Metode Bagi Dua (Biseksi)
Metode Bagi Dua (Biseksi)
Ide awal metode ini adalah metode tabel, dimana area dibagi menjadi N bagian. Hanya
saja metode biseksi ini membagi range menjadi dua bagian, dari dua bagian ini dipilih bagian
mana yang mengandung dan bagian yang tidak mengandung akar dua. Hal ini dilakukan
berulang-ulang hingga diperoleh akar persamaan.
Untuk menggunakan metode biseksi, terlebih dahulu ditentukan batas bawah (a) dan batas (b).
kemudian dihitung nilai tengahnya:
=
Dari nilai x ini perlu dilakukan pe ngecekan keberadaan akar. Secara matematik, suatu range
terdapat akar persamaan bila f(a) dan f(b) berlawanan tanda atau dituliskan:
.
−
C>
@A
@A
@A
%
2
/%
7 ln 2 > ln b − a − ln ε
7>
<
ket: in adalah logaritma natura
@A B
@A B
Yang dalam hal ini R adalah jumlah lelaran (jumlah pembagian selang) yang dibutuhkan untuk
menjamin bahwa c adalah hampiran akar yang memiliki galat kurang dari %.
Contoh : 1
=2−
1. Hitung D2, misalkan
Maka:
1) = 1 dan (2) = −2
Jadi akar terletak antara
(1) = 1
.
= 1.5 dan (
(2) = −2
) = −2.5
=
(
+
2
)
(
1
1
2
1.5
-0.25
2
1
1.5
1.25
0.4375
3
1.25
1.5
1.375
0.109375
4
1.375
1.5
1.4375
-0.06641
5
1.375
1.4375
1.40625
0.022461
6
1.40625
1.4375
1.421875
-0.02173
7
1.40625
1.421875
1.4140625
0.000427
8
1.414063
1.421875
1.41796875
-0.01064
9
1.414063
1.417969
1.416015875
-0.0051
10
1.414063
1.416016
1.415039438
-0.00234
Akar dari 2 adalah 1,4150
Sampai pada kesalahan relative terkecil.
)
Contoh : 2
Cari akar dari ( ) =
+ 3 − 5 ,yang ada dalam interval
:
= 1,
= 2! dengan % = 0,001
i
A
X
B
f(a)
f(x)
f(b)
1
1
1,5
2
-1
2,875
9
2
1
1,25
1,5
-1
0,703125
2,875
3
1
1,125
1,25
-1
-0,201171875
0,703125
4
1,125
1,1875
1,25
-0,201171875
0,237060546875
0,703125
5
1,125
1,15625
1,1875
-0,201171875
0,014556884765625
0,237060546875
6
1,125
1,140625
1,15625
-0,201171875
-0,0941429138183594
0,0145568847656
7
1,140625
1,1484375
1,15625
-0,09414291381835
-0,0400032997131348
0,01455688476562
8
1,1484375
1,15324375
1,15625
-0,04000329971313
-0,0127759575843811
0,01455688476562
9
1,16324375
1,154296875
1,15625
-0,01277559575843
0,000877253711223
0,0145568847656
Jadi akar 2 adalah 1,154297
Contoh : 3
Carilah salah satu akar dari persamaan berikut :
G=
:
+
−3 −3
Disyaratkan bahwa batas kesalahan relatif % < 0,01 %
Hasil hitungan ditabelkan dalam tabel berikut :
Iterasi
H
)
( H)
(
f(xu)
f(xr)
( H ). f(xr)
% (%)
1
1
2
1,5
-4
3
-1,875
7,5
2
1,5
2
1,75
-1,875
3
0,171875
-0,3222656
14,285714
3
1,5
1,75
1,625
-1,875
0,171875
-0,9433594
1,7687988
-7, 6923077
4
1,625
1,75
1,6875
-0,9433594
0,171875
-0,4094238
0,3862338
3,7037037
5
1,6875
1,75
1,71875
-0,4094238
0,171875
-0,1247864
0,0510905
1,8181818
6
1,71875
1,75
1,734375
-0,1247864
0,171875
0,02202299
-0,002749
0.9009009
7
1,71875
1,734375
1,7265625
-0,1247864
0,0220299
-0,0517554
0,0064584
-0,4524887
8
1,7265625
1,734375
1,7304688
-0,0517554
0,0220299
-0,0149572
0,0007741
0,2257336
9
1,7304688
1,734375
1,7324219
-0,0149572
0,0220299
0,0035127
-5,254E-05
0,11227396
10
1,7304688
1,7324219
1,7314453
-0,0149572
0,0035217
-0,00557282
8,568E-05
-0,0564016
11
1,7314453
1,7324219
1,7319336
-0,0057282
0,0035217
-0,0011092
6,354E-06
0,0281928
12
1,7319336
1,7324219
1,7321777
-0,0011092
0,0035217
0,0012013
-1,333E-06
0,0140944
13
1,7319336
1,7321777
1,7320557
-0,0011092
0,0011092
4,596E-05
-5,098E-08
-0,0070477
Dari hasil hitungan tampak bahwa akar persamaan adalah 1,7320557 bandingkan dengan
akar eksaknya yang bernilai D3 = 1,73205080756…..
