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    1

    CISE301: Numerical Methods

    Unit 1:Introduction to Numerical methods and

    Review of Taylor Series

    Lectures 1-3:

    Dr.Sayyid Anas Vaqar

    Dr.Sayyid Anas Vaqar OfficeHours

    UTR8:009:00AMand11:0011:11:30AM ,

    orbyappointment

    Office 221481

    Tel 8607876

    Email:

    usetheBlackboardemail

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    2

    GradingPolicy Standard

    Grading

    policy

    will

    be

    adapted

    Attendance 5%(1%perabsence)

    Quizzes 10%

    HWs/CompWork 10%

    MajorI 20% topics1,2,3(Date: Monday24th ofJune2013

    Time: 4:30PM 6:30PM)

    MajorII 25% topics4,5,6(Date: Saturday13th ofJuly 2013

    Time: 9:30PM 11:30PM)

    Final

    30%

    topics

    7,8,9

    (DatesetbyRegistrar,Duration=2hoursTime: 7:00PM9:00PM)

    Range Grade

    [95,100] A+

    A

    B+

    B

    C+

    C

    D+

    D

    [0,50 ) F9/9/2013 3

    RulesandRegulations

    Nomakeupquizzes/exams

    DNgrade 4 unexcusedabsences

    HomeworkAssignments

    are

    due

    to

    the

    beginningofthelectures.Absenceisnotanexcusefornotsubmitting

    theHomework. Latesubmissionmaynotbeaccepted

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    3

    LearningObjectives

    DefinenumericalmethodsandlisttopicscoveredinthecourseCompareanalyticsolutionsandnumericalsolutions;DiscusstheneedfornumericalmethodsDiscusstheneedforanalyzingthequalityofnumericalsolutions;

    Reading

    Assignment:pages310oftextbook

    AnalyticalSolutions Analyticalmethodsgiveexactsolutions

    Example:Analyticalmethodtoevaluatetheintegral

    Numericalmethodsaremathematicalprocedurestocalculateapproximatesolution

    (TrapezoidMethod)

    3

    1

    3

    0

    3

    1

    3

    1

    0

    31

    0

    2 x

    dxx

    21)1()0(

    201 221

    0

    2 dxx

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    Numerical

    MethodsNumericalMethods:Algorithmsthatareusedtoobtainapproximatesolutions ofamathematicalproblem.

    Whydoweneedthem?

    1.Noanalyticalsolutionexists,

    2.Ananalyticalsolutionisdifficulttoobtain

    ornotpractical.

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    WhatdoweneedBasicNeedsintheNumericalMethods:

    Practical:canbecomputedinareasonableamountoftime.

    Accurate:

    Goodapproximatetothetruevalue

    Informationabouttheapproximation error

    (Bounds,error

    order,

    )is

    available

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    5

    Outlines

    of

    the

    Course1.TaylorTheorem,NumberRepresentation,Errors

    2.SolutionofnonlinearEquations

    3.SolutionoflinearEquations

    4.LeastSquarescurvefitting

    5.Interpolation

    6.NumericalDifferentiation

    7.NumericalIntegration

    8.Solution

    of

    ordinary

    differential

    equations

    9.SolutionofPartialdifferentialequations

    9/9/2013 9

    SolutionofNonlinearEquations Somesimpleequationscanbesolvedanalytically

    Many

    other

    equations

    have

    no

    analytical

    solution

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    31

    )1(2

    )3)(1(444:solutionAnalytic

    034

    2

    2

    xandx

    roots

    xx

    solutionanalyticNo052 29

    xex

    xx

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    6

    MethodsforsolvingNonlinearEquationso BisectionMethod

    o NewtonRaphsonMethod

    o SecantMethod

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    SolutionofSystemsofLinearEquations

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    unknowns?1000inequations1000

    haveweifdoWhat to

    123,2

    523,3

    asitsolvecanWe

    52

    3

    12

    2221

    21

    21

    xx

    xxxx

    xx

    xx

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    Cramers

    Rule

    is

    not

    practical

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    compute.toyears10thanmoreneedscomputersuperA

    needed.aretionsmultiplica102.38system,30by30asolveTo

    tions.multiplica1)N!1)(N(NneedweunknownsNinequationsNsolveTo

    2

    21

    11

    51

    31

    ,1

    21

    11

    25

    13

    52

    3

    systemthesolvetousedbecanRulesCramer'

    20

    35

    21

    21

    21

    xx

    xx

    xx

    Cramers Ruleis not practicalfor largeproblems

    MethodsforsolvingSystemsofLinearEquations

    o NaiveGaussianElimination

    o GaussianEliminationwithScaledPartialpivoting

    o AlgorithmforTridiagonalEquations

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    8

    CurveFitting

    Givenasetofdata

    Selectacurvethatbestfitthedata.

