روش ژاکوبی

در ${\displaystyle Ax=b}$ با داشتن مقدار اولیه (حدس اولیه)، جواب را بدست می آوریم.

${\displaystyle A={\begin{bmatrix}2&1\\5&7\\\end{bmatrix}},\ b={\begin{bmatrix}11\\13\\\end{bmatrix}}\quad {\text{and}}\quad x^{(0)}={\begin{bmatrix}1\\1\\\end{bmatrix}}.}$

با استفاده از رابطۀ ${\displaystyle x^{(k+1)}=D^{-1}(b-Rx^{(k)})}$، که در بالا توضیح داده شد، به تخمین ${\displaystyle x}$ می پردازیم تا جواب بدست آید.

با بازنویسی رابطۀ اخیر به فرم ساده‌تر ${\displaystyle D^{-1}(b-Rx^{(k)})=Tx^{(k)}+C}$،که در آن ${\displaystyle T=-D^{-1}R}$ و ${\displaystyle C=D^{-1}b}$.

توجه کنید که ${\displaystyle R=L+U}$ که ${\displaystyle L}$ و ${\displaystyle U}$ به ترتیب قسمت پایینی و بالایی ${\displaystyle A}$ هستند.

با استفاده از مقادیر داده شده:

${\displaystyle D^{-1}={\begin{bmatrix}1/2&0\\0&1/7\\\end{bmatrix}},\ L={\begin{bmatrix}0&0\\5&0\\\end{bmatrix}}\quad {\text{and}}\quad U={\begin{bmatrix}0&1\\0&0\\\end{bmatrix}}.}$

we determine ${\displaystyle T=-D^{-1}(L+U)}$ as

${\displaystyle T={\begin{bmatrix}1/2&0\\0&1/7\\\end{bmatrix}}\left\{{\begin{bmatrix}0&0\\-5&0\\\end{bmatrix}}+{\begin{bmatrix}0&-1\\0&0\\\end{bmatrix}}\right\}={\begin{bmatrix}0&-1/2\\-5/7&0\\\end{bmatrix}}.}$

Further, C is found as

${\displaystyle C={\begin{bmatrix}1/2&0\\0&1/7\\\end{bmatrix}}{\begin{bmatrix}11\\13\\\end{bmatrix}}={\begin{bmatrix}11/2\\13/7\\\end{bmatrix}}.}$

With T and C calculated, we estimate ${\displaystyle x}$ as ${\displaystyle x^{(1)}=Tx^{(0)}+C}$:

${\displaystyle x^{(1)}={\begin{bmatrix}0&-1/2\\-5/7&0\\\end{bmatrix}}{\begin{bmatrix}1\\1\\\end{bmatrix}}+{\begin{bmatrix}11/2\\13/7\\\end{bmatrix}}={\begin{bmatrix}5.0\\8/7\\\end{bmatrix}}\approx {\begin{bmatrix}5\\1.143\\\end{bmatrix}}.}$

با تکرار بعدی داریم:

${\displaystyle x^{(2)}={\begin{bmatrix}0&-1/2\\-5/7&0\\\end{bmatrix}}{\begin{bmatrix}5.0\\8/7\\\end{bmatrix}}+{\begin{bmatrix}11/2\\13/7\\\end{bmatrix}}={\begin{bmatrix}69/14\\-12/7\\\end{bmatrix}}\approx {\begin{bmatrix}4.929\\-1.714\\\end{bmatrix}}.}$

این فرایند تا جایی که همگرایی مشهود باشد تکرار می‌شود. (به عبارت دقیق تر ${\displaystyle \|Ax^{(n)}-b\|}$ کوچک باشد). جواب پس از ۲۵ بار تکرار خواهد بود:

${\displaystyle x={\begin{bmatrix}7.111\\-3.222\end{bmatrix}}.}$

فرض کنید معادلات زیر را بخواهیم حل کنیم:

${\displaystyle {\begin{aligned}10x_{1}-x_{2}+2x_{3}&=6,\\-x_{1}+11x_{2}-x_{3}+3x_{4}&=25,\\2x_{1}-x_{2}+10x_{3}-x_{4}&=-11,\\3x_{2}-x_{3}+8x_{4}&=15.\end{aligned}}}$

با انتخاب (0, 0, 0, 0) به عنوان تقریب اولیه:

${\displaystyle {\begin{aligned}x_{1}&=(6-0-0)/10=0.6,\\x_{2}&=(25-0-0)/11=25/11=2.2727\\x_{3}&=(-11-0-0)/10=-1.1,\\x_{4}&=(15-0-0)/8=1.875.\end{aligned}}}$

پاسخ‌ها پس از ۵ بار تکرار در جدول زیر آورده شده است:

