Title: HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR Pertemuan 3
1HAMPIRAN NUMERIK SOLUSI PERSAMAAN
NIRLANJARPertemuan 3
- Matakuliah K0342 / Metode Numerik I
- Tahun 2006
2Pertemuan
3
HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR
3PERSAMAAN NIRLANJAR (N0N LINIER)
Yaitu persamaan yang mengandung variabel
berpangkat lebih dari satu dan/atau yang
mengandung fungsi-fungsi transenden
Contoh
1.
2.
3.
dsb
4Numerical method for finding roots of non linear
equations
Newton-Raphson method
Bisecton method
Fixed point method
False position method
Secant method
5Bracketing Methods - At least two guesses are
required - Require that the guesses bracket the
root of an equation - More robust that
open methods
Open 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 !!
6Bracketing Methods
- Bisection method
- Method of False position
- 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 ?
7- Bila f(xu) dan f(xl) berlainan tanda
- maka pasti akar, xr, diantara xu dan xl.
- i.e. xl lt xr lt xu.
- Atau terdapat akar yang banyaknya ganjil.
8- Bila f(xu) dan f(xl) mempunyai tanda
- yang sama, maka kemungkinan tidak
- terdapa akar diantara xl and xu.
xl
xu
- Atau kemungkinan terdapat banyaknya
- akar genap diantara xl and xu.
9There are exceptions to the rules
When the function is tangential to the x-axis,
multiple roots occur
Functions with discontinuities do 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) - Choose upper and lower limits (a and b)
- 2. Make sure a lt 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) lt 0) - 4. Calculate the midpoint of a and b (mid
(ab)/2) - 5. if f(mid)f(a) lt 0 then the root lies
between mid and a (set bmid), otherwise it lies
between b and mid (set amid) - 6. if f(mid) is greater than epsilon then loop
back to step 4, otherwise report the value of
mid as the root.
11Metoda Bisection
12Bisection 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
13Iterasi Metoda bagi dua
Double Click disini
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15(No Transcript)
16Metode Bisection
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18Metoda Posisi Salah
Metoda 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) lt 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
n0,1,2,3,
19Metoda Posisi Salah
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21Metoda Terbuka
1. Metoda titik tetap
Pada metoda tetap, rumus iterasi diperoleh dari
f(x) 0 yaitu dengan mengubah f(x) 0 menjadi
atau
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24Contoh f(x) 1 x x30
Rumus iterasi diperoleh dengan xx f(x)
yaitu 1-2x-x3 -x, kemudian diubah menjadi
Jawab
Jadi akar pendekatan adalah
25Hitung f(x) 3 x2
X kx 3 x2 x kx
Jawab
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
x6 (1.732056) 1-(1.732056)2/3 1.732051
Jadi akar pendekatan adalah 1.732051
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27Metode Newton
28Double click disini
29Terima kasih