Tugas Metode Numerik Pendidikan Matematika UMT
-
Upload
rukmono-budi-utomo -
Category
Science
-
view
101 -
download
0
Transcript of Tugas Metode Numerik Pendidikan Matematika UMT
1.1 Barisan Fibonacci
METODE NUMERIK FIBONACCI
Amelia Noviasari 1384202071Denny Hardi 1384202110
Mona Yulinda Santika 1384202115Risti Apriani Dewi 1384202141Rudi Alviansyah 1384202100
March 11, 2016
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Barisan Fibonacci
1 1.1 Barisan Fibonacci
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Definisi
Barisan f0, f1, f2, f3, ..., fn − 2, fn − 1, fn disebut Fibonacci jika untukf0 = 1, f1 = 0 + f0, f2 = f0 + f1,f3 = f2 + f1,...,fn = fn − 2 + fn − 1
contoh barisan fibonacci1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Definisi
Barisan f0, f1, f2, f3, ..., fn − 2, fn − 1, fn disebut Fibonacci jika untukf0 = 1, f1 = 0 + f0, f2 = f0 + f1,f3 = f2 + f1,...,fn = fn − 2 + fn − 1
contoh barisan fibonacci1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Algoritma nilai optimal dengan Metode Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
dibentukL0 = Fn+1Ln
dibentuk
λi = ai +F(n+1)−i−1
F(n+1)−i+1(bi − ai )
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Algoritma nilai optimal dengan Metode Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
dibentukL0 = Fn+1Ln
dibentuk
λi = ai +F(n+1)−i−1
F(n+1)−i+1(bi − ai )
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Algoritma nilai optimal dengan Metode Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
dibentukL0 = Fn+1Ln
dibentuk
λi = ai +F(n+1)−i−1
F(n+1)−i+1(bi − ai )
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan
dicari
µi = ai +F(n+1)−i
F(n+1)−i+1(bi − ai )
jikaf (µi ) > f (λi )
ambil µi dan ai , masing-masing sebagai bi+1 dan ai+1
iterasi berhenti ketika bi − ai < 2δ
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan
dicari
µi = ai +F(n+1)−i
F(n+1)−i+1(bi − ai )
jikaf (µi ) > f (λi )
ambil µi dan ai , masing-masing sebagai bi+1 dan ai+1
iterasi berhenti ketika bi − ai < 2δ
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan
dicari
µi = ai +F(n+1)−i
F(n+1)−i+1(bi − ai )
jikaf (µi ) > f (λi )
ambil µi dan ai , masing-masing sebagai bi+1 dan ai+1
iterasi berhenti ketika bi − ai < 2δ
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Soal
minimalkanf (x) = 2x3 − 3x2
dengan δ = 0, 1 pada selang −2 <= x <= 3
dengan cara analitik, diperoleh nilai x yang meminimalkanf (x) adalah x = 1
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Soal
minimalkanf (x) = 2x3 − 3x2
dengan δ = 0, 1 pada selang −2 <= x <= 3
dengan cara analitik, diperoleh nilai x yang meminimalkanf (x) adalah x = 1
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
1
F7+1<
2δ
5
1
34<
1
25
dibentukL0 = Fn+1Ln
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
1
F7+1<
2δ
5
1
34<
1
25
dibentukL0 = Fn+1Ln
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Dicari nilai n terkecil
1
Fn+1<
2δ
L
1
F7+1<
2δ
5
1
34<
1
25
dibentukL0 = Fn+1Ln
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi I
λ1 = a1 +F(7+1)−1−1
F(7+1)−1+1(b1 − a1)
λ1 = −2 +F(6)F(8)
(3 − (−2))
λ1 = −2 +13
34(5)
λ1 = −0, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi I
λ1 = a1 +F(7+1)−1−1
F(7+1)−1+1(b1 − a1)
λ1 = −2 +F(6)F(8)
(3 − (−2))
λ1 = −2 +13
34(5)
λ1 = −0, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi I