Contoh 4
Carilah akar dari persamaan f(x) =
:
−
−1 =0
Jawab
Ambil a = 1 dan b = 2
F(1) = 1: − 1 − 1 = −1
F(2) = 2: − 2 − 1 = 5
Karena f(1) . f(2) = (-1) . (5) = -5
Maka f(1).f(2) < 0
JKLMLNOP
Kesalahan relative semu = I
JML5QROS TU
JML5QROS TU
Iterasi 1
(1) = −1
=1
(2) = 5
=2
=
=
= 1,5
f(1,5) = 0,875
Karena f( ). f( ) = (-1). 0,875
= -0,875
Maka f( ). f( ) < 0
Iterasi 2
=
= 1,5
=
X
(1) = −1
=1
X
=
,Y
(1,5) = 0,875
= 1,25
I
f(1,25) = -0,296875
). f( ) = (-1).(-0,296875)
Karena f(
= 0,296875
,Y
Z77[7 =
, Y
, Y
= 0,2
). f( ) > 0
Jadi f(
Iterasi 3
:
=
:
=
:
=
(1,25) = −0,296875
= 1,25
= 1,5
^
^
, Y
=
(1,5) = 0,875
,Y
= 1,375
f(1,375) = 0,22460938
f(
: ).
f( : ) = (-0,296875).( 0,22460938)
= -0,0666809
,:Y
Z77[7 =
: ).
Jadi f(
,:_Y
= 0,0909091
,:_Y
f( : ) < 0
Iterasi 4
`
=
`
=
`
=
:
:
b
(1,25) = −0,296875
= 1,25
= 1,375
b
, Y
=
,:_Y
(1,375) = 0,22460938
= 1,3125
f(1,3125) = -0,0515137
f(
` ).
f( ` ) = (-0,296875).( -0,0515137)
= -0,0152931
,:_Y
Z77[7 =
` ).
Jadi f(
,:
,:
Y
Y
= 0,04761905
f( ` ) < 0
Iterasi 5
Y
Y
=
=
`
`
= 1,25
= 1,3125
(1,25) = −0,296875
(1,3125) = −0,0515137
Y
=
c
c
, Y
=
,:
Y
= 1,28125
f(1,28125) = -0,177948
f(
Y ).
f( Y ) = (-0,296875).( -0,177948)
= 0,0528283
,:
Z77[7 =
Y ).
Jadi f(
Y
, d
, d
Y
Y
= 0,02439024
f( Y ) > 0
Iterasi 6
e
=
e
=
e
=
Y
Y
(1,28125) = −0,177948
= 1,28125
= 1,3125
f
f
, d
=
Y
,:
(1,3125) = −0,0515137
Y
= 1,296875
f(1,296875) = -0,1156807
f(
e ).
f( e ) = (-0,296875).( -0,1156807)
= 0,0343427
, d
Z77[7 =
e ).
Jadi f(
Y
, ged_Y
, ged_Y
= 0,0121482
f( e ) > 0
Iterasi 7
_
=
_
=
_
=
e
e
(1,296875) = −0,1156807
= 1,296875
= 1,3125
h
h
=
, ged_Y
,:
(1,3125) = −0,0515137
Y
= 1, 3046875
f(1, 3046875) = -0,0838361
f(
_ ).
f( _ ) = (-0,1156807).( −0,0838361)
= 0,0096982 > 0
Z77[7 =
Jadi f(
, ged_Y
,:9`ed_Y
,:9`ed_Y
_ ).
f( _ ) > 0
= 0,005988
Iterasi 8
d
=
d
=
d
=
_
_
(1,3046875) = −0,0838361
= 1,3046875
= 1,3125
i
i
=
,:9`ed_Y
,:
Y
(1,3125) = −0,0515137
= 1,30859375
f(1,30859375 ) = -0,0677348
f(
d ).
f( d ) = (−0,0838361).( −0,0677348)
= 0,0056786 > 0
,:9`ed_Y
Z77[7 =
d ).
Jadi f(
,:9dYg:_Y
,:9dYg:_Y
= 0,0029851
f( d ) > 0
Iterasi 9
g
=
g
=
g
=
d
d
(1,30859375) = −0,0677348
= 1,30859375
= 1,3125
j
j
=
,:9dYg:_Y
,:
Y
(1,3125) = −0,0515137
= 1,31054688
f(1,31054688 ) = -0,0596392
f( g ). f( g ) = (−0,0677348 ).( −0,0596392)
= 0,0040396 > 0
,:9`ed_Y
Z77[7 =
,:9dYg:_Y
,:9dYg:_Y
= 0,0014903
Jadi f( g ). f( g ) > 0
Iterasi 10
9
=
9
=
9
=
g
g
= 1,3125
k
(1,31054688) = −0,0596392
= 1,31054688
k
=
,: 9Y`edd
,:
Y
(1,3125) = −0,0515137
= 1,31152344
f(1,31152344 ) = -0,0555802
f(
9 ).
9)
f(
= (−0,0596392).( −0,0555802)
= 0,0033148 > 0
,: 9Y`edd
Z77[7 =
9 ).
