Deret taylor

1. Fungsi Pembangkit
DeretTaylor
Ada 2 fungsi yaitu:
1) f(x) = ex
2) f(x) =
1
(1−𝑥)
RumusDeretTaylor:
𝑓( 𝑥) = ∑
1
𝑛!
𝑓 𝑛(0) 𝑥 𝑛
~
𝑛=0
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
(5𝑥 + 2)10 = 1(5𝑥 + 2)−10
𝑓′( 𝑥) = −10(5𝑥 + 2)−11 .5
𝑓′(𝑥) = −50(5𝑥 + 2)−11 =
−50
(5𝑥 + 2)11
𝑓( 𝑥) =
1
(𝑥2 + 4𝑥)6 = 1(𝑥2 + 4𝑥)−6
𝑓′( 𝑥) = −6( 𝑥2 + 4𝑥)−7.2𝑥 + 4
𝑓′( 𝑥) =
−6(2𝑥 + 4)
(𝑥2 + 4𝑥)7 =
−12𝑥 − 24
(𝑥2 + 4𝑥)7
Derettaylor
1) f(x) = ex
2) f(x) =
1
(1−𝑥)
TentukanDeretTaylordari f(x) = ex
gunakan:
𝑓( 𝑥) ≈ ∑
1
𝑛!
𝑓 𝑛(0) 𝑥 𝑛
~
𝑛=0
Contoh:0! = 1 , 1!=1, 2! = 2x1=2, 3! = 3x2x1=6
Fn
(0) = turunanke n
F(x) = ex
→ f’(x) = 1ex
= ex

2. F(x) = e2x
→ f’(x) =2e2x
.
F(x) = 10e-3x
→ f’(x) = -30 e-3x
𝑓( 𝑥) = 𝑒 𝑥2+4𝑥 → 𝑓′( 𝑥) = (2𝑥 + 4) 𝑒 𝑥2+4𝑥
F(x) = e-5x + 1
→ f’(x) = -5 e-5x+1
Tentukanderettaylordari f(x) = ex
gunakan:
𝑓( 𝑥) ≈ ∑
1
𝑛!
𝑓 𝑛(0) 𝑥 𝑛
~
𝑛=0
𝑓(𝑥) = 𝑒 𝑥 → 𝑓(0) = 𝑒0 = 1
𝑓’(𝑥) = 𝑒 𝑥 → 𝑓’(0) = 𝑒0 = 1
𝑓’’(𝑥) = 𝑒 𝑥 → 𝑓’’(0) = 𝑒0 = 1
𝑓’’’(𝑥) = 𝑒 𝑥 → 𝑓’’’(0) = 𝑒0 = 1
𝑑𝑒𝑟𝑒𝑡 𝑓( 𝑥) = 𝑒 𝑥 ≈ ∑
1
𝑛!
.1𝑥 𝑛 = ∑
1
𝑛!
𝑥 𝑛
~
𝑛=0
𝑛
𝑛=0
∶ 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) = 𝑒2𝑥 ≈ ∑
1
𝑛!
. 2 𝑛 𝑥 𝑛
~
𝑛=0
= 1 + 2𝑥 +
4𝑥2
2
+
8𝑥3
6
+ ⋯
Derettaylordari𝑓( 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
Derettayloruntuk 𝑓( 𝑥) =
1
(1−𝑥)
= (1 − 𝑥)−1 gunakan:
𝑓( 𝑥) ≈ ∑
1
𝑛!
𝑓 𝑛(0). 𝑥 𝑛
~
𝑛=0
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!

3. Derettaylor
(1 − 𝑥)−1 ≈ ∑
1
𝑛!
~
𝑛=0
𝑛1.𝑥 𝑛
∑ 𝑥 𝑛
~
𝑛=0
= 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
= -1
F’’(x) =2 (1+x)-3
. 1 = 2 (1+x)-3
→ f’’(0) = 2(1+0)-3
= 2
F’’’(x) =-6 (1+x)-4
. 1 = -6(1+x)-4
→ f’’’(0) = -6(1+0)-4
= 6
F’’’’(x) =24 (1+x)-5
. 1 = 24(1+x)-5
→ f’’’’(0) = 24(1+0)-5
= 24
:
Fn
(-1)n
.n
Fungsi Pembangkit
1) Kombinasi 𝑐 𝑟𝑛 = 𝑘 𝑟 = ( 𝑛
𝑟
) =
𝑛!
( 𝑛−𝑟)!𝑟!𝑛
2) Permutasi 𝑝𝑟𝑛 =
𝑛!
( 𝑛−𝑟)!
Contoh:
𝐾25 = (
5
2
) =
5!
(5 − 2)! 2!
=
5.4.3.2.1
3.2.1.2.1
=
20
2
= 10
Deret
1
(1−𝑥) 𝑛
≅ ∑ ( 𝑛+𝑘−1
𝑘
)𝑛
𝑘=0 𝑥 𝑘
1
(1 − 𝑥)3 ≈ ∑ (
3 + 𝑘 − 1
𝑘
)
3
𝑘=0
𝑥 𝑘 = ∑ (
𝑘 + 2
𝑘
)
3
𝑘=0
𝑥0
𝑑𝑒𝑟𝑒𝑡 (
2
0
) 𝑥0 + (
3
1
) 𝑥1 + (
4
2
) 𝑥2 + (
5
3
) 𝑥3 = 1 + 3𝑥 + 6𝑥2 + 10𝑥3
Fungsi Pembangkit
1) Fungsi PembangkitBiasa(FPB)
2) Fungsi PembangkitExporter(FPE)
𝐹𝑃𝐵 → 𝑝( 𝑥) = ∑ 𝑎 𝑛
~
𝑛=0
𝑥 𝑛
𝐹𝑃𝐸 → 𝑝( 𝑥) = ∑ 𝑎 𝑛
~
𝑛=0
𝑥 𝑛
𝑛!
An barisanbilangandari suatuderetan = a0,a1, a2,a3, ...
Contohtentukanfungsi pembangkit (FPB)dari FPEjikaan diketahui
𝑎 𝑛 {
0, 𝑛 ≤ 3
1, 𝑛 > 3
→𝑎 𝑛 = 𝑎0, 𝑎1, 𝑎2,𝑎3, 𝑎4, 𝑎5,…
= 0, 0,0, 0,1,1, …
Catatan
𝑒 𝑥:1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+
𝑥4
4!
+ ⋯
1
1 − 𝑥
∶ 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯

4. 𝐹𝑃𝐵 → 𝑝( 𝑥) ∑ 𝑎 𝑛
~
𝑛=0
𝑥 𝑛: 𝑎4 𝑥4 + 𝑎5 𝑥5 + 𝑎6 𝑥6 + ⋯
P(x) = 1x4
+ 1x5
+ 1x6
+ ...
P(x) = X4
+ X5
+ X6
+ ....
= x4
(1 + x + x2
+ x3
+ ....)
= 𝑥4.
1
1 − 𝑥
=
𝑥4
1 − 𝑥
→∴ 𝑝( 𝑥) =
𝑥4
1 − 𝑥
𝐹𝑃𝐸 → 𝑝( 𝑥) = ∑ 𝑎 𝑛
~
𝑛=0
𝑥 𝑛
𝑛!
= 𝑎4
𝑥4
4!
+ 𝑎5
𝑥5
5!
+ 𝑎6
𝑥6
6!
+ ⋯
𝑝( 𝑥) = 1
𝑥4
4!
+ 1
𝑥5
5!
+ 1
𝑥6
6!
+ ⋯
𝑝( 𝑥) =
𝑥4
4!
+
𝑥5
5!
+
𝑥6
6!
+ ⋯
𝑝( 𝑥) = 𝑒 𝑥 − 1 − 𝑥 −
𝑥2
2!
−
𝑥3
3!
MenentukanAn dari fungsi Pembangkit
Contoh:TentukanAn jikap(x) = X2
ex
Catatan:
𝑒 𝑥 = 1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+ ⋯ ∑
𝑥 𝑛
𝑛!
~
𝑛=0
1
1 − 𝑥
= 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ ∑ 𝑥 𝑛
~
𝑛=0
1) 𝑝( 𝑥) = 𝑥2 𝑒 𝑥 = 𝑥2 ∑
𝑥 𝑘
𝑘!
𝑛
𝑘=0 = ∑
𝑥 𝑘+2
𝑘!
𝑛
𝑘=0 = ∑
𝑥 𝑛
( 𝑛−2)!
𝑛
𝑛−2
𝑑𝑖𝑚𝑎𝑛𝑎 𝑚𝑖𝑠𝑎𝑙𝑛𝑦𝑎 𝑘 + 2 = 𝑛, 𝑘 = 𝑛 − 2
𝑎5. 𝑎2 = 𝑎7
𝑎 𝑛 {
0, 𝑛 < 2
1
( 𝑛 − 2)!
, 𝑛 ≥ 2
, 𝑎 𝑛 𝑠𝑒ℎ𝑖𝑛𝑔𝑔𝑎 𝐹𝑃𝐵
𝑎 𝑛 {
0, 𝑛 < 2
𝑛!
( 𝑛 − 2)!
, 𝑛 ≥ 2
, 𝑎 𝑛 𝑠𝑒ℎ𝑖𝑛𝑔𝑔𝑎 𝐹𝑃𝐸
2) 𝑝( 𝑥) =
𝑥
(1−𝑥)
= 𝑥1∑ 𝑥 𝑘𝑛
𝑘=0 = ∑ 𝑥 𝑘+1𝑛
𝑘=0 = ∑ 1𝑥 𝑛𝑛
𝑛−1
𝑑𝑖𝑚𝑎𝑛𝑎 𝑚𝑖𝑠𝑎𝑙𝑛𝑦𝑎 𝑘 + 1 = 𝑛, 𝑘 = 𝑛 − 1
𝑎 𝑛 {
0, 𝑛 < 1
1, 𝑛 ≥ 1
, 𝑎 𝑛 𝑠𝑒ℎ𝑖𝑛𝑔𝑔𝑎 𝐹𝑃𝐵
𝑎 𝑛 {
0, 𝑛 < 1
𝑛!, 𝑛 ≥ 1
, 𝑎 𝑛 𝑠𝑒ℎ𝑖𝑛𝑔𝑔𝑎 𝐹𝑃𝐸