HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR Pertemuan 3

1

HAMPIRAN NUMERIK SOLUSI HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJARPERSAMAAN NIRLANJAR

Pertemuan 3

Matakuliah : K0342 / Metode Numerik I Tahun : 2006

2

PertemPertemuan uan

33HAMPIRAN NUMERIK HAMPIRAN NUMERIK SOLUSI PERSAMAAN SOLUSI PERSAMAAN

NIRLANJARNIRLANJAR

3

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

4

Numerical method for finding roots of non linear equations

Bracketing methods

Open methods

Bisecton method

False positionmethod

Fixed pointmethod

Newton-Raphson method

Secant method

5

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 !!

6

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

7

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.

8

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.

9

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

10

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.

11

Metoda Bisection

12

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

13

Fixed Point Iteration Java Applet.mht

Double Click disini

Iterasi Metoda bagi dua

14

15

16

17

18

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,…

19

Metoda Posisi Salah

20

21

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

22

23

24

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:

25

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 :

26

27

Metode Newton

28

Equations of One Variable.mht

Double click disini

29

Terima kasih