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A survey of the development of daubechies' scaling functions

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A survey of the development of daubechies' scaling functions
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Age, Amber
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Daubechies
Cascade Algorithm
Scaling Functions
Multiresolution Analysis
Wavelet
Dissertations, Academic -- Mathematics and Statistics -- Masters -- USF   ( lcsh )
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non-fiction   ( marcgt )

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Abstract:
ABSTRACT: Wavelets are functions used to approximate data and can be traced back to several different areas, including seismic geology and quantum mechanics. Wavelets are applicable in many areas, including fingerprint and data compression, earthquake prediction, speech discrimination, and human vision. In this paper, we first give a brief history on the origins of wavelet theory. We will then discuss the work of Daubechies, whose construction of continuous, compactly supported scaling functions resulted in an explosion in the study of wavelets in the 1990's. These scaling functions allow for the construction of Daubechies' wavelets. Next, we shall use the algorithm to construct the Daubechies D4 scaling filters associated with the D4 scaling function. We then explore the Cascade Algorithm, which is a process that uses approximations to get possible representations for the D2N scaling function of Daubechies. Lastly, we will use the Cascade Algorithm to get a visual representation of the D4 scaling function.
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Thesis (MA)--University of South Florida, 2010.
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Includes bibliographical references.
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by Amber Age.
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ABSTRACT: Wavelets are functions used to approximate data and can be traced back to several different areas, including seismic geology and quantum mechanics. Wavelets are applicable in many areas, including fingerprint and data compression, earthquake prediction, speech discrimination, and human vision. In this paper, we first give a brief history on the origins of wavelet theory. We will then discuss the work of Daubechies, whose construction of continuous, compactly supported scaling functions resulted in an explosion in the study of wavelets in the 1990's. These scaling functions allow for the construction of Daubechies' wavelets. Next, we shall use the algorithm to construct the Daubechies D4 scaling filters associated with the D4 scaling function. We then explore the Cascade Algorithm, which is a process that uses approximations to get possible representations for the D2N scaling function of Daubechies. Lastly, we will use the Cascade Algorithm to get a visual representation of the D4 scaling function.
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ASurveyoftheDevelopmentofDaubechiesScalingFunctions by AmberE.Age Athesissubmittedinpartialfulllment oftherequirementsforthedegreeof MasterofArts Departmentof Mathematics&Statistics Collegeof ArtsandSciences UniversityofSouthFlorida MajorProfessor: CatherineBeneteau,Ph.D. DmitryKhavinson,Ph.D. SherwinKouchekian,Ph.D. DateofApproval: July6th,2010 Keywords:Daubechies,CascadeAlgorithm,ScalingFunctions,MultiresolutionAnalysis,Wavelet c Copyright 2010 ,AmberE.Age

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Dedication Iwouldliketodedicatethistoallmymathematicsteachers andinstructorswhohavetaughtmeovertheyears.Itis becauseoftheirguidanceandenthusiasmthatInotonly startedmyquestformathematicaleducation,butalsokept mypathsteadfastandtrue.Toallofthem,Igivethanks.

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Acknowledgments Theauthorwouldliketothankthefollowingindividualsforalltheirsupportinthewriting ofthisthesis.Iwouldliketothankmymother,forinspiringmetoachievegreatnessandfor alltheencouragementbeforeexams.Iwouldalsoliketothankmyfatherforhisloveand support.IwouldliketothankmyhusbandTomforconstantloveandencouragement,andfor hisinvaluableLatexhelp.IwouldalsoliketothankthegraduatestudentsofPhysics107for providingmomentsoflevityduringotherwisechallengingtimes.Iwouldalsoliketothank ShaneStadlerforhissubtle(andnotsosubtle)reminderstokeepatit.Iwouldliketothank theDepartmentofMathematicsandStatisticsattheUniversityofSouthFloridaforgiving metheopportunitytostudymathematics.Lastly,Iwouldliketothankmyadvisor,Catherine B en eteaunotonlyforherguidance,encouragement,andsuggestions,butformakingmea bettermathematician.

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TableofContents ListofFigures.......................................... ii Abstract............................................. iii Chapter1Wavelets:ABriefHistoryandBasicDenitions.................1 Chapter2AnAlgorithmfortheConstructionofDaubechiesScalingFunctions......10 2.1TheDaubechiesAlgorithm.............................. 10 2.2TheRootsof F ( z ) .................................. 16 2.3TheConstructionofthe D 4 scalinglter....................... 19 Chapter3TheCascadeAlgorithm..............................22 References............................................ 37 i

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ListofFigures Figure1IteratesoftheCascadeAlgorithmforD4scalingfunctionwithn=1,2,3,4..... 35 Figure2IteratesoftheCascadeAlgorithmforD4scalingfunctionwithn=5,6,7,8..... 36 ii

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ASurveyoftheDevelopmentofDaubechiesScalingFunctions AmberE.Age ABSTRACT Waveletsarefunctionsusedtoapproximatedataandcanbetracedbacktoseveraldifferentareas,includingseismicgeologyandquantummechanics.Waveletsareapplicableinmanyareas,including ngerprintanddatacompression,earthquakeprediction,speechdiscrimination,andhumanvision. Inthispaper,werstgiveabriefhistoryontheoriginsofwavelettheory.Wewillthendiscuss theworkofDaubechies,whoseconstructionofcontinuous,compactlysupportedscalingfunctions resultedinanexplosioninthestudyofwaveletsinthe1990's.Thesescalingfunctionsallowforthe constructionofDaubechies'wavelets.Next,weshallusethealgorithmtoconstructtheDaubechies D 4 scalingltersassociatedwiththe D 4 scalingfunction.WethenexploretheCascadeAlgorithm, whichisaprocessthatusesapproximationstogetpossiblerepresentationsforthe D 2 N scaling functionofDaubechies.Lastly,wewillusetheCascadeAlgorithmtogetavisualrepresentationof the D 4 scalingfunction. iii

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Chapter1 Wavelets:ABriefHistoryandBasicDenitions Thestudyofwaveletscanbetracedbacktoseveraldifferentelds,includingmathematics,quantum physics,andelectricalengineering.Whileseveralareasareresponsibleforindependentlydevelopingwavelets,perhapsthemostimportantcontributioncamefromseismicgeologyandthework ofJeanMorlet.Morletneededawaytoanalyzeseismicsignalswhichcarriedinformationabout geologicallayers.BuildingofftheworkofDennisGabor,Morlet,alongwithAlexGrossman,explicitlydenedandbeganusingthewordwavelet.St ephaneMallatandYvesMeyerthenused these Grossman-Morlet waveletsasbuildingblocksfor multiresolutionanalysis ,anotionbuiltoff previousworkdonebyBurtandAdelson[1]andthatusestheconceptof orthonormalwavelet bases .Theorthonormalwaveletbases,asrstdenedbyMeyerin[7],werenotperfectedforuse inapplicationsuntiltheworkofIngridDaubechiesin1988.ItwasatthistimethatDaubechies presentedaconstructionthatresultedinasetoforthonormalwaveletbasesthatwereofcompact supportandcontinuouslydifferentiable[2].Thisdevelopmentresultedinanexplosioninthestudy ofwaveletsandtheirapplications.Currently,waveletsareusedinavarietyofareas,including humanvision,speechdiscrimination,ngerprintcompression,earthquakeprediction,andnuclear engineering.Forfurtherinformation,thereadershouldsee[3]or[6]. Waveletsarefunctionsthatcanbeusedtoapproximatedata.Whileothertypesoffunctions, suchastheFouriertransform,canalsobeusedtoapproximatedata,thewavelettransformhasthe additionalabilitytoanalyzedataatdifferentresolutions.Dataanalyzedatalargerresolutiongivesa roughapproximationwhiledataanalyzedatsmallerresolutionsprovidesmoredetailedinformation. Itisthisabilityofthewavelettransformtozoominandoutofthedatathatmakesitsuperiortothe Fouriertransformforcertainapplications.Asstatedin[3],waveletalgorithmsallowustosee"both theforestandthetrees." Whileseveraldifferentapproachesweretakeninthedevelopmentoftheanalysisofwavelets,for thepurposesofthispaper,weshallfocusontheworkdonebyHaarandtheideaofamultiresolution 1

