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PENYELESAIAN SOAL-SOAL FUNGSI PEMBANGKIT Tugas ini dibuat untuk memenuhi salah satu tugas mata kuliah
MATEMATIKA DISKRIT
Dosen Pembimbing : Prof. I KETUT BUDAYASA, Ph.D
Oleh : Hariyadi (10715013) Agus Nur Sofan (10715015) Yohanes Nova Probo W (10715018)
PROGRAM STUDI PENDIDIKAN MATEMATIKA PROGRAM PASCASARJANA UNIVERSITAS NEGERI SURABAYA 2010
1. Tulis FPB dari barisan-barisan berikut, dan sederhanakan jika mungkin. a. ( 0, 0, 0, 1, 1, 1, . . . ) Jawab: P(x) = x 3 + x 4 + x 5 + β― = π₯ 3 (1 + π₯ + π₯ 2 + π₯ 3 + β― ) 1
= π₯3 (
1βπ₯
=
)
π₯3 1βπ₯
1
1
1
b. (0, 0, , , , β¦ ) 2! 3! 4! Jawab: π(π₯) =
1 2 1 3 1 4 π₯ + π₯ + π₯ +β― 2! 3! 4!
= (1 + π₯ +
1 2!
π₯2 +
1 3!
π₯3 +
1 4!
π₯4 + β― ) β π₯ β 1
= ππ₯ β π₯ β 1 1
1
1
c. ( + + + β― ) 3! 4! 5! Jawab: π(π₯) =
1 1 1 π₯ + π₯2 + π₯3 + β― 3! 4! 5!
= (1 + π₯ +
1 2!
π₯2 +
1 3!
π₯3 +
1 4!
π₯4 +
1 5!
1
= π π₯ β 1 β π₯ β π₯2 2!
1
1
1
1
d. (1, β1, , β , , β , β¦ ) 2! 3! 4! 5! Jawab: π(π₯) = 1 β π₯ + = π βπ₯
1
π₯5 + β― ) β 1 β π₯ β π₯2
1 2 1 3 1 4 1 5 π₯ β π₯ + π₯ β π₯ +β― 2! 3! 4! 5!
2!
e. (0,1,0,1,0,1,0,1, β¦ ) π±ππππ: π(π₯) = 0 + π₯ + 0π₯ 2 + π₯ 3 + 0π₯ 4 + π₯ 5 + 0π₯ 6 + π₯ 7 + β― = π₯ + π₯3 + π₯5 + π₯7 + β― = π₯(1 + π₯ 2 + π₯ 4 + π₯ 6 + β― ) 1
= π₯(
1βπ₯ 2
2
)
2
f. (2,0, , 0, , β¦ ) 3 5 π±ππππ: 2 2 P(x) = 2 + 0x + x 2 + 0x 3 + x 4 + β― 3 5 2
2
3
5
= 2 + π₯2 + π₯4 + β― 1
1
3
5
= 2 (1 + x 2 + x 4 + β― ) = = =
1 3
1 5
2(x+ x3 + x5 +β― ) x 1 2
2( ππ(
1+π₯ )) 1βπ₯
π₯ 1 π₯
(ππ(1 + π₯) β ππ(1 β π₯))
2. Tulislah fungsi pembangkit eksponensial dari barisan berikut. a. (3, 3,3,3, β¦ ) Jawab: π(π₯) = 3 + 3π₯ +
3 2 3 3 3 4 π₯ + π₯ + π₯ +β― 2! 3! 4!
= 3 (1 + π₯ + = 3(π π₯ ) = 3π π₯
1 2!
π₯2 +
1 3!
π₯3 +
1 4!
π₯4 + β― )
b. (0,1,0,1,0, β¦ ) Jawab: π₯2 π₯3 π₯4 π(π₯) = 0 + π₯ + 0 + + 0 + β― 2! 3! 4! =π₯+ =
1 2
π₯2 3!
+
π₯5 5!
+β―
(π π₯ β π βπ₯ )
c. (3,1,3,1,3,1, β¦ ) Jawab: 3π₯ 2 π₯ 3 3π₯ 4 π₯ 5 π(π₯) = 3 + π₯ + + + + +β― 2! 3! 4! 5! = (3 +
3π₯ 2 2!
= 3 (1 + =
3 2
π₯2 2!
+ +
3π₯ 4 4! π₯4 4!
+ β― ) + (π₯ +
+ β― ) + (π₯ + 1
(π π₯ + π βπ₯ ) + (π π₯ β π βπ₯ ) 2
= 2π π₯ + π βπ₯
d. ππ = 3π π½ππ€ππ: π P(π₯) = β~ π=0 3
= β~ π=0 = π 3π₯
π₯π π!
(3π₯)π π!
π₯3 3!
π₯3 3!
+
+
π₯5 5!
π₯5 5!
