exmple 7: If f(x) = x + x x +, fid f[, -,, ], f[, b, c, d], f[4, 6, 7, 8]. Solutio: Sice f(x) is cubic polyomil, the rd order divided differeces of f(x) with y set of rgumet re costt d equl to, the coefficiet of x i f(x). Thus, it follows tht f[, -,, ], f[, b, c, d], d f[4, 6, 7, 8] re ech equl to. I the ext sectio, we re goig to discuss bout bouds o the iterpoltio error..5 Further Results o Iterpoltio Error We hve derived error formul E (x) = f(x) P (x) = ( x x ) i= i f ( + ) ς ( x) ( + )!. 9

41 MTH MODULE We ssume tht f(x) is ( + ) times cotiuously differetible i the itervl of iterest [, b] = I tht cotis x,..., x d x. sice ς ( x) is kow we my replce f (+) mx ( + ) ( ς ( x) ) by x Î I f (x). If we deote (x - x ) (x x )...(x x ) by (x) the we hve E (x) = f(x) P (x) mx f ( + ) (t) x Î I mx ( + )! x Î I y (t) (4) Cosider ow the cse whe the odes re eqully spced, tht is (m x j = x + jh), j =,...,N, d h is the spcig betwee cosecutive odes. For the cse = we hve lier iterpoltio. If x Î [x i-, x i ], the we pproximte f(x) by P (x) which iterpoltes t x i-, d x i. From Eq. (4) we hve E (x) where (x) = (x x i- ) (x - x i ). Now, dy = x x - = dx gives x = (x i- - x i )/. mx f "(t) t Î I mx y (t) t Î I Hece, the mximum vlue of (x x i- ) (x - x i ) occurs t x = x * = (x i- - x i )/. The mximum vlue is give by (xi - x i- ) h (x*) = = 4 4. Thus, we hve for lier iterpoltio, for y x I (xi - x i- ) E (x) = f(x) P (x) 4 mx f "(x) x Î I h = M. (5) 8 where f (x) M o I. For the cse =, it c be show tht for y x [x i-, x i+ ]. 9

42 MTH NUMERICAL ANALYSIS E (x) h M 9 where f (x) M o I. (6) 9

43 MTH MODULE Exmple 8: Determie the spcig h i tble of eqully spced vlues of the fuctio of f(x) = x betwee d, so tht iterpoltio with first degree polyomil i this tble will yield seve plce ccurcy. Solutio: Here f"(x) = - 4 x-/ mx f "(x) x d E (x) = 4. h. For seve plce ccurcy, h is to be chose such tht h < or h < (6) -8 tht is h <.. 4. CONCLUSION This uit shll be cocluded by givig summry of wht we hve covered i it. 5. SUMMARY I this uit we hve derived form of iterpoltig polyomil clled Newto s geerl form, which hs some dvtge over the Lgrge s form discussed i Uit. This form is useful i derivig some other iterpoltig formuls. We hve itroduced the cocept of divided differeces d discussed some of its importt properties before derivig Newto s geerl form. The error term hs lso bee derived d utilizig the error term we hve estblished reltioship betwee the divided differece d the derivtive of the fuctio f(x) for which the iterpoltig polyomil hs bee obtied. The mi formul derived re listed below: ) f[x,...,x j ] = f[x,..., x j] - f[x,..., x j- ] x - x j 9

44 MTH NUMERICAL ANALYSIS j ) P (x) = f [ x,..., xi ] ( x x j ) i= j= ) E (x) = f [ x,..., x, x] ( x x j ) j= 4) f[x,...,x ] = f ( ) ( ) ξ, ξ ]mi xi, mx i [! 6. TUTOR-MARKED ASSIGNMENT ) Fid the Lgrge s iterpoltig polyomil of f(x) from the tble of vlues give below d show tht it is the sme s the Newto s divided differeces iterpoltig polyomil. x 4 5 f(x) 8 68 ) Form the tble of vlues give below, obti the vlue of y whe x =.5 usig ) divided differeces iterpoltio formul. b) Lgrge s iterpoltio formul. x 4 5 f(x) ) Usig Newto s divided differece iterpoltio formul, fid the vlues of f(8) d f(5) from the followig tble. x f(x) ) If f(x) = x x + 7x +, wht is the vlue of f[,,, 4]? 5) If f(x) = x x + 5, fid f[, ], f[, ] d f[,, ]. 6) If f(x) tkes the vlues -, 5, d respectively whe x ssumes the vlues -,, d, fid the polyomil which pproximtes f(x). 7) Fid the polyomil which pproximte f(x), tbulted below 94

