r are all of its complex roots We will look at how to 64257nd roots or zeros of polynomials in one variable The solution of multivariate polynomials can often be transformed into a problem that requires the solution of singlevariate polynomials 1 R ID: 22799
Download Pdf The PPT/PDF document "Roots of Polynomials Com S Notes YanBin..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
RootsofPolynomials(ComS477/577Notes)Yan-BinJiaSep22,2020Adirectcorollaryofthefundamentaltheoremofalgebra[9,p.247]isthatp(x)canbefactorizedoverthecomplexdomainintoaproductn(x 1)(x 2)(x n),wherenistheleadingcoecientand1;r2;:::;rnareallofitsncomplexroots.Wewilllookathowto ndroots,orzeros,ofpolynomialsinonevariable.Intheory,root ndingformulti-variatepolynomialscanbetransformedintothatforsingle-variatepolynomials.1RootsofLowOrderPolynomialsWewillstartwiththeclosed-formformulasforrootsofpolynomialsofdegreeuptofour.Forpolynomialsofdegreesmorethanfour,nogeneralformulasfortheirrootsexist.Root ndingwillhavetoresorttonumericalmethodsdiscussedlater.1.1QuadraticsAquadraticequationax2+bx+c=0;a=0;hastworoots:x= p 2 4ac 2:Ifthecoecientsa;b;carereal,itfollowsthatif2 4ac�0therootsarerealandunequal;if2 4ac=0therootsarerealandequal;if2 4ac0therootsareimaginary.1.2CubicsThecubicequationx3+px2+qx+=0canbereducedbythesubstitutionx=y p 31 tothenormalformy3+ay+=0;(1)where=1 3(3q p2);=1 27(2p3 9pq+27):Equation(1)hasthreeroots:y1=A+B;y2= 1 2(A+B)+ip 3 2(A B);y3= 1 2(A+B) ip 3 2(A B);wherei2= 1andA=3s 2+r 2 4+3 27;B=3s 2 r 2 4+3 27:Thethreerootscanbeveri edbelow:(y y1)(y y2)(y y3)=(y A B)y2+(A+B)y+A2 AB+B2=y3 3ABy (A+B)(A2 AB+B2)=y3 3ABy A3 B3=y3+ay+b:Supposep;q;rarereal(andhenceandarereal).Threecasesexist:b2 4+a3 27�0:Thereareonerealrooty=y1andtwoconjugateimaginaryroots.b2 4+a3 27=0:Therearethreerealyrootsofwhichatleasttwoareequal: 2p a 3;p a 3;p a 3ifb�0,2p a 3; p a 3; p a 3ifb0,0;0;0if=0.b2 4+a3 270:Therearethreerealandunequalroots:yk=2r 3cos 3+2k 3;k=0;1;2;wherecos =8�&#x-2.9;领: q b2=4 a3=27ifb�0;q b2=4 a3=27ifb0.Everyrootykthusobtainedcorrespondstoarootxk=yk p=3oftheoriginalcubicequation.2 1.3QuarticsThequarticequationx4+px3+qx2+rx+s=0canbereducedtotheformy4+ay2+by+c=0(2)bythesubstitutionx=y p 4;where=q 3p2 8;=+p3 8 pq 2;c=s 3p4 256+p2q 16 pr 4:Thequartic(2)canbefactorizedundersomecondition.Theequationthatmustbesolvedtomakeitfactorizableiscalledtheresolventcubic:z3 qz2+(pr 4s)z+(4qs 2 p2s)=0:Letz1bearealrootoftheabovecubic.Thenthefourrootsoftheoriginalquarticarex1= p 4+1 2(+D);x2= p 4+1 2( D);x3= p 4 1 2( E);x4= p 4 1 2(+E);where=r 1 4p2 q+z1;D=8�&#x-300;:q 3 4p2 2 2q+1 4(4pq 8 p3) 1if=0,q 3 4p2 2q+2p z21 4sif=0,E=8�&#x-300;:q 3 4p2 2 2q 1 4(4pq 8 p3) 1if=0;q 3 4p2 2q 2p z21 4sif=0.Formoredetailssee[1]or[10].3 2RootCountingConsiderapolynomialofdegreen:p(x)=nxn+n 1xn 1++1x+0;an=0.Thefundamentaltheoremofalgebrastatesthatphasnrealorcomplexroots,countingmultiplic-ities.Ifthecoecients0;a1;:::;anareallreal,thenthecomplexrootsoccurinconjugatepairs,thatis,intheformcdi,wherec;d2Randi2= 