Part 3
2nd Iteration
For x1 = [x1,y1,z1] = [-0.72254, 0.86392, 0.02569]
|
|
1.467279449
|
1.709282902
|
-0.6242208
|
Df(x0)
|
=
|
0.8639224733
|
-0.72254268
|
-0.0513797
|
|
|
0.4855161696
|
2.372448332
|
1
|
f1 = -1.131742062
f2 = - 0.7148808356
f3 = 2.473654395
the system of equations, in matrix format, is Df(x0)∆x = −f(x0).
1.467279449
|
1.709282902
|
-0.6242208
|
∆x
|
=
|
1.131742062
|
0.8639224733
|
-0.72254268
|
-0.0513797
|
∆y
|
=
|
0.7148808356
|
0.4855161696
|
2.372448332
|
1
|
∆z
|
=
|
-2.473654395
|
or,
1.467279449∆x + 1.709282902∆y – 0.6242208∆z = 1.131742062
0.8639224733∆x – 0.72254268∆y – 0.0513797∆z = 0.7148808356
0.4855161696∆x + 2.372448332∆y + ∆z = -2.473654395
Solving this system of equations gives us,
∆x = 0.4174047065, ∆y = -0.360889953, ∆z = -1.820118362
From which,
x2 = x1 + ∆x, y2 = y1 + ∆y, z2 = z1 + ∆z,
x2 = -0.72254 + 0.4174047, y1 = 0.86392 - 0.36089, z2 = 0.02569 - 1.82012,
x2 = -0.305135, y2 = 0.50303, z2 = -1.79443,
x2 = [-0.305135, 0.50303, -1.79443]
3rd Iteration
For x2 = [x2,y2,z2] = [-0.305135, y2 = 0.50303, z2 = -1.79443]
|
|
-0.292379716
|
1.553613689
|
-0.1534944
|
Df(x0)
|
=
|
0.5030325661
|
-0.30513813
|
3.58885701
|
|
|
0.7370215630
|
1.653728716
|
1
|
f1 = -0.9046327596
f2 = -3.463468076
f3 = 0. 186321774
the system of equations, in matrix format, is Df(x0)∆x = −f(x0).
-0.292379716
|
1.553613689
|
-0.1534944
|
∆x
|
=
|
0.9046327596
|
0.5030325661
|
-0.30513813
|
3.58885701
|
∆y
|
=
|
3.463468076
|
0.7370215630
|
1.653728716
|
1
|
∆z
|
=
|
-0.186321774
|
or,
-0.292379716∆x + 1.553613689∆y – 0.1534944∆z = 0.9046327596
0.5030325661∆x – 0.30513813∆y + 3.58885701∆z = 3.463468076
0.737021563∆x + 1.653728716∆y + ∆z = -0.186321774
Solving this system of equations gives us,
∆x = -2.592141381, ∆y = 0.2276073932, ∆z = 1.347741436
From which,
x3 = x2 + ∆x, y3 = y2 + ∆y, z3 = z2 + ∆z,
x3 = -0.305135 - 2.592141381, y3 = 0.50303 + 0.2276073932, z3 = -1.79443 + 1.347741436,
x3 = -2.897279, y3 = 0.730637, z3 = -0.446688,
x3 = [-2.897279, 0.730637, -0.446688]
This 3d iteration is far from accurate, although it is converging. 10 iterations will give 4 d.p. accuracy with the solution:
X10 = [-0.3287, -1.3027. -0.5816]