45 MTH MODULE x -4-5 f(x) Also fid pproximte vlue of f(x) t x = d x = -. 8) From the followig tble, fid the vlue of y whe x = x y REFERENCES/FURTHER READINGS. Egieerig Mthemtics P.D.S. Verm. Geerlized Fuctios i Mthemticl Physics by V.S. Vidimirov. Fudmetls of the Fiite Elemet Method. Hrtley Grdi, Fr. 95

46 MTH NUMERICAL ANALYSIS UNIT INTERPOLATION AT EQUALLY SPACED POINTS CONTENTS. Itroductio. Objectives. Mi Cotet. Differeces.. Forwrd Differeces.. Bckwrd Differeces.. Cetrl Differeces. Differece Formuls.. Newto s Forwrd-Differece Formul.. Newto s Bckwrd-Formul 4. Coclusio 5. Summry 6. Tutor Mrked Assigmet 7. Refereces/Further Redigs. INTRODUCTION Suppose tht y is fuctio of x. The exct fuctiol reltio y = f(x) betwee x d y my or my ot be kow. But, the vlues of y t ( + ) eqully spced of x re supposed to be kow, i.e., (x i, y i ); i =,..., re kow where x i x i- = h (fixed), i =,,...,. Suppose tht we re required to determie pproximte vlue of f(x) or its derivtive f (x) for some vlues of x i the itervl of iterest. The methods for solvig such problems re bsed o the cocept of fiite differeces. We hve itroduced the cocept of forwrd, bckwrd d cetrl differeces d discussed their iterreltioship i the previous uit We hve lredy itroduced two importt forms of the iterpoltig polyomil i Uits d. These forms simply whe the odes re equidistt. For the cse of equidistt odes, we hve derived the Newto s forwrd, bckwrd differece forms d Stirlig s cetrl differece form of iterpoltig, ech suitble for use uder specific situtio. We hve derived these methods i the previous uit d lso give the correspodig error term. 96

47 MTH MODULE. OBJECTIVES After redig this uit, you should be ble to: write forwrd differece i terms of fuctio vlues from tble of forwrd differeces d locte differece of give order t give poit write bckwrd differece i terms of fuctio vlues from tble of bckwrd differeces d idetify differeces of vrious orders t y give poit from the tble expd cetrl differece i terms of fuctio vlues d form tble of cetrl differeces estblish reltios betwee V,, d d divided differece obti the iterpoltig polyomil of f(x) for give dt by pplyig y oe of the iterpoltig formuls compute f(x) pproximtely whe x lies er the begiig of the tble d estimte the error compute f(x) pproximtely whe x lies er the ed of the tble d estimte the error estimte the vlue of f(x) whe x lies er the middle of the tble d estimte the error.. MAIN CONTENTS. Differeces Suppose tht we re give tble of vlues (x i, y i ), i =,,,..., N where y i = f(x i ) = f j. Let the odl poits be equidistt. Tht is x i = + ih, i =,..., N, with N = (b )/h () For simplicity we itroduce lier chge of vribles s = s(x) = x - x h, so tht x = x(s) = x + sh () d itroduce the ottio f(x) = f(x + sh) = f s () The lier chge of vribles i Eq. () trsforms polyomils of degree i x ito polyomils of degree is s. we hve lredy itroduced the divided-differece tble to clculte polyomil of 97

