Fungsi Gamma Beta (1).pdf

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    KALKULUS 4

    Dr. D. L. Crispina Pardede, SSi., DEA.

    SARMAG TEKNIK MESIN

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    1. Deret Fourier

    1.1. Fungsi Periodik

    1.2. Fungsi Genap dan Ganjil,1.3. Deret Trigonometri,1.4. Bentuk umum Deret Fourier,1.5. Kondisi Dirichlet,1.6. Deret Fourier sinus atau cosinus separuh jangkauan.

    KALKULUS 4 - SILABUS

    .

    3. Transformasi Laplace

    3.1. Definisi dan sifat Transformasi Laplace3.2. Invers dari transformasi Laplace3.3. Teorema Konvolusi

    3.4. Penerapan transformasi Laplace dalam penyelesaian P. D.dengan syarat batas.

    4. Fungsi Gamma dan Fungsi Beta

    4.1. Fungsi Gamma

    4.2. Fungsi Beta4.3. Penerapan fungsi Gamma dan fungsi Beta

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    Tabel Nilai Fungsi Gamma

    4.1. Fungsi Gamma

    n (n)

    1,00 1,0000

    1,10 0,9514

    1 20 0,9182

    1,30 0,89751,40 0,88731,50 0,8862

    1,60 0,8935

    1,70 0,90861,80 0,9314

    1,90 0,9618

    2,00 1,0000

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    Grafik Fungsi Gamma

    4.1. Fungsi Gamma

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    FUNGSI GAMMA : (n)

    4.1. FUNGSI GAMMA

    ==

    bx1nx1n

    konvergen untuk n>0

    0

    b0

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    Contoh:

    4.1. Fungsi Gamma

    dxex)1(

    bx11

    x

    0

    11

    =

    [ ] [ ] 1eelimelimdxelim

    xexm

    0b

    b

    b

    0

    x

    b

    b

    0

    x

    b

    0b

    =+==

    =

    =

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    Contoh:

    4.1. Fungsi Gamma

    dxex)2(

    bx1

    x

    0

    12

    =

    ..........

    xexm0

    b

    =

    =

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    Rumus Rekursi dari Fungsi Gamma

    (n+1) = n (n)dimana (1) = 1

    4.1. Fungsi Gamma

    Contoh:

    1. (2) = (1+1) = 1 (1) = 1.

    2. (3) = (2+1) = 2 (2) = 2.

    3. (3/2) = ( + 1 ) = ().

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    Bila n bilangan bulat positif

    (n+1) = n!dimana (1) = 1

    4.1. Fungsi Gamma

    Contoh:

    1. (2) = (1+1) = 1! = 1.

    2. (3) = (2+1) = 2! = 2.

    3. (4) = (3+1) = 3! = 6.

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    Contoh:

    Hitunglah4. (6)

    4.1. Fungsi Gamma

    . (3)

    6. (6)/2(3)

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    Bila n bilangan pecahan positif

    (n) = (n-1) . (n-2) . ( )dimana 0 < < 1

    4.1. Fungsi Gamma

    Contoh:

    1. (3/2) = (1/2) (1/2)

    2. (7/2) = (5/2)(3/2)(1/2)(1/2)

    3. (5/3) = (2/3)(2/3).

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    Bila n bilangan pecahan negatif

    4.1. Fungsi Gamma

    n

    )1n()n( +=

    atau

    )...1n(n

    )mn()n(

    +=

    m bilangan

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    Contoh:

    4.1. Fungsi Gamma

    +

    =

    =

    +

    =

    121

    211

    23

    3

    =

    =

    4

    32

    1

    2

    1

    2

    32

    1

    2222

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    Contoh:

    4.1. Fungsi Gamma

    +

    =

    =

    +

    =

    2

    3

    2

    5

    1

    2

    3

    2

    52

    3

    2

    5

    1

    2

    5

    25

    =

    =

    +

    =

    =

    8

    15

    2

    1

    2

    1

    2

    3

    2

    5

    2

    1

    2

    1

    2

    3

    2

    5

    12

    2

    3

    2

    5

    2

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    Beberapa hubungan dalam fungsi gamma

    4.1. Fungsi Gamma

    = )( 21

    =

    n

    )1n()n(

    +=

    =

    nsin)n1()n(

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    Contoh Soal:

    4.1. Fungsi Gamma

    ( )

    21 2

    5

    .3

    25.1

    ( ) ( )( )

    ( )( )

    3

    25

    386

    .5

    5,55,23.4

    ( )( )2

    12

    1

    .3

    2.