Jadi f(
,:
9)
f(
,:
Y :``
Y :``
= 0,0007446
>0
Iterasi 11
=
=
9
9
(1,31152344) = −0,0555802
= 1,31152344
= 1,3125
=
,:
=
Y :``
,:
Y
(1,3125) = −0,0515137
= 1,31201172
F(1,31201172 ) = -0,0535479
f(
). f(
) = (−0,0555802 ).( −0,0535479)
= 0,00297762 > 0
,:
Z77[7 =
Y :``
,:
). f(
Jadi f(
9
,:
_
9
_
= 0,0003722
)>0
Iterasi 12
=
=
=
(1,31201172) = −0,0535479
= 1,31201172
= 1,3125
X
X
,:
=
9
_
,:
Y
(1,3125) = −0,0515137
= 1,31225586
f( 1,31225586 ) = -0, 052531
f(
). f(
) = (−0,0535479 ).( −0, 052531)
= 0,0028129 > 0
Z77[7 =
Jadi f(
,:
). f(
9
_
,:
,:
YYde
)>0
YYde
= 0,000186
Iterasi 13
:
=
:
=
:
=
(1,31225586) = −0,052531
= 1,31225586
= 1,3125
^
^
,:
=
YYde
,:
Y
(1,3125) = −0,0515137
= 1,31237793
f(1,31237793 ) = -0,0519603
: ).
f(
:)
f(
= (−0,052531 ).( −0,0519603)
= 0,0027264 > 0
: ).
Jadi f(
:)
f(
,:
Z77[7 =
>0
YYde
,:
,:
:__g:
:__g:
= 0,000093
Iterasi 14
`
=
`
=
`
=
:
:
(1,31237793) = −0,0519003
= 1,31237793,
= 1,3125 ,
b
b
,:
=
:__g:
,:
Y
(1,3125) = −0,0515137
= 1,312243897
f(1,312243897 ) = -0,0515849
` ).
f(
`)
f(
= (−0,0519003 ).( −0,0515849)
= 0,0026773 > 0
,:
Z77[7 =
Jadi f(
` ).
:__g:
,:
f(
,:
`:dg_
`)
>0
(
`)
`:dg_
= 0,0000465
Iterasi 15
Karena (
Y
=
Y
=
Y
=
`
Y
c
` ).
>0
(1,31243897) = −0,0515849
= 1,31243897
= 1,3125
c
=
,:
`:dg_
,:
Y
(1,3125) = −0,0515137
= 1,31246949
(1,31246949) = (1,31246949): − 1,31246949 − 1
= 2,26082865 − 1,31246949 − 1 = −0,0516408
(
Y)
(
Y)
= (−0,0515849)(−0,0516408) = 0,00266389
,:
Z77[7 =
` ).
Jadi f(
`:dg_
,:
f(
,:
`eg`g
`)
>0
(
Y)
`eg`g
= 0,0000232
Iterasi 16
Karena (
e
=
Y
=
Y
=
Y
Y
Y ).
>0
(1,31246949 ) = −0,0513137
= 1,31246949
= 1,3125
c
c
=
,:
`eg`g
,:
Y
(1,3125) = −0,0515137
= 1,31248475
(1,31248475) = −0,0494923
(
e)
(
Z77[7 =
Jadi f(
e)
= (−0,0513137)(−0,0494923) = 0,0025558
,:
`eg`g
` ).
f(
,:
`)
,:
`d`_Y
>0
`d`_Y
= 0,0000117
Hasil hitungan ditabelkan dalam tabel berikut:
Iterasi
an
bn
xn
f(an)
f(bn)
f(xn)
f(an).f(xn)
1
1
2
1,5
-1
5
0,875
-0,875
2
1
1,5
1,25
-1
0,875
-0,296875
0,296875
3
1,25
1,5
1,375
-0,296875
0,875
0,22460938
-0,0666809
4
1,25
1,375
1,3125
-0,296875
0,22460938
-0,0515137
-0,015293
5
1,25
1.3125
1,28125
-0,296875
-0,0515137
-0,177948
0,0528283
6
1,28125
1,3125
1,296875
-0,177948
-0,0515137
-0,1156807
0,0343427
7
1,296875
1,3125
1,3046875
-0,1156807
-0,0515137
-0,0838361
0,0096982
8
1,3046875
1,3125
1,30859375
-0,0838361
-0,0515137
-0,0677348
0,0056786
9
1,30859375
1,3125
1,31054688
-0,0677348
-0,0515137
-0,0596392
0,0040396
10
1,31054688
1,3125
1,31152344
-0,0596392
-0,0515137
-0,0555802
0,0033148
11
1,31152344
1,3125
1,31201172
-0,0555802
-0,0515137
-0,0555802
0,0029762
12
1,31201172
1,3125
1,31225586
-0,0535479
-0,0515137
-0,052531
0,0028129
13
1,31225586
1,3125
1,31237793
-0,052531
-0,0515137
-0,0519003
-0,0027264
14
1,31237793
1,3125
1,31243897
-0,0519003
-0,0515137
-0,0515849
0,0026773
15
1,31243897
1,3125
1,31246949
-0,0515849
-0,0515137
-0,0516408
0,00266389
16
1,31246949
1,3125
1,31248475
-0,0516408
-0,0515137
-0,0494923
0,002558
Iterasi dihentikan Sampai pada kesalahan relative terkecil.