    Onechoiceisfindthecurvesothat thesumofthe

    square

    of

    the

    error

    is

    minimized.

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    xi 0 1 2

    yi 0.5 10.3 15.3

    Interpolation

    Givenasetofdata

    findapolynomial

    P(x)whosegraphpasses

    throughalltabulatedpoints.

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    xi 0 1 2

    yi 0.5 10.3 15.3

    tablein theis)( iii xifxPy

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    Methodsfor

    Curve

    Fitting

    o LeastSquares

    o LinearRegression

    o NonlinearleastSquaresProblems

    o Interpolation

    o Newtonpolynomialinterpolation

    o Lagrangeinterpolation

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    Integration

    Somefunctionscanbeintegratedanalytically

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    ?

    solutionsanalyticalnohavefunctionsmanyBut

    dxe

    xxdx

    a

    x

    0

    3

    1

    23

    1

    2

    42

    1

    2

    9

    2

    1

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    Methodsfor

    Numerical

    Integration

    o UpperandLowerSums

    o TrapezoidMethod

    o RombergMethod

    o GaussQuadrature

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    SolutionofOrdinaryDifferentialEquations

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    onlycasesspecial

    foravailablearesolutionsAnalytical*

    equationsthesatisfiesthatfunctionais

    0)0(;1)0(

    0)(2)(3)(

    equationaldifferentitheosolution tA

    x(t)

    xx

    txtxtx

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    SolutionofPartialDifferential

    EquationsPartialDifferentialEquationsaremoredifficulttosolve

    thanordinarydifferentialequations

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    )sin()0,(,0),1(),0(

    022

    2

    2

    2

    xxututu

    t

    u

    x

    u

    Summary

    NumericalMethods:

    Algorithmsthatareusedtoobtainnumericalsolutionofamathematicalproblem.

    Weneedthemwhen

    Noanalyticalsolutionexistor itisdifficulttoobtain

    analyticalsolution.

    1. Introduction

    2. SolutionofnonlinearEquations

    3. SolutionoflinearEquations

    4. LeastSquaresCurvefitting

    5. Interpolation

    6. NumericalIntegration

    7. Numerical

    Differentiation

    8. Solutionofordinarydifferentialequations

    9. SolutionofPartialdifferentialequations

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    Topics Covered in the Course

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    NumberRepresentation NormalizedFloatingPointRepresentation SignificantDigits AccuracyandPrecision RoundingandChopping

    Readingassignment:Chapter3

    RepresentingRealNumbers

    Youarefamiliarwiththedecimalsystem

    DecimalSystem Base=10,Digits(0,1,9)

    StandardRepresentations

    21012 10510410210110345.312

    partpart

    fractionintegralsign

    54.213

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    NormalizedFloatingPointRepresentation

    NormalizedFloating

    Point

    Representation

    Nointegralpart,

    Advantage Efficientinrepresentingverysmallorverylargenumbers

    integer:,0

    exponentmantissasign

    10.0

    1

    4321

    nd

    ddddn

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    CalculatorExample supposeyouwanttocompute

    3.578*2.139

    usingacalculatorwithtwodigitfractions

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    3.57 * 2.13 7.60=

    7.653342True answer

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    BinarySystem

    BinarySystem Base=2,Digits{0,1}

    10 11 bb

    exponentmantissasign

    21.0 432n

    bbb

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    1010

    321

    2 )625.0()212021()101.0(

    Fact

    Numberthathavefiniteexpansioninonenumberingsystemmayhaveaninfiniteexpansioninanothernumberingsystem

    Youcanneverrepresent0.1exactlyinanycomputer

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    210 ...)011000001100110.0()1.0(

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    ConvertingDecimalFractionstoBinary

    A simple,stepbystepmethodforcomputingthebinaryexpansionontherighthandsideofthepoint.Wewillillustratethemethodbyconvertingthedecimalvalue.625toabinaryrepresentation..

    Step1:Beginwiththedecimalfractionandmultiplyby2.Thewholenumberpartoftheresultisthefirstbinarydigittotherightofthepoint.

    Because.625x2= 1.25,thefirstbinarydigittotherightofthepointisa 1.Sofar,wehave.625=.1???...(base2).

    Step2:Nextwedisregardthewholenumberpartofthepreviousresult(the1inthiscase)andmultiplyby2onceagain.Thewholenumberpartofthisnewresultisthe secondbinarydigittotherightofthepoint.Wewillcontinuethisprocessuntilwegetazeroasourdecimalpartoruntilwerecognizeaninfiniterepeatingpattern.Because.25x2= 0.50,thesecondbinarydigittotherightofthepointisa 0.Sofar,wehave.625=.10??...(base2).

    Step3:Disregardingthewholenumberpartofthepreviousresult(thisresultwas.50sothereactuallyisnowholenumberparttodisregardinthiscase),wemultiplyby2onceagain.Thewholenumberpartoftheresultisnowthenextbinarydigittotherightofthepoint.

    Because.50x2= 1.00,thethirdbinarydigittotherightofthepointisa 1.Sonowwehave.625=.101??...(base2).

    Step4:Infact,wedonotneedaStep4.WearefinishedinStep3,becausewehad0asthefractionalpartofourresultthere.

    Hencetherepresentationof.625=.101(base2).

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    ConvertingDecimalFractionstoBinary InfiniteBinaryFractions

    Themethodwejustexploredcanbeusedtodemonstrate howsomedecimalfractionswillproduceinfinite binaryfractionexpansions.Weillustrate byusingthatmethodtoseethatthebinaryrepresentation ofthedecimalfraction1/10is,infact,infinite.

    Step1:Beginwiththedecimalfractionandmultiplyby2.Thewholenumberpartoftheresultisthefirstbinarydigittotherightofthepoint.Because.1x2= 0.2,thefirstbinarydigittotherightofthepointisa 0.Sofar,wehave.1(decimal)=.0???...(base2).

    Step2:Nextwedisregardthewholenumberpartofthepreviousresult(0inthiscase)andmultiplyby2onceagain.Thewholenumberpartofthisnewresultisthe secondbinarydigittotherightofthepoint.Wewillcontinuethisprocessuntilwegetazeroasourdecimalpartoruntilwerecognizeaninfiniterepeatingpattern. Because.2x2= 0.4,thesecondbinarydigittotherightofthepointisalsoa 0.Sofar,wehave.1(decimal)=.00??...(base2).

    Step3:Disregardingthewholenumberpartoftheprevious result(againa0),wemultiplyby2onceagain.Thewholenumberpartoftheresultisnowthenextbinarydigittotherightofthepoint.Because.4x2= 0.8,thethirdbinarydigittotherightofthepointisalsoa 0.

    So

    now

    we

    have

    .1

    (decimal)

    =

    .000??

    .

    .

    .

    (base

    2)

    . Step4:Wemultiplyby2onceagain,disregarding thewholenumberpartofthepreviousresult(againa0inthiscase).Because

    .8x2= 1.6,thefourthbinarydigittotherightofthepointisa 1.Sonowwehave.1(decimal)=.0001??...(base2).

    Step5:Wemultiplyby2onceagain,disregarding thewholenumberpartofthepreviousresult(a1inthiscase).Because.6x2= 1.2,thefifthbinarydigittotherightofthepointisa 1.Sonowwehave.1(decimal)=.00011??...(base2).

    Step6:Wemultiplyby2onceagain,disregarding thewholenumberpartofthepreviousresult.Let'smakeanimportantobservationhere.Noticethatthisnextsteptobeperformed(multiply2.x2)is exactlythesameactionwehadinstep2.Wearethenboundtorepeatsteps25,thenreturntoStep2againindefinitely.Inotherwords,wewillnevergeta0asthedecimalfractionpartofourresult.Insteadwewilljustcyclethroughsteps25forever.Thismeanswewillobtainthesequence ofdigitsgeneratedinsteps25,namely0011,overandover.Hence,thefinalbinaryrepresentation willbe.

    .1(decimal)=.00011001100110011...(base2).