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analysispresentedbyMallat[4]andMeyer[7].WewillthenbrieymentiontheShannonmultiresolutionanalysisanditsshortcomingsintermsofapplications.Theseshortcomingswillbecorrected byDaubechies'constructionofcontinuous,compactlysupportedscalingfunctions.Thesescaling functionsaretheessentialtoolsneededtobuildorthonormalwaveletbases.Inthispaper,wewill focusonthedevelopmentofthesescalingfunctions.Muchofthefollowingdiscussionismotivated bytheworksof[6]and[8]. Whilethestudyofwaveletsstartedinthe1930's,therootsofwaveletsgobackto1807andthe ideasofFourierregardingtheconvergenceoffunctions.Fourierclaimedthatevery 2 periodic function f ( ) couldbeexpressedasasumofitsFourierSeries.Forthepurposesofthispaper,we shallonlybeconsideringfunctions f ( ) inthespace L 2 ( R ) Denition1.1 Wedenethespace L 2 ( R ) asthesetofallLebesguemeasurable,complex-valued functions f suchthat || f || 2 2 := R | f ( ) | 2 d < # Wedenethespace L 2 ([ ! ]) inananalogous way. Denition1.2 WedenetheFourierSeriesforafunctionfor f ( ) $ L 2 ([ ! ]) by k = "! c k e ik wherethecoefcient c k isgivenby 1 2 # " f ( ) e ik d Notethatif f $ L 2 ([ ! ]) theabove-mentionedseriesexistsa.e.andconvergesto f inthe L 2 norm,andsoweoftenwrite f ( )= k = "! c k e ik However,itmustbepointedoutthatgivenacontinuousfunction f, itisnotalwaysthecasethat itsFourierserieswillconvergeto f pointwise.Infact,problemswithconvergenceofFourierseries cameintheformofanexampleconstructedbyP.DuBois-Reymond.Inhisexample,DuBoisReymondfounda 2 periodicfunctionwhoseFourierseriesdivergedatagivenpoint.Itwasthen concludedthatcertainrestrictionswouldneedtobemetinorderforFourier'soriginaltheoryto hold. 2

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MathematiciansbegantryingtojustifyFourier'stheoryinoneofthreeways.Therstwaywas toattempttore-evaluatethedenitionofafunctionandtrytomodifyitinsuchawaythatitwould coincidewiththeideaofFourierseries.Thesecondideawastore-evaluatethenotionofconvergenceforFourierseries.Itwasthoughtthatbyputtingcertainconditionson f ( x ) convergenceof theFourierserieswouldfollow.Forexample,itcanbeshownthatif f ( x ) isatwice-differentiable functiononthecircle,thentheFourierSeriesof f convergesabsolutelyanduniformlyto f (see [10]).Thethirdideawastouseorthogonalsystemsintheplaceoftrigonometricpolynomials.In otherwords,usingorthogonalsystems,couldoneachieveconvergenceoftheFourierseriestoa function?TherstsignicantresultwasgivenbyHaarin1801[6]. Recallthatthe innerproduct oftwofunctions f ( t ) ,g ( t ) $ L 2 ( R ) isdenedas % f ( t ) ,g ( t ) & = # R f ( t ) g ( t ) dt. Forasetoffunctions { # k ( t ) } k # Z $ W where W isasubspaceof L 2 ( R ) ,wesaythat { # k ( t ) } k # Z isa basis for W ifeveryfunction f $ W canbeexpressedasalimitofnitelinearcombinationsof { # k ( t ) } k # Z andif { # k ( t ) } k # Z islinearlyindependent.Abasis { # k ( t ) } k # Z isan orthonormalbasis for W if % # k ( t ) # j ( t ) & = $ % & 1 j = k 0 otherwise Haarbeganbylookingforanorthonormalsystemoffunctions { # j ( t ) } $ R suchthatfora continuousfunction f ( t ) $ L 2 ( R ) theseriesgivenby j # Z % f ( t ) # j ( t ) & # j ( t ) wouldconvergeuniformlyto f ( t ) Haarcreatedavectorspacethatcontainedallstep-functionsthathadbreakpointsattheintegers. Hedenedaninitialfunctionas # ( t )= $ % & 10 t< 1 0 otherwise Thisfunctionisknownasthe Haarfunction .ItshouldbenotedthattheHaarfunctionissometimes referredtoasthe boxfunction .TheHaarfunctionanditsintegertranslateswereusedtodenethe 3

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space V 0 where V 0 = span { # ( t k ) } k # Z ( L 2 ( R ) Thismeansthateveryfunction f $ V 0 canbeexpressedasan L 2 limitofnitelinearcombinations of { # ( t k ) } k # Z .V 0 isknownasthe HaarSpace .Notethatthefunctions { # ( t k ) } k # Z forman orthonormalbasisfor V 0 Haarprojectedfunctions f $ L 2 ( R ) onto V 0 inthestandardway: P 0 f ( t ) ( = k # Z % f ( t ) # ( t k ) & # ( t k ) Thisprojectionof f ( t ) $ L 2 ( R ) into V 0 givesaroughapproximationof f. Toaltertheapproximation,forbetterorworse,Haarsimplychangedwherethebreakpointsofthefunctionoccurred.Haar denedthesespaces V j as V j = span { # (2 j t k ) } k # Z ( L 2 ( R ) Asbefore,for j $ Z bydeningthefunction # j,k ( t ) as # j,k ( t )=2 j 2 # (2 j t k ) onecanshowthat { # j,k ( t ) } isanorthonormalbasisforthespace V j Onceagain,wecanproject functions f $ L 2 ( R ) onto V j bydeningourprojection P j f ( t ) ( as P j f ( t ) ( = k # Z % # j,k ( t ) ,f ( t ) & # j,k ( t ) =2 j k # Z % # (2 j t k ) ,f ( t ) & # (2 j t k ) TheHaarSpaces V j obeysomenicepropertiesthatwillbeneededlater,soweshallpresentthem here.First,the V j spacesare nested spaces.Thatis,theysatisfy ... ) V 2 ) V 1 ) V 0 ) V 1 ) V 2 .... Wecanalsomovefromonespacetoanotherwithrelativeease: f ( t ) $ V j f (2 t ) $ V j +1 Thereasonwecan"zoom"betweenspacesisbecauseourfunction # ( t ) satiseswhatiscalleda dilationequation givenby # ( t )= + 2 2 # 1 0 ( t )+ + 2 2 # 1 1 ( t )= # (2 t )+ # (2 t 1) 4

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Because # ( t ) satisessuchanequation, # ( t ) isusuallycalleda scalingfunction .Byconsidering theunionandintersectionoftheHaarspaces,weobtaintwomoreusefulproperties,namelythat thesespacessatisfyaseparationconditionandaredense(see[8]): ) j # Z V j = { 0 } and j # Z V j = L 2 ( R ) Weshalllaterseethatthesepropertiesareessentialinourdenitionofamultiresolutionanalysis. Haar'sapproachhasitslimitations,asitsscalingfunction # ( t ) isnotcontinuousanditsrstderivativeiszeroa.e..Forpurposesofapplication,wedesireascalingfunctionthatisbothcontinuous andhasanumberofderivatives.DaubechieswasabletobuildoffofHaar'soriginalconstructionto obtainascalingfunctionthathasbothoftheseproperties.BeforewecandiscusshowDaubechies builtsuchscalingfunctions,wemustdescribetheideaof multiresolutionanalysis ,aresultofthe workofMeyer[7]andMallat[4]builtupontheearlierworkofBurtandAdelson[1]. Denition1.3 Let { V j } j # Z beasequenceofsubspacesof L 2 ( R ) Wesaythat { V j } j # Z isa multiresolutionanalysis of L 2 ( R ) if 1. V j ) V j +1 2. + j # Z V j = L 2 ( R ) 3. j # Z V j = { 0 } 4. f ( t ) $ V 0 f (2 j t ) $ V j andthereexistsafunction # ( t ) $ V 0 calledascalingfunction,with R # ( t ) dt =0 suchthatthe set { # ( t k ) } k # Z isanorthonormalbasisfor V 0 Notethatsincewearerequiringthat R # ( t ) dt =0 wecannormalizethescalingfunctionso that # R # ( t ) dt =1 5

