Afnita Sandini (16029054) Tugas 10 Metode Secant.docx
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Afnita Sandini (16029054) Tugas 10 Metode Secant.docx as PDF for free.
More details
- Words: 1,586
- Pages: 9
TUGAS 10 METODE SECANT oleh AFNITA SANDINI 16029054
Dari fungsi 1a sampai 1f kerjakan dengan metode secant? Penyelesaian : a.
f ( x) x cos( x)
1. Iterasi 1 Ambil x0 1 dan x1 0 f 1 1 cos( 1) 0,4597 f 0 0 cos(0) 1
x n 1 x n
x n 1 0
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
(1)(0 (1)) 0,68507 1 (1)
2. Iterasi 2 Ambil x1 0 dan x2 0,68507 f 0 0 cos(0) 1 f 0,68507 (0,68507) cos( 0,68507) 0,089299
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 (0,68507)
(0,089299)(( 0,68507) 0) 0,75225 0,089299 1
3. Iterasi 3 Ambil x2 0,68507 dan x3 0,75225 f 0 (0,68507) cos( 0,68507) 0,089299 f 0 (0,75225) cos( 0,75225) 0,02209
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 (0,75225)
(0,02209)(( 0,75225) (0,68507)) 0,73892 (0,02209) (0,089299)
Tabel iterasi :
f xn1
f xn
Iterasi
x n 1
1
-1
-0.4597
0
1
2
0
1
-0.68507
0.089299
3
-0.68507
0.089299
-0.75225
-0.02209
4
-0.75225
-0.02209
-0.73892
0.000268
b.
f ( x) x 2 sin( x 2) 1. Iterasi 1 Ambil x0 2 dan x1 1
f 2 2 2 sin( 2 2) 4,756802 f 1 12 sin( 1 2) 0,85888
xn
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 1
(0.85888)( 1 (2)) 0,77966 (0,85888) (4,756802)
2. Iterasi 2 Ambil x1 1 dan x2 0,77966
f 1 12 sin( 1 2) 0,85888 f 0,77966 0,77966 2 sin( 0,77966 2) 0,25378 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 0,77966
(0,25378)( 0,77966 (1)) 0,68724 (0,25378) (0,85888)
3. Iterasi 3 Ambil x 2 0,77966 dan x3 0,68724
f 0,77966 0,77966 2 sin( 0,77966 2) 0,25378 f 0,68724 0,68724 2 sin( 0,68724 2) 0,033429 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 0,68724
(0,033429)(( 0,68724) (0,77966)) 0,68724 (0,033429) (0,25378)
Tabel iterasi :
f xn1
xn
f xn
-2
4.756802
-1
0.85888
2
-1
0.85888
-0.77966
0.25378
3
-0.77966
0.25378
-0.68724
0.033429
4
-0.68724
0.033429
-0.67323
0.001802
Iterasi
x n 1
1
c.
f ( x) e x sin( x) 1. Iterasi 1
Ambil x0 3 dan x 1 4
f 3 e 3 sin( 3) 0,190907 f 4 e 4 sin( 4) 0,73849 x n 1 x n x n 1 4
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
(0,73849)( 4 3) 3,20541 (0,73849) (0,190907)
2. Iterasi 2 Ambil x1 4 dan x2 3,20541
f 4 e 4 sin( 4) 0,73849 f 3,20541 e 3, 20541 sin( 3,20541) 0,02323 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 3,20541
(0,02323)(3,20541 4) 3,179601 (0,02323) (0,73849)
3. Iterasi 3 Ambil x2 3,20541 dan x3 3,20541
f 3,20541 e 3, 20541 sin( 3,20541) 0,02323 f 3,179601 e 3,179601 sin( 3,179601) 0,003603
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 3,20541
(0,003603)(3,179601 3,20541) 3.183066 (0,003603) (0,02323)
Tabel iterasi :
f xn1
f xn
Iterasi
x n 1
1
3
0.190907
4
-0.73849
2
4
-0.73849
3.20541
-0.02323
3
3.20541
-0.02323
3.179601
0.003603
4
3.179601
0.003603
3.183066
-3.4E-06
d.