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    7BitRepresentation(sign:1bit,Mantissa3bits,exponent3bits)

    9/9/2013 31

    exponentmantissasign

    21.0 432n

    bbb

    7Bit

    Representation

    (sign:1bit,Mantissa3bits,exponent3bits)

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    Representation

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    Hypothetical Machine (real computers use 23 bit mantissa)

    Mantissa 2 bits exponent 2 bit sign 1 bit

    Possible machine numbers

    .25 .3125 .375 .4375 .5 .625 .75 .875

    1 1.25 1.5 1.75

    Remarks

    Numbersthatcanbeexactlyrepresentedarecalledmachinenumbers

    Differencebetweenmachinenumbersisnotuniform sumofmachinenumbersisnotnecessarilyamachinenumber

    0.25+.3125=0.5625

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    Machinenumber

    Machinenumber

    NOTMachinenumber

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    Significant

    Digits

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    Significant digits are those digits that can beused with confidence.

    0 1 2 3 4

    Length of green rectangle = 3.45

    significant

    LossofSignificance Mathematicaloperationsmayleadtoreducingthe

    numberofsignificantdigits

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    0.123466 E+02 6 significant digits

    0.123445 E+02 6 significant digits

    0.000021E+02 2 significant digits

    0. 210000E-02

    Subtracting nearly equal numbers causes loss of significance

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    Accuracy

    and

    Precision

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    Accuracy is related to closeness to the true value

    Precision is related to the closeness to other estimatedvalues

    AccuracyandPrecision

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    Accuracy isrelated tocloseness to thetrue value

    Precision isrelated to thecloseness to

    other estimatedvalues

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    AccuracyandPrecision

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    Better

    Precision

    Betteraccuracy

    Accuracy isrelated tocloseness to thetrue value

    Precision isrelated to thecloseness toother estimatedvalues

    RoundingandChopping Rounding: Replacethenumberbythenearest

    machinenumber

    Chopping:Throwallextradigits

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    0 1 2

    True 1.1681

    Rounding (1.2) Chopping (1.1)

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    ErrorDefinitions

    TrueError

    canbe

    computed

    if

    the

    true

    value

    is

    known

    100*valuetrue

    ionapproximatvaluetrue

    ErrorelativePercent RAbsolute

    ionapproximatvaluetrue

    ErrorTrueAbsolute

    t

    tE

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    100*valuetrue

    ionapproximatvaluetrue

    ErrorRelativeTrue

    ionapproximatvaluetrue

    ErrorTrue

    t

    tE

    ErrorDefinitionsEstimatederrorUsedwhenthetruevalueisnotknown

    100*estimatecurrent

    estimateprevoiusestimatecurrent

    ErrorelativePercent RAbsoluteEstimated

    estimateprevoiusestimatecurrent

    ErrorAbsoluteEstimated

    a

    aE

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    100*estimatecurrent

    estimateprevoiusestimatecurrent

    ErrorRelativePercenteApproximat

    estimateprevoiusestimatecurrent

    ErrorA

    a

    aE

    pproximate

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    Errorasstoppingcriterion

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    Itis

    often

    useful

    to

    employ

    the

    absolute

    value

    error

    as

    astoppingcriterion.

    Weareinterestedinwhetherthepercentabsoluteerrorislowerthanaprespecifiedpercenttolerance s

    sa

    Notation

    Wesaytheestimateiscorrecttondecimaldigitsif

    Wesaytheestimateiscorrecttondecimaldigitsrounded if

    n 1021Error

    n 10Error

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    Wesaytheestimateiscorrecttondecimaldigitsif

    Wesaytheestimateiscorrecttondecimaldigitsrounded if

    n 10Error

    n 102

    1Error

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    Notation

    Wesay

    the

    estimate

    is

    correct

    to

    ndecimal

    digits

    if

    Wesaytheestimateiscorrecttondecimaldigitsrounded if

    n 102

    1Error

    n 10Error

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    Wesaytheestimateiscorrecttoatleastnsignificantdigitsif:

    Summary

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    Number Representation

    Number that have finite expansion in one numbering system may

    have an infinite expansion in another numbering system.

    Normalized Floating Point Representation Efficient in representing very small or very large numbers

    Difference between machine numbers is not uniform Representation error depends on the number of bits used in the

    mantissa.

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    Summary Accuracy Precision

    Rounding Chopping

    ErrorDefinitions:

    Absolutetrueerror

    TruePercentrelativeerror

    Estimatedabsoluteerror

    Estimatedpercentrelativeerror

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