Kuliah 4 Interpolasi-NewtonDividedDifference

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    KAEDAH

    TERBAHAGI

    NEWTON(Newtons Divided Difference Polynomial Method)(Newtons Divided Difference Polynomial Method)

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    Bentuk umum:

    0 1 0 2 0 1 0 1 1( ) ( ) ( )( ) ... ( )( )...( )n n nf x b b x x b x x x x b x x x x x x = + + +

    0 1, ,... nb b b

    Data yang diberikan digunakan untuk mendapatkan nilai-nilaipekali;

    2

    Data yang diberi adalah

    0 0 1 1( , ( )),( , ( ))...( , ( ))n nx f x x f x x f x n + 1 data

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    NEWTONS DIIDED DI!!E"EN#E $ET%OD

    Daripada data& diper'leh:

    0 0 0

    1 0

    1 1 1 01 0

    1 02 1

    ( )

    ( ) ( )[ , ]

    ( ) ( )( ) ( )

    x x b f x

    f x f xx x b f x x

    x x

    f x f xf x f x

    = =

    = = =

    3

    2 1 1 02 2 2 1 0

    2 0

    1 1 0

    [ , , ]

    [ , , , ]n n n n

    x x b f x x xx x

    x x b f x x x x

    = = =

    = =

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    Diberi yang mana),,( 00 yx ),,( 11 yx

    )()( 0101 xxbbxf +=

    )( 00 xfb =

    INTERPOLASI NEWTON LINEAR

    4

    01

    01

    1)()(

    xxxfxfb

    =

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    CONTO !(")

    %ala(u menaik )ebuah r'ket diberikan )ebagai *ung)i ma)a

    )eperti dalam +adual 1, #ari hala(u pada t#!$ )aat menggunakan

    kaedah bea terbahagi Ne.t'n /interp'la)i linear0 ,

    t v(t)

    s m/s

    %

    0 0

    10 227.04

    15 362.78

    20 517.3522.5 602.97

    30 901.67

    +adual 1

    el'ity 2), time data *'r the r'ket e3ample

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    ,15=t 78.362=tv

    0 1 0 2 0 1 0 1 1( ) ( ) ( )( ) ... ( )( )...( )n n nf x b b x x b x x x x b x x x x x x = + + +

    $

    ,201 =t 35.517)( 1 =tv

    )( 00 tvb = 78.362=

    01

    01

    1

    )()(

    tt

    tvtv

    b

    = 914.30=

    )()( 010 ttbbtv +=

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    10 12 14 16 18 20 22 24350

    400

    450

    500

    550517.35

    362.78

    y s

    f range( )

    f x desired( )

    x s1

    10+x s0

    10 x s range, x desired,

    )()( 010 ttbbtv +=

    ),15(914.3078.362 += t 2015 t

    At 16=t

    )1516(914.3078.362)16( +=v

    69.393= m/s

    7

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    INTERPOLASI NEWTON KUADRATIK

    Diberi ),,( 00 yx ),,( 11 yx dan ),,( 22 yx memenuhi interpolasi kuadratik.

    ))(()()( 1020102 xxxxbxxbbxf ++=

    )( 00 xfb =

    8

    01

    011

    )()(xx

    xfxfb

    =

    02

    01

    01

    12

    12

    2

    )()()()(

    xx

    xx

    xfxf

    xx

    xfxf

    b

    =

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    CONTO !(&)

    %ala(u menaik )ebuah r'ket diberikan )ebagai *ung)i ma)a

    )eperti dalam +adual 1, #ari hala(u pada t#!$ )aat menggunakan

    kaedah bea terbahagi Ne.t'n /interp'la)i kuadratik0 ,

    t v(t)

    s m/s

    '

    0 0

    10 227.04

    15 362.78

    20 517.3522.5 602.97

    30 901.67

    +adual 1

    el'ity 2), time data *'r the r'ket e3ample

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    200

    250

    300

    350

    400

    450

    500

    550517.35

    227.04

    y s

    f range( )

    f x desired( )

