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1. INTERPOLASI (METODE NUMERIK)
Erik Pebrinasyah
Ali Martun Pulungan
Nyella Kenanga Andini
Dwi Yulia Ningsih
Ario Amanda

2. Tentukakanan lah 5 titik Interpolasi (sebarang titik) dari
data berikut dengan mengugunakan 4 metode yaitu
1. Interpolasi Liner x y
2. Interpolasi kuadrat 15 14944
20 19867
3. Interpolasi Newton
25 24740
4. Interpolasi Lagrange 30 29552
35 34290
40 38945
45 43497
50 47943
55 52269
60 56464
(Sumber: Nasution;54)

3. Penyelesaian :
Misalkan diambil sembarang lima titik interpolasi yaitu 18,
23,33,43, dan 53.
1. Interpolasi Linier
 Untuk x = 18
 X0 = 15 → f(x0) = 14944
 X1 = 20 → f(x1) = 19867

4.  Untuk x = 23
 X0 = 20 → f(x0) = 19867
 X1 = 25 → f(x1) = 24740
 Untuk x = 33
 X0 = 30 → f(x0) = 29552
 X1 = 35 → f(x1) = 34290

5.  Untuk x = 43
 X0 = 40 → f(x0) = 38945
 X1 = 45 → f(x1) = 43497
 Untuk x = 53
 X0 = 50→ f(x0) = 47934
 X1 = 55 → f(x1) = 52269

6. Tabel Interpolasi linier

7. 2. Interpolasi Kuadrat
 Untuk x = 18
 X0 = 15 → f(x0) = 14944
 X1 = 20 → f(x1) = 19867
 X2 = 25 → f(x2) = 24740
 b0 = f(x0)
= 14944
 b1 = = 984,6
 b2 = = =-1

8.  F (x) = b0 + b1 ( x – x0) + b2 (x-x0)(x-x1)
= 14944 + 984,6(18-15) + (-1)(18-15)(18-20)
= 14944 + 984,6 (3) + (-1)(3)(-2)
= 14944 + 2953,8 + 6
= 17903,8
 Untuk x = 23
 X0 = 20 ,f(x0 ) =19867
 X1 = 25 ,f(x1 ) =24740
 X2 = 30 ,f(x2 ) =29552
 b0 = f(x0 )
= 19867

9.  b1 = = = 974,6
 b2 = = = = -1,22
 F (x) = b0 + b1(x- x0) + b2(x- x0) (x- x1)
= 19867 + 974.6 (23-20) + (-1.22)(23-20)(23-25)
= 19867 + 2923.8 + 7.32
= 22798.12
 Untuk x = 33
 X0 = 30 , f(x0 ) =29552
 X1 = 35 ,f(x1 ) =34290
 X2 = 40 ,f(x2 ) =38945
 b0 = f(x0 )
= 29552

10.  b1 = = 947,6
 b2 = = = -1,66
 F (x) = b0 + b1(x- x0) + b2(x- x0) (x- x1)
= 29552 + 947.6 (33-30) + (-1.66)(33-30)(33-35)
= 29552 + 2842.8 + 9.96
= 32404.76
 untuk x = 43
 X0 = 40 → f(x0) = 38945
 X1 = 45 → f(x1) = 43497
 X2 = 50 → f(x2) = 47943
 b0 = f(x0)
= 38945

11.  b1 = = = 910,4
 b2 = = = -2,12
 F (x) = b0 + b1 ( x – x0) + b2 (x-x0)(x-x1)
= 38945 + 910,4(43-40) + (-2,12)(43-40)(43-45)
= 38945 + 910,4 (3) + (-2,12)(3)(-2)
= 38945 + 2731,2 + 12,72
= 41688,92
 Untuk x =53
 X0 = 50 → f(x0) = 47943
 X1 = 55 → f(x1) = 52269
 X2 = 60 → f(x2) = 56464

12.  b0 = f(x0)
= 47943
 b1 = = 865,2
 b2 = = -13,1
 F (x) = b0 + b1 ( x – x0) + b2 (x-x0)(x-x1)
= 47943 + 865,2(53-50) + (-13,1)(53-50)(53-55)
= 47943 + 2595,6 + 78,6
= 50617,2

13. Tabel Interpolasi Kuadrat

14. 3. Interpolasi newton.

19. Tabel Hasil Interpolasi Newton

20.  4. Interpolasi Lagrange

30. Tabel Hasil Interpolasi
Lagrange