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    Penggunaan Fungsi Gamma

    ==

    dydxy2xMisalkan

    :awabJ

    21

    4.1. Fungsi Gamma

    0

    x26dxexHitung.1

    ==

    ==

    ymaka,xbila

    ,

    8452

    6!(7))(

    dyey)(dyey)(

    dyey)(dye)y(dxex

    7

    7

    21

    0

    y1-77

    21

    0

    y67

    21

    0

    y67

    21

    0

    21

    y6

    21

    0

    x26

    ===

    ==

    ==

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    4.1. Fungsi Gamma

    =

    0

    xysubstitusidengandyeyHitung.233y

    ==

    ==

    ==

    ymaka,xbila

    0ymaka0,xbilady3ydxxyMisalkan:awabJ

    23

    ===

    ==

    =

    =

    3

    1)(3

    1dxex3

    1

    dxex3

    1dxex

    3

    1

    dxx3

    1exdx

    x3

    1exdyey

    21

    x-

    0

    121

    x-

    0

    21

    x-

    0

    32

    61

    x-

    0

    61

    23

    1

    x-

    0

    31

    03

    2

    3y

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    4.1. Fungsi Gamma

    uln xsubstitusidenganln x-

    dx

    Hitung.3

    1

    0

    =

    due-xdexuln xMisalkan

    :awabJu-u-

    ===

    ,, ====

    ===

    ==

    )(dueu

    du)e(u

    u

    due-

    ln x-

    dx

    21

    0

    001

    0

    u-

    u--u

    21

    21

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    FUNGSI BETA dinyatakan sebagai

    4.2. FUNGSI BETA

    =1

    1n1mdx)x1(x)n,m(B

    konvergen untuk m > 0 dan n > 0.

    Sifat: B(m,n) = B(n,m)

    Bukti:

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    Bukti:

    4.2. Fungsi Beta

    dx)x1(x)n,m(B

    11n1m

    1

    0

    1n1m

    =

    =

    Terbukti

    )m,n(B

    dx)y1()y(

    1

    0

    1m1n

    0

    =

    =

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    HUBUNGAN

    Fungsi Beta dengan Fungsi Gamma

    4.2. Fungsi Beta

    )nm(

    )n,m(B +=

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    Contoh:

    1. Hitung B(3,5).Jawab:

    4.2. Fungsi Beta

    ......)8(

    )5()3()53()5()3()5,3(B =

    =

    +=

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    Contoh:

    2. Hitung B(5 , 2).Jawab:

    4.2. Fungsi Beta

    . 2 , .Jawab:

    4. Hitung B(1/3 ,2/3).

    Jawab:

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    Penggunaan Fungsi Beta

    4.2. Fungsi Beta

    1

    0

    34

    dx)x1(xHitung.1

    :Jawab

    280

    1

    !8!3!4

    4)(5(4)(5)

    )4,5(B

    dx)x1(xdx)x1(x0

    141-5

    0

    34

    =

    =+=

    =

    =

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    4.2. Fungsi Beta

    2

    0

    2

    x2

    dxxHitung.2

    du2dx2uxMisalkan:Jawab ==

    ...)(3,B24

    du)u1(u24u1

    duu2

    8

    u12

    duu8

    u22

    du2)u2(

    x2

    dxx

    21

    1

    0

    2

    1

    0

    2

    1

    0

    21

    0

    22

    0

    2

    21

    ==

    =

    =

    =

    =

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    Penggunaan Fungsi Beta

    4.2. Fungsi Beta

    dyyayHitung.3

    a

    0

    224

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    LATIHAN

    dxexHitung.20

    x4

    )(

    )()3(b).

    )3()4(2

    )7(a).Hitung.1

    29

    23

    dxexHitung.30

    duu)-4(uHitung.6

    dx)x1(xHitung.5

    42/53/2

    1

    32

    0

    0

    )32,3

    12

    3 (Bb).2),(Ba).Hitung.4