hampiran numerik solusi persamaan nirlanjar pertemuan 3

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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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