Contoh 5
Dik : pers ( ) =
:
+2
+ 10 − 20 = 0 dengan selang 1; 1,5!
Penyelesaian :
= 1 dan
= 1,5
(1) = (1): + 2(1) + 10(1) − 20 = 1 + 2 + 10 − 20 = −7
(1,5) = (1,5): + 2(1,5) + 10(1,5) − 20 = 2,875
• Iterasi 1
=1
, (1) = −7
= 1,5 , (1,5) = 2,875
=
=
,Y
= 1,25
( ) = −2,421875
( ). ( ) = (−7). (−2,421875) = 16,953125
( ). ( ) > 0
• Iterasi 2 , karena ( ). ( ) > 0 , maka
= 1,25 , (1,25) = −2,421875
=
= 1,5 , (1,5) = 2,875
=
=
X
X
, Y
=
,Y
= 1,375
( ) = 0,1308594
(
). ( ) = (−2,421875). (0,1308594) = −0,31692511
Jadi , ( ). ( ) < 0
Z77[7 =
X
, Y
=
X
= 1,25 , (1,25) = −2,421875
=
:
=
:
=
= −0,0909091
). ( ) < 0 , maka
• Iterasi 3 , karena (
:
,:_Y
,:_Y
= 1,3475 , (1,375) = 0,1308594
^
^
( :) =
, Y
=
:
+2
,:_Y
= 1,3125
+ 10 − 20 = 0
= (1,3125): + 2(1,3125) + 10(1,3125) − 20
= 2,2609863 + 3,4453215 + 13,125 − 20
(
= −1,1687012
: ).
Jadi, (
( : ) = (−2,421875). (−1,1687012) = 2,3804822
: ).
Z77[7 =
X
( :) > 0
^
^
=
,:_Y
,:
,:
Y
Y
= 0,04761905
• Iterasi 4
`
=
`
=
`
=
:
:
,:
= 1,3125
= 1,375
Y
,:_Y
(1,3125) = −1,1687012
(1,375) = 0,1308594
= 1,34375
( ` ) = (1,34375): + 2(1,34375) + 10(1,34375) − 20
= 2,42636108 + 3,61132812 + 13,4375 − 20 = −0,5248108
(
` ).
( ` ) = −1,1687012 . (−0,5248108)
(
` ).
( `) > 0
= 0,613347
Z77[7 =
^
b
,:
=
b
Y
,:`:_Y
,:`:_Y
= 0,0232558
• Iterasi 5
Y
=
Y
=
Y
=
`
`
(1,34375) = −0,5248108
= 1,34375
(1,375) = 0,1308594
= 1,375
,:`:_Y
,:_Y
= 1,359375
(1,359375) = (1,359375): + 2(1,359375) + 10(1,359375) − 20
= 2,51198959 + 3,69580078 + 13,59375 − 20
= − 0,1984596
(
Y ).
( Y ) = (−0,5248108). (−0,1984596)
(
Y ).
( Y) > 0
= 0,1041537
Z77[7 =
b
c
=
c
,:`:_Y
,:Yg:_Y
,:Yg:_Y
= 0,0114943
• Iterasi 6
e
=
e
=
e
=
Y
Y
= 1,359375
= 1,375
,:Yg:_Y
,:_Y
(1,359375) = −0,1984596
(1,375) = 0,1308594
= 1,3671875
(1,3671875) = (1,3671875): + 2(1,3671875) + 10(1,3671875) − 20
= 2,55554914 + 3,73840332 + 13,671875 − 20
= −0,0341275
(
(
e ).
e ).
( e ) = (−0,1984596). (−0,0341275) = 0,00677293
( e) > 0
Z77[7 =
,:Yg:_Y
,:e_ d_Y
,:e_ d_Y
= 0,0057143
• Iterasi 7
_
=
_
=
_
=
e
e
= 1,3671875
= 1,375
,:e_ d_Y
,:_Y
(1,3671875) = −0,0341275
(1,375) = 0,1308594
= 1,37109375
( _ ) = (1,37109375): + 2(1,37109375) + 10(1,37109375) − 20 = 0,0482501
(
(
_ ).
( _ ) = (−0,0341275). (0,0482501) = − 0,0016467
_ ). ( _ )
0
,:eg `9e:
Z77[7 =
,:ed e`9_
= 0,00071377
,:ed e`9_
• Iterasi 10
9
9
9
(
=
=
g
g
= 1,36914063
,:ed e`9_
=
9)
(
= 1,36816407
,:eg `9e:
(
9)
9)
= −0,0135842
= 0,0070156
= 1,36865235
= (1,36865235): + 2(1,36865235) + 10(1,36865235) − 20
= 2,56377226 + 3,74641852 + 13,6865235 − 20
= −0,0032857
,:ed e`9_
Z77[7 =
,:edeY :Y
= 0,0003568
,:edeY :Y
• Iterasi 11
=
=
=
(
(
(
9
9
(
= 1,36865235
(
= 1,36914063
,:edeY :Y
,:eg `9e:
) = −0,0032857
9)
= 0,0070156
= 1,36889649
) = (1,36889649): + 2(1,36889649) + 10(1,36889649) − 20
= 2,56514447 + 3,7477552 + 13,6889649 − 20
= 0,0018646
). (
). (
Z77[7 =
) = (−0,0032857). (0,0018646) = −0,00000613
)
Ide awal metode ini adalah metode tabel, dimana area dibagi menjadi N bagian. Hanya
saja metode biseksi ini membagi range menjadi dua bagian, dari dua bagian ini dipilih bagian
mana yang mengandung dan bagian yang tidak mengandung akar dua. Hal ini dilakukan
berulang-ulang hingga diperoleh akar persamaan.