xn
f ( x) 1 x e 2 x
1.Iterasi 1 Ambil x0 1 dan x1 2
f 1 1 1 e 2*1 0,13534 f 2 1 2 e 2*2 1,01832 xn1 xn xn1 2
f ( xn )( xn xn 1 ) f ( xn ) f ( xn 1 )
(1,01832)( 2 1) 0,846729 (1,01832) (0,13534)
2.Iterasi 2 Ambil x1 2 dan x2 0,846729
f 2 1 2 e 2*2 1,01832 f 0,846729 1 0,846729 e 2*0,846729 0,03061
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 0,846729
(0,03061)(0,846729 2) 0,810986 (0,03061) (1,01832)
3.Iterasi 3 Ambil x2 0,846729 dan x3 0,810986
f 0,846729 1 0,846729 e 2*0,846729 0,03061 f 0,810986 1 0,810986 e 2*0,810986 0,00849 xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 0,810986
(0,00849)(0,810986 0,846729) 0,797257 (0,00849) (0,03061)
Tabel iterasi
f xn1
f xn
Iterasi
x n 1
1
1
-0.13534
2
-1.01832
2
2
-1.01832
0.846729
-0.03061
3
0.846729
-0.03061
0.810986
-0.00849
4
0.810986
-0.00849
0.797257
-0.00026
e.
f ( x) 2 x tan x
1.Iterasi 1 Ambil x0 1,9 dan x1 1,8 f 1,9 2(1,9) tan( 1,9) 0,8729 f 1,8 2(1,8) tan( 1,8) 0,686262
xn
xn1 xn
f ( xn )( xn xn 1 ) f ( xn ) f ( xn1 )
xn1 1,8
(0,686262)(( 1,8) (1,9)) 1,84401 (0,686262) (0,8729)
2.Iterasi 2 Ambil x1 1,8 dan x2 1,84401 f 1,8 2(1,8) tan( 1,8) 0,686262 f 1,84401 2(1,84401) tan( 1,84401) 0,11948
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 1,84401
(0,11948)(( 1,84401) (1,8)) 1,83749 (0,11948) (0,686262)
3.Iterasi 3 Ambil x2 1,84401 dan x3 1,83749 f 1,84401 2(1,84401) tan( 1,84401) 0,11948 f 1,83749 2(1,83749) tan( 1,83749) 0,01465
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 1,83749
(0,01465)(( 1,83749) (1,84401)) 1,83658 (0,01465) (0,11948)
Tabel iterasi iterasi
x n 1
f xn1
xn
1
-1.9
-0.8729
-1.8
0.686262
2
-1.8
0.686262
-1.84401
-0.11948
3
-1.84401
-0.11948
-1.83749
-0.01465
f xn
-1.83749
4
f.
-0.01465
-1.83658
0.00035
f ( x) 2 x 2 e x
1.Iterasi 1 Ambil x0 2 dan x1 1 xn 1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn 1 )
xn 1 (1)
(0,71828)(( 1) (2)) 1,54038 f 2 2(2) 2 e ( 2) 0,610944 (0,71828) (0,610944) f 1 2(1) 2 e ( 1) 0,71828
2.Iterasi 2 Ambil x1 1 dan x2 1,54038
f 1 2(1) 2 e ( 1) 0,71828 f 1,54038 2(1,54038) 2 e ( 1,54038) 0,079172 xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 (1,54038)
(0,079172)(( 1,54038) (1)) 1,48673 (0,079172) (0,71828)
3.Iterasi 3 Ambil x2 1 dan x3 1,54038
f 1,54038 2(1,54038) 2 e ( 1,54038) 0,079172 f 1,48673 2(1,48673) 2 e ( 1, 48673) 0,00188
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 (1,48673)
(0,00188)(( 1,48673) (1,54038)) 1,48797 (0,00188) (0,079172)
Table iterasi
f xn1
xn
-2
0.610944
-1
-0.71828
2
-1
-0.71828
-1.54038
0.079172
3
-1.54038
0.079172
-1.48673
-0.00188
4
-1.48673
-0.00188
-1.48797
1.65E-05
iterasi
x n 1
1
f xn
Dari fungsi 1a sampai 1f kerjakan dengan metode secant? Penyelesaian : a.