${\displaystyle x_{1}}$	${\displaystyle x_{2}}$	${\displaystyle x_{3}}$	${\displaystyle x_{4}}$
${\displaystyle 0.6}$	${\displaystyle 2.27272}$	${\displaystyle -1.1}$	${\displaystyle 1.875}$
${\displaystyle 1.04727}$	${\displaystyle 1.7159}$	${\displaystyle -0.80522}$	${\displaystyle 0.88522}$
${\displaystyle 0.93263}$	${\displaystyle 2.05330}$	${\displaystyle -1.0493}$	${\displaystyle 1.13088}$
${\displaystyle 1.01519}$	${\displaystyle 1.95369}$	${\displaystyle -0.9681}$	${\displaystyle 0.97384}$
${\displaystyle 0.98899}$	${\displaystyle 2.0114}$	${\displaystyle -1.0102}$	${\displaystyle 1.02135}$

جواب دقیق (1, 2, −1, 1) است.

import numpy as np

ITERATION_LIMIT = 1000
# initialize the matrix
A = np.array([[10., -1., 2., 0.],
              [-1., 11., -1., 3.],
              [2., -1., 10., -1.],
              [0.0, 3., -1., 8.]])
# initialize the RHS vector
b = np.array([6., 25., -11., 15.])
# prints the system
print("System:")
for i in range(A.shape[0]):
    row = ["{}*x{}".format(A[i, j], j + 1) for j in range(A.shape[1])]
    print(" + ".join(row), "=", b[i])
print()

x = np.zeros_like(b)
for it_count in range(ITERATION_LIMIT):
    print("Current solution:", x)
    x_new = np.zeros_like(x)

    for i in range(A.shape[0]):
        s1 = np.dot(A[i, :i], x[:i])
        s2 = np.dot(A[i, i + 1:], x[i + 1:])
        x_new[i] = (b[i] - s1 - s2) / A[i, i]

    if np.allclose(x, x_new, rtol=1e-10):
        break

    x = x_new

print("Solution:")
print(x)
error = np.dot(A, x) - b
print("Error:")
print(error)

خروجی به این شکل خواهد بود:

System:
10.0*x1 + -1.0*x2 + 2.0*x3 + 0.0*x4 = 6.0
-1.0*x1 + 11.0*x2 + -1.0*x3 + 3.0*x4 = 25.0
2.0*x1 + -1.0*x2 + 10.0*x3 + -1.0*x4 = -11.0
0.0*x1 + 3.0*x2 + -1.0*x3 + 8.0*x4 = 15.0

Current solution: [ 0.  0.  0.  0.]
Current solution: [ 0.6         2.27272727 -1.1         1.875     ]
Current solution: [ 1.04727273  1.71590909 -0.80522727  0.88522727]
Current solution: [ 0.93263636  2.05330579 -1.04934091  1.13088068]
Current solution: [ 1.01519876  1.95369576 -0.96810863  0.97384272]
Current solution: [ 0.9889913   2.01141473 -1.0102859   1.02135051]
Current solution: [ 1.00319865  1.99224126 -0.99452174  0.99443374]
Current solution: [ 0.99812847  2.00230688 -1.00197223  1.00359431]
Current solution: [ 1.00062513  1.9986703  -0.99903558  0.99888839]
Current solution: [ 0.99967415  2.00044767 -1.00036916  1.00061919]
Current solution: [ 1.0001186   1.99976795 -0.99982814  0.99978598]
Current solution: [ 0.99994242  2.00008477 -1.00006833  1.0001085 ]
Current solution: [ 1.00002214  1.99995896 -0.99996916  0.99995967]
Current solution: [ 0.99998973  2.00001582 -1.00001257  1.00001924]
Current solution: [ 1.00000409  1.99999268 -0.99999444  0.9999925 ]
Current solution: [ 0.99999816  2.00000292 -1.0000023   1.00000344]
Current solution: [ 1.00000075  1.99999868 -0.99999899  0.99999862]
Current solution: [ 0.99999967  2.00000054 -1.00000042  1.00000062]
Current solution: [ 1.00000014  1.99999976 -0.99999982  0.99999975]
Current solution: [ 0.99999994  2.0000001  -1.00000008  1.00000011]
Current solution: [ 1.00000003  1.99999996 -0.99999997  0.99999995]
Current solution: [ 0.99999999  2.00000002 -1.00000001  1.00000002]
Current solution: [ 1.          1.99999999 -0.99999999  0.99999999]
Current solution: [ 1.  2. -1.  1.]
Solution:
[ 1.  2. -1.  1.]
Error:
[ -2.81440107e-08   5.15706873e-08  -3.63466359e-08   4.17092547e-08]