Bab 3_Interpolasi Part 2 (Pertemuan 5).pptx

Metode Numerik

1. METODE NUMERIK 1 INTERPOLASI DWI HARTINI TEKNIK DIRGANTARA ITDA TD036

2. Interpolasi Polinomial Newton 2 Secara umum: f 1 (x) = b + b 1 (x- x ) f 2 (x) = b + b 1 (x- x ) + b 2 (x-x )(x- x 1 ) f 3 (x) = b + b 1 (x- x ) + b 2 (x-x )(x- x 1 ) + b 3 (x-x )(x-x 1 )(x- x 2 ) … f n (x) = b + b 1 (x- x ) + b 2 (x-x )(x- x 1 ) + b 3 (x-x )(x-x 1 )(x- x 2 ) + … + b n (x-x 1 )(x-x 2 )…(x- x n- 1 )

3. Interpolasi Polinomial Newton 3 Dengan: b = f(x ) b 1 = f[x 1 , x ] b 2 = f[x 2 , x 1 , x ] … b n = f[x n , x n- 1 , x n- 2 , . . . ., x ]

4. 4 Diketahui: n titik ( x 1 , y 1 ), ( x 2 , y 2 ), …, ( x n , y n ) ( y i = f ( x i ), i =1,2,…, n ) Ditanya : f n ( x ) = a + a 1 x + a 2 x 2 + … + a n x n yang melewati n titik tersebut . Interpolasi Polinomial Newton

5. dengan 5 Interpolasi Polinomial Newton

6. Interpolasi Polinomial Newton Diketahui : (1, 0), (4, 1.386294), (6, 1.791759), (5, 1.609438) (dari fungsi ln x ) Ditanya : Perkirakan ln 2 dengan interpolasi Newton orde ke- 3 6

7. Interpolasi Polinomial Newton 7 Interpolasi Newton Orde 2:  butuh 3 titik x = 1 x 1 = 4 x 2 = 6 f(x ) = f(x 1 ) = 1.386294 f(x 2 ) = 1.791759 b = f(x ) = f 2 (x) = b + b 1 (x- x ) + b 2 (x-x )(x- x 1 )

8. Interpolasi Polinomial Newton 8 x0 = 1 x1 = 4 x2 = 6 f(x0) = f(x1) = 1.386294 f(x2) = 1.791759 f 2 (x) = b + b 1 (x- x ) + b 2 (x-x )(x- x 1 ) f 2 (x) = + 0,462(2- 1) – 0,052 (2-1)(2- 4 ) f 2 (x) = 0,566

9. Interpolasi Polinomial Newton 9 Interpolasi Newton Orde 3:  butuh 4 titik x = 1 x 1 = 4 x 2 = 6 x 3 = 5 f(x ) = f(x 1 ) = 1.386294 f(x 2 ) = 1.791759 f(x 3 ) = 1.609438 b = f(x ) =

10. Interpolasi Polinomial Newton f 3 (2) = 0.629 10 x = 1 x 1 = 4 x 2 = 6 x 3 = 5 f(x ) = f(x 1 ) = 1.386294 f(x 2 ) = 1.791759 f(x 3 ) = 1.609438

11. Interpolasi Polinomial Newton  𝑎 = ( 0,6931472−0,566 0,6931472 )100% = 18,343% Kesalahan perkiraan terbaik Orde 2 Orde 3  𝑎 = ( 0,6931472−0,629 11 0,6931472 )100% = 9,254%

12. Interpolasi Polinomial Newton (Ex.) 12 Hitung nilai tabel distribusi ‘Student t’ pada derajat bebas dengan  = 4%, jika diketahui: t 10% = 1,476 t 5% = 2,015 t 2,5% = 2,571 t 1% = 3,365 dengan interpolasi Newton orde 2 dan orde 3!

13. Interpolasi Polinomial Newton(Ex.) Interpolasi Newton Orde 2:  butuh 3 titik f(x ) = 2,015 f(x 1 ) = 2,571 f(x 2 ) = 3,365 x = 5 x 1 = 2,5 x 2 = 1 b = f(x ) = 2,015 13

14. Interpolasi Polinomial Newton (Ex.) 14 f 2 (x) = b + b 1 (x- x ) + b 2 (x- x )(x- x 1 ) = 2,015 + (- 0,222) (4- 5) + 0,077 (4-5)(4- 2,5) = 2,121

15. Interpolasi Polinomial Newton(Ex.) 15 Interpolasi Newton Orde 3:  butuh 4 titik x = 5 x 1 = 2,5 x 2 = 1 x 3 = 10 f(x ) = 2,015 f(x 1 ) = 2,571 f(x 2 ) = 3,365 f(x3) = 1,476

