fungsi gamma beta (1).pdf
TRANSCRIPT
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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