Untuk menggunakan metode biseksi, terlebih dahulu ditentukan batas bawah (a) dan batas (b).
kemudian dihitung nilai tengahnya:
=
Dari nilai x ini perlu dilakukan pe ngecekan keberadaan akar. Secara matematik, suatu range
terdapat akar persamaan bila f(a) dan f(b) berlawanan tanda atau dituliskan:
.
−
C>
@A
@A
@A
%
2
/%
7 ln 2 > ln b − a − ln ε
7>
<
ket: in adalah logaritma natura
@A B
@A B
Yang dalam hal ini R adalah jumlah lelaran (jumlah pembagian selang) yang dibutuhkan untuk
menjamin bahwa c adalah hampiran akar yang memiliki galat kurang dari %.
Contoh : 1
=2−
1. Hitung D2, misalkan
Maka:
1) = 1 dan (2) = −2
Jadi akar terletak antara
(1) = 1
.
= 1.5 dan (
(2) = −2
) = −2.5
=
(
+
2
)
(
1
1
2
1.5
-0.25
2
1
1.5
1.25
0.4375
3
1.25
1.5
1.375
0.109375
4
1.375
1.5
1.4375
-0.06641
5
1.375
1.4375
1.40625
0.022461
6
1.40625
1.4375
1.421875
-0.02173
7
1.40625
1.421875
1.4140625
0.000427
8
1.414063
1.421875
1.41796875
-0.01064
9
1.414063
1.417969
1.416015875
-0.0051
10
1.414063
1.416016
1.415039438
-0.00234
Akar dari 2 adalah 1,4150
Sampai pada kesalahan relative terkecil.
)
Contoh : 2
Cari akar dari ( ) =
+ 3 − 5 ,yang ada dalam interval
:
= 1,
= 2! dengan % = 0,001
i
A
X
B
f(a)
f(x)
f(b)
1
1
1,5
2
-1
2,875
9
2
1
1,25
1,5
-1
0,703125
2,875
3
1
1,125
1,25
-1
-0,201171875
0,703125
4
1,125
1,1875
1,25
-0,201171875
0,237060546875
0,703125
5
1,125
1,15625
1,1875
-0,201171875
0,014556884765625
0,237060546875
6
1,125
1,140625
1,15625
-0,201171875
-0,0941429138183594
0,0145568847656
7
1,140625
1,1484375
1,15625
-0,09414291381835
-0,0400032997131348
0,01455688476562
8
1,1484375
1,15324375
1,15625
-0,04000329971313
-0,0127759575843811
0,01455688476562
9
1,16324375
1,154296875
1,15625
-0,01277559575843
0,000877253711223
0,0145568847656
Jadi akar 2 adalah 1,154297
Contoh : 3
Carilah salah satu akar dari persamaan berikut :
G=
:
+
−3 −3
Disyaratkan bahwa batas kesalahan relatif % < 0,01 %
Hasil hitungan ditabelkan dalam tabel berikut :
Iterasi
H
)
( H)
(
f(xu)
f(xr)
( H ). f(xr)
% (%)
1
1
2
1,5
-4
3
-1,875
7,5
2
1,5
2
1,75
-1,875
3
0,171875
-0,3222656
14,285714
3
1,5
1,75
1,625
-1,875
0,171875
-0,9433594
1,7687988
-7, 6923077
4
1,625
1,75
1,6875
-0,9433594
0,171875
-0,4094238
0,3862338
3,7037037
5
1,6875
1,75
1,71875
-0,4094238
0,171875
-0,1247864
0,0510905
1,8181818
6
1,71875
1,75
1,734375
-0,1247864
0,171875
0,02202299
-0,002749
0.9009009
7
1,71875
1,734375
1,7265625
-0,1247864
0,0220299
-0,0517554
0,0064584
-0,4524887
8
1,7265625
1,734375
1,7304688
-0,0517554
0,0220299
-0,0149572
0,0007741
0,2257336
9
1,7304688
1,734375
1,7324219
-0,0149572
0,0220299
0,0035127
-5,254E-05
0,11227396
10
1,7304688
1,7324219
1,7314453
-0,0149572
0,0035217
-0,00557282
8,568E-05
-0,0564016
11
1,7314453
1,7324219
1,7319336
-0,0057282
0,0035217
-0,0011092
6,354E-06
0,0281928
12
1,7319336
1,7324219
1,7321777
-0,0011092
0,0035217
0,0012013
-1,333E-06
0,0140944
13
1,7319336
1,7321777
1,7320557
-0,0011092
0,0011092
4,596E-05
-5,098E-08
-0,0070477
Dari hasil hitungan tampak bahwa akar persamaan adalah 1,7320557 bandingkan dengan
akar eksaknya yang bernilai D3 = 1,73205080756…..