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Weshallassumethatthescalingfunctionsatisesthispropertyfortheremainderofthispaper.Itis alsoimportanttonotethatwhileDenition1.3holdsforcomplex-valuedscalingfunctions,forthe purposesofthispaper,weshallassumethatallscalingfunctionsarereal-valuedfunctions. OneshouldnoticeimmediatelythatthepreviouslydiscussedHaarspaces V j satisfytherequirementsofamultiresolutionanalysis.Infact,thestrengthofamultiresolutionanalysisisthatitgives usawaytodecomposethespace L 2 ( R ) intonestedsubspacesthatwecanusetoapproximatefunctions,muchliketheHaarspaces.SeveralofthepropertiesthatholdfortheHaarspaceswillalso holdforamoregeneralsetofsubspaces V j satisfyingthepropertiesofamultiresolutionanalysis. Usingthescalingfunction # ( t ) ofamultiresolutionanalysis V j $ L 2 ( R ) alongwithitsinteger translates,wedenethefunction # j,k ( t ) as # j,k ( t )=2 j 2 # (2 j t k ) Oneshouldrstobservethat # j,k ( t ) $ V j duetothenestedproperty.Itcanalsobeshownthat || # j,k ( t ) || 2 =1 .Thesetwofactscanbeusedtoshowthat,justasinthecaseoftheHaarSpaces, { # j,k ( t ) } k # Z formsanorthonormalbasisforthespace V j Asadirectresult,wecanrepresentany function f $ V j as f ( t )= k # Z % f ( t ) # j,k ( t ) & # j,k ( t ) Thescalingfunction # ( t ) ofamultiresolutionanalysis V j satisesageneraldilationequation givenby # ( t )= + 2 k # Z h k # (2 t k ) where h k = % # ( t ) # 1 ,k ( t ) & Thecoeffcients h k ,k $ Z formwhatiscalledthescalinglter. Inmanyapplications,itiseasiertoconsiderthescalingfunction # ( t ) inthetransformdomain, asrstdevelopedbyFourier.Thisshouldcomeasnosurprisewhenoneconsidersthattherootsof waveletanalysislieinHarmonicAnalysis!Beforewecandescribetheprocess,itisnecessaryto denetheFouriertransform. Denition1.4 Wedenethe FourierTransform f ( ) of f ( t ) as f ( )= 1 + 2 # R f ( t ) e it dt. 6

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Onecanshowthatif f $ L 2 ( R ) andsatisessomeadditionalconditions(see[8]or[10]),then f ( t )= 1 + 2 # R f ( ) e i t d Thisintegralisknownasthe InverseFourierTransform .Fortheremainderofthispaper,weshall refertothedomainoffunctions f ( t ) $ L 2 ( R ) asthetimedomain,whiletheFouriertransforms ofthesefunctions,denotedby f ( ) shallbeconsideredtolieinthetransformdomain.Oneofthe reasonstheFouriertransformisusefulinsolvingproblemsisthat convolution inthetimedomain becomesmultiplicationintheFourierdomain: Denition1.5 For f,g $ L 2 ( R ) ,wedenethe convolution of f ( t ) and g ( t ) as ( f g )( t )= # R f ( u ) g ( t u ) du. Notethat f g $ L 1 ( R ) Usingthisdenition,wehavethefollowingtheorem,asstatedin[10]: Theorem1.1 For f,g $ L 2 ( R ) ( f g )= f g. Wecantranslatethedilationequation # ( t )= + 2 k # Z h k # (2 t k ) intothetransformdomainbyrstobservingthatifwelet g ( t )= # (2 t k ) theFouriertransform ofgis g ( )= 1 2 e ik 2 # 2 . UsingTheorem1.1andbasicpropertiesoftheFouriertransform(see[10]),weseethatthedilation equationsatisedby # ( t ) $ L 2 ( R ) becomes,inthetransformdomain: # ( )= 1 + 2 k # Z h k e ik 2 # 2 . (1.1) Thisgivesusaveryimportantpropertyofthescalingfunction # ( t ) asitstatesthat # ( t ) $ L 2 ( R ) satisesthedilationequationinthetimedomainifandonlyifitsFouriertransform, # ( ) satises (1.1).Let H ( ) bethetrigonometricseries H ( )= 1 $ 2 / k # Z h k e ik where h k arethescaling ltercoefcients.Weshallreferto H ( ) asthe symbol of # ( t ) andwecanuseittore-write(1.1) as # ( )= H 2 # 2 . 7

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Weshallrefertothisequationasthedilationequationinthetransformdomain. Ifwethinkofoursymbolbeingoftheform H ( z )= 1 $ 2 / k # Z h k z k ,whichcanbeobtained fromouroriginaldenitionof H ( ) byletting z = e i wecanmakeadditionalremarksabout thesupportofourscalingfunction # ( t ) Forourpurposes,weshalldenethe support of f, denoted bysupp(f),asthesetofallvalues t $ R suchthat f ( t ) =0 Wesay f is compactlysupported ifsupp( f )iscontainedinaclosedintervalofnitelength.Inthiscase,wesaythatthe compact supportof f isthesmallestclosedinterval [ a,b ] suchthatsupp ( f ) ) [ a,b ] Thisintervalisdenoted by supp ( f ) Itshouldbenotedthatisadifferentnotionofcompactsupport,andmoredetailscan befoundin[8]. Itcanbeshownthatfor # ( t ) generatingamultiresolutionanalysis V j $ L 2 ( R ) if # ( t ) hascompactsupport,then supp ( # )=[0 ,N ] providedthatthesymbol H ( z ) isapolynomialoftheform H ( z )= 1 $ 2 / N k =0 h k z k for N $ Z Itcanalsobeshownthat H (0)=1 andthat H ( ) is 2 periodic.ThesetwopropertiesareinfactnecessaryfortheresultsobtainedusingtheCascadeAlgorithm,whichwillbediscussedinmoredetailinChapter3.Perhapsthemostimportantpropertyof thesymbol H ( ) isthatissatiseswhatweshallhereafterrefertoasthe orthonormalitycondition : | H ( ) | 2 + | H ( + ) | 2 =1 (1.2) Ifthesymbol H ( ) satisesthiscondition,theresultisthatthescalingfunction # ( t ) andits translates,areorthonormalinthetimedomain.Conversely,if H ( ) satisescertainconditions,we canguaranteetheexistenceofascalingfunction # ( t ) thatwillgenerateamultitresolutionanalysis: Inotherwords,if H ( ) hasnitedegreeandsatisestheorthonormalitycondition,andif H (0)=1 and H ( z ) isoftheform H ( z )= 1+ z 2 N S ( z ) where S ( z ) satises max | z | =1 | S ( z ) | 2 N 1 with z = e i thenthereexistsascalingfunction # ( t ) thatgeneratesamultiresolutionanalysis. Asweshallsee,thisideaiscrucialintheconstructionofDaubechies'scalingfunctions. Inorderforamultiresolutionanalysistobeusefulinapplications,wewouldlikeforittosatisfy threeproperties.Wewouldlikethescalingfunctiontohavecompactsupport,inordertosimplify computations.Wewouldalsolikeourscalingfunction # ( t ) tobesufcientlysmooth.Thatis,we 8

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wouldlike # ( t ) tohaveanitenumberofcontinuousderivatives.Finally,wewouldalsolikeour scalingfunction # ( t ) tohaveorthogonaltranslates.Thatis,wewouldlikeourscalingfunctionto satisfytheorthonormalityconditiongivenby(1.2).Whilethereareseveralexamplesofmultiresolutionanalyses,upuntiltheworkofDaubechies',therewasnotasinglemultiresolutionanalysis thatsatisedallthreeoftheaboveproperties.TheShannonmultiresolutionanalysisissuchan example.TheShannonmultiresolutionanalysishasascalingfunction,namely sin( t ) t := sinc(t) whoseFouriertransformisthecharacteristicfunctionofaninterval,andcanthereforebethought ofastheHaarmultiresolutionanalysisinthetransformdomain.ThescalingfunctionoftheShannonmultiresolutionanalysisdoesnothavecompactsupport,thusmakingithardtouseinmany applications.Inthenextchapter,weshalldiscusstheDaubechiesconstructionofasetoforthonormalscalingfunctionsthatwerebothcompactlysupportedandpossessedasufcientnumberof continuousderivatives. 9

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Chapter2 AnAlgorithmfortheConstructionofDaubechiesScalingFunctions 2.1TheDaubechiesAlgorithm WewillnowexploreanalgorithmpresentedbyIngridDaubechiesin1988toobtainamultiresolutionanalysis.Thefollowingconstructionfollowsfromanoutlinewhichcanbefoundin[8].As mentionedinthepreviouschapter,wewouldliketoworkwithascalingfunctionthathascompact supportandhasorthogonaltranslates.Fromourdiscussioninthepreviouschapter,weknowthat thisisequivalenttosayingthatwedesiretohavethedilationequationbeoftheform # ( )= H ( 2 ) # ( 2 ) (2.1) where H ( )= 1 $ 2 / N k =0 h k e ik andsatisestheorthonormalityequationgivenby | H ( ) | 2 + | H ( + ) | 2 =1 $ R (2.2) Finally,wewouldlikeourscalingfunctionstobesufcientlysmooth.Inotherwords,wewould likeourscalingfunctiontohavecontinuous N 1 derivatives.Daubechies'approachwastorequire thesymbol H ( ) tobeoftheform H ( )= 1+ e i 2 N S ( ) (2.3) where S ( )= A k =0 a k e ik withrealcoeffcients a k Byrequiringoursymbol H ( ) tosatisfytheseconditions,wecanguarantee theexistenceofascalingfunction # ( t ) thatwillgenerateamultiresolutionanalysis { V j } j # Z of L 2 R thatwillhave H ( ) asitssymbol. Webeginbyre-writingtheorthonormalitycondition: 10

