matematika diskrit: fungsi pembangkit part 3
TRANSCRIPT
Fungsi Pembangkit
Deret Taylor
Ada 2 fungsi yaitu:
1) f(x) = ex
2) f(x) = 1
(1−x )
Rumus Deret Taylor:
f ( x )=∑n=0
1n!
f n (0 ) xn
F(x) = (3x + 5)5
F’(x) = 5(3x +2)4 . 3 = 15(3x+2)4
F(x) = 4 (x2 + 4x)4
F’(x) = 16(x2 + 4x)3.(2X +4)
f ( x )= 1(5 x+2 )10
=1 (5 x+2 )−10
f ' ( x )=−10 (5 x+2 )−11 .5
f '( x)=−50 (5 x+2)−11= −50(5 x+2)11
f ( x )= 1(x2+4 x)6
=1(x2+4 x )−6
f ' ( x )=−6 (x2+4 x)−7.2 x+4
f ' ( x )= −6 (2 x+4 )
(x¿¿2+4 x)7= −12x−24(x¿¿2+4 x)7¿
¿
Deret taylor
1) f(x) = ex
2) f(x) = 1
(1−x )Tentukan Deret Taylor dari f(x) = ex gunakan:
f ( x )≈∑n=0
1n!
f n (0 ) xn
Contoh: 0! = 1 , 1!=1, 2! = 2x1=2, 3! = 3x2x1=6Fn(0) = turunan ke nF(x) = ex → f’(x) = 1ex = ex
F(x) = e2x→ f’(x) = 2e2x.F(x) = 10e-3x→ f’(x) = -30 e-3x
f ( x )=ex2+4 x→f ' ( x )=(2 x+4 ) ex2+ 4x
F(x) = e-5x + 1 → f’(x) = -5 e-5x+1
Tentukan deret taylor dari f(x) = ex gunakan:
f ( x )≈∑n=0
1n!
f n (0 ) xn
f (x)=ex→f (0)=e0=1f ’ (x )=ex→f ’ (0)=e0=1f ’ ’(x )=ex→f ’’ (0)=e0=1f ’ ’ ’(x)=ex→f ’’’(0)=e0=1
deret f ( x )=ex≈∑n=0
n 1n !.1xn=∑
n=0
1n!
xn:1+x+ x2
2+ x3
6+…
F(x) = e2x →f(0) =e0 =1 →20
F’(x) = 2e2x → f’(0) = 2e0 = 2 →21
F’’(x) = 4e2x → f’’(0) = 4e0 = 4 →22
F’’’(x) = 8e2x →f’’’(0) = 8e0 = 8 → 23
:Fn(0) = 2n
f ( x )=e2 x≈∑n=0
1n !
.2n xn=1+2 x+ 4 x2
2+ 8 x
3
6+…
Deret taylor darif ( x )= 1(1−x )
f ( x )= 1(1−x )
=(1−x )−1→f ' (x )=−1 (1−x )−2.−1=1 (1−x )−2= 1(1−x )2
f ' ' ( x )=−2(1−x )−3 .−1=2 (1−x)−3= 2(1−x )3
f ' ' ' ( x )=−6 (1−x )−4 .−1=6 (1−x )−4= 6(1−x )4
f ' ' ' ' ( x )=−24 (1−x )−5 .−1=24 (1−x )−5= 24(1−x)5
Deret taylor untuk f ( x )= 1(1−x)
=(1−x )−1 gunakan:
f ( x )≈∑n=0
1n!
f n (0 ) . xn
F(x) = (1-x)-1 →f(0) = (1-0)-1= 1 → 0!F’(x) = -1(1-x)-2. (-1) = 1(1-x)-2 →f’(0) = 1(1-0)-2 = 1 → 1! F’’(x) = -2(1-x)-3.(-1) = 2(1-x)-3 →f’’(0) = 2(1-0)-3 = 2 → 2!
F’’’(x) = -6(1-x)-4.(-1) = 6(1-x)-4 → f’’’(0) = 6(1-0)-4 = 6 → 3!F’’’’(x) = -24(1-x)-5. (-1) = 24(1-x)-5 →f’’’’(0) =24(1-0)-5 = 24 → 4!Fn(0) = n!
