ROOTS OF NON LINIER EQUATIONS - kuliah.ftsl.itb.ac.id · ROOTS OF NON LINIER EQUATIONS • Metode...
Transcript of ROOTS OF NON LINIER EQUATIONS - kuliah.ftsl.itb.ac.id · ROOTS OF NON LINIER EQUATIONS • Metode...
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ROOTS OF NON LINIER EQUATIONS
• Metode Bagi dua (Bisection Method)• Metode Regula Falsi (False Position Method)• Metode Grafik• Iterasi Titik-Tetap (Fix Point Iteration)• Metode Newton-Raphson• Metode Secant
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SOLUSI PERSAMAAN KUADRAT TINGKAT 2
aacbbx
cbxaxxf
24
0)(2
2
Persamaan di atas memberi akar-akar penyelesaian untuk fungsi aljabarf(x)
Yaitu nilai-nilai x yang memberikan f(x) = 0
Kalau persaamaannya f(x) = e-x - x?
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OVERVIEW OF METHODS
• Bracketing methodsGraphing methodBisection methodFalse position
• Open methodsOne point iterationNewton-RaphsonSecant method
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SPECIFIC STUDY OBJECTIVES
• Memahami konsep konvergensi dan divergensi.
• Memahami bahwa metode tertutup selalu konvergen, sedangkan metode terbuka kadang-kadang divergen.
• Konvergensi pada metode terbuka biasanya didapat jika initial guess –nya dekat dengan akar sebenarnya.
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METODE TERTUTUP
• Graphical• Bisection method• False position method
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CARA GRAFIK
• Plotkan fungsinya dan tentukan dimana memotong sumbu x.
• Lacks precision• Trial and error f(x)=e-x-x
-2
0
2
4
6
8
10
-2 -1 0 1 2
x
f(x)
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CARA GRAFIK (LIMITED PRACTICAL VALUE)
x
f(x)
x
f(x)
x
f(x)
x
f(x)
Pembatas atas dan Bawah memiliki tanda sama. Akar tidak ada atau banyak akar
Tanda berbeda, jumlah akar-akar ganjil
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BISECTION METHOD
• Memanfaatkan beda tanda dua nilai batas• f(xl)f(xu) < 0 dimana l=lower (batas bawah) dan
u=upper (batas atas)• Minimal ada satu akar
x
f(x)
x
f(x)
x
f(x)
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ALGORITHM• Pilih xu dan xl. Cek beda tanda nilai fungsi
keduanyaf(xl)f(xu) < 0
• Perkirakan akarxr = (xl + xu) / 2
• Tentukan interval berikut ada di subinterval atas atau subinterval bawahf(xl)f(xr) < 0 then xu baru = xr RETURN f(xl)f(xr) >0 then xl baru = xr RETURN f(xl)f(xr) =0 then root equals xr - COMPLETE
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METODE BAGI DUA
Asumsi: Fungsi f(x) kontinu dalam interval 00 ,ba
0)()( 00 bfaf
do n = 0,1,…2/)( nn bam
if ,0)()( mfaf n then ,1 nn aa mbn 1
else ,1 ma n nn bb 1
if 11 nn ab exitend do
or 0)( mf
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BISECTION METHOD
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ERROR
100
akhir
awalakhira perkiraan
perkiraanperkiraan
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•f(x) = e-x - x•xl = -1 xu = 1
CONTOHGunakan bisection method untuk mencari akar-akar persamaan
3.7 18282
-0.63212
-2
0
2
4
6
8
10
-2 -1 0 1 2
x
f(x)
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SOLUTION
3.7 18282
-0.63212
1
-2
0
2
4
6
8
10
-2 -1 0 1 2
x
f(x)
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-0.63212
1
0.106531
-2
0
2
-1 0 1 2
x
f(x)
SOLUTION
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FALSE POSITION METHOD
• “Brute Force” dari metode bagi dua kurang efisien• Menghubungkan dua nilai batas dengan garis lurus• Mengganti kurva menjadi garis lurus memberikan
“false position”• Mempercepat perkiraan
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xl
xuf(xl)
f(xu)next estimate, xr
f xx x
f xx x
x xf x x xf x f x
l
r l
u
r u
r uu l u
l u
Based on similar triangles
Nilai f(xr) dicek tandanya, kemudian tentukan xu dan xl yang baruberdasarkan perbedaan tanda seperti pada metode bagi dua
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REGULA FALSI
Asumsi: Fungsi f(x) kontinu dalam interval 00 ,ba
0)()( 00 bfaf
do n = 0,1,…)]()(/[])()([ nnnnnn afbfbafabfw
if ,0)()( wfaf n then ,1 nn aa wbn 1
else ,1 wa n nn bb 1
if 11 nn ab exit
end do
or 0)( wf
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REGULA FALSI
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CONTOH
Tentukan akar persamaan dari persamaan berikut menggunakan false position method, mulai dengan initial estimate xl=4.55 and xu=4.65
f(x) = x3 - 98-40
-30
-20
-10
0
10
20
30
4 4.5 5
x
f(x)
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OPEN METHODS
• Simple one point iteration• Newton-Raphson method• Secant method
• Pada metode tertutup, akar terdapat di antara kedua interval yang dibatasi batas atas dan bawah
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OPEN METHODS• Metode terbuka diharapkan konvergen
solution moves closer to the root as the computation progresses
• Metode terbuka;• single starting value, atau• two starting values that do not necessarily bracket the
root• Ada kemungkinan metode ini divergen
solution moves farther from the root as the computation progresses
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The tangentgives next estimate.xi
f(x)
x
f(xi)
xi+1
f(xi+1 )