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Lemma2.1 If H ( )= 1+ e i 2 N S ( ) with S ( )= A k =0 a k e ik thentheorthonormalitycondition | H ( ) | 2 + | H ( + ) | 2 =1 canbere-writtenas 0 cos 2 0 2 11 N | S ( ) | 2 + 0 sin 2 0 2 11 N | S ( + ) | 2 =1 (2.4) Proof. Byrstconsideringthesymbol H ( )= 1+ e i 2 N S ( ) foraxedNandsubstitutingitinto(2.2)weobtain 2 2 2 2 1+ e i 2 N S ( ) 2 2 2 2 2 + 2 2 2 2 1+ e i ( + ) 2 N S ( + ) 2 2 2 2 2 =1 Byobservingthat 2 2 2 2 1+ e i 2 2 2 2 2 2 =cos 2 0 2 1 andthat 2 2 2 2 1+ e i ( + ) 2 2 2 2 2 2 =sin 2 0 2 1 weobtain 0 cos 2 0 2 11 N | S ( ) | 2 + 0 sin 2 0 2 11 N | S ( + ) | 2 =1 (2.5) Wenowwanttond S ( ) Webeginbyletting L ( )= | S ( ) | 2 Wewouldliketotransform L ( ) intoapolynomial P ( y ) whichwillbeeasiertoworkwith. Lemma2.2 If S ( )= / A k =0 a k e ik wherethe a k arereal,thenbydening y = sin 2 2 ( there existsapolynomial P ( y ) suchthat | S ( ) | 2 = P ( y ) 11

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Proof. Byrstobservingthat L ( )= | S ( ) | 2 = S ( ) S ( ) andbynotingthatthecoefcientsof S ( ) arerealweobtain L ( )= S ( ) S ( " ) whichthenmakes L ( ) aproductoftrigonometricpolynomials.Wecanrewrite L ( ) as L ( )= S ( ) S ( " )= A k =0 a k e ik .A k =0 a k e ik = A j = A c j e ij where c j = c j ,whichcanthenbewrittenas L ( )= c 0 +2 A j =1 c j cos( j ) (2.6) since e ij + e ij =2cos( j ) Wecanre-write cos( j ) as cos ( j )= j k =0 t k cos k ( ) byrstnotingthat ( e i ) j = cos( )+ i sin( ) ( j = j k =0 3 4 j k 5 6 (cos( )) k ( i sin( )) j k (2.7) =cos( j )+ i sin( j ) Notethatthecontributiontotherealpartof(2.7)occurswhen j k =2 n iseven,andthenwecan write ( i sin( )) ( j k ) =( 1) n (1 cos 2 ( )) n Thus,therealpartof(2.7)canbewrittenaslinearcombinationsofpowersof cos( ) ofdegreeat most j : cos( j )= j k =0 t k cos k ( ) Thus,wecanre-write L ( ) as L ( )= A k =0 d k cos k ( ) 12

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Nowbecause cos( )=1 2sin 2 0 2 1 wehave L ( )= A k =0 d k 0 1 2sin 2 0 2 11 k Let y =sin 2 ( 2 ) Then | S ( ) | 2 = L ( )= P ( y )= A k =0 d k (1 2 y ) k (2.8) Sonow 0 cos 2 0 2 11 N | S ( ) | 2 + 0 sin 2 0 2 11 N | S ( + ) | 2 =1 hasbeentranslatedinto (1 y ) N P ( y )+ y N L ( + )=1 with L ( + )= A k =0 d k 0 1 2sin 2 0 2 + 2 11 k = A k =0 d k 0 1 2(1 sin 2 0 2 11 k = A k =0 d k (1 2(1 y )) k = P (1 y ) ThuswewanttondapolynomialP(y)satisfying (1 y ) N P ( y )+ y N P (1 y )=1 (2.9) Tondtheexplicitformof P ( y ) weshallfollow[11].Notethatforaxed N $ N wecanwrite 1= (1 y )+ y ( 2 N 1 = 2 N 1 k =0 2 N 1 k (1 y ) k y 2 N 1 k = N 1 k =0 2 N 1 k (1 y ) k y 2 N 1 k + 2 N 1 k = N 2 N 1 k (1 y ) k y 2 N 1 k 13

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Observethat 3 4 2 N 1 k 5 6 = (2 N 1)! k !(2 N 1 k )! = 3 4 2 N 1 2 N 1 k 5 6 Letting m =2 N 1 k, wecanre-writethesecondsumas N 1 m =0 2 N 1 m (1 y ) 2 N 1 m y m giving 1= N 1 k =0 2 N 1 k (1 y ) k y 2 N 1 k + N 1 m =0 2 N 1 m (1 y ) 2 N 1 m y m = y N N 1 k =0 2 N 1 k (1 y ) k y N 1 k +(1 y ) N N 1 m =0 2 N 1 m (1 y ) N 1 m y m = y N P (1 y )+(1 y ) N P ( y ) where P ( y )= N 1 k =0 2 N 1 k y k (1 y ) N 1 k (2.10) Clearly, P ( y ) / 0 for 0 y 1 Also,since y = sin 2 ( 2 ) ,allvaluesofyfallintheinterval[0,1]. WeshallstatetheaboveresultsasaTheorem: Theorem2.1 ThereexistapolynomialP(y),ofdegreeN-1,suchthat (1 y ) N P ( y )+ y N P (1 y )=1 wherethepolynomial P ( y )= N 1 k =0 2 N 1 k y k (1 y ) N 1 k (2.11) Tosummarize,wehaveaccomplishedthefollowing:Inordertocreateasufcientlysmooth scalingfunctionofcompactsupportwhoseintegertranslatesareorthogonal,theassociatedsymbol H ( ) canbewrittenintermsofatrigonometricpolynomial S ( ) thatsatisesLemma2.1.Wecan express | S ( ) | 2 asapolynomial P ( y ) thatsatises(2.9).Theorem2.1guaranteestheexistenceof thispolynomial. 14

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Havingnowfoundpolynomial P ( y ) wewishtoworkbackwardstondtheexplicitformof L ( ) Wecandothisbysimplysubstituting y =sin 2 ( 2 )= 1 cos 2 ( ) 2 into(2.11),weobtain P ( y )= N 1 k =0 2 N 1 k .1 cos( ) 2 k 1 1 cos( ) 2 .. N 1 k = N =1 k =0 2 N 1 k .1 cos 2 k 1+ cos ( ) 2 N 1 k Thuswehavefound L ( ) using P ( y ) Wenowwishtondanexplicitformulafor S ( ) using L ( ) Todothis,weneedtofactor L ( ) insuchawaythatbybuilding S ( ) usinghalfofthefactorsfrom L ( ) weobtain | S ( ) | 2 = L ( ) Since P ( y )= L ( ) / 0 $ [ ! ] andsince L ( ) isatrigonometricpolynomialofdegree N 1 ,wecanapplyawell-knownresultfromharmonicanalysisinordertofactor L ( ) Theorem2.2(Fej er-RieszTheorem,[9]) Atrigonometricpolynomial L ( )= / A j = A c j e ij thatsatises L ( ) / 0 forall $ [ ! ] isexpressibleintheform L ( )= | F ( z ) | 2 forsomepolynomial F ( z ) with z = e i whichtakestheform F ( z )= c A 7 j =1 ( z $ j ) where $ j satisfy | $ j | 1 Adirectapplicationofthistheoremallowsustofactor L ( ) as L ( )= | F ( z ) | 2 where F ( z )= c N 1 7 j =1 ( z $ j ) Wecanobtain S ( ) bysimplysubstituting z = e i into F ( z ): S ( )= F ( e i )= N 1 k =0 a k e ik 15