Deret taylor
(1−x )−1≈∑n=0
1n!
n1 . xn
∑n=0
xn=1+x+x2+x3+…
f ( x )= 1(1+x )
=(1+x )−1→f (0 )=(1+0 )−1=1
F’(x) = -1 (1+x)-2 . 1 = -1(1+x)-2 →f’(0) = -1(1+0)-2 = -1F’’(x) = 2 (1+x)-3 . 1 = 2 (1+x)-3 → f’’(0) = 2(1+0)-3 = 2F’’’(x) = -6 (1+x)-4 . 1 = -6(1+x)-4 → f’’’(0) = -6(1+0)-4 = 6F’’’’(x) = 24 (1+x)-5 . 1 = 24(1+x)-5 → f’’’’(0) = 24(1+0)-5 = 24:Fn(-1)n. nFungsi Pembangkit
1) Kombinasi crn❑ = kr=(nr )= n !
(n−r )!r !n
❑
2) Permutasi prn❑ = n !
(n−r )!Contoh:
K 25❑ =(52)= 5 !
(5−2 )!2 !=5.4 .3 .2.13.2 .1.2 .1
=202
=10
Deret 1
(1−x )n≅∑
k=0
n
(n+k−1k )xk
1(1−x )3
≈∑k=0
3
(3+k−1k ) xk=∑k=0
3
(k+2k ) x0deret (20)x0+(31)x1+(42) x2+(53)x3=1+3 x+6 x2+10 x3Fungsi Pembangkit
1) Fungsi Pembangkit Biasa (FPB)2) Fungsi Pembangkit Exporter (FPE)
FPB→p ( x )=∑n=0
an xnFPE→ p (x )=∑
n=0an
xn
n!An barisan bilangan dari suatu deret an = a0,a1, a2, a3, ...Contoh tentukan fungsi pembangkit (FPB) dari FPE jika an diketahui
an{0 , n≤31, n>3→an=a0 , a1 , a2 , a3 , a4 ,a5 ,…
¿0 ,0 ,0 ,0 ,1 ,1 ,…
Catatan
ex :1+ x+ x2
2 !+ x3
3 !+ x4
4 !+…
11−x
:1+x+ x2+x3+x4+…
FPB→p ( x )∑n=0
an xn: a4 x
4+a5 x5+a6 x
6+…
P(x) = 1x4 + 1x5 + 1x6 + ...P(x) = X4 + X5 + X6 + ....
= x4 (1 + x + x2 + x3 + ....)
¿ x4 . 11−x
= x4
1−x→∴ p (x )= x4
1−x
FPE→ p (x )=∑n=0
anxn
n!=a4 x
4
4 !+a5 x
5
5 !+a6 x
6
6 !+…
p ( x )=1 x4
4 !+1 x
5
5 !+1 x
6
6 !+…
p ( x )= x4
4 !+ x5
5!+ x6
6 !+…
p ( x )=ex−1−x− x2
2 !− x3
3 !Menentukan An dari fungsi PembangkitContoh: Tentukan An jika p(x) = X2ex
Catatan:
ex=1+x+ x2
2!+ x3
3 !+…∑
n=0
xn
n!
11−x
=1+ x+x2+x3+…∑n=0
xn
1) p ( x )=x2ex=x2∑k=0
n xk
k !=∑
k=0
n xk+2
k !=∑
n−2
n xn
(n−2 )!dimanamisalnya k+2=n , k=n−2a5 . a2=a7
an{ 0 , n<21
(n−2 ) !,n≥2
,an sehinggaFPB
an{ 0 , n<2n !
(n−2 ) !,n≥2
,an sehingga FPE
2) p ( x )= x(1−x )
=x1∑k=0
n
xk=∑k=0
n
xk +1=∑n−1
n
1 xn
dimanamisalnya k+1=n , k=n−1
an{0 , n<11 ,n≥1, an sehinggaFPB
an{ 0 , n<1n! , n≥1, an sehinggaFPE