f ( x) x cos( x)
1. Iterasi 1 Ambil x0 1 dan x1 0 f 1 1 cos( 1) 0,4597 f 0 0 cos(0) 1
x n 1 x n
x n 1 0
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
(1)(0 (1)) 0,68507 1 (1)
2. Iterasi 2 Ambil x1 0 dan x2 0,68507 f 0 0 cos(0) 1 f 0,68507 (0,68507) cos( 0,68507) 0,089299
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 (0,68507)
(0,089299)(( 0,68507) 0) 0,75225 0,089299 1
3. Iterasi 3 Ambil x2 0,68507 dan x3 0,75225 f 0 (0,68507) cos( 0,68507) 0,089299 f 0 (0,75225) cos( 0,75225) 0,02209
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 (0,75225)
(0,02209)(( 0,75225) (0,68507)) 0,73892 (0,02209) (0,089299)
Tabel iterasi :
f xn1
f xn
Iterasi
x n 1
1
-1
-0.4597
0
1
2
0
1
-0.68507
0.089299
3
-0.68507
0.089299
-0.75225
-0.02209
4
-0.75225
-0.02209
-0.73892
0.000268
b.
f ( x) x 2 sin( x 2) 1. Iterasi 1 Ambil x0 2 dan x1 1
f 2 2 2 sin( 2 2) 4,756802 f 1 12 sin( 1 2) 0,85888
xn
x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 1
(0.85888)( 1 (2)) 0,77966 (0,85888) (4,756802)
2. Iterasi 2 Ambil x1 1 dan x2 0,77966
f 1 12 sin( 1 2) 0,85888 f 0,77966 0,77966 2 sin( 0,77966 2) 0,25378 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 0,77966
(0,25378)( 0,77966 (1)) 0,68724 (0,25378) (0,85888)
3. Iterasi 3 Ambil x 2 0,77966 dan x3 0,68724
f 0,77966 0,77966 2 sin( 0,77966 2) 0,25378 f 0,68724 0,68724 2 sin( 0,68724 2) 0,033429 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 0,68724
(0,033429)(( 0,68724) (0,77966)) 0,68724 (0,033429) (0,25378)
Tabel iterasi :
f xn1
xn
f xn
-2
4.756802
-1
0.85888
2
-1
0.85888
-0.77966
0.25378
3
-0.77966
0.25378
-0.68724
0.033429
4
-0.68724
0.033429
-0.67323
0.001802
Iterasi
x n 1
1
c.
f ( x) e x sin( x) 1. Iterasi 1
Ambil x0 3 dan x 1 4
f 3 e 3 sin( 3) 0,190907 f 4 e 4 sin( 4) 0,73849 x n 1 x n x n 1 4
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
(0,73849)( 4 3) 3,20541 (0,73849) (0,190907)
2. Iterasi 2 Ambil x1 4 dan x2 3,20541
f 4 e 4 sin( 4) 0,73849 f 3,20541 e 3, 20541 sin( 3,20541) 0,02323 x n 1 x n
f ( x n )( x n x n 1 ) f ( x n ) f ( x n 1 )
x n 1 3,20541
(0,02323)(3,20541 4) 3,179601 (0,02323) (0,73849)
3. Iterasi 3 Ambil x2 3,20541 dan x3 3,20541
f 3,20541 e 3, 20541 sin( 3,20541) 0,02323 f 3,179601 e 3,179601 sin( 3,179601) 0,003603
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 3,20541
(0,003603)(3,179601 3,20541) 3.183066 (0,003603) (0,02323)
Tabel iterasi :
f xn1
f xn
Iterasi
x n 1
1
3
0.190907
4
-0.73849
2
4
-0.73849
3.20541
-0.02323
3
3.20541
-0.02323
3.179601
0.003603
4
3.179601
0.003603
3.183066
-3.4E-06
d.
