HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR Pertemuan 3

1. HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJARPertemuan 3 Matakuliah : K0342 / Metode Numerik I Tahun : 2006

2. Pertemuan 3 HAMPIRAN NUMERIK SOLUSI PERSAMAAN NIRLANJAR

3. PERSAMAAN NIRLANJAR (N0N LINIER) Yaitu persamaan yang mengandung variabel berpangkat lebih dari satu dan/atau yang mengandung fungsi-fungsi transenden Contoh: 1. 2. 3. dsb

4. Bracketing methods Open methods Numerical method for finding roots of non linear equations Newton-Raphson method Bisecton method Fixed point method False position 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 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 !!

6. Bracketing 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. xr xr xr xu xl f(x) f(x) xu x x xl xr • Bila f(xu) dan f(xl) berlainan tanda • maka pasti akar, xr, diantara xudan xl. • i.e. xl < xr< xu. • Atau terdapat akar yang banyaknya ganjil.

8. xr xr xl xu f(x) f(x) x x • Bila f(xu) dan f(xl) mempunyai tanda • yang sama, maka kemungkinan tidak • terdapa akar diantara xland xu. xl xu • Atau kemungkinan terdapat banyaknya • akar genap diantara xland xu.

9. Multiple roots occur here f(x) f(x) x x There 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 < 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. Iterasi Metoda bagi dua Double Click disini

16. Metode Bisection

18. Metoda 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) < 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: n=0,1,2,3,…

19. Metoda Posisi Salah

21. Metoda Terbuka 1. Metoda titik tetap Pada metoda tetap, rumus iterasi diperoleh dari f(x) =0 yaitu dengan mengubah f(x) = 0 menjadi: atau

24. Contoh: f(x) = 1 – x – x^3=0 Rumus iterasi diperoleh dengan x=x +f(x) yaitu: 1-2x-x^3 = -x, kemudian diubah menjadi: Jawab : Jadi akar pendekatan adalah

25. Hitung 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

27. Metode Newton

28. Double click disini

29. Terima kasih