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Notethatthisisthedesiredformfor S ( ) Toobtainoursymbol H ( ) wesubstitute S ( ) into (2.3): H ( )= 1+ e i 2 N S ( )= 1+ e i 2 N N 1 k =0 a k e ik Expanging H ( ) givesatrigonometricpolynomialofdegree 2 N 1 oftheform H ( )= 1 + 2 2 N 1 k =0 h k e ik wherethecoefcients h k arereal.Byconstruction, H ( ) satisesthedesiredorthonormalitycondition(2.2).Thereforeweknowthatthereisascalingfunction # ( t ) thatgeneratesamultiresolution analysis { V j } j # Z of L 2 ( R ) ADaubechiesscalingfunction # ( t ) whosesymbol H ( ) foragiven N obtainedthroughthisconstructioniscalledthe D 2 N scalingfunction .Thecorrespondinglter h k has 2 N nonzerocomponents,andisreferredtoasthe D2NDaubechiesscalinglter .Weshall lateruseDaubechies'algorithmtoconstructthe D 4 scalinglterscorrespondingtothe D 4 scaling function.Firstwewouldliketogiveamoredetaileddescriptionoftherootsofthepolynomial F ( z ) 2.2TheRootsof F ( z ) Inher1988paper,DaubechiespresentedanalternativeproofoftheFej er-Reisztheorem[2].Her proofwasbasedonthestructureoftherootsofthepolynomial F ( z ) Inthissection,wediscussher constructionofthepolynomial F ( z ) basedontheclassicationoftheseroots.Weshallfollowher constructionasdescribedin[8]. Recallthat P ( y )= N 1 k =0 2 N 1 k y k (1 y ) N 1 k (2.12) and L ( )= P ( y ) where y =sin 2 2 ( Werstbeginbywriting L ( ) as T ( z ) Writing L ( ) inthisformallowsustore-grouptheroots basedontheirlocationintermsoftheunitcircle.Classifyingtherootsinthiswaywillleadusto theconstructionof F ( z ) Byletting z = e i andbynotingthatwecanwrite 1+cos 2 = 1+ z 2 .1+ 1 z 2 16

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and 1 cos 2 = 1 z 2 .1 1 z 2 wecanre-write L ( ) as: L ( )= N 1 k =0 2 N 1 k .1 cos 2 k 1+ cos ( ) 2 N 1 k = N 1 k =0 -1 z 2 .1 1 z 2 .. k -1+ z 2 .1+ 1 z 2 .. N 1 k =: T ( z ) Fromthisform,itiseasilyseenthat T ( z )= T ( 1 z ) andthat z =1 and z = 1 arenotrootsofT(z). Byexpanding -1 z 2 .1 1 z 2 .. and -1+ z 2 .1+ 1 z 2 .. weobtain T ( z )= N 1 k =0 2 N 1 k ( z +2 z 1 ) k 4 k ( z +2+ z 1 ) N 1 k 4 N 1 k = 1 4 N 1 N 1 k =0 2 N 1 k ( z +2 z 1 ) k ( z +2+ z 1 ) N 1 k = $ z ( N 1) 2 N 2 7 k =1 ( z z k ) where $ istheleadingcoefcientand z k aretherootsof T ( z ) Inordertondapolynomial F ( z ) suchthat T ( z )= | F ( z ) | 2 = F ( z ) F ( z ) itwouldbeidealtogrouptherootsof T ( z ) usingconjugatepairs.NotethattheFej er-Riesztheorem guaranteestheexistenceofthepolynomial F ( z ) Therootsof T ( z ) canbereal,ontheunitcircle, ornotrealandnotontheunitcircle.Letusconsidereachofthesecasesinturn. For z k $ R wehavethat z k = z k Fromearlier,wenotedthat T ( z )= T ( 1 z ) Thus,if z k =0 is arootof T ( z ) thensois 1 z k Notethatsinceweknowthat 1 arenotrootsof T ( z ) thisputseach realrooteitherinsideoroutsideoftheunitcircle,i.e., | z k | < 1 or | z k | > 1 Weshalldenotethe numberofpairsofsuchrootsby K andweshalldenotetherootsinsidetheunitcircleby r k For z k $ C \ R either z k liesontheunitcircleor z k liesofftheunitcircle.Letusdenoteroots ontheunitcircleby z u j anddenoterootsofftheunitcircleby z c i 17

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Foreachroot z u j ontheunitcircle, z u j = 1 z u j isalsoaroot.Itcanbeshownthatthemultiplicityof theserootsiseven(see[9]),sowewillgrouptherootsinpairsoftwo.Weshalldenotethenumber ofthesepairsby J Foreachroot z c i offtheunitcircle,thenbecause T ( z )= T ( 1 z ) ,wehavethat 1 z c i isalsoaroot. Since T ( z ) hasrealcoeffcients,if z c i isaroot,thensois z c i Inotherwords,if z c i isroot,thensoare 1 z c i z c i and 1 z c i Nowagain,sincerootsofthistypeareeitherinsideoroutsideoftheunitcircle,we havethateither | z c i | < 1 or | 1 z c i | < 1 whichresultsineither | z c i | > 1 or | 1 z c i | > 1 respectively.We shallcreategroupsof4anddenotethenumberofsuchgroupsby L Lemma2.3 Let z = e i .If T ( z )= 1 4 N 1 N 1 k =0 2 N 1 k ( z +2 z 1 ) k ( z +2+ z 1 ) N 1 k thenthereexistsapolynomial F ( z ) suchthat | F ( z ) | 2 = T ( z ) and F ( z ) isoftheform F ( z )= 8 | $ | 9 L 7 i =1 | z c i | 1 K 7 k =1 | r k | 1 2 : L 7 i =1 ( z z c i )( z z c i ) J 7 j =1 ( z z u j )( z 1 /z u j ) K 7 k =1 ( z r k ) where r k arerealwith | r k | < 1 ,z c i arein C \ R with | z c i | < 1 and | z c u | =1 Proof. Weshallrstwrite T ( z ) asaproductofitsfactors,whereeachofitsfactorsisgrouped accordingtothelocationoftheroots: T ( z )= $ z ( N 1) 2 N 2 7 k =1 ( z z k ) = $ z N 1 L 7 i =1 ( z z c i )( z 1 /z c i )( z z c i )( z 1 / z c 1 ) J 7 j =1 ( z z u i ) 2 ( z 1 /z u i ) 2 K 7 k =1 ( z r k )( z 1 /r k ) Byobservingthatbecause L ( )= | L ( ) | wehavethat T ( z )= | T ( z ) | andbyregrouping complexrootsonandofftheunitcircle,weobtain | T ( z ) | = | $ | L 7 i =1 | ( z z c i )( z 1 / z c i ) || ( z z c i )( z 1 /z c i ) | J 7 j =1 | ( z z u j )( z 1 /z u j ) | 2 K 7 k =1 | ( z r k )( z 1 /r k ) | Nowsince | z | =1 ,wecanwrite | ( z z c i )( z 1 /z c i ) | = | z c i | 1 | z z c i | 2 18

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Similarly, | ( z z c i )( z 1 /z c i ) | = | z c i | 1 | z z c i | 2 and | ( z r k )( z 1 /r k ) | = | r k | 1 | z r k | 2 Makingtheabovesubstitutionsgives | T ( z ) | = | $ | L 7 i =1 | z c i | 1 | z z c i | 2 | z c i | 1 | z z c i | 2 J 7 j =1 | ( z z u j )( z 1 /z u j ) | 2 K 7 k =1 | r k | 1 | z r k | 2 = | $ | L 7 i =1 | z c i | 2 | z z c i | 2 | z z c i | 2 J 7 j =1 | ( z z u j )( z 1 /z u j ) | 2 K 7 k =1 | r k | 1 | z r k | 2 CreatethesquarerootofT(z),denotedF(z),bychoosingallfactorswithrootsinsidetheunitcircle andonefactorfromeachdoubleroot z u j ontheunitcircle.Thus F ( z ) isdenedas F ( z )= 8 | $ | 9 L 7 i =1 | z C i | 1 K 7 k =1 | r k | 1 2 : L 7 i =1 ( z z c i )( z z C i ) J 7 j =1 ( z z u j )( z 1 /z u j ) K 7 k =1 ( z r k ) Notethat | F ( z ) | 2 = | T ( z ) | = T ( z ) asdesired,andthatthedegreeof F ( z ) is N 1 since thedegreeofT(z)is 2 N 2 NotealsothatF(z)hasrealcoefcients.Thuswehavethedesired polynomial F ( z ) 2.3TheConstructionofthe D 4 scalinglter Inthissection,wewillconstructtheD4scalinglterassociatedwiththeD4scalingfunctionusing thealgorithmofDaubechies.Forclarity,wewillmakeuseoftheconstructionpresentedinthe previoussection.Weshallbeginbyndingthepolynomial P ( y ) Recallthat P ( y ) hastheform P ( y )= N 1 k =0 2 N 1 k y k (1 y ) N 1 k (2.13) Sinceweknowthat N =2 weobtain P ( y )= 1 k =0 3 k y k (1 y ) 1 k =1+2 y. (2.14) 19