Contoh 4
Carilah akar dari persamaan f(x) =
:
−
−1 =0
Jawab
Ambil a = 1 dan b = 2
F(1) = 1: − 1 − 1 = −1
F(2) = 2: − 2 − 1 = 5
Karena f(1) . f(2) = (-1) . (5) = -5
Maka f(1).f(2) < 0
JKLMLNOP
Kesalahan relative semu = I
JML5QROS TU
JML5QROS TU
Iterasi 1
(1) = −1
=1
(2) = 5
=2
=
=
= 1,5
f(1,5) = 0,875
Karena f( ). f( ) = (-1). 0,875
= -0,875
Maka f( ). f( ) < 0
Iterasi 2
=
= 1,5
=
X
(1) = −1
=1
X
=
,Y
(1,5) = 0,875
= 1,25
I
f(1,25) = -0,296875
). f( ) = (-1).(-0,296875)
Karena f(
= 0,296875
,Y
Z77[7 =
, Y
, Y
= 0,2
). f( ) > 0
Jadi f(
Iterasi 3
:
=
:
=
:
=
(1,25) = −0,296875
= 1,25
= 1,5
^
^
, Y
=
(1,5) = 0,875
,Y
= 1,375
f(1,375) = 0,22460938
f(
: ).
f( : ) = (-0,296875).( 0,22460938)
= -0,0666809
,:Y
Z77[7 =
: ).
Jadi f(
,:_Y
= 0,0909091
,:_Y
f( : ) < 0
Iterasi 4
`
=
`
=
`
=
:
:
b
(1,25) = −0,296875
= 1,25
= 1,375
b
, Y
=
,:_Y
(1,375) = 0,22460938
= 1,3125
f(1,3125) = -0,0515137
f(
` ).
f( ` ) = (-0,296875).( -0,0515137)
= -0,0152931
,:_Y
Z77[7 =
` ).
Jadi f(
,:
,:
Y
Y
= 0,04761905
f( ` ) < 0
Iterasi 5
Y
Y
=
=
`
`
= 1,25
= 1,3125
(1,25) = −0,296875
(1,3125) = −0,0515137
Y
=
c
c
, Y
=
,:
Y
= 1,28125
f(1,28125) = -0,177948
f(
Y ).
f( Y ) = (-0,296875).( -0,177948)
= 0,0528283
,:
Z77[7 =
Y ).
Jadi f(
Y
, d
, d
Y
Y
= 0,02439024
f( Y ) > 0
Iterasi 6
e
=
e
=
e
=
Y
Y
(1,28125) = −0,177948
= 1,28125
= 1,3125
f
f
, d
=
Y
,:
(1,3125) = −0,0515137
Y
= 1,296875
f(1,296875) = -0,1156807
f(
e ).
f( e ) = (-0,296875).( -0,1156807)
= 0,0343427
, d
Z77[7 =
e ).
Jadi f(
Y
, ged_Y
, ged_Y
= 0,0121482
f( e ) > 0
Iterasi 7
_
=
_
=
_
=
e
e
(1,296875) = −0,1156807
= 1,296875
= 1,3125
h
h
=
, ged_Y
,:
(1,3125) = −0,0515137
Y
= 1, 3046875
f(1, 3046875) = -0,0838361
f(
_ ).
f( _ ) = (-0,1156807).( −0,0838361)
= 0,0096982 > 0
Z77[7 =
Jadi f(
, ged_Y
,:9`ed_Y
,:9`ed_Y
_ ).
f( _ ) > 0
= 0,005988
Iterasi 8
d
=
d
=
d
=
_
_
(1,3046875) = −0,0838361
= 1,3046875
= 1,3125
i
i
=
,:9`ed_Y
,:
Y
(1,3125) = −0,0515137
= 1,30859375
f(1,30859375 ) = -0,0677348
f(
d ).
f( d ) = (−0,0838361).( −0,0677348)
= 0,0056786 > 0
,:9`ed_Y
Z77[7 =
d ).
Jadi f(
,:9dYg:_Y
,:9dYg:_Y
= 0,0029851
f( d ) > 0
Iterasi 9
g
=
g
=
g
=
d
d
(1,30859375) = −0,0677348
= 1,30859375
= 1,3125
j
j
=
,:9dYg:_Y
,:
Y
(1,3125) = −0,0515137
= 1,31054688
f(1,31054688 ) = -0,0596392
f( g ). f( g ) = (−0,0677348 ).( −0,0596392)
= 0,0040396 > 0
,:9`ed_Y
Z77[7 =
,:9dYg:_Y
,:9dYg:_Y
= 0,0014903
Jadi f( g ). f( g ) > 0
Iterasi 10
9
=
9
=
9
=
g
g
= 1,3125
k
(1,31054688) = −0,0596392
= 1,31054688
k
=
,: 9Y`edd
,:
Y
(1,3125) = −0,0515137
= 1,31152344
f(1,31152344 ) = -0,0555802
f(
9 ).
9)
f(
= (−0,0596392).( −0,0555802)
= 0,0033148 > 0
,: 9Y`edd
Z77[7 =
9 ).