48 MTH NUMERICAL ANALYSIS degree which iterpoltes f(x) t x, x,..., x. For eqully spced odes, we shll del with three types of differeces, mely, forwrd, bckwrd d cetrl d discuss their represettio i the form of tble. We shll lso derive the reltioship of these differeces with divided differeces d their iterreltioship... Forwrd Differeces We deote the forwrd differeces of f(x) if ith order t x = x + sh by i f s d defie it s follows: i f s = { s f i = V( V f ) = V f - V f, i >. i- i- i- s s+ s Where V deotes forwrd differece opertor. Whe s = k, tht is, x = x k, we hve for i = for i = f k = f k+ - f k f k = f k+ - f k = f k+ - f k+ [f k+ - f k ] = f k+ - f k+ + f k Similrly f k = f k+ - f k+ + f k+ - f k We recll the biomil theorem ( + b) s s = j= j j j b r (4) where s is rel o-egtive iteger. We give below i Lemm the reltioship betwee the forwrd d divided differeces. This reltio will be utilized to derive the Newto s forwrd differece formul which iterpoltes f(x) t x k + ih, i =,,...,. Lemm : For ll i f[x k,..., k k+ ] = i i!h i f k (5) 98

49 MTH MODULE Proof: We prove the result by iductio. For i =, both sides of reltio (5) re sme by covetio, tht is, f[x k ] = f(x k ) = f k = f k. Assumig tht reltio (5) holds for i =, we hve for i = + f[x k +,..., x k + + ] - f[x k,..., x k + ] f[x k, x k+,..., k k++ ] = x - x k + + k [ ] [ ] f k + /! h f k /! h = = x f + k+ ( k + + ) h x kh f k = + + ( + )! h ( + )! h This shows tht reltio (5) holds for i = + lso. Hece (5) is proved. We ow give result which immeditely follows from this theorem i the followig corollry. Corollry: If P (x) is polyomil of degree with ledig coefficiet, d x is rbitrry poit, the P (x ) =! h + d + P (x ) =, i.e., ll higher differeces re zero. Proof: Tkig k = i reltio (5) we hve f k f[x,..., x i ] = i!h i f. (6) Let us recll tht f[x,..., x i ] = f ( ) (! ξ ) (7) where f(x) is rel-vlued fuctio defied o [, b] d i times differetible i ], b[ d ξ ], b[. Tkig i = d f(x) = P (x) i Eqs. (6) d (7), we get 99

50 MTH NUMERICAL ANALYSIS () i P (x ) =!h P [x,..., x ] =!h P ( x)! = h!. Sice i + P (x ) = i P (x ) - i P (x ) This completes the proof = h! - h! =. The shift opertor E is defied s Ef i = f i+ (8) I geerl Ef(x) = f(x + h). We hve E s f i = f i+s For exmple, E f i = f i+, E / f i = f i+/ d E -/ f i = f i-/ Now, i f i = f i+ - Ef i f i = (E )f i Hece the shift d forwrd differece opertios re relted by or = E E = + Opertig s times, we get s = (e ) s j = E ( ) j= s j Mkig use of reltio (8) i Eq. (9), we get s f i = ( ) j j= s j r f + r (9) We ow give i Tble, the forwrd differeces of vrious orders usig 5 vlues.

51 MTH MODULE Tble : Forwrd Differece Tble x f(x) f f f 4 f x f f x f f f f x f f 4 f f f x f f x 4 f 4 f Note tht the forwrd differece k f lie o stright lie slopig dowwrd to the right... Bckwrd Differeces Let f be rel-vlued fuctio of x. let the vlues of f(x) t + eqully spced poits x, x,..., x be f, f,..., f respectively. The bckwrd differeces of f(x) of ith order t x k = x + kh re deoted by i f k. They re defied s follows: i f k, = f k = { i - i - ( f ) k = [fk - f k- ], i ³ () where deotes bckwrd differece opertor. Usig (), we hve for i = ; f k = f k f k- i = ; f k = ( f k f k- ) = f k f k- = f k f k- + f k- i = ; f k = [f k f k- ] = f k - f k- = [f k ] - [f k- ] = [f k - f k- ] - [f k- - f k- ] = f k f k- - f k- + f k- = f k f k- [f k- + f k- ] + f k- - f k- = f k f k- + f k- - f k-