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Wenowneedtore-write P ( y ) as L ( ) Substituting y = 1 cos ( ) 2 gives P ( y )=1+2 y =1+2 1 cos ( ) 2 = L ( ) Byusingtheidentity 1 cos ( 2 = 1 z 2 1 1 z 2 wecanconvert L ( ) into T ( z ): L ( )=1+2 1 z 2 (' 1 1 z 2 ( = 1 2 4 z 1 z ( = 1 2 1 z z 2 4 z +1 ( = T ( z ) Directcalculationgivesthattherootsof T ( z ) are z =2+ + 3 and z =2 + 3 Weshallchoosethe rootinsidetheunitcircletobuildthepolynomial F ( z ) Usingtheformulafor F ( z ) asdescribedin theprevioussection,wehave F ( z )= ; 1 2(2 + 3) z (2 + 3) ( Substituting z = e i into F ( z ) weobtain S ( ): F ( e i )= ; 1 2(2 + 3) e i (2 + 3) ( = S ( ) Wecanndoursymbol H ( ) byrstrecallingthatwewantedoursymboltobeoftheform H ( )= 1+ e i 2 ( N S ( ) Substitutionof S ( ) gives: H ( )= 1+ e i 2 2 < 1 2 .1 8 2 + 3 e i (2 + 3) ( (2.15) 20

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Furthersimplicationof H ( ) gives H ( )= 1 + 3 4 + 2 + 3 + 3 4 + 2 e i + 3+3 + 3 4 + 2 e 2 i + 1+ + 3 4 + 2 e 3 i (2.16) Thus,the D 4 scalingltercoefcientsare h 0 = 1+ + 3 4 + 2 h 1 = 3+3 + 3 4 + 2 h 2 = 3 + 3 4 + 2 and h 3 = 1 + 3 4 + 2 21

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Chapter3 TheCascadeAlgorithm Itturnsoutthatfor N / 3 itisnotpossibletocomeupwithaclosed-formfortheDaubechies 2 N scalingfunctions.TheCascadeAlgorithmwasdevelopedbyDaubechiesandLagariasinan attempttoobtaingoodapproximationsofthesescalingfunctions.Itwasproposedthatbytaking sequencesofapproximationsgivenbyiteratesbasedonarstguess,say # 0 ( t ) where # 0 ( t ) isthe characteristicfunctionontheinterval [0 1) andwitheachsuccessiveapproximationgivenby: # n +1 = + 2 M k =0 h k # n (2 t k ) (3.1) thatwewouldultimatelyhaveconvergenceoftheapproximationstotheactualscalingfunction # ( t ) associatedwiththesymbol H ( ) describedintheprevioussection.Infact,thisalgorithmproduces thescalingfunctiongivenanumberofotherrstguesses,butforthepurposesofthispaper,we shallonlyconsidertherstguessof # 0 ( t ) Inthischapter,weshallshowthatwehaveconvergenceoftheiteratesinthetimedomain,given # 0 ( t ) Byrstshowingthat,foreach $ R wehaveconvergenceoftheiteratesinthetransform domaintoacontinuousfunction g ( ) andthenshowingthatthis g ( ) satisesthetransformdomain dilationequation,weshallbeabletoshowpointwiseconvergenceoftheiterates { # n ( t ) } tothe scalingfunction # ( t ) Thefollowingargumentisbasedontheoutlineaspresentedin[8]. Proposition3.1 [see[8],p.258]Supposethesymbolgivenby H ( )= 1 + 2 M k =0 h k e ik (3.2) satises H (0)=1 Thenthecascadealgorithmiterates { = # n ( ) } denedby > # n ( )= n 7 k =1 H 2 k 1 + 2 e i 2 n +1 sinc 2 n +1 (3.3) convergesforeach $ R inthetransformdomaintoacontinuousfunction g ( ) 22

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Beforeprovingtheaboveproposition,recallthatsinc(t) = sin( t ) t andnotethatthefunctionappearing ontherighthandsideof(3.3)isanaturalchoice,sincetheFouriertransformfor # 0 ( ) is > # 0 ( )= 1 2 e i 2 sinc 2 ) Proof. Weshallprovethepropositionviaaseriesoftwoclaims.Theresultoftherstclaimis neededtogiveanexplicitformfor g ( ) Oncewehaveour g ( ) oursecondclaimwillshowthat wehaveconvergenceofthecascadealgorithmiteratestothis g ( ) inthetransformdomain. Claim3.1 k =1 2 2 2 2 H 2 k 1 2 2 2 2 convergestoanitevalue. Proof. Let H ( )= 1 $ 2 / M k =0 h k e ik Weshallre-write H ( ) as H ( )= 1 + 2 M k =0 h k e ik 1 + 2 M k =0 h k + 1 + 2 M k =0 h k = 1 + 2 M k =0 h k e ik h k ( + 1 + 2 M k =0 h k = 1 + 2 M k =0 h k ( e ik 1)+ 1 + 2 M k =0 h k Now,since H (0)= 1 $ 2 / M k =0 h k =1 ,byourhypothesis,wehave H ( )= 1 + 2 M k =0 h k ( e ik 1)+ H (0) = 1 + 2 M k =0 h k ( e ik 1)+1 23

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Weknowthat 1 e i =2 ie i 2 sin( 2 ) Thisgivesusthat 2 2 2 2 1 + 2 M k =0 h k ( e ik 1) 2 2 2 2 = 2 2 2 2 1 + 2 M k =0 h k 2 ie ik 2 sin( k 2 ) 2 2 2 2 = 2 2 2 2 2 i + 2 M k =0 h k e ik 2 sin( k 2 ) 2 2 2 2 + 2 M k =0 | h k || e ik || sin( k 2 ) | = + 2 M k =0 | h k || sin( k 2 ) | Fromtheabove,wenowhavethat 2 2 2 2 H ( ) 1 2 2 2 2 = 2 2 2 2 1 + 2 M k =0 h k ( e ik 1) 2 2 2 2 + 2 M k =0 | h k | 2 2 2 2 sin( k 2 ) 2 2 2 2 + 2 M k =0 | h k | 2 2 2 2 k 2 2 2 2 2 = + 2 2 M k =0 | h k || k | = + 2 2 M k =0 | h k | k | | = | | + 2 2 M k =0 | h k | k Let % = $ 2 2 / M k =0 | h k | k. Thentheabovegives 2 2 2 2 H ( ) 1 2 2 2 2 % | | Byrelabelingandusinggeometricseries,weobtain k =1 2 2 2 2 H ( 2 k ) 1 2 2 2 2 k =1 % 2 2 2 2 2 k 2 2 2 2 = k =1 % | | 2 2 2 2 1 2 k 2 2 2 2 = % | | 24

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Therefore,wehavethat / k =1 | H ( 2 k ) 1 | convergesforeach in R and,infact,converges uniformlyoncompactsubsetsof R Nowsince / k =1 | H ( 2 k ) 1 | convergesabsolutely,byatheoremincomplexanalysis(see[11], Theorem8.33,p.238),theproductgivenby ? k =1 H ( 2 k ) convergesabsolutely,andtherefore,there existsa g ( ) suchthat g ( )= 1 + 2 7 k =1 H 0 2 k 1 Nowthatwehavegivenanexplicitformulafor g ( ) wecannowshowthatwehaveconvergence ofthecascadealgorithmiteratesto g ( ) Claim3.2 Thesequenceofcascadealgorithmiterates > # n ( ) convergesto g ( ) Proof. Dene > # n ( )= n 7 k =1 H 0 2 k 1 1 + 2 e i 2 n +1 sinc 0 2 n +1 1 Then lim n %! > # n ( )=lim n %! n 7 k =1 H 0 2 k 1 ( 1 + 2 e i 2 n +1 sinc( 2 n +1 ) = 1 + 2 lim n %! n 7 k =1 H ( 2 k ) = g ( ) since e i 2 n +1 sinc( 2 n +1 ) 0 1 as n 0# Itremainstoshowthat g ( ) iscontinuous.Notethateach > # n ( ) iscontinuousbyconstruction. since,fromclaims, > # n ( ) 0 g ( ) uniformlyoncompactsets,wehavethat g ( ) iscontinuous. ThiscompletestheproofofProposition3.1. Nextweshallshowthat g ( ) asfoundabove,satisesthetransformdomaindilationequation whenthesymbol H ( z ) isobtainedfromaDaubechiesscalingfunction. Proposition3.2 [see[8],p.258]Supposethesymbol H ( z )= 1 + 2 M k =0 h k e ik 25

