Delet Tayor
3
4 3 2 1 f (x) = -0.1 X - 0.15 X - 0.5 X - 0.25 X + 1.2 Xi = 0 Xi+1 = 1 Solusi : Orde 0 : Xi = 1 f(1) = 0.2 Xi = 0 f (0) = 1.2 Et1 = -1 Orde 1 : f (Xi+1) = f (xi) + f'(xi).h 3 2 1 f' (Xi) = -0.4 X - 0.45 X - 1 X - 0.25 xi = 0 f" (xi) = -0.25 h = 1 f (Xi+1) = 0.95 Et2 = -0.75 Orde 2 : 2 1 f"(Xi) = -1.2 X - 0.9 X - 1 Xi = 0 f"(Xi) = -1 f (Xi+1) = (Xi) + f'(Xi).h+f"(Xi)/2!.h
description
Metode Numerik
Transcript of Delet Tayor
4 3 2 1f (x) = -0.1 X - 0.15 X - 0.5 X - 0.25 X + 1.2
Xi = 0Xi+1 = 1
Solusi :
Orde 0 :
Xi = 1
f(1) = 0.2
Xi = 0
f (0) = 1.2
Et1 = -1
Orde 1 :
f (Xi+1) = f (xi) + f'(xi).h
3 2 1f' (Xi) = -0.4 X - 0.45 X - 1 X - 0.25
xi = 0
f" (xi) = -0.25 h = 1
f (Xi+1) = 0.95
Et2 = -0.75
Orde 2 :2 1
f"(Xi) = -1.2 X - 0.9 X - 1
Xi = 0
f"(Xi) = -1
f (Xi+1) = f (Xi) + f'(Xi).h+f"(Xi)/2!.h2
f (Xi+1) = 0.45
Et3 = -0.25
Orde 3 :
3 2 1f (x) = 0.25 X + 0.5 X + 0.25 X + 0.5
Xi = 0Xi+1= 1
Gunakan Deret Taylor Untuk mengaproksimasi di atas :
Orde 0 :
untuk x = 0
f (xi) = f(0) = 0.5
Untuk x = 1
f (xi+1) = f(1) = 1.5
Et1 = 1
Orde 1 :
f (xi+1) = f (xi) + f'(xi).h2 1
f'(xi) = 0.75 X + 1.0 X + 0.3
xi = 0
f'(xi) = 0.25
f(xi+1) = 0.75
Et2 = 0.75
Orde 2 :