1.Ifthecoecientsarecomplex,thecomplexrootsneednotberelated.UsingDescartes'rulesofsign,wecancountthenumberofrealpositivezerosthatp(x)has.Morespeci cally,letbethenumberofvariationsinthesignofthecoecientsn;an 1;:::;a0(ignoringcoecientsthatarezero).Letnpbethenumberofrealpositivezeros.Then(i)np,(ii) npisaneveninteger.Anegativezeroofp(x),ifexists,isapositivezeroofp( x).Thenumberofrealnegativezerosofp(x)isrelatedtothenumberofsignchangesinthecoecientsofp( x).Example1.Considerthepolynomialp(x)=x4+2x2 x 1.Thenv=1,sonpiseither0or1byrule(i).Butbyrule(ii)v npmustbeeven.Hencenp=1.Nowlookatp( x)=x4+2x2+x 1.Again,thecoecientshaveonevariationinsign,sop( x)hasonepositivezero.Inotherwords,p(x)hasonenegativezero.Tosummarize,simplybylookingatthecoecients,weconcludethatp(x)hasonepositiverealroot,onenegativerealroot,andtwocomplexrootsasaconjugatepairDescartes'ruleofsignstillleavesanuncertaintyastotheexactnumberofrealzerosofapolynomialwithrealcoecients.Anexacttestwasgivenin1829bySturm,whoshowedhowtocounttherealrootswithinanygivenrangeofvalues.Letf(x)bearealpolynomial.Denoteitbyf0(x)anditsderivativef0(x)byf1(x).ProceedasinEuclid'salgorithmto ndf0(x)=q1(x)f1(x) f2(x);f1(x)=q2(x)f2(x) f3(x);...fk 2(x)=qk 1(x)fk 1(x) fk(x);fk 1(x)=qk(x)fk(x);wheredeg(fi(x))deg(fi 1(x)),for1ik.ThesignsoftheremaindersarenegatedfromthoseintheEuclidalgorithm.Notethatthedivisorfkthatyieldszeroremainderisagreatestcommondivisoroff(x)andf0(x).Thesequencef0;f1;:::;fkiscalledaSturmsequenceforthepolynomialf.Theorem1(Sturm'sTheorem)Thenumberofdistinctrealzerosofapolynomialf(x)withrealcoecientsin(a;b)isequaltotheexcessofthenumberofchangesofsigninthesequencef0();:::;fk 1();fk()overthenumberofchangesofsigninthesequencef0();:::;fk 1();fk().4 Infact,wecanmultiplyfbyapositiveconstant,orafactorinvolvingx,providedthatthefactorremainspositivethroughout(a;b),andthemodi edfunctioncanbeusedforcomputingallfurthertermsfiofthesequence.Example2.UseSturm'stheoremtoisolatetherealrootsofx5+5x4 20x2 10x+2=0We rstcomputetheSturmfunctions:f0(x)=x5+5x4 20x2 10x+2f1(x)=x4+4x3 8x 2f2(x)=x3+3x2 1f3(x)=3x2+7x+1f4(x)=17x+11f5(x)=1Bysettingx= 101,weseethatthereare3negativerootsand2positiveroots.Allrootsliebetween 10and10,infact,between 5and5.Wethentryallintegralvaluesbetween 5and5.Thefollowingtablerecordsthework: 1 10 5 4 3 2 1 0 1 2 5 10 1 f0 + + + + + + f1 + + + + + + + + f2 + + + + + + + f3 + + + + + + + + + + + f4 + + + + + + f5 + + + + + + + + + + + + + var. 5 5 5 5 4 3 3 2 1 0 0 0 0Thus,thereisarootin( 4 3),arootin( 3 2),arootin( 10),arootin(01),andarootin(12).3MoreBoundsonRootsItisknownthatp(x)musthaveatleastoneroot,realorcomplex,insidethecircleofradius1abouttheoriginofthecomplexplane,where1=min(n 0 1 ;ns 0 n ):(3)Inthecase1=0,wehave a0 a1 =1and1=nr a0 an .Meanwhile,ifwelet2=1+max0kn 1 k n ;(4)thenallzerosofp(x)mustlieinsidethecircleofradius2abouttheorigin.5 Theorem2(Cauchy'sTheorem)Givenapolynomialp(x)=nxn+n 1xn 1++0,n=0,de nethepolynomialsP(x)=jnjxn+jn 