xn
f ( x) 1 x e 2 x
1.Iterasi 1 Ambil x0 1 dan x1 2
f 1 1 1 e 2*1 0,13534 f 2 1 2 e 2*2 1,01832 xn1 xn xn1 2
f ( xn )( xn xn 1 ) f ( xn ) f ( xn 1 )
(1,01832)( 2 1) 0,846729 (1,01832) (0,13534)
2.Iterasi 2 Ambil x1 2 dan x2 0,846729
f 2 1 2 e 2*2 1,01832 f 0,846729 1 0,846729 e 2*0,846729 0,03061
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 0,846729
(0,03061)(0,846729 2) 0,810986 (0,03061) (1,01832)
3.Iterasi 3 Ambil x2 0,846729 dan x3 0,810986
f 0,846729 1 0,846729 e 2*0,846729 0,03061 f 0,810986 1 0,810986 e 2*0,810986 0,00849 xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 0,810986
(0,00849)(0,810986 0,846729) 0,797257 (0,00849) (0,03061)
Tabel iterasi
f xn1
f xn
Iterasi
x n 1
1
1
-0.13534
2
-1.01832
2
2
-1.01832
0.846729
-0.03061
3
0.846729
-0.03061
0.810986
-0.00849
4
0.810986
-0.00849
0.797257
-0.00026
e.
f ( x) 2 x tan x
1.Iterasi 1 Ambil x0 1,9 dan x1 1,8 f 1,9 2(1,9) tan( 1,9) 0,8729 f 1,8 2(1,8) tan( 1,8) 0,686262
xn
xn1 xn
f ( xn )( xn xn 1 ) f ( xn ) f ( xn1 )
xn1 1,8
(0,686262)(( 1,8) (1,9)) 1,84401 (0,686262) (0,8729)
2.Iterasi 2 Ambil x1 1,8 dan x2 1,84401 f 1,8 2(1,8) tan( 1,8) 0,686262 f 1,84401 2(1,84401) tan( 1,84401) 0,11948
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 1,84401
(0,11948)(( 1,84401) (1,8)) 1,83749 (0,11948) (0,686262)
3.Iterasi 3 Ambil x2 1,84401 dan x3 1,83749 f 1,84401 2(1,84401) tan( 1,84401) 0,11948 f 1,83749 2(1,83749) tan( 1,83749) 0,01465
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 1,83749
(0,01465)(( 1,83749) (1,84401)) 1,83658 (0,01465) (0,11948)
Tabel iterasi iterasi
x n 1
f xn1
xn
1
-1.9
-0.8729
-1.8
0.686262
2
-1.8
0.686262
-1.84401
-0.11948
3
-1.84401
-0.11948
-1.83749
-0.01465
f xn
-1.83749
4
f.
-0.01465
-1.83658
0.00035
f ( x) 2 x 2 e x
1.Iterasi 1 Ambil x0 2 dan x1 1 xn 1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn 1 )
xn 1 (1)
(0,71828)(( 1) (2)) 1,54038 f 2 2(2) 2 e ( 2) 0,610944 (0,71828) (0,610944) f 1 2(1) 2 e ( 1) 0,71828
2.Iterasi 2 Ambil x1 1 dan x2 1,54038
f 1 2(1) 2 e ( 1) 0,71828 f 1,54038 2(1,54038) 2 e ( 1,54038) 0,079172 xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 (1,54038)
(0,079172)(( 1,54038) (1)) 1,48673 (0,079172) (0,71828)
3.Iterasi 3 Ambil x2 1 dan x3 1,54038
f 1,54038 2(1,54038) 2 e ( 1,54038) 0,079172 f 1,48673 2(1,48673) 2 e ( 1, 48673) 0,00188
xn1 xn
f ( xn )( xn xn1 ) f ( xn ) f ( xn1 )
xn1 (1,48673)
(0,00188)(( 1,48673) (1,54038)) 1,48797 (0,00188) (0,079172)
Table iterasi
f xn1
xn
-2
0.610944
-1
-0.71828
2
-1
-0.71828
-1.54038
0.079172
3
-1.54038
0.079172
-1.48673
-0.00188
4
-1.48673
-0.00188
-1.48797
1.65E-05
iterasi
x n 1
1
f xn
Related Documents
Afnita Sandini (16029054) Tugas 10 Metode Secant.docx
September 2019 5
Tugas Metode Penelitian.docx
May 2020 24
Tugas Metode Penelitian.docx
June 2020 13
Tugas Metode Statistika Ii
May 2020 25
Makalah Metode Tugas Besar.docx
November 2019 28More Documents from "Made Yoga Pradnyana"
Cluster Rs Kelompok 3.docx
December 2019 0
Afnita Sandini (16029054) Tugas 10 Metode Secant.docx
September 2019 5
Ppt%20%20afni.pptx
April 2020 0
Fluida.docx
April 2020 0
Sap Sadari Fix.docx
December 2019 1