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HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR Pertemuan 3. Matakuliah: K0342 / Metode Numerik I Tahun: 2006. Pertemuan . 3. HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR. PERSAMAAN NIRLANJAR (N0N LINIER). Yaitu persamaan yang mengandung variabel berpangkat lebih - PowerPoint PPT Presentation
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HAMPIRAN NUMERIK SOLUSI HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJARPERSAMAAN NIRLANJAR
Pertemuan 3
Matakuliah : K0342 / Metode Numerik I Tahun : 2006
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PertemPertemuan uan
33HAMPIRAN NUMERIK HAMPIRAN NUMERIK SOLUSI PERSAMAAN SOLUSI PERSAMAAN
NIRLANJARNIRLANJAR
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PERSAMAAN NIRLANJAR (N0N LINIER)
Yaitu persamaan yang mengandung variabel berpangkat lebih dari satu dan/atau yang mengandung fungsi-fungsi transendenContoh:
0...)( 33
221
0
00
nn
n
k
kk xaxaxaxaxaxaxf1.
02)( 2 xexf x2.
0sin)( xxexf x3.
dsb
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Numerical method for finding roots of non linear equations
Bracketing methods
Open methods
Bisecton method
False positionmethod
Fixed pointmethod
Newton-Raphson method
Secant method
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Bracketing Methods:- At least two guesses are required- Require that the guesses bracket the
root of an equation
- More robust that open methodsOpen Methods:
- Most of the time, only one initial guess is required
- Do that require that the guesses bracket the root of the equation
- More computationally efficient than bracketing methods but they do not always work…..may blow up !!
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Bracketing Methods
• These methods are known as bracketing methods because they rely on having two initial guesses. - xl - lower bound and
- xu - upper bound.
• The guesses must bracket (be either side of) the root. WHY ?
• Bisection method• Method of False position
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f(x)
xxl
xuxr xr xr
• Atau terdapat akar yang banyaknya ganjil.
f(x)
xxu
xl xr
• Bila f(xu) dan f(xl) berlainan tanda maka pasti akar, xr, diantara xu dan xl. i.e. xl < xr < xu.
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f(x)
xxl xu
xrxr
f(x)
xxl xu
• Bila f(xu) dan f(xl) mempunyai tanda yang sama, maka kemungkinan tidak terdapa akar diantara xl and xu.
• Atau kemungkinan terdapat banyaknya akar genap diantara xl and xu.
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There are exceptions to the rulesf(x)
x
Multipleroots occur here
f(x)
x
When the function is tangentialto the x-axis, multiple roots occur
Functions with discontinuitiesdo not obey the rules above
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The Bisection Method can be used to solve the roots for such an equation. The method can be described by the following algorithm to solve for a root for the function f(x):
1. Choose upper and lower limits (a and b) 2. Make sure a < b, and that a and b lie within the
range for which the function is defined. 3. Check to see if a root exists between a and b
(check to see if f(a)*f(b) < 0) 4. Calculate the midpoint of a and b (mid = (a+b)/2) 5. if f(mid)*f(a) < 0 then the root lies between mid
and a (set b=mid), otherwise it lies between b and mid (set a=mid)
6. if f(mid) is greater than epsilon then loop back to step 4, otherwise report the value of mid as the root.
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Metoda Bisection
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Bisection method…
• This method converges to any pre-specified tolerance when a single root exists on a continuous function
• Example Exercise: write a function that finds the square root of any positive number that does not require programmer to specify estimates
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Fixed Point Iteration Java Applet.mht
Double Click disini
Iterasi Metoda bagi dua
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Metoda Posisi SalahMetoda posisi salah (Regula Falsi) tetap menggunakan dua
titik perkiraan awal seperti pada metoda bagi dua yaitu a0 dan b0 dengan syarat f(a0).f(b0) < 0. Metoda Regula Falsi dibuat untuk mempecepat konvergensi iterasi pada metoda bagi dua yaitu dengan melibatkan f(a) dan f(b)
Rumus iterasi Regula Falsi:
)())()((
)(1 n
nn
nnnn bf
afbfab
bc
n=0,1,2,3,…
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Metoda Posisi Salah
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Metoda Terbuka
,...3,2,1,0)(1 nxgx nn
Pada metoda tetap, rumus iterasi diperoleh dari f(x) =0 yaitudengan mengubah f(x) = 0 menjadi:
1. Metoda titik tetap
atau
,...3,2,1,0)(1 nxfxx nnn
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Contoh:f(x) = 1 – x – x^3=0
3
2
^n x- 1nx
1nx
0 0 x Jawab :
2
30 - 10 x1
0 .5
00 2
3.5 - 1.5 x2
0 .6875
6875.
230.6875 - 10 x3
0 .6813
6813.
230.6813 - 10 x4
0.6825
6825.
230.6825 - 10 x5
0 .6823
00
23 .6823 - 1 .6823 x6
0 .6823
Jadi akar pendekatan adalah .68230
Rumus iterasi diperolehdengan x=x +f(x) yaitu:1-2x-x^3 = -x, kemudian diubah menjadi:
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Hitung f(x) = 3 – x2
X + kx + 3 – x2 = x + kx
x6= (1.732056) + 1-(1.732056)2/3 =1.732051
Jadi akar pendekatan adalah 1.732051
x0 = 1
x1= (1) + 1-(1)2/3 =1.666667
x2= (1.666667) + 1-(1.666667)2/3 =1.740741
x3= (1.740741) + 1-(1.740741)2/3 =1.730681
x4= (1.730681) + 1-(1.730681)2/3 =1.732018
x5= (1.732018) + 1-(1.732018)2/3 =1.732056
Jawab :
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Metode Newton
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Equations of One Variable.mht
Double click disini
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Terima kasih
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