Jadi f(
,:
9)
f(
,:
Y :``
Y :``
= 0,0007446
>0
Iterasi 11
=
=
9
9
(1,31152344) = −0,0555802
= 1,31152344
= 1,3125
=
,:
=
Y :``
,:
Y
(1,3125) = −0,0515137
= 1,31201172
F(1,31201172 ) = -0,0535479
f(
). f(
) = (−0,0555802 ).( −0,0535479)
= 0,00297762 > 0
,:
Z77[7 =
Y :``
,:
). f(
Jadi f(
9
,:
_
9
_
= 0,0003722
)>0
Iterasi 12
=
=
=
(1,31201172) = −0,0535479
= 1,31201172
= 1,3125
X
X
,:
=
9
_
,:
Y
(1,3125) = −0,0515137
= 1,31225586
f( 1,31225586 ) = -0, 052531
f(
). f(
) = (−0,0535479 ).( −0, 052531)
= 0,0028129 > 0
Z77[7 =
Jadi f(
,:
). f(
9
_
,:
,:
YYde
)>0
YYde
= 0,000186
Iterasi 13
:
=
:
=
:
=
(1,31225586) = −0,052531
= 1,31225586
= 1,3125
^
^
,:
=
YYde
,:
Y
(1,3125) = −0,0515137
= 1,31237793
f(1,31237793 ) = -0,0519603
: ).
f(
:)
f(
= (−0,052531 ).( −0,0519603)
= 0,0027264 > 0
: ).
Jadi f(
:)
f(
,:
Z77[7 =
>0
YYde
,:
,:
:__g:
:__g:
= 0,000093
Iterasi 14
`
=
`
=
`
=
:
:
(1,31237793) = −0,0519003
= 1,31237793,
= 1,3125 ,
b
b
,:
=
:__g:
,:
Y
(1,3125) = −0,0515137
= 1,312243897
f(1,312243897 ) = -0,0515849
` ).
f(
`)
f(
= (−0,0519003 ).( −0,0515849)
= 0,0026773 > 0
,:
Z77[7 =
Jadi f(
` ).
:__g:
,:
f(
,:
`:dg_
`)
>0
(
`)
`:dg_
= 0,0000465
Iterasi 15
Karena (
Y
=
Y
=
Y
=
`
Y
c
` ).
>0
(1,31243897) = −0,0515849
= 1,31243897
= 1,3125
c
=
,:
`:dg_
,:
Y
(1,3125) = −0,0515137
= 1,31246949
(1,31246949) = (1,31246949): − 1,31246949 − 1
= 2,26082865 − 1,31246949 − 1 = −0,0516408
(
Y)
(
Y)
= (−0,0515849)(−0,0516408) = 0,00266389
,:
Z77[7 =
` ).
Jadi f(
`:dg_
,:
f(
,:
`eg`g
`)
>0
(
Y)
`eg`g
= 0,0000232
Iterasi 16
Karena (
e
=
Y
=
Y
=
Y
Y
Y ).
>0
(1,31246949 ) = −0,0513137
= 1,31246949
= 1,3125
c
c
=
,:
`eg`g
,:
Y
(1,3125) = −0,0515137
= 1,31248475
(1,31248475) = −0,0494923
(
e)
(
Z77[7 =
Jadi f(
e)
= (−0,0513137)(−0,0494923) = 0,0025558
,:
`eg`g
` ).
f(
,:
`)
,:
`d`_Y
>0
`d`_Y
= 0,0000117
Hasil hitungan ditabelkan dalam tabel berikut:
Iterasi
an
bn
xn
f(an)
f(bn)
f(xn)
f(an).f(xn)
1
1
2
1,5
-1
5
0,875
-0,875
2
1
1,5
1,25
-1
0,875
-0,296875
0,296875
3
1,25
1,5
1,375
-0,296875
0,875
0,22460938
-0,0666809
4
1,25
1,375
1,3125
-0,296875
0,22460938
-0,0515137
-0,015293
5
1,25
1.3125
1,28125
-0,296875
-0,0515137
-0,177948
0,0528283
6
1,28125
1,3125
1,296875
-0,177948
-0,0515137
-0,1156807
0,0343427
7
1,296875
1,3125
1,3046875
-0,1156807
-0,0515137
-0,0838361
0,0096982
8
1,3046875
1,3125
1,30859375
-0,0838361
-0,0515137
-0,0677348
0,0056786
9
1,30859375
1,3125
1,31054688
-0,0677348
-0,0515137
-0,0596392
0,0040396
10
1,31054688
1,3125
1,31152344
-0,0596392
-0,0515137
-0,0555802
0,0033148
11
1,31152344
1,3125
1,31201172
-0,0555802
-0,0515137
-0,0555802
0,0029762
12
1,31201172
1,3125
1,31225586
-0,0535479
-0,0515137
-0,052531
0,0028129
13
1,31225586
1,3125
1,31237793
-0,052531
-0,0515137
-0,0519003
-0,0027264
14
1,31237793
1,3125
1,31243897
-0,0519003
-0,0515137
-0,0515849
0,0026773
15
1,31243897
1,3125
1,31246949
-0,0515849
-0,0515137
-0,0516408
0,00266389
16
1,31246949
1,3125
1,31248475
-0,0516408
-0,0515137
-0,0494923
0,002558
Iterasi dihentikan Sampai pada kesalahan relative terkecil.
Contoh 5
Dik : pers ( ) =
:
+2
+ 10 − 20 = 0 dengan selang 1; 1,5!