+ β―)
+ β―)
3. P(x) adalah fungsi pembangkit biasa dari barisan ππ . Carilah barisan (ππ ). a. π(π₯) = 1 +
1 1βπ₯
π½ππ€ππ: P(π₯) = 1 + = =
1 1βπ₯
1βπ₯+1 1βπ₯ 2βπ₯ 1βπ₯
π = (2 β π₯) β~ π=0 π₯ ~ π π+1 = 2 β~ π=0 π₯ β βπ=0 π₯ ~ π π = 2 β~ π=0 π₯ β βπ=1 π₯
Maka barisannya adalah: (ππ ) = {
b. π(π₯) =
2; π = 0 atau (ππ ) = ( 2,1,1,1,1, β¦ ) 1; π ο³ 1
π₯5 1+8π₯
Jawab: π₯5 π(π₯) = 1 + 8π₯ = π₯5 (
1
1+8π₯
= π₯5 (
1
) )
1β(β8π₯)
π = π₯ 5 β~ π=0(β8π₯) π π+5 = β~ π=0(β8) π₯ πβ5 π = β~ π₯ π=5(β8)
Maka barisannya adalah: (ππ ) = {
(β8)πβ5 ; π ο³ 5 0 ; 0ο£π < 5
Atau (ππ ) = (0,0,0,0,0,1, β8,64, β¦ )
c. π(π₯) =
2 1βπ₯
+ 3π₯ 2 + 6π₯ + 1
Jawab: π(π₯) =
= = =
2 + 3π₯ 2 + 6π₯ + 1 1βπ₯ 2+3π₯ 2 (1βπ₯)+6π₯(1βπ₯)+(1βπ₯) 1βπ₯ 2+3π₯ 2 β3π₯ 3 +6π₯β6π₯ 2 +1βπ₯ 1βπ₯ 3+5π₯β3π₯ 2 β3π₯ 3 1βπ₯
π = (3 + 5π₯ β 3π₯ 2 β 3π₯ 3 ) β~ π=0 π₯ ~ π π+1 π+2 π+3 = 3 β~ β 3 β~ β 3 β~ π=0 π₯ + 5 βπ=0 π₯ π=0 π₯ π=0 π₯ ~ ~ ~ π π π π = 3 β~ π=0 π₯ + 5 βπ=1 π₯ β 3 βπ=2 π₯ β 3 βπ=3 π₯
Jadi barisannya adalah: ππ =
3, 8, {5, 2,
π’ππ‘π’π π=0 π’ππ‘π’π π=1 π’ππ‘π’π π=2 π’ππ‘π’π πβ₯3
atau ππ = (3,8,5,2,2,2, β¦ ) d. π(π₯) = 2π₯ + π βπ₯ Jawab: π(π₯) = 2π₯ + π βπ₯
=
2π₯ 1!
π + β~ π=0(β1)
~ = 2π₯ + βπ=0
(β1)π π!
π₯π π!
π₯π
1, π’ππ‘π’π π = 0 Jadi barisannya adalah : ππ = { β1,π π’ππ‘π’π π = 1 (β1) , π’ππ‘π’π π β₯ 2 π!
1
1
1
1
2!
3! 4!
5!
Atau ππ = (1, β1, , β , , β , β¦ )
1
e. π(π₯) =
2
(π π₯ + π βπ₯ )
Jawab: π(π₯) = =
1 2 1 2
(π π₯ + π βπ₯ ) [β~ π=0
1 π!
π₯ π + β~ π=0
1
(β1)π
1
= β~ π=0 2 (π! + = β~ π=0 (
1+(β1)π 2π!
π!
(β1)π π!
) π₯π
) π₯π
Jadi barisannya adalah : ππ =
1+(β1)π 2π!
Atau ππ = (1, 0, f. P(π₯) =
1 1β3π₯
+
π₯π]
, π’ππ‘π’π π β₯ 0
1 2!
, 0,
1 4!
, 0,
1 6!
, . . .)
4π₯ 1βπ₯
Jawab: P(π₯) =
1 1β3π₯
+
4π₯ 1βπ₯
~ π π = β~ π=0(3π₯) + 4π₯ βπ=0 π₯ ~ π π π+1 = β~ π=0 3 π₯ + βπ=0 4π₯ ~ π π π = β~ π=0 3 π₯ + βπ=1 4π₯ ~ π π π = 1 + β~ π=1 3 π₯ + βπ=1 4π₯ π π = 1 + β~ π=1(3 + 4)π₯
1, π’ππ‘π’π π = 0 Jadi barisannya adalah: ππ = { π 3 + 4, π’ππ‘π’π π β₯ 1 Atau
ππ = (1, 7, 13, 31, 85, β¦ )
4. Tulis barisan (ππ ) yang fungsi pembangkit eksponensialnya sebagai berikut! a. π(π₯) = 5 + 5π₯ + 5π₯ 2 + 5π₯ 3 + β¦ Jawab: π(π₯) = 5 + 5π₯ + 5π₯ 2 + 5π₯ 3 + β¦ = 5 (1 + π₯ + π₯ 2 + π₯ 3 + β― ) π = 5 β~ π=0 π₯
= 5 β~ π=0 π!
π₯π π!
Jadi barisannya adalah ππ = 5π! atau ππ = (5, 5, 10, 30, β¦ )
b. π(π₯) =
1 1β4π₯
Jawab: π(π₯) =
1 1 β 4π₯
π = β~ π=0(4π₯) π π = β~ π=0 4 π₯ π = β~ π=0 4 π!