λ1 = a1 +F(7+1)−1−1
F(7+1)−1+1(b1 − a1)
λ1 = −2 +F(6)F(8)
(3 − (−2))
λ1 = −2 +13
34(5)
λ1 = −0, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi I
λ1 = a1 +F(7+1)−1−1
F(7+1)−1+1(b1 − a1)
λ1 = −2 +F(6)F(8)
(3 − (−2))
λ1 = −2 +13
34(5)
λ1 = −0, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
µ1 = a1 +F(7+1)−1
F(7+1)−1+1(b1 − a1)
µ1 = −2 +F(7)F(8)
(3 − (−2))
µ1 = −2 +21
34(5)
µ1 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
µ1 = a1 +F(7+1)−1
F(7+1)−1+1(b1 − a1)
µ1 = −2 +F(7)F(8)
(3 − (−2))
µ1 = −2 +21
34(5)
µ1 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
µ1 = a1 +F(7+1)−1
F(7+1)−1+1(b1 − a1)
µ1 = −2 +F(7)F(8)
(3 − (−2))
µ1 = −2 +21
34(5)
µ1 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
µ1 = a1 +F(7+1)−1
F(7+1)−1+1(b1 − a1)
µ1 = −2 +F(7)F(8)
(3 − (−2))
µ1 = −2 +21
34(5)
µ1 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (λ1) = 2λ13 − 3λ1
2
f (λ1) = 2(−0, 088)3 − 3(−0, 088)2
f (λ1) = −0, 001 − 0, 023
f (λ1) = −0, 024
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (λ1) = 2λ13 − 3λ1
2
f (λ1) = 2(−0, 088)3 − 3(−0, 088)2
f (λ1) = −0, 001 − 0, 023
f (λ1) = −0, 024
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (λ1) = 2λ13 − 3λ1
2
f (λ1) = 2(−0, 088)3 − 3(−0, 088)2
f (λ1) = −0, 001 − 0, 023
f (λ1) = −0, 024
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (λ1) = 2λ13 − 3λ1
2
f (λ1) = 2(−0, 088)3 − 3(−0, 088)2
f (λ1) = −0, 001 − 0, 023
f (λ1) = −0, 024
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi I
f (µ1) = 2µ13 − 3µ1
2
f (µ1) = 2(1, 088)3 − 3(1, 088)2
f (µ1) = 2, 576 − 3, 551
f (µ1) = −0, 975
f (λ1) > f (µ1)
λ1 = −0, 088(a2)
b1 = 3(b2)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi II
λ2 = a2 +F(7+1)−2−1
F(7+1)−2+1(b2 − a2)
λ2 = −0, 088 +F(5)F(7)
(3 − (−0, 088))
λ2 = −0, 088 +8
21(3, 088)
λ2 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi II
λ2 = a2 +F(7+1)−2−1
F(7+1)−2+1(b2 − a2)
λ2 = −0, 088 +F(5)F(7)
(3 − (−0, 088))
λ2 = −0, 088 +8
21(3, 088)
λ2 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi II
λ2 = a2 +F(7+1)−2−1
F(7+1)−2+1(b2 − a2)
λ2 = −0, 088 +F(5)F(7)
(3 − (−0, 088))
λ2 = −0, 088 +8
21(3, 088)
λ2 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi II
λ2 = a2 +F(7+1)−2−1
F(7+1)−2+1(b2 − a2)
λ2 = −0, 088 +F(5)F(7)
(3 − (−0, 088))
λ2 = −0, 088 +8
21(3, 088)
λ2 = 1, 088
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
µ2 = a2 +F(7+1)−2
F(7+1)−2+1(b2 − a2)
µ2 = −0, 088 +F(6)F(7)
(3 − (−0, 088))
µ2 = −0, 088 +13
21(3, 088)
µ2 = 1, 744
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
µ2 = a2 +F(7+1)−2
F(7+1)−2+1(b2 − a2)
µ2 = −0, 088 +F(6)F(7)
(3 − (−0, 088))
µ2 = −0, 088 +13
21(3, 088)
µ2 = 1, 744
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
µ2 = a2 +F(7+1)−2
F(7+1)−2+1(b2 − a2)
µ2 = −0, 088 +F(6)F(7)
(3 − (−0, 088))
µ2 = −0, 088 +13
21(3, 088)
µ2 = 1, 744
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
µ2 = a2 +F(7+1)−2
F(7+1)−2+1(b2 − a2)
µ2 = −0, 088 +F(6)F(7)
(3 − (−0, 088))
µ2 = −0, 088 +13
21(3, 088)
µ2 = 1, 744
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (λ2) = 2λ23 − 3λ2
2
f (λ2) = 2(1, 088)3 − 3(1, 088)2
f (λ2) = 2, 576 − 3, 551
f (λ2) = −0, 975
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (λ2) = 2λ23 − 3λ2
2
f (λ2) = 2(1, 088)3 − 3(1, 088)2
f (λ2) = 2, 576 − 3, 551
f (λ2) = −0, 975