Penyelesaian :
= 1 dan
= 1,5
(1) = (1): + 2(1) + 10(1) − 20 = 1 + 2 + 10 − 20 = −7
(1,5) = (1,5): + 2(1,5) + 10(1,5) − 20 = 2,875
• Iterasi 1
=1
, (1) = −7
= 1,5 , (1,5) = 2,875
=
=
,Y
= 1,25
( ) = −2,421875
( ). ( ) = (−7). (−2,421875) = 16,953125
( ). ( ) > 0
• Iterasi 2 , karena ( ). ( ) > 0 , maka
= 1,25 , (1,25) = −2,421875
=
= 1,5 , (1,5) = 2,875
=
=
X
X
, Y
=
,Y
= 1,375
( ) = 0,1308594
(
). ( ) = (−2,421875). (0,1308594) = −0,31692511
Jadi , ( ). ( ) < 0
Z77[7 =
X
, Y
=
X
= 1,25 , (1,25) = −2,421875
=
:
=
:
=
= −0,0909091
). ( ) < 0 , maka
• Iterasi 3 , karena (
:
,:_Y
,:_Y
= 1,3475 , (1,375) = 0,1308594
^
^
( :) =
, Y
=
:
+2
,:_Y
= 1,3125
+ 10 − 20 = 0
= (1,3125): + 2(1,3125) + 10(1,3125) − 20
= 2,2609863 + 3,4453215 + 13,125 − 20
(
= −1,1687012
: ).
Jadi, (
( : ) = (−2,421875). (−1,1687012) = 2,3804822
: ).
Z77[7 =
X
( :) > 0
^
^
=
,:_Y
,:
,:
Y
Y
= 0,04761905
• Iterasi 4
`
=
`
=
`
=
:
:
,:
= 1,3125
= 1,375
Y
,:_Y
(1,3125) = −1,1687012
(1,375) = 0,1308594
= 1,34375
( ` ) = (1,34375): + 2(1,34375) + 10(1,34375) − 20
= 2,42636108 + 3,61132812 + 13,4375 − 20 = −0,5248108
(
` ).
( ` ) = −1,1687012 . (−0,5248108)
(
` ).
( `) > 0
= 0,613347
Z77[7 =
^
b
,:
=
b
Y
,:`:_Y
,:`:_Y
= 0,0232558
• Iterasi 5
Y
=
Y
=
Y
=
`
`
(1,34375) = −0,5248108
= 1,34375
(1,375) = 0,1308594
= 1,375
,:`:_Y
,:_Y
= 1,359375
(1,359375) = (1,359375): + 2(1,359375) + 10(1,359375) − 20
= 2,51198959 + 3,69580078 + 13,59375 − 20
= − 0,1984596
(
Y ).
( Y ) = (−0,5248108). (−0,1984596)
(
Y ).
( Y) > 0
= 0,1041537
Z77[7 =
b
c
=
c
,:`:_Y
,:Yg:_Y
,:Yg:_Y
= 0,0114943
• Iterasi 6
e
=
e
=
e
=
Y
Y
= 1,359375
= 1,375
,:Yg:_Y
,:_Y
(1,359375) = −0,1984596
(1,375) = 0,1308594
= 1,3671875
(1,3671875) = (1,3671875): + 2(1,3671875) + 10(1,3671875) − 20
= 2,55554914 + 3,73840332 + 13,671875 − 20
= −0,0341275
(
(
e ).
e ).
( e ) = (−0,1984596). (−0,0341275) = 0,00677293
( e) > 0
Z77[7 =
,:Yg:_Y
,:e_ d_Y
,:e_ d_Y
= 0,0057143
• Iterasi 7
_
=
_
=
_
=
e
e
= 1,3671875
= 1,375
,:e_ d_Y
,:_Y
(1,3671875) = −0,0341275
(1,375) = 0,1308594
= 1,37109375
( _ ) = (1,37109375): + 2(1,37109375) + 10(1,37109375) − 20 = 0,0482501
(
(
_ ).
( _ ) = (−0,0341275). (0,0482501) = − 0,0016467
_ ). ( _ )
0
,:eg `9e:
Z77[7 =
,:ed e`9_
= 0,00071377
,:ed e`9_
• Iterasi 10
9
9
9
(
=
=
g
g
= 1,36914063
,:ed e`9_
=
9)
(
= 1,36816407
,:eg `9e:
(
9)
9)
= −0,0135842
= 0,0070156
= 1,36865235
= (1,36865235): + 2(1,36865235) + 10(1,36865235) − 20
= 2,56377226 + 3,74641852 + 13,6865235 − 20
= −0,0032857
,:ed e`9_
Z77[7 =
,:edeY :Y
= 0,0003568
,:edeY :Y
• Iterasi 11
=
=
=
(
(
(
9
9
(
= 1,36865235
(
= 1,36914063
,:edeY :Y
,:eg `9e:
) = −0,0032857
9)
= 0,0070156
= 1,36889649
) = (1,36889649): + 2(1,36889649) + 10(1,36889649) − 20
= 2,56514447 + 3,7477552 + 13,6889649 − 20
= 0,0018646
). (
). (
Z77[7 =
) = (−0,0032857). (0,0018646) = −0,00000613
)