    10

    2010 xs

    range, xdesired

    ,

    0 0

    1 1

    2 2

    10, ( ) 227.04

    15, ( ) 362.78

    20, ( ) 517.35

    t v t

    t v t

    t v t

    = =

    = =

    = =

    ))(()()( 1020102 xxxxbxxbbxf ++=

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    )( 00 tvb =

    04.227=

    01

    01

    1

    )()(

    tt

    tvtvb

    =

    1015

    04.22778.362

    =

    148.27=

    0112 )()()()( tvtvtvtv

    04.22778.36278.36235.517

    11

    02

    01122

    ttttttb

    =

    102010151520

    =

    10

    148.27914.30 =

    37660.0=

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    ))(()()( 102010 ttttbttbbtv ++=

    ),15)(10(37660.0)10(148.2704.227 ++= ttt 2010 t

    Pada ,16=t

    )1516)(1016(37660.0)1016(148.2704.227)16( ++=v 19.392= m/s

    Ralat mutlak relatif, di antara keputusan polinomial peringkat pertamaa

    12

    dan peringkat kedua adalah

    a 100x

    19.392

    69.39319.392 =

    = 0.38247 %

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    !O"$456 B67I INTE"8O56SI NEWTON

    946D"6TI9))(()()( 1020102 xxxxbxxbbxf ++=

    yang mana

    )(][ 000 xfxfb ==

    01 )()( xfxf

    !3

    +ika dituli) )emula& maka

    ))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf ++=

    01

    011 ,

    xx

    xx

    ==

    02

    01

    01

    12

    12

    02

    0112

    0122

    )()()()(

    ],[],[],,[

    xx

    xx

    xfxf

    xx

    xfxf

    xx

    xxfxxfxxxfb

    =

    ==

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    Polinomial peringkat ketiga, diberi ),,( 00 yx ),,( 11 yx ),,( 22 yx dan ),,( 33 yx

    adalah

    ))()(](,,,[

    ))(](,,[)](,[][)(

    2100123

    1001200103

    xxxxxxxxxxf

    xxxxxxxfxxxxfxfxf

    +

    ++=

    !O"$456 B67I INTE"8O56SI NEWTON 94BI9

    14

    0b

    0x )( 0xf 1b

    ],[ 01 xxf 2b

    1

    x )(1

    xf ],,[012

    xxxf

    3

    b

    ],[ 12 xxf ],,,[ 0123 xxxxf

    2x )( 2xf ],,[ 123 xxxf

    ],[ 23 xxf

    3x

    )( 3xf

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    CONTO !(C)

    %ala(u menaik )ebuah r'ket diberikan )ebagai *ung)i ma)a

    )eperti dalam +adual 1, #ari hala(u pada t#!$ )aat menggunakan

    kaedah bea terbahagi Ne.t'n /interp'la)i kubik0 ,

    t v(t)