π₯π π!
Jadi barisannya adalah: ππ = 4π π! atau ππ = (1, 4, 32, 384, β¦ ) c. π(π₯) = π π₯ + π 4π₯ Jawab: π(π₯) = π π₯ + π 4π₯ = β~ π=0
π₯π π!
π + β~ π=0 4
π = β~ π=0(1 + 4 )
π₯π π!
π₯π π!
Jadi barisannya adalah: ππ = 1 + 4π Atau ππ = (2, 15, 17, 65, β¦ )
d. π(π₯) = (1 + π₯ 2 )π Jawab: π(π₯) = (1 + π₯ 2 )π π 2 π = β~ π=0[π ](π₯ ) π₯π
π = β~ π=0[π ](2π)! π!
Misal: n = 2k 1
K= π 2
π = β~ π=0 [1π ] π! 2
π₯π π!
2, π’ππ‘π’π π ππππππ Jadi barisannya adalah: ππ = { π (1π) π! , π’ππ‘π’π π πππππ 2
5. Cari konvolusi dari pasangan barisan berikut! a. (1, 1, 1, 1, β¦ β¦ β¦ ) πππ (1, 1, 1, 1, β¦ β¦ β¦ ) Jawab: Misalkan konvolusi dari pasangan barisan tersebut adalah ππ = (ππ ) β (ππ ) Maka ππ = βππ=0 ππ ππβπ Karena ππ = 1, ππ = 1 , untuk setiap I ,maka: ππ = βππ=0 1 . 1 = βππ=0 1 = 1 + 1 + 1 + 1 + β¦β¦β¦.. n + 1 suku = n + 1 , untuk setiap n b. ( 1, 1, 1, 1, β¦β¦β¦β¦. ) dan ( 0, 1, 2, 3, β¦β¦β¦.... ) Jawab: Misalkan ππ = ( 1, 1, 1, 1, β¦β¦β¦β¦. )
ππ = 1
ππ = ( 0, 1, 2, 3, β¦β¦β¦.... )
ππ = π
Konvolusi dari ππ πππ ππ adalah: ππ = (ππ ) β (ππ ) = βππ=0 ππ ππβπ = βππ=0 1. (π β π) = βππ=0 π β βππ=0 π 1
= n ( n+1 ) - π ( n+1 ) 2
1
= π ( n+1 ) 2
c. ( 1, 1, 1, 0, 0, 0, . . . ) dan ( 0, 1, 2, 3, . . . ) Jawab : ππ = (1, 1, 1, 0, 0, 0, β¦ ) β¦ β¦ β¦ β¦ β¦ β¦ β¦ (1) ππ = (0, 1, 2, 3, β¦ ) β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (2) FPB dari (1) adalah A(x) = 1 + x + π₯ 2 FPB dari (2) adalah B(x) = π₯ + 2π₯ 2 + 3π₯ 3 + β¦ π = β~ π=0 ππ₯ π Misalkan : A(x) B(x) = β~ π=0 ππ π₯
ππ adalah konvolusi dari (ππ ) πππ (ππ ) π A(x) B(x) = (1 + π₯ + π₯ 2 ) β~ π=0 ππ₯ ~ π π 2 ~ π = β~ π=0 ππ₯ + π₯ βπ=0 ππ₯ + π₯ βπ=0 ππ₯ ~ π π+1 π+2 = β~ + β~ π=0 ππ₯ + βπ=0 ππ₯ π=0 ππ₯ ~ ~ π π π = β~ π=0 ππ₯ + βπ=1(π β 1)π₯ + βπ=2(π β 2)π₯
οπΆπ=
0 { π + (π β 1) π+πβ1+πβ2
ππππ π = 0 ππππ π = 1 ππππ π β₯ 2
Atau
οπΆπ=
0 { 2π β 1 3(π β 1)
ππππ π = 0 ππππ π = 1 ππππ π β₯ 2
d. ( 0, 0, 0, 1, 0, 0, 0, . . . ) dan ( 6, 7, 8, 9, . . . ) Jawab: Misalkan :
ππ = (0, 0, 0, 1, 0, 0, 0, β¦ ) β¦ β¦ β¦ β¦ β¦ β¦ β¦ (1) ππ = (6, 7, 8, 9, β¦ ) β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (2) FPB dari (1) adalah A(x) = 0 + 0x + 0π₯ 2 + π₯ 3 + 0π₯ 4 + β― = π₯3 FPB dari (2) adalah B(x) = 6 + 7π₯ + 8π₯ 2 + 9π₯ 3 + β¦ π = β~ π=0(π + 6)π₯ π Misalkan : A(x) B(x) = β~ π=0 ππ π₯
ππ adalah konvolusi dari (ππ ) πππ (ππ ) π A(x) B(x) = π₯ 3 β~ π=0(π + 6)π₯ π+3 = β~ π=0(π + 6)π₯ π = β~ πβ3=0(π + 6 β 3)π₯ π = β~ π=3(π + 3)π₯
Jadi ππ = {
0 π+3
ππππ 0 β€ π β€ 3 ππππ π β₯ 3