![Page 1: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/1.jpg)
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
![Page 2: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/2.jpg)
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
![Page 3: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/3.jpg)
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
![Page 4: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/4.jpg)
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
![Page 5: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/5.jpg)
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
![Page 6: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/6.jpg)
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
![Page 7: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/7.jpg)
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
![Page 8: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/8.jpg)
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
![Page 9: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/9.jpg)
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
![Page 10: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/10.jpg)
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
![Page 11: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/11.jpg)
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
![Page 12: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/12.jpg)
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
![Page 13: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/13.jpg)
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
![Page 14: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/14.jpg)
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
![Page 15: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/15.jpg)
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
![Page 16: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/16.jpg)
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
![Page 17: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/17.jpg)
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
![Page 18: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/18.jpg)
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
![Page 19: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/19.jpg)
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
![Page 20: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/20.jpg)
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
![Page 21: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/21.jpg)
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
![Page 22: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/22.jpg)
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
![Page 23: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/23.jpg)
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
![Page 24: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/24.jpg)
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
![Page 25: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/25.jpg)
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
![Page 26: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/26.jpg)
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
![Page 27: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/27.jpg)
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
![Page 28: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/28.jpg)
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
![Page 29: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/29.jpg)
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
![Page 30: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/30.jpg)
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
![Page 31: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/31.jpg)
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
![Page 32: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/32.jpg)
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
![Page 33: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/33.jpg)
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
![Page 34: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/34.jpg)
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
![Page 35: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/35.jpg)
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
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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
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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
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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
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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
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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
![Page 41: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/41.jpg)
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
![Page 42: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/42.jpg)
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
![Page 43: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/43.jpg)
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
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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
![Page 45: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/45.jpg)
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
![Page 46: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/46.jpg)
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
![Page 47: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/47.jpg)
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
![Page 48: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/48.jpg)
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
![Page 49: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/49.jpg)
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
![Page 50: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/50.jpg)
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
![Page 51: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/51.jpg)
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
![Page 52: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/52.jpg)
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
![Page 53: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/53.jpg)
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
![Page 54: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/54.jpg)
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
![Page 55: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/55.jpg)
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
![Page 56: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/56.jpg)
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
![Page 57: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/57.jpg)
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
![Page 58: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/58.jpg)
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
![Page 59: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/59.jpg)
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
![Page 60: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/60.jpg)
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
![Page 61: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/61.jpg)
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
![Page 62: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/62.jpg)
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
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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
![Page 64: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/64.jpg)
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
![Page 65: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/65.jpg)
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
![Page 66: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/66.jpg)
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
![Page 67: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/67.jpg)
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
![Page 68: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/68.jpg)
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
![Page 69: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/69.jpg)
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
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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
![Page 71: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/71.jpg)
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
![Page 72: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/72.jpg)
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
![Page 73: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/73.jpg)
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
![Page 74: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/74.jpg)
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
![Page 75: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/75.jpg)
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
![Page 76: Tugas Metode Numerik Pendidikan Matematika UMT](https://reader031.vdocuments.mx/reader031/viewer/2022022202/587f36ce1a28ab121d8b7155/html5/thumbnails/76.jpg)
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