    s m/s

    !%

    0 010 227.04

    15 362.78

    20 517.3522.5 602.97

    30 901.67

    +adual 1

    el'ity 2), time data *'r the r'ket e3ample

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    8r'*il hala(u adalah

    ))()(())(()()( 2103102010 ttttttbttttbttbbtv +++=

    Data yang menghampiri nilai adalah16=t

    ,100 =t 04.227)( 0 =tv

    ,151 =t 78.362)( 1 =tv

    !$

    ,2 .2

    ,5.223 =t 97.602)( 3 =tv

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    0b

    100 =t 04.227 1b

    148.27 2b

    ,151 =t 78.362 37660.0 3b

    914.30 3104347.5 x

    ,202 =t 35.517 44453.0

    248.34

    17

    b0 = 227.04; b1= 27.148; b2= 0.37660; b3= 5.4347*10-3

    ,5.223 =t 97.602

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    Maka

    ))()(())(()()( 2103102010 ttttttbttttbttbbtv +++=

    )20)(15)(10(10*4347.5

    )15)(10(37660.0)10(148.2704.2273

    +

    ++=

    ttt

    ttt

    Pada t = 16

    )2016)(1516)(1016(10*4347.5)1516)(1016(37660.0)1016(148.2704.227)16(

    3+

    ++=

    v

    06.392= m/s

    18

    Ralat mutlak relatif adalah

    a 100x

    06.392

    19.39206.392 =

    = 0.033158 %

    a

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    JADUAL PERBANDINGAN

    Peringkat

    Polinomial

    1 2 3

    v(t=16) 393.69 392.19 392.06

    m s

    Ralat MutlakRelatif

    ---------- 0.38247 % 0.033158 %

    19

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    ""*""* D"+P"D"D"+P"D" PPO,+-O,+- "-"."-".

    Dapatkan (arak yang dilalui 'leh r'ket ter)ebut pada ma)a

    t#!!s hin//a t#!$s

    t v(t)

    s m/s

    20

    10 227.04

    15 362.78

    20 517.35

    22.5 602.97

    30 901.67

    +adual 1

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    )20)(15)(10(10*4347.5

    )15)(10(37660.0)10(148.2704.227)(

    3 +

    ++=

    ttt

    ttttv

    5.2210 t

    32 0054347.013204.0265.212541.4 ttt +++= 5.2210 t

    Maka

    ( ) ( ) ( )=16

    11

    1116 dttvss

    21

    dtttt )0054347.013204.0265.212541.4( 32

    16

    11

    +++= 16

    11

    432

    40054347.0

    313204.0

    2265.212541.4

    +++=

    tttt

    m1605=

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    32 0054347.013204.0265.212541.4)( ttttv +++=

    ( )32 0054347.013204.0265.212541.4)()( tttdt

    dtv

    dt

    dta +++==

    P0C.T"N D"+P"D" PO,+- "-".

    Dapatkan peutan r'ket pada t1

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    BBENTUKENTUK UUMUMMUM

    Diberi )1( +n data iaitu ( ) ( ) ( ) ( )nnnn yxyxyxyx ,,,,......,,,, 111100 sebagai

    ))...()((....)()( 110010 +++= nnn xxxxxxbxxbbxf

    yang mana

    xb =

    ],[ 011 xxfb =

    ],,[ 0122 xxxfb =

    ],....,,[ 0211 xxxfb nnn =

    ],....,,[ 01 xxxfb nnn =

    23

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    NEWTONS DIIDED DI!!E"EN#E $ET%ODdan

    ( ) ( )[ , ]

    ( ) ( ) ( ) ( )

    [ , ] [ , ]

    i j

    i j

    i j

    i j j k

    i j j k i j j k

    f x f xf x x

    x x

    f x f x f x f x

    x x x x f x x f x x

    =

    24

    1 1 1 2 1 0

    1 1 00

    , ,

    [ , , ] [ , , ]

    [ , , , ]

    i j k

    i k i k

    n n n n

    n nn

    x x x x

    f x x x f x x x xf x x x x

    x x

    = =

    ==

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    NEWTONS DIIDED DI!!E"EN#E $ET%OD

    0 1 0 0 2 1 0 0 1

    1 1 0 0 1 1

    maka diperoleh:

    ( ) ( ) [ , ]( ) [ , , ]( )( )

    [ , , , ]( )( ) ( )

    n

    n n n

    n

    f x f x f x x x x f x x x x x x x

    f x x x x x x x x x x

    = + +

    +

    2%

    0 1 1 0 0 1 1

    1

    , , ,k k kk

    =

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    0

    1

    ( )

    ( )

    ( )

    f x

    f x

    f x

    0

    1

    x

    x 1 0[ , ]f x x

    [ , ]f x x x

    x [ , ]i jf x x [ , , ]i j kf x x x

    26

    2

    3

    ( )

    ( )

    ( )n

    f x

    f x

    f x

    2

    3

    n

    x

    x

    x

    2 1

    3 2

    1

    ,

    [ , ]

    [ , ]n n

    f x x

    f x x

    3 2 1

    1 2

    [ , ]

    [ , , ]n n n

    f x x x

    f x x x

    1 0[ , , ]n nf x x x