52 MTH NUMERICAL ANALYSIS By iductio we c prove the followig lemm which coects the divided differece with the bckwrd differece. Lemm : The followig reltio holds f[x -k,..., x ] = k k!h k f(x ) () The reltio betwee the bckwrd differece opertor d the shift opertor E is give by = E - or E = (i - ) - Sice f k = f k f k- = f k E - f k = [ E]f k. Opertig s times, we get s f k = [ E] s m m f k = E ( ) f k s m s m f km () m j= = ( ) j= We c exted the biomil coefficiet ottio to iclude egtive umbers, by lettig s = i s ( s )( s )...( s i + ) = i! (-) i s(s + )...(s + i - ) i! The bckwrd differeces of vrious orders with 5 odes re give i Tble. Tble : Bckwrd Differece Tble x f(x) f f f 4 f x f f x f f f f x f f 4 f 4 f f 4 x f f 4 f 4 x 4 f 4 Let us cosider the followig exmple:

53 MTH MODULE Exmple : Evlute the differeces () [ x + x + ] (b) [ x + x + x + ]. Solutio: () [ x + x + ] = (b) [ x + x + x + ]. = (x ) + [ x + x + ] =.! h Note tht the bckwrd differeces k f 4 lie o stright lie slopig upwrd to the right. Also ote tht V f k = f k+ = f k+ f k. Try to show tht V 4 f = 4 f 4. Let us ow discuss bout the cetrl differeces... Cetrl Differeces The first order cetrl differece of f t x k, deoted by df k, is defied s df = f(x + h/) f(x h/) = f k+/ f k-/. Opertig with d, we obti the higher order cetrl differeces s d s f k = f k whe s =. The secod order cetrl differece is give by d f k = d[f k+/ f k-/ ] = d[f k+/ ] - d[f k-/ ] = f k+ f k f k + f k- = f k+ f k + f k- Similrly, d f k = f k+/ - f k+/ + f k-/ - f k-/ d d 4 f k = f k+ - 4f k+ + 6f k - 4f k- + f k-.

54 MTH NUMERICAL ANALYSIS Notice tht the eve order differeces t tbulr vlue x k re expressed i terms of tbulr vlues of f d odd order differeces t tbulr vlue x k re expressed i terms of o-tbulr vlue of f. lso ote tht the coefficiets of d s f k re the sme s those of the biomil expsio of ( x) s, s =,,,.... Sice df k f k+/ f k-/ = (E / E -/ )f k We hve the opertio reltio d = E / E -/ (4) The cetrl differeces t o-tbulr poit x k+/ c be clculted i similr wy. For exmple, df k+/ = f k+ - f k d f k+/ = f k+/ - f k+/ + f k-/ d f k+/ = f k+ - f k+ + f k - f k- (5) d 4 f k+/ = f k+/ - 4f k+/ + 6f k+/ - 4f k-/ + f k-/ Reltio (5) c be obtied esily by usig the reltio (4) We hve d s f k = [E / E -/ ] s f k i / ( i) / i = E E ( ) f k i= s i i = ( ) f k+ ( / ) i= s i (6) The followig formuls c lso be estblished: f[x,..., x m ] = m d m f m (7) (m)!h f[x,..., x m+ ] = m d m+ f (m + )!h + m+/ (8) f[x -m,...,x,..., x m ] = m d m f (9) (m)!h 4

55 MTH MODULE f[x -m,...,x,..., x m+ ] = m d m+ f (m + )!h + / () f[x -(m+),...,x,..., x m ] = m d m+ f (m + )!h + -/ () We ow give below the cetrl differece tble with 5 odes. Tble : Cetrl Differece Tble x f df d f d f d 4 f x - f - df -/ x - f - d f - df -/ d f -/ x f d f d 4 f df / d f / x f d f df / x f Note tht the differece d m f lie o horizotl lie show by the dotted lies. Tble 4: Cetrl Differece Tble x f df d f d f d 4 f x f df / x f d f df / d f / x f d f d 4 f df 5/ d f 5/ x f d f df 7/ x 4 f 4 Note tht the differece d m f lie o horizotl lie. We ow defie the me opertor ms follows mf k = [f k+/ + f k-/ ] = [E/ + E -/ ]f k. 5