1jxn 1++j1jx j0j;Q(x)=jnjxn jn 1jxn 1 j0j:ByDescartes'rules,P(x)hasexactlyonerealpositivezero1andQ(x)hasexactlyonerealpositivezero2.Thenallzeroszofp(x)lieintheannularregion1jzj2:(5)Example3.Letp(x)=x5 37x4+74x3 108x2+108x 68.Applyingbound(3),weknowthatatleastonezeroofpisinsidethecirclexj1,where1=min(568 1085r 68 1)=minf314815146724g=146724Applyingbound(4),allzeroslieinsidethecirclexj2,where2=1+maxf377410810868g=118LetustrytoapplyCauchy'stheorem.First,wehavethatP(x)=x5 37x4 74x3 108x2 108x 68Q(x)=x5+37x4+74x3+108x2+108x 68whosepositivezerosarer1=063r2=56Thereforeallzerosofp(x)lieintheregion063jxj56Howdoesoneusetheaboverootbounds?Usethemasheuristicstogiveusawayoflocalizingthepossiblezerosofapolynomial.Bylocalizingthezeros,wecanguidetheinitialguessesofournumericalroot nders.4De\rationThee ortofroot ndingcanbesigni cantlyreducedbytheuseofde\ration.Onceyouhavefoundarootofapolynomialp(x),considernextthede\ratedpolynomialq(x)whichsatis esp(x)=(x )q(x):Therootsofqareexactlytheremainingrootsofp(x).Becausethedegreedecreases,thee ortof ndingtheremainingrootsdecreases.Moreimportantly,withde\rationwecanavoidtheblunder6 ofhavingouriterativemethodconvergetwicetothesamerootinsteadofseparatelytotwodi erentroots.Wecanobtainq(x)byevaluatingp(x)atx=usingHornerscheme.Recalltheintermediatequantities0;:::;bnfromtheevaluation,wheren=n,andi 1=i 1+rbi,fori=n;:::;1.Weknowthatp(x)=0+(x )(nxn 1++2x+1):(6)Since0=p()=0,weobtainq(x)=nxn 1++2x+1:(7)De\rationcanalsobecarriedoutbysyntheticdivisionofp(x)byq(x)whichactsonthearrayofpolynomialcoecients.Itmust,however,beutilizedwithcare.Becauseeachrootisknownwithonly niteaccuracy,errorscanbuildupintherootsasthepolynomialsarede\rated.Theorderinwhichrootsarefoundcana ectthestabilityofthede\ratedcoecients.Forexample,supposethenewpolynomialcoecientsarecomputedinthedecreasingorderofthecorrespondingpowersofx.Themethodisstableiftherootofsmallestabsolutevalueisdividedoutateachstage.Rootsofde\ratedpolynomialsarereallyjust\goodsuggestions"oftherootsofp.Oftenweneedtousetheoriginalpolynomialp(x)topolishtherootsintheend.5Newton'sMethodThe rstmethodweintroduceonpolynomialroot ndingisNewton'smethod.Supposewehaveanestimatexkofarootatiterationstepk.Newton'smethodyieldstheestimateatstepk+1:xk+1=xk p(xk) p0(xk):(8)Recallthatwecanevaluatep(t)andp0(t)togetherecientlyusingHornerscheme,generatingintermediatequantitiesn;bn 1;:::;b0=p(t)(andthusthepolynomialq(x)in(7)).Equations(6)and(7)arecombinedintop(x)=p(t)+(x t)q(x);(9)whichimpliesp0(t)=q(t).Iftisactuallyarootofp(x),itfollowsfromequation(9)thatq(x)=p(x) x tisalreadythede\ratedpolynomial,whichcanbeviewedasabyproductoftheNewton'smethod.To ndazeroofp(x),Newton'smethodtakesaninitialguessx0ofarootanditeratesaccordingto(8)untilsometerminationconditionissatis ed,forinstance,untilp(xi)orjxi+1 xijiscloseenoughtozero.OnesituationthatNewton'smethoddoesnotworkwelliswhenthepolynomialp(x)hasadoubleroot,thatis,whenp(x)=(x )2h(x):Inthiscasep(x)sharesafactorwithitsderivativewhichisoftheformp0(x)=2(x )h(x)+(x )2h0(x):7 Asxapproaches,bothpandp0approachzero.InapplyingNewton'smethod,machineimprecisionwilldominateevaluationofthetermp(x) p0(x)asxtendsto.Inthiscaseonetinynumberisdividedbyoneverysmallnumber.Onecanavoidthedouble-rootproblembyseekingquadraticfactorsdirectlyinde\ration.Thisisalsousefulwhenlookingforapairofcomplexconjugaterootsofapolynomialwithrealcoecients.Forhigherorderroots,onecandetecttheirpossibilitiestosomeextentandapplyspecialtechniquestoeither ndorruleoutthem.Polynomialsareoftensensitivetovariationsintheircoecients.Consequently,afterseveralde\rations,theremainingrootsmaybeveryinaccurate.Onesolutionistopolishtherootsus-ingaveryaccuratemethod,onceapproximationstotheserootshavebeenfoundfromde\ratedpolynomials.Newton'smethodisgenerallygoodforpolishingbothrealandcomplexroots.6Muller'sMethodNewton'smethodisalocalmethod,thatis,itmayfailtoconvergeiftheinitialestimateistoofarfromaroot.Nowweintroduceaglobalroot ndingtechnique.Aftertherootsarefound,wecanpolishthemaswedesire(oftenusingNewton'smethod).Muller'smethodcan ndanynumberofzeros,realorcomplex,oftenwithglobalconvergence.Sinceitisalsoapplicabletofunctionsotherthanpolynomials,weherepresentitfor ndingazeroofageneralfunctionf(x),notnecessarilyapolynomial.Themethodmakesuseofquadraticinterpolation.Supposethethreepriorestimatesofazerooff(x)inthecurrentiterationarethepointsxk 2;xk 1;xk.Tocomputethenextestimatewewillconstructthepolynomialofdegree2thatinterpolatesf(x)atxk 2;xk 1;xk,then ndoneofitsroots.WebeginwiththeinterpolatingpolynomialintheNewtonformp(x)=f(xk)+f[xk 1;xk](x xk)+f[xk 2;xk 1;xk](x xk)(x xk 1);wheref[xk 1;xk]andf[xk 2;xk 1;xk]aredivideddi erencesde nedbelow:f[xk 1;xk]=f(xk 1) f(xk) xk 1 xk;f[xk 2;xk 1;xk]=f[xk 2;xk 1] f[xk 1;xk] xk 2 xk=f(xk 2) f(xk 1) xk 2 xk 1 f(xk 1) f(xk) xk 1 xk xk 2 xk:Usingtheequality(x xk)(x xk 1)=(x xk)2+(x xk)(xk xk 1);wecanrewritep(x)asp(x)=f(xk)+(x xk)+(x xk)2;where=f[xk 2;xk 1;xk];=f[xk 1;xk]+f[xk 2;xk 1;xk](xk xk 1)8 Nowletp(x)=0andsolveforxasthenextapproximationxk+1=xk 2f(xk) p 2 4af(xk):Intheabovewechoosetheformofzerosofthequadraticax2+bx+ctobex= 2c p 2 4acinsteadofx= p 2 4ac 2fornumericalstability.Thisismainlybecausec=f(xk)getsverysmallneararootoff.Alsofornumericalstability,the`'signinthedenominatorp 2 4ac,ischosensoastomaximizethemagnitudeofthedenominator.Notethatcomplexestimatesareintroducedautomaticallyduetothesquare-rootoperation.Algorithm1 ndsallroots,withorwithoutmultiplicities,ofapolynomialthathasonlyrealco-ecients,bycombiningMuller'smethodwithde\rationandrootpolishing(usingNewton'smethod).De\rationmaystopwhenthedegreeofthepolynomialreducestofourorthree,andclosed-formrootsinSections1.2and1.3canbecomputed.Example2.Findtherootsofp(x)=x3 x 1usingMuller'smethod.Foreachroottheiterationstartwithinitialguessesx 2=1;x 1=15,andx0=20.Weterminatetheiterationatstepiwhenxi xi 1j510 5Themethodestimatesthe rstroottober1=1324718withanaccuracyabout3410 7 ixip(xi) 21.0 1 