1) The document discusses Taylor series and functions. It provides examples of calculating the Taylor series for functions like ex, 1/(1-x), and others. 2) It also discusses function generators in basic form (FPB) and exported form (FPE). An example is worked out to determine the coefficients for the FPB and FPE of a given function. 3) Determining the coefficients for different function generators is demonstrated, such as for p(x)=x2ex and p(x)=x/(1-x).

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