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where z = e i canbefactoredas H ( z )= 1+ z 2 N S ( z ) where S ( z ) satises S (1)=1 and max | z | =1 | S ( z ) | 2 N 1 Then g ( ) satisesthetransform domaindilationequation g ( )= H ( 2 ) g ( 2 ) (3.4) and | g ( ) | C 1+ | | forsomeconstantC. Proof. First,notethat g ( ) satisesthetransformdomaindilationequation g ( )= H ( 2 ) g ( 2 ) sincebyProposition3.1,wehave g ( )= 1 + 2 lim n %! n 7 k =1 H ( 2 k ) andthereforewecanre-writetherighthandsideof(3.4)as H ( 2 ) g ( 2 )= H ( 2 ) 1 + 2 lim n %! n 7 k =2 H ( 2 k ) = 1 + 2 lim n %! n 7 k =1 H 0 2 k 1 = g ( ) Thus g ( ) satisesthedilationequationinthetransformdomain.Itremainstoshowthat g ( ) C 1+ | | Claim3.3 7 k =1 1+ e i 2 k 2 N = 1 e i i N 26

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Proof. Observethat 7 k =1 1+ e i 2 k 2 N =lim n %! n 7 k =1 1+ e i 2 k 2 N Now lim n %! n 7 k =1 1+ e i 2 k 2 N =lim n %! n 7 k =1 1+ e i 2 k 2 1 e i 2 k 1 e i 2 k N =lim n %! n 7 k =1 1 e i 2 k 1 2(1 e i 2 k ) N =lim n %! 1 2 n 1 e i 1 e i 2 n N Now 1 e i 2 n =2 ie i 2 n +1 sin( 2 n +1 ) sowehavethat lim n %! 1 2 n 1 e i 1 e i 2 n N =lim n %! 1 2 n 1 e i 2 ie i 2 n +1 sin( 2 n +1 ) N =lim n %! 1 2 n 1 e i 2 ie i 2 n +1 sin( 2 n +1 ) N 2 n +1 2 n +1 N =lim n %! ( 2 n +1 ) N (1 e i ) N (sin( 2 n +1 )) N ( i e i 2 n +1 ) N Nowweknowthat lim n %! ( 2 n +1 ) sin( 2 n +1 ) N =1 since lim x % 0 x sin x =1 Soweareleftwith lim n %! (1 e i ) N ( i e i 2 n +1 ) N = 1 e i i N since e i 2 n +1 0 1 as n 0# Thuswehave 7 k =1 1+ e i 2 k 2 N = 1 e i i N andClaim3.3isshown,asrequired. Now g ( )= 1 + 2 ! 7 k =1 H 2 k = 1 + 2 ! 7 k =1 1+ e i 2 k 2 N S 2 k 27

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bythefactorizationof H ( z ) fromourhypothesisandfromourpreviousclaim,sowehavethat g ( )= 1 + 2 ! 7 k =1 1+ e i 2 k 2 N S 2 k = 1 + 2 1 e i i N 7 k =1 S 2 k . Thusinordertoshowthat | g ( ) | C 1+ | | weneedtoobtainestimatesforboth | 1 e i i | N and 2 2 2 2 ? k =1 S ( 2 k ) 2 2 2 2 Claim3.4 | 1 e i i | N 2 N min(1 | | N ) Proof. Notethatwereallyneedtoshowthat 2 2 2 2 1 e i i 2 2 2 2 N 2 N and 2 2 2 2 1 e i i 2 2 2 2 N 2 N | | N Bythetriangleinequality,wehavethat | 1+ e i | 1+ | e i | =2 Thus, | 1 e i | | i | = | 1 e i | | | 2 | | Therefore 2 2 2 2 1 e i i 2 2 2 2 N 2 N | | N Soif 1 | | N 1 then min(1 1 | | N )= 1 | | N andfromtheargumentabove,wehave 2 2 2 2 1 e i i 2 2 2 2 N 2 N | | N =2 N min(1 | | N ) If 1 | | N / 1 then | | N 1 whichimplies | | 1 So min(1 | | N )=1 Notethat | 1 e i | = | 2 ie i 2 sin( 2 ) | =2 | sin( 2 ) | So 2 2 2 2 1 e i i 2 2 2 2 = 2 | sin( 2 ) | | | =2 2 2 2 2 sin( 2 ) 2 2 2 2 Notethat 2 2 2 2 sin( 2 ) 2 2 2 2 1 $ R Thuswehavethat 2 2 2 2 1 e i i 2 2 2 2 =2 2 2 2 2 sin( 2 ) 2 2 2 2 2 28

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Therefore 2 2 2 2 1 e i i 2 2 2 2 N 2 N andwehavethat 2 2 2 2 1 e i i 2 2 2 2 N 2 N min(1 1 | | N ) asrequiredforClaim3.5. Nowletusgiveanestimatefor ? k =1 S ( 2 k ) Forsimplicity,let T ( )= ? k =1 S ( 2 k ) FixMand suchthat 2 M 1 | | 2 M Then T ( )= 7 k =1 S ( 2 k )= M 7 k =1 S ( 2 k ) 7 M +1 S ( 2 k ) Let l = k M. Thisgives T ( )= M 7 k =1 S 2 k 7 M +1 S 2 k = M 7 k =1 S 2 k 7 l =1 S 2 l + M = M 7 k =1 S 2 k 7 l =1 S 2 M 2 l = M 7 k =1 S 2 k T (2 M ) Let U =max | | & 1 | T ( ) | andnotethatsince | | 2 M 1 wehavethat | T (2 M ) | U. Sowehave,byourhypothesison S ( z ) | T ( ) | = 2 2 2 2 M 7 k =1 S 2 2 2 2 2 | T (2 M ) | (2 N 1 ) M | T (2 M ) | =(2 M ) N 1 | T (2 M ) | =(2 2 M 1 ) N 1 | T (2 M ) | (2 2 M 1 ) N 1 U. 29

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Becausewexed andM,weknowthat 2 M 1 | | Thentheabovegivesthat | T ( ) | 2 N 1 | | N 1 U =(2 | | ) N 1 U. ThisresultsholdsindependentofourchoiceofMandthuswillholdforall Therefore 2 2 2 2 7 k =1 S ( 2 k ) 2 2 2 2 (2 | | ) N 1 U Wehavenowshownthat | g ( ) | = 1 + 2 2 2 2 2 1 e i i 2 2 2 2 N 2 2 2 2 7 k =1 S 2 k 2 2 2 2 1 + 2 2 N min(1 | | N ) 2 N 1 | | N 1 U. Claim3.5 1 + 2 2 N min(1 | | N ) 2 N 1 | | N 1 U 2 2 N 1 U + 2 2 1+ | | Proof. Notethatwereallyonlyneedtoshowthat min(1 | | N ) | | N 1 2 1+ | | ,sincetherestof theinequalitycomesfromrearrangingterms.Firstassume | | 1 Then min(1 | | N )=1 and | | N 1 + | | N 2 1 | | N 1 + | | N 1 | | 2 1 | | N 1 (1+ | | ) 2 1 | | N 1 2 (1+ | | ) 1 min(1 | | N ) | | N 1 2 (1+ | | ) Nextassume | | > 1 Then min(1 | | N )= | | N and | | 1 2 1 | | 1 + | | 0 2 1 | | 1 (1+ | | ) 2 1 | | 1 2 (1+ | | ) 1 | | N | | N 1 2 (1+ | | ) 1 min(1 | | N ) | | N 1 2 (1+ | | ) Claim3.6isthusshown. 30

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Nowfromourseriesofclaims,wehavethat | g ( ) | = 1 + 2 2 2 2 2 1 e i i 2 2 2 2 N 2 2 2 2 7 k =1 S 2 k 2 2 2 2 1 + 2 2 N min(1 | | N ) 2 N 1 | | N 1 U 2 2 N 1 U + 2 2 1+ | | Let C = 2 2 N 1 U $ 2 2 Then | g ( ) | C 1+ | | Tosummarize,wehavenowshownthatbyputtingtheconditionsonoursymbol H ( z ) aslisted inProposition3.2,wehaveconvergenceofthecascadealgorithmiterates { = # n ( ) } toacontinuous function g ( ) foreach $ R andthatthis g ( ) satisesthedilationequationinthetransform domain.Nextweshallshowconvergenceoftheiterates { # n ( t ) } toafunction { # ( t ) } inthetime domain. Theorem3.1 [see[8],p.261]Supposethesymbol H ( z )= 1 + 2 M k =0 h k e ik canbefactoredas H ( z )= 1+ z 2 N S ( z ) where S ( z ) satises S (1)=1 and max | z | =1 | S ( z ) | 2 N 1 Thenthesequenceoffunctions { # n ( t ) } denedby # n +1 = + 2 M k =0 h k # n (2 t k ) with # 0 ( t ) asourinitialguess,convergesin L 2 ( R ) toafunction # ( t ) thatsatisesthedilation equationgivenby # ( t )= + 2 M k =0 h k # (2 t k ) Moreover, supp ( # n )=[0 ,M 2 n ( M 1)] for n / 1 and supp ( # )=[0 ,M ] 31