11.50.87502.0511.333330.03703721.32447 00010531.32471814410 641.32471871510 13Next,weworkwiththede\ratedpolynomialq(x)=p(x)=(x r1)and ndthesecondroottober2= 066236+056228iwithanaccuracyof2210 5ixiq(xi) 1 06623697+05622605i2135510 5 1210 5i2 066235898+05622795i10 11Thethirdrootmustbeaconjugateofthesecondroot.Sowehaver3= 066236 056228iLetusdoamorecompleteexample.Example3.Findtherootsofthepolynomialp(x)=x4+2x2 x 1Earlierbycheckingthesignchangesinthecoecientsequenceweknewthatp(x)hasonepositiveroot,onenegativeroot,andonepairofcomplexconjugateroots.9 Algorithm1Muller'smethod Input:p(x)=nxn+1x+0,wherei2Rfor0inandn=0Output:itsroots0;r1;:::;rn 1ifdeg(p)4thenusetheclosedformsgiveninSection1elsep0(x) p(x)l 0whiledeg(pl)5doonerootofplmustlieinsidethecircleofradius1in(3)inSection3generatethreerootestimateswithinthecircle.useMuller'smethodto ndonerootlofpl(x)iflisrealthenpl+1(x) pl=(x l)(applyHornerschemeforde\ration)l l+1whilepl(l 1)=0(amultipleroot)dol l 1pl+1(x) pl=(x l)l l+1endwhileelsel=l+iblanditscomplexconjugate l=l iblaretworootspl+2(x) pl=((x l)(x l))=pl=(x2 2lx+2l+2l)l l+2whilepl(l 2)=0dol l 2l+1 l 1pl+2(x) pl=(x2 2lx+2l+2l)l l+2endwhileendifendwhile ndtherootsofpl(withdegreesatmostfour)usingtheclosedformsinSection1.2or1.3.polishalltheroots0;:::;rn 1usingNewton'smethodonp(x)endif 10 Nowletuslookattheboundheuristics:=min(na0 a1ns a0 an)=minf41g=1Thusthereisatleastonezeroinsidethecomplexcircleofradiusoneabouttheorigin.Furthermore,allzerosofp(x)lieinsidethecircleofradiusr=1+max0kn 1 ak an =1+maxf1120g=3Oursearchforzeroneedonlyfocusonthecomplexdiskofradius3.ForeachrootwestartMuller'smethodwiththeinitialguessesx0= 1 2;x1=0;x2=1 2andterminatethesearchoncexi510 5roots#iterationsaccuracy 0.825109852410 7 0481815641410 6 0171647+1576686i25110 6 0171647 1576686iNotethatthetwocomplexrootsliebetweenthecircleofradius1andthecircleofradius3.De\rationswereusedinseekingthesecondandthethirdroots.Weshouldnextpolishthefourrootestimatesusingtheoriginalpolynomialp(x),whichcanbedoneautomaticallywithinMuller'smethod.References[1]M.AbramowitzandI.A.Stegun.HandbookofMathematicalFunctions:withFormulas,Graphs,andMathematicalTables.NewYork:Dover,pp.17-18,1972.[2]R.S.Burington.HandbookofMathematicalTablesandFormulas.McGraw-Hill,Inc.,5thedition,1973.[3]S.D.ConteandCarldeBoor.ElementaryNumericalAnalysis.McGraw-Hill,Inc.,2ndedition,1972.[4]M.Erdmann.Lecturenotesfor16-811MathematicalFundamentalsforRobotics.TheRoboticsInstitute,CarnegieMellonUniversity,1998.[5]C.C.MacDu ee.TheoryofEquations.JohnWiley&Sons,Inc.,NewYork,1954.[6]W.H.Press,etal.NumericalRecipesinC++:TheArtofScienti cComputing.CambridgeUniversityPress,2ndedition,2002.11 [7]J.StoerandR.Bulirsch.IntroductiontoNumericalAnalysis.Springer-VerlagNewYork,Inc.,2ndedition,1993.[8]H.W.Turnbull.TheoryofEquations,5thedition.OliverandBoydLtd.,Edinburgh,1952.[9]B.L.vanderWaerden.Algebra,VolumeI.Springer-VerlagNewYork,Inc.,1991.[10]WolframMathWorld.http://mathworld.wolfram.com/QuarticEquation.html.12