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Solution can “overshoot”the root and potentiallydiverge
x0
f(x)
x
x1x2
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SIMPLE ONE POINT ITERATION /
METODE TITIK TETAP• Merubah formula untuk memperkirakan akar• Re-arrange fungsi f(x) sehingga ada satu nilai x
pada sebelah kiri dari persamaan Contoh, untuk f(x) = x2 - 2x + 3 = 0Ubah menjadix = (x2 + 3) / 2
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SIMPLE ONE POINT ITERATION
• Contoh lain, untuk f(x) = sin x = 0,menjadix = sin x + x
• Hitung nilai x = g(x)• Perkiraan nilai berikut berdasar pada
x i+1 = g(xi)
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ITERASI TITIK TETAP
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CONTOH
• Untuk f(x) = e-x -3x• Ubah menjadi g(x) = e-x / 3• Initial guess x = 0
-6-4-202468
10121416
-2 -1 0 1 2
x
f(x)
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Initial guess 0.000
g(x) f(x) a
0.333 -0.283
0.239 0.071 39.561
0.263 -0.018 9.016
0.256 0.005 2.395
0.258 -0.001 0.612
0.258 0.000 0.158
0.258 0.000 0.041
-6-4-202468
10121416
-2 -1 0 1 2
x
f(x)
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METODE NEWTON RAPHSON
tangent
dydx
f
f xf xx x
rearrange
x xf xf x
ii
i i
i ii
i
'
'
'
0
1
1
f(xi)
xi
tangent
xi+1
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METODE NEWTON-RAPHSON
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NEWTON RAPHSONPITFALLS
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CONTOH
Gunakan metode Newton Raphson untuk mencari akar-akar dari f(x) = x2 - 11 memakai initial guess xi = 3
-20
0
20
40
60
80
100
0 2 4 6 8 10
x
f(x)
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NEWTON RHAPSON SECANT
• Include an upper limit on the number of iterations
• Establish a tolerance, s
• Check to see if a is increasing
Bagaimana jika turunan fungsinya sulit dipecahkan?SECANT METHOD
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SECANT METHOD
f x f x f xx xi i
i i
'
1
1
Memperkirakan turunan menggunakan finite divided difference
APAKAH finite divided difference? HINT: dy / dx = y / x
Masukkan FDD pada rumus untuk Newton Raphson
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SECANT METHOD
ii
iiiii
i
iii
xfxfxxxfxx
xfxfxx
1
11
1 '
Masukkan perkiraan dengan finite difference pada rumus untuk Newton Raphson
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METODE SECANT
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SECANT METHOD
• Membutuhkan dua nilai perkiraan awal• f(x) tidak harus berbeda tanda,
membedakan dengan metode tertutup, false position method.
ii
iiiii xfxf
xxxfxx
1
11
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x
f(x)
1
2
new est.
x
f(x)
1
new est.
2
FALSE POSITION
SECANT METHOD
Perkiraan baru dipilih dari potongangaris dengan sumbu x
Perkiraan baru bisa diluar batas kurva
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SYSTEMS OF NON-LINEAR EQUATIONS
• Kita telah mengenal sistem persamaan linierf(x) = a1x1 + a2x2+...... anxn - C = 0dimana a1 , a2 .... an dan C adalah konstanta
• Maka, perhatikan sistem persamaan non-liniery = -x2 + x + 0.5y + 5xy = x3
• Selesaikan x dan y
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SYSTEMS OF NON-LINEAR EQUATIONS
• Buat persamaan sama dengan nolu = x2 + xy – 10v = y + 3xy2 – 57
• u(x,y) = x2 + xy – 10 = 0• v(x,y) = y + 3xy2 – 57 = 0• Solusi adalah nilai-nilai x dan y yang akan
memberikan nilai fungsi u dan v sama dengan nol.
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METODE TITIK TETAP
• Mulai dengan nilai awal x0 = 1.5 dan y0 = 3.5
1
01
001
357
10
xyy
yxx
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METODE NEWTON RHAPSON
xv
yu
yv
xu
xuv
xvu
yy
xv
yu
yv
xu
yuv
yvu
xx
iiii
ii
i
ii
iiii
ii
i
ii
1
1
Versi dua persamaan untuk Newton-Raphson
u(x,y) dan v(x,y)
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DETERMINAN JACOBIAN (TAMBAHAN SAJA)
xv
yu
yv
xu
xuv
xvu
yy
xv
yu
yv
xu
yuv
yvu
xx
iiii
ii
i
ii
iiii
ii
i
ii
1
1THE DENOMINATOROF EACH OF THESEEQUATIONS ISFORMALLYREFERRED TOAS THE DETERMINANTOF THEJACOBIAN
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JACOBIAN (TAMBAHAN JUGA)
• The general definition of the Jacobian for nfunctions of n variables is the following set of partial derivatives:
n
nnn
n
n
n
n
xf
xf
xf
xf
xf
xf
xf
xf
xf
xxxfff
...............
...
...
),...,,(),...,,(
21
2
2
2
1
2
1
2
1
1
1
21
21
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JACOBIAN (INI JUGATAMBAHAN)
• The Jacobian can be used to calculate derivatives from a function in one coordinate sytem from the derivatives of that same function in another coordinate system.
• Equations u=f(x,y), v=g(x,y), then x and y can be determined as functions of u and v (possessing first partial derivatives) as follows:
• With similar functions for xv and yv.• The determinants in the denominators are examples of
the use of Jacobians.
yx
yx
x
yx
yx
y
yx
yx
ggff
guy
ggff
gux
yggxggyxgvxffyffyxfu
/;/);,(/;/);,(
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CONTOH
u = 2x3 + 2xy – 2v = 2y + 4xy2 – 3Mulai dengan nilai awal x0 = 0.5 dan y0 = 1.5
yxxux
yuy
xvxy
yv 26;2;4;82 22