Fungsi Gamma Dan Beta
March 31, 2019 | Author: ANas NazRoe DHeen | Category: N/A
Short Description
kalkulus lanjut...
Description
KALK KA LKULU ULUS S LANJUT
MODUL 11 Fungs un gsii Gama dan Be B eta
Zuhair Jurusan Jur usan Te Teknik kni k Info Informatik rmatika a Universitas Mercu Buana Jakarta 2009.12.20
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Fungsi Gama dan Beta
I.
Fungsi Gamma ∞
Γ (n) = ∫
Definisi:
x
n−1
e
−x
dx
0
Konvergen untuk n > 0 Rumus rekursi fungsi gama:
Γ (n + 1) = n Γ (n) Contoh:
•
Γ (2)
=1
Γ ( 3 2 )
=
1
Γ 2
………………………………………. (1)
(1)
Γ (12 )
Untuk n = bilangan bulat positif → n = 1, 2, 3, ...
Γ
(n + 1) = n!
...................................................... (2)
Contoh:
Γ (2) = 1! = 1 Γ (3) = 2! = 2 •
Γ (5) = 4! Γ (6) = 5!
Untuk n = bilangan pecahan positif
Γ (n) = (n – 1) (n – 2) … α Γ (α), 0 < α < 1
Γ (7 2 ) =
Contoh:
Γ (5 3 ) = •
………………… (3)
2
3
.
5
2
.
3
2
.
1
2
.
Γ ( 1 2 )
Γ (2 3 )
Untuk n < 0, n ≠ -1, -2, …
Γ(n + 1)
Γ (n)
=
atau
Γ (n) =
………………… (4)
n
Γ (n + m) n (n + 1)...
……………….… (5)
1 4 24 3
m bilangan
m = bilangan bulat positif
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1 = Γ ( 2 ) = −2 1
Γ (− 1 2 )
−
2
Γ (− 5 2 ) = Γ ( − 3 2 ) = −
5
Γ ( 1 2 )
Γ (− −
2
5
2
1
2)
.−
3
= 2
=
−
5
2
Γ ( 12) .− 32 .−
1
2
− 815 Γ( 1 2 )
Jika pakai rumus (5)
Γ(− 5 2 + 3) = − 815 Γ ( 1 2 ) 5 3 1 − 2 .− 2 .− 2
Γ (− 5 2 ) =
Γ ( 12 ) = π
Ingatlah:
Γ (1)
• Γ (x) Γ (1-x) =
cari buktinya!
=1
π
………………..… (6)
sin x π
0 0 Buktikan !
Bukti:
Misalkan: y
= ln
1 x
= ln
x −1
= − ln
x
→ x = e− y dx = −e Jika x = 0
−y
dy
→ y = ∞
x = 1 → y = 0
⎛ 1 ⎞ ∫ ⎜ ln ⎟ ⎝ x ⎠
n −1
1 0
dx
= ∫ ∞ y n −1 . − e − y dy 0
∞
= ∫ 0 y n −1 e − y dy = Γ (n ) terbukti ! Contoh:
1
∫0
(ln ) 1 x
Ι = ∫01 =
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1 3
dx 4
(ln ) 1
x
3
−1
Γ (4 3 ) = 1 3 Γ ( 1 3 )
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II. Fungsi Beta Definisi: β (m, n) =
1 m-1 n-1 ∫ 0 x (1-x) dx
→ konvergen untuk m > 0, n > 0 Hubungan fungsi beta dengan fungsi gamma :
Γ
β (m, n) =
( m) . Γ ( n )
Γ ( m + n)
Contoh: 1
1) ∫ 0 x3 (1 - x)2 dx = β (4,3) =
Γ (4) . Γ (3) = Γ (4 + 3)
3 !. 2 ! 6!
=
6.2 6.5 !
=
2 1 = 120 60
2
2) ∫
3 0
x dx
3 − x
Penyelesaian: Misalkan
dx 9 u 2 3 du
Ι = ∫10
3 − 3u
= 3 du 27
=
27 3
27 3
=
=
∫0
1 − x x
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1
x
− 13
→
u=0
u=1
β (3, 1 2 )
Γ(3) .Γ( 12) Γ(7 2 ) 3
=
3
x=3
→
27
=
3) ∫ 10
= 3u x = 0
1 2 -1/2 ∫ 0 u (1 - u) du
3
=
x
.
