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