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (λ2) = 2λ23 − 3λ2
2
f (λ2) = 2(1, 088)3 − 3(1, 088)2
f (λ2) = 2, 576 − 3, 551
f (λ2) = −0, 975
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (λ2) = 2λ23 − 3λ2
2
f (λ2) = 2(1, 088)3 − 3(1, 088)2
f (λ2) = 2, 576 − 3, 551
f (λ2) = −0, 975
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi II
f (µ2) = 2µ23 − 3µ2
2
f (µ2) = 2(1, 744)3 − 3(1, 744)2
f (µ2) = 10, 609 − 9, 125
f (µ2) = 1, 484
f (µ2) > f (λ2)
µ2 = −1, 744(b3)
a2 = −0, 088(a3)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi III
λ3 = a3 +F(7+1)−3−1
F(7+1)−3+1(b3 − a3)
λ3 = −0, 088 +F(4)F(6)
(1, 744 − (−0, 088))
λ3 = −0, 088 +5
13(1, 832)
λ3 = −0, 617
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi III
λ3 = a3 +F(7+1)−3−1
F(7+1)−3+1(b3 − a3)
λ3 = −0, 088 +F(4)F(6)
(1, 744 − (−0, 088))
λ3 = −0, 088 +5
13(1, 832)
λ3 = −0, 617
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi III
λ3 = a3 +F(7+1)−3−1
F(7+1)−3+1(b3 − a3)
λ3 = −0, 088 +F(4)F(6)
(1, 744 − (−0, 088))
λ3 = −0, 088 +5
13(1, 832)
λ3 = −0, 617
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Iterasi III
λ3 = a3 +F(7+1)−3−1
F(7+1)−3+1(b3 − a3)
λ3 = −0, 088 +F(4)F(6)
(1, 744 − (−0, 088))
λ3 = −0, 088 +5
13(1, 832)
λ3 = −0, 617
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
µ3 = a3 +F(7+1)−3
F(7+1)−3+1(b3 − a3)
µ3 = −0, 088 +F(5)F(6)
(1, 744 − (−0, 088))
µ3 = −0, 088 +8
13(1, 832)
µ3 = 1, 039
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
µ3 = a3 +F(7+1)−3
F(7+1)−3+1(b3 − a3)
µ3 = −0, 088 +F(5)F(6)
(1, 744 − (−0, 088))
µ3 = −0, 088 +8
13(1, 832)
µ3 = 1, 039
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
µ3 = a3 +F(7+1)−3
F(7+1)−3+1(b3 − a3)
µ3 = −0, 088 +F(5)F(6)
(1, 744 − (−0, 088))
µ3 = −0, 088 +8
13(1, 832)
µ3 = 1, 039
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
µ3 = a3 +F(7+1)−3
F(7+1)−3+1(b3 − a3)
µ3 = −0, 088 +F(5)F(6)
(1, 744 − (−0, 088))
µ3 = −0, 088 +8
13(1, 832)
µ3 = 1, 039
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (λ3) = 2λ33 − 3λ3
2
f (λ3) = 2(0, 617)3 − 3(0, 617)2
f (λ3) = 0, 470 − 1, 142
f (λ3) = −0, 672
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (λ3) = 2λ33 − 3λ3
2
f (λ3) = 2(0, 617)3 − 3(0, 617)2
f (λ3) = 0, 470 − 1, 142
f (λ3) = −0, 672
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (λ3) = 2λ33 − 3λ3
2
f (λ3) = 2(0, 617)3 − 3(0, 617)2
f (λ3) = 0, 470 − 1, 142
f (λ3) = −0, 672
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (λ3) = 2λ33 − 3λ3
2
f (λ3) = 2(0, 617)3 − 3(0, 617)2
f (λ3) = 0, 470 − 1, 142
f (λ3) = −0, 672
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
lanjutan iterasi III
f (µ3) = 2µ33 − 3µ3
2
f (µ3) = 2(1, 039)3 − 3(1, 039)2
f (µ3) = 2, 243 − 3, 239
f (µ3) = −1, 086
f (λ3) > f (µ3)
λ3 = 0, 617(a4)
b3 = 1, 744(b4)
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
Tabel perhitungan dengan Metode Fibonacci
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
x∗ = a8 +
(b8 − a8
2
)
x∗ = 1, 039 +
(1, 180 − 1, 039
2
)x∗ = 1, 109
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
x∗ = a8 +
(b8 − a8
2
)
x∗ = 1, 039 +
(1, 180 − 1, 039
2
)
x∗ = 1, 109
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI
1.1 Barisan Fibonacci
x∗ = a8 +
(b8 − a8
2
)
x∗ = 1, 039 +
(1, 180 − 1, 039
2
)x∗ = 1, 109
Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI