Metode posisi palsu
Contoh:
Tentukan salah satu akar dari f(x) = exp(x) – 4x dengan menggunakan metode posisi palsu.
Penyelesaian:
Seperti pada metode bagi dua, karena tidak diberitahu batas intervalnya, dan dari gambar 3.4, misalkan diambil a = 0 dan b = 1 (alasannya f(a)*f(b) < 0).
Dengan menggunakan metode posisi palsu (secara manual) selesaiannya adalah sebagai berikut.
Iterasi 1
a=0, b = 1;
cek = 2*b – a = 2;
f(a) = exp(a) – 4a = exp(0) – 4*0 = 1
f(b) = exp(b) – 4b = exp(1) – 4*1 = -1.2817
c=b-f(b)(b-a)/(f(b)-f(a))=1- (-1.2817)*(1 – 0) / (-1.2817 – 1) = 0.4383
f(c) = exp(c) – 4c = exp(0.4383) – 4*0.4383 = -0.203047
f(a)*f(c) = -0.203047 < 0 maka b = c = 0.4383
Iterasi 2
a = 0, b = 0.4383
f(a) = 1
f(b) = -0.203047
c=b-f(b)(b-a)/(f(b)-f(a))= 0.4383 – (-0.203047) * (0.4383 – 0)/( -0.203047- 1) = 0.3643
f(c) = exp(c) – 4c = exp(0.3643) – 4*0.3643 = -0.017686
f(a)*f(c) = -0.017686 < 0 maka b = c = 0.3643
Iterasi 3
a = 0, b = 0.3643
f(a) = 1
f(b) = -0.017686
c=b-f(b)(b-a)/(f(b)-f(a))= 0.3643 – (-0.017686)* (0.3643 – 0) /(-0.017686 – 1)= 0.3580
f(c) = exp(c) – 4c = exp(0.3580) – 4*0.3580 = -0.001447
f(a)*f(c) = -0.001447 < 0 maka b = c = 0.3580
dst.
Sampai iterasi ke tiga, maka hampiran akar = c = 0.3580
| | Iterasi 1 | Iterasi 2 | Iterasi 3 |
| a | 0 | 0 | 0 |
| +1b | 1 | 0.4383 | 0.3643 |
| c | 0.4383 | 0.3643 | 0.3580 |
| f(a)*f(c) | -0.203047<0 | -0.017686< 0 | |
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Contoh
Diketahui persamaan f(x) = ex – 4x; Tentukan hampiran akar menggunakan x0 = 0 dan tol = 0.00001.
Penyelesaian:
Turunan pertama f’(x) = ex – 4.
x0 = 0,
f(x0) = e0 – 4*0 = 1, f’(x0) = e0 – 4 = -3,
x1 = x0 – f(x0)/f’(x0) = 0 – 1/(-3) = 0.333333
f(x1) = f(0.333333) = 0.062279 , f’(x1) = f’(0.333333) = -2.604388
x2 = x1 – f(x1)/f’(x1) = 0.357246
f(x2) = f(0.357246) = 0.000402, f’(x2) = f’(0.357246) = -2.570612
x3 = x2 – f(x2)/f’(x2) = 0.357403
f(x3) = f(0.357403) = 0.000000, f’(x3) = f’(0.357403) = -2.570388
x4 = x3 – f(x3)/f’(x3) = 0.357403
Pada iterasi ketiga, dapat disimpulkan bahwa akarnya adalah 0.357403.
Soal no 2
Soal f(x)=x^3-3x^2-x+3
Carilah akar-akar persamaan dari fungsi diatas dengan menggunakan beberapa metode sebagai berikut :
a. Metode Bisection, dengan nilai awal Xn = 2 dan Xn+1 =5.
b. Metode Interpolasi, dengan nilai awal Xn = 2 dan Xn+1 = 5.
c. Metode Newton – Rapshon, dengan nilai Xi = -4
Pembahasan :
a. Metode Bisection dengan nilai awal Xn = 2 dan Xn+1 =5.