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Proof. Inordertotranslate g ( ) intothetimedomain,werstneedtoshowthat g ( ) $ L 2 ( R ) Thiscanbeseeneasily,sincebyProposition3.2,weknow # R | g ( ) | 2 d # R C 1+ | | 2 d anditisclearthat # R C 1+ | | 2 d < # Thuswehavethat g ( ) $ L 2 ( R ) Wecannowdene @ g ( )= " # ( ) astheFouriertransformof g ( ) ByProposition3.2,weknowthat g ( ) satisesthetransform domaindilationequationgivenby g ( )= H ( 2 ) g ( 2 )= 1 + 2 M k =0 h k e ik 2 g ( 2 ) SofromanearlierresultfromChapter1(see(1.1)),wehavethat # ( t ) satisesthedilationequation, givenby # ( t )= + 2 k # Z h k # (2 t k ) inthetimedomain.NowinProposition3.1,weshowedthat = # n ( ) 0 = # ( )= g ( ) Usingthis fact,onecanshow(see[11],Theorem8.36,p.243),that # n ( t ) 0 # ( t ) in L 2 ( R ) Itremainstoshowthat supp ( # n )=[0 ,M 2 n ( M 1)] for n / 1 andhence supp ( # )=[0 ,M ] Recallthat supp ( # ) isthecompactsupportof # ( t ) aspreviouslydenedinChapter1.Weshall showthat supp ( # n )=[0 ,M 2 n ( M 1)] usinginduction.Firstobservethat # 1 ( t )= + 2 M k =0 h k # 0 (2 t k )= + 2 M k =0 h k B 0 (2 t k ) since,byourhypothesis, # 0 ( t ) isourinitialguess.Recallthat # 0 ( t ) isdenedas # 0 ( t )= $ % & 10 t< 1 0 otherwise Weknowthat # 0 ( t ) hasasupportof [0 1) bythedenitionofthefunction.Thus,weknowthat thesupportof # 0 (2 t k ) hasasupportof [ k 2 k +1 2 ) Now,thesesupportintervalsdonotoverlap,so 32

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wehavethat supp ( # 1 ( t ))= M k =0 9 k 2 k +1 2 : = 9 0 M +1 2 : =[0 ,M 2 1 ( M 1)] Nowassumethatfor # n ( t ) supp ( # n ( t ))=[0 ,M 2 n ( M 1)]=[0 ,A ] andconsider # n +1 ( t )= + 2 M k =0 h k # n (2 t k ) Byourinductionhypothesis, # n ( t ) hascompactsupport [0 ,A ] sothecompactsupportof # n (2 t k ) is [ k 2 A + k 2 ] Taking k =0 ontheleftsideoftheintervalgives 0 Taking k = M ontherightsideof theinterval,wehavethat A + k 2 = M 2 n ( M 1)+ M 2 = 2 M 2 n ( M 1) 2 = M ( M 1) 2 n +1 = M 2 ( n +1) ( M 1) Thuswehaveshownthat supp ( # n +1 )=[0 ,M 2 ( n +1) ( M 1)] Therefore, supp ( # n )=[0 ,M 2 n ( M 1)] for n / 1 Toshowthat supp ( # )=[0 ,M ] wesimplyneedtotakethelimitsofthesupportsoftheiterates, lim n %! supp ( # n ( t ))=lim n %! [0 ,M 2 n ( M 1)]=[0 ,M ] TheproofofTheorem3.1isnowcomplete. TheaboveresultsholdforDaubechiesscalingfunctions,asdescribedintheprevioussection.In fact,asadirectresultofTheorem2.9,wehavethefollowingcorollary: Corollary3.1 Suppose # ( t ) istheDaubechies D 2 N scalingfunctionwhereNisapositiveinteger. Then supp # ( t )=[0 2 N 1] Let # ( t ) betheDaubechies D 4 scalingfunction.Weknowthatthe D 4 scalingltercoeffcients are h 0 = 1+ + 3 4 + 2 33

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h 1 = 3+3 + 3 4 + 2 h 2 = 3 + 3 4 + 2 and h 3 = 1 + 3 4 + 2 FromTheorem3.1,weknowthatthescalingfunctioniteratesarerepresentedby # n +1 ( t )= + 2 2 N 1 k =0 h k # n (2 t k ) (3.5) Weknowthatthesupportof B 0 ( t )= # 0 ( t )=[0 1] Using(3.5),onecanshowthatthecompact supportof # 1 ( t ) is [0 2] Furthermore,thecompactsupportof # 2 ( t ) is [0 2 5] thecompactsupport of # 3 ( t ) is [0 2 75] andthecompactsupportof # 4 ( t ) is [0 2 875] Takingfurtheriteratesshows that supp # ( t )=[0 3] verifyingtheabovecorollaryfor N =2 PlottedonthefollowingpagesarethersteightiteratesofthecascadealgorithmfortheDaubechies D 4 scalingfunction,withaninitialguessof # 0 ( t )= B 0 ( t ) Onecanseefromthegraphsthatthe supportforthescalingfunctionis [0 3] andthatwithinaveryfewnumberofiterations,theCascade Algorithmgivesagoodestimateofthescalingfunction. Inconclusion,wehaveexploredDaubechies'constructionofscalingfunctionsthatarebothcompactlysupportedandsmooth,whicharedesiredpropertiesformanyapplications.Thesescaling functionsareassociatedwithamultiresolutionanalysis, { V j } j # Z ,whichisasequenceofsubspaces of L 2 ( R ) thatsatisesanumberofdifferentproperties.Whilenoclosed-formformulaforascaling functionexistsforvaluesof N / 3 wecanapproximatethescalingfunctionsusingtheCascade Algorithm,asrstpresentedbyDaubechiesandLagarias.Daubechiesresultsledtoanexplosion inthestudyofwaveletsinthe1990'sandresultedinmanymodern-dayapplicationsofwavelet theory,includingspeechdiscriminationandearthquakeprediction,aswellasngerprintanddata compression. 34

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0.5 1.0 1.5 2.0 2.5 3.0 0.2 0.2 0.4 0.6 0.8 1.0 1.2 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 Figure1. :IteratesoftheCascadeAlgorithmforD4scalingfunctionwithn=1,2,3,4 35

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0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 Figure2. :IteratesoftheCascadeAlgorithmforD4scalingfunctionwithn=5,6,7,8 36

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References [1] Burt,P.J.;E.H.Adelson, TheLaplacianpyramidasacompactimagecode ,IEEETrans. Comm.,COM-31(1983),pp.532-540. [2] Daubechies,I. Orthonormalbasesofcompactlysupportedwavelets ,Comm.PureAppl.Math., Vol.41(1988),pp.909-996. [3] Graps,Amara. AnIntroductiontoWavelets ,IEEEComp.ScienceandEngineering,Vol.2, No.2,1995. [4] Mallat,St ephane. Multiresolutionapproximationsandwaveletorthonormalbasesof l 2 ( R ) Trans.Amer.Math.Soc.,Vol.315(1989),pp.69-87. [5] Meyer,Yves. FundamentalPapersinWaveletTheory ;EditedbyChristopherHeilandDavid Walnut;pp.265-999.PrincetonUniversityPress:Princeton,NJ.2006. [6] Meyer,Yves. Wavelets:AlgorithmsandApplications ;TranslatedandRevisedbyRobertD. Ryan.;SocietyforIndustrialandAppliedMathematics:Philadelphia,PA.1993. [7] Meyer,Yves. WaveletsandOperators ;AdvancedMathematics.CambridgeUniversityPress, 1992. [8] Ruch,DavidK.;VanFleet,PatrickJ. WaveletTheory:AnElementaryApproachWithApplications ;JohnWileyandSons,Inc.:Hoboken,NJ.2009. [9] Sheil-Small,Terry. ComplexPolynomials ;CambridgeUniversityPress:NewYork,NY.2002. [10] Stein,EliasM.;Shakarachi,Rami. FourierAnalysis:AnIntroduction ;PrincetonUniversity Press:Princeton,NJ.2003. [11] Walnut,David. AnIntroductiontoWaveletAnalysis ;Birkhauser:Cambridge,MA.2002. 37