48 5
2! 5
2
3
2
Γ ( 12 ) 1 1 2 Γ ( 2)
3
dx
(1 − x )
1
3
dx
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= β
(2 3 , 4 3 )
=
=
Γ (2 3 ) . Γ ( 4 3 ) Γ (2) Γ (2 3 ). 13 Γ ( 13 ) 1!
Γ ( 1 3 ) Γ (2 3 )
= 13 =
1
=
1
=
2
π 3
sin
1
3 π
π 3
1
2
3
3 π
9
•
2 m −1 θ . cos 2 n −1 θ dθ = ∫ 0 sin
•
p p ∫ 0 sin θ dθ = ∫ 0 cos θ dθ
π
π
2
π
2
1
2
β (m, n)
2
=
1 . 3 . 5 ... ( p − 1) π . 2 . 4 . 6 ... p 2
=
2 . 4 . 6 ... ( p − 1)
Jika p genap positif
Jika p ganjil positif
1 . 3 . 5 ... p
Contoh-contoh Soal π
1) ∫ 0 2 sin 4
θ . cos 2 θ dθ = Ι 2m – 1 = 4 → m = 5
cara 1:
2n – 1 = 2
Ι=
1
2 β (
5
2
,3
2 ) =
→ n=
3
2 2
Γ (5 2 ) . Γ (3 2 ) 2 Γ ( 4)
3
= 3
π
( 12 ). 12 Γ ( 1 2 )
8
π
12
=
π 32
4 2 ∫ 0 sin θ . cos θ dθ 2
π
= ∫ 0 2 sin
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. 1 2 .Γ
2.3 !
= cara 2:
2
4
θ . (1 – sin 2 θ) dθ
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π
= ∫ 0 2 sin 1. 3
=
=
π
π
= 2 ∫ 0 2 sin =2.
3π
.
−
16
π 2
1. 3 . 5
−
15π 96
2.4.6
=
π
.
2
18π − 15π 96
=
π 32
θ dθ
7
2) ∫ 0 sin
2.4
θ - sin 6 θ dθ
4
7
θ dθ
2.4.6 1. 3 . 5 . 7
=
96 105
=
32 35
Boleh juga dilakukan dengan cara berikut: π
2 ∫ 0 2 sin 7
θ dθ, 2m – 1 = 7 → 2n – 1 = 0
=2.
1
2
β (m, n) = β (m, n)
m=4
→
n = 1
= β (4, =
=
=
1
2
2
)
Γ (4) . Γ ( 12) Γ ( 9 2) 3 ! . Γ ( 12) 7
2
.
6 105 16
5
2
.
3
2
=6.
.
1
2
16 105
Γ ( 12 )
=
32 35
Mana cara yang lebih mudah, bisa dipilih dari cara penyelesaiannya. _____________________________________________________________________
SOAL-SOAL
1.
Γ (3) . Γ ( 3 2 ) =? Γ (9 2 )
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16
105
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∞
6
−3 x
2.
∫0
3.
∫0
4.
Tunjukkan ∫ 0
∞
x e 2
x e
80
dx =
− 2 x 2
2π
= ∞
16 e
− st
t
dt =
Petunjuk: Misalkan 5.
(ln x )
5
1
∫0
1
6.
∫ ( x
7.
Γ (− 3 2 ) =
8.
Γ (− 9 2 ) =
9.
β (3, 5) =
10.
0
π s
,s>0
y = st -120
dx
ln x ) dx = 3
− 3128 4
−
π
3
32 π 945 1 105
β ( 3 2 , 2) =
4
β ( 1 3 , 2 3 ) =
2π
∫ 0 (4 − x 2
2
)
3
2
15
3 dx
3π
(Petunjuk: misalkan x2 = 4y) 11.
243
4 3/2 5/2 ∫ 0 u (4 – u) du
12π
(Petunjuk: misalkan u = 4x) π
12.
4 4 ∫ 0 sin θ cos θ dθ =
13.
2π 6 ∫ 0 cos θ dθ =
2
3 π 256 5 π 8
_____________________________________________________________________
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