16. Interpolasi Polinomial Newton (Ex.) b = f(x ) = 2,015 b 1 = - 0,222  f[x 1 ,x ] b 2 = 0,077  f[x 2 ,x 1 ,x ] b 3 = ?  f[x 3 ,x 2 ,x 1 ,x ] 16 x = 5 f(x ) = 2,015 x 1 = 2,5 f(x 1 ) = 2,571 x 2 = 1 f(x 2 ) = 3,365 x 3 = 10 f(x 3 ) = 1,476

17. Interpolasi Polinomial Newton(Ex.) 17 f 3 (x) = b + b 1 (x- x ) + b 2 (x- x )(x- x 1 ) + b 3 (x- x )(x-x 1 )(x- x 2 ) = 2,015 + (- 0,222)(4- 5) + 0,077 (4-5)(4- 2,5) + (-0,007)(4-5)(4-2,5)(4- 1) = 2,015 + 0,222 + 0,1155 + 0,0315 = 2,153

18. Kesalahan Interpolasi Polinomial 18 R n = |f[x n+1 ,x n ,x n-1 ,…,x ](x- x )(x- x 1 )…(x-x n )| Menghitung R 1 Perlu 3 titik (karena ada x n+1 ) R 1 = |f[x 2 ,x 1 ,x ](x- x )(x- x 1 )| Menghitung R 2 Perlu 4 titik sebagai harga awal R 2 = |f[x 3 ,x 2 ,x 1 ,x ](x-x )(x- x 1 )(x- x 2 )|

19. Kesalahan Interpolasi Polinomial (Ex.) 19 Berdasarkan contoh: R 1 = |f[x 2 ,x 1 ,x ](x-x )(x-x 1 )| = |0.077 (4-5)(4-2.5)| = 0.1155 R 2 = |f[x 3 ,x 2 ,x 1 ,x ](x- x )(x-x 1 )(x- x 2 )| = |- 0.007 (4-5)(4-2.5)(4- 1)| = 0.0315

20. Interpolasi Lagrange Interpolasi Lagrange pada dasarnya dilakukan untuk menghindari perhitungan dari differensiasi terbagi hingga (Interpolasi Newton) Rumus: dengan 20

21. Interpolasi Lagrange Pendekatan orde ke- 1 f 1 (x) = L (x)f(x ) + L 1 (x)f(x 1 ) i=0, j=1, n=1, j≠i i=1, j=0, n=1 21

22. Interpolasi Lagrange Pendekatan orde ke- 2 f 2 (x) = L (x)f(x ) + L 1 (x)f(x 1 ) + L 2 (x)f(x 2 ) 22

23. Interpolasi Lagrange Pendekatan orde ke- 3 f 3 (x) = L (x)f(x ) + L 1 (x)f(x 1 ) + L 2 (x)f(x 2 ) + L 3 (x)f(x 3 ) 3 3 2 3 2 2 3 2 2  f  x      3 1   x  x  x  x  x  x   x  x  x  x 1  x  x 2  f  x      x  x   1     x  x  x  x  x  x  x  x 1  x  x 3 23

24. Interpolasi Lagrange (Ex.) 24 Berapa nilai distribusi t pada  = 4 %?  = 2,5 %  x = 2,5  f(x ) = 2,571  = 5 %  x 1 = 5  f(x 1 ) = 2,015  = 10 %  x 2 = 10  f(x 2 ) = 1,476

25. Interpolasi Lagrange (Ex.) Pendekatan orde ke- 1 f 1 (x) = L (x)f(x ) + L 1 (x)f(x 1 ) 1 1 1 f  x  f  x   x  x 1 x  x x  x 1 x  x f  x    25     2,237  2,5  5   5  2,5    4  5   2,571    4  2,5   2,015  x = 2,5  f(x ) = 2,571 x 1 = 5  f(x 1 ) = 2,015 x 2 = 10  f(x 2 ) = 1,476

26. Interpolasi Lagrange (Ex.) Pendekatan orde ke- 2 f 2 (x) = L (x)f(x ) + L 1 (x)f(x 1 ) + L 2 (x)f(x 2 ) 26 x = 2,5  f(x ) = 2,571 x 1 = 5  f(x 1 ) = 2,015 x 2 = 10  f(x 2 ) = 1,476

27. Interpolasi Lagrange (Ex.) 27 Latihan Gunakan interpolasi polinomial Lagrange order dua untuk menghitung ln 2 ? x = 1 x 1 = 4 x 2 = 6 f(x ) = f(x 1 ) = 1,3862944 f(x 2 ) = 1,791759 5