| iterasi | Xn | Xn+1 | Xt | f(Xn) | f(Xn+1) | f(Xt) |
| 1 | 2 | 5 | 3.5 | -3 | 48 | 5.625 |
| 2 | 2 | 3.5 | 2.75 | -3 | 5.625 | -1.64063 |
| 3 | 2.75 | 3.5 | 3.125 | -1.64063 | 5.625 | 1.095703 |
| 4 | 2.75 | 3.125 | 2.9375 | -1.64063 | 1.095703 | -0.47681 |
| 5 | 2.9375 | 3.125 | 3.03125 | -0.47681 | 1.095703 | 0.25589 |
| 6 | 2.9375 | 3.03125 | 2.984375 | -0.47681 | 0.25589 | -0.12354 |
| 7 | 2.984375 | 3.03125 | 3.007813 | -0.12354 | 0.25589 | 0.062867 |
| 8 | 2.984375 | 3.007813 | 2.996094 | -0.12354 | 0.062867 | -0.03116 |
| 9 | 2.996094 | 3.007813 | 3.001953 | -0.03116 | 0.062867 | 0.015648 |
| 10 | 2.996094 | 3.001953 | 2.999023 | -0.03116 | 0.015648 | -0.00781 |
| 11 | 2.999023 | 3.001953 | 3.000488 | -0.00781 | 0.015648 | 0.003908 |
| 12 | 2.999023 | 3.000488 | 2.999756 | -0.00781 | 0.003908 | -0.00195 |
| 13 | 2.999756 | 3.000488 | 3.000122 | -0.00195 | 0.003908 | 0.000977 |
| 14 | 2.999756 | 3.000122 | 2.999939 | -0.00195 | 0.000977 | -0.00049 |
b. Metode Interpolasi Linear dengan nilai awal Xn = 2 dan Xn+1 =5.
| iterasi | Xn | Xn+1 | X* | f(Xn) | f(Xn+1) | f(X*) |
| 1 | 2 | 5 | 2.176471 | -3 | 48 | -3.07755 |
| 2 | 2.176471 | 5 | 2.346595 | -3.07755 | 48 | -2.94457 |
| 3 | 2.346595 | 5 | 2.499961 | -2.94457 | 48 | -2.62511 |
| 4 | 2.499961 | 5 | 2.629598 | -2.62511 | 48 | -2.19085 |
| 5 | 2.629598 | 5 | 2.733067 | -2.19085 | 48 | -1.72697 |
| 6 | 2.733067 | 5 | 2.811795 | -1.72697 | 48 | -1.29978 |
| 7 | 2.811795 | 5 | 2.869487 | -1.29978 | 48 | -0.94413 |
| 8 | 2.869487 | 5 | 2.910584 | -0.94413 | 48 | -0.66807 |
| 9 | 2.910584 | 5 | 2.939266 | -0.66807 | 48 | -0.46397 |
| 10 | 2.939266 | 5 | 2.958994 | -0.46397 | 48 | -0.31803 |
| 11 | 2.958994 | 5 | 2.972428 | -0.31803 | 48 | -0.21604 |
| 12 | 2.972428 | 5 | 2.981513 | -0.21604 | 48 | -0.14586 |
| 13 | 2.981513 | 5 | 2.987627 | -0.14586 | 48 | -0.09806 |
| 14 | 2.987627 | 5 | 2.99173 | -0.09806 | 48 | -0.06575 |
| 15 | 2.99173 | 5 | 2.994477 | -0.06575 | 48 | -0.044 |
| 16 | 2.994477 | 5 | 2.996314 | -0.044 | 48 | -0.02941 |
| 17 | 2.996314 | 5 | 2.997541 | -0.02941 | 48 | -0.01964 |
| 18 | 2.997541 | 5 | 2.99836 | -0.01964 | 48 | -0.01311 |
| 19 | 2.99836 | 5 | 2.998906 | -0.01311 | 48 | -0.00874 |
| 20 | 2.998906 | 5 | 2.999271 | -0.00874 | 48 | -0.00583 |
| 21 | 2.999271 | 5 | 2.999514 | -0.00583 | 48 | -0.00389 |
| 22 | 2.999514 | 5 | 2.999676 | -0.00389 | 48 | -0.00259 |
| 23 | 2.999676 | 5 | 2.999784 | -0.00259 | 48 | -0.00173 |
| 24 | 2.999784 | 5 | 2.999856 | -0.00173 | 48 | -0.00115 |
| 25 | 2.999856 | 5 | 2.999904 | -0.00115 | 48 | -0.00077 |
c. Metode Newton Raphson dengan nilai awal Xi = -4.
Turunan pertama dari f(x)=x^3-3x^2-x+3 adalah f’(x) = 3x^2-6x-1
| iterasi | Xi | Xi+1 | f(Xi) | f(Xi+1) | f’(X) |
| 1 | -4 | -2.52113 | -105 | -29.5716 | 71 |
| 2 | -2.52113 | -1.63028 | -29.5716 | -7.67617 | 33.195 |
| 3 | -1.63028 | -1.17214 | -7.67617 | -1.56005 | 16.75515 |
| 4 | -1.17214 | -1.01851 | -1.56005 | -0.15018 | 10.15463 |
| 5 | -1.01851 | -1.00025 | -0.15018 | -0.00201 | 8.223197 |
| 6 | -1.00025 | -1 | -0.00201 | -3.8E-07 | 8.00302 |
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