• 1 Постановка задачи
• 2 Метод
• 3 Условие сходимости
  • 3.1 Теорема
• 4 Условие окончания
• 5 Пример реализации на Python
• 6 См. также
• 7 Примечания
• 8 Ссылки

${\displaystyle \left\{{\begin{array}{lcr}a_{11}x_{1}&=&-a_{12}x_{2}-a_{13}x_{3}-\ldots -a_{1n}x_{n}+b_{1}\\a_{21}x_{1}+a_{22}x_{2}&=&-a_{23}x_{3}-\ldots -a_{2n}x_{n}+b_{2}\\\ldots &&\\a_{(n-1)1}x_{1}+a_{(n-1)2}x_{2}+\ldots +a_{(n-1)(n-1)}x_{n-1}&=&-a_{(n-1)n}x_{n}+b_{n-1}\\a_{n1}x_{1}+a_{n2}x_{2}+\ldots +a_{n(n-1)}x_{n-1}+a_{nn}x_{n}&=&b_{n}\end{array}}\right.}$

${\displaystyle (\mathrm {L} +\mathrm {D} ){\vec {x}}=-\mathrm {U} \,{\vec {x}}+{\vec {b}},}$

${\displaystyle (\mathrm {L} +\mathrm {D} ){\vec {x}}^{(k+1)}=-\mathrm {U} {\vec {x}}^{(k)}+{\vec {b}},\quad k=0,1,2,\ldots }$

${\displaystyle \left\{{\begin{array}{ccccccccccc}{x_{1}}^{(k+1)}&=&c_{12}{x_{2}^{(k)}}&+&c_{13}x_{3}^{(k)}&+&{\ldots }&+&c_{1n}{x_{n}}^{(k)}&+&d_{1}\\{x_{2}}^{(k+1)}&=&c_{21}{x_{1}^{(k+1)}}&+&c_{23}x_{3}^{(k)}&+&{\ldots }&+&c_{2n}{x_{n}}^{(k)}&+&d_{2}\\\ldots &&&&&&&&&&\\{x_{n}}^{(k+1)}&=&c_{n1}{x_{1}^{(k+1)}}&+&c_{n2}{x_{2}^{(k+1)}}&+&{\ldots }&+&c_{n(n-1)}{x_{n-1}}^{(k+1)}&+&d_{n}\end{array}}\right.,}$

${\displaystyle c_{ij}={\begin{cases}-{\frac {a_{ij}}{a_{ii}}},&j\neq i,\\0,&j=i.\end{cases}}\quad d_{i}={\frac {b_{i}}{a_{ii}}},\quad i=1,\ldots ,n.}$

${\displaystyle x_{i}^{(k+1)}=\sum _{j=1}^{i-1}c_{ij}x_{j}^{(k+1)}+\sum _{j=i}^{n}c_{ij}x_{j}^{(k)}+d_{i},\quad i=1,\ldots ,n.}$

${\displaystyle x_{1}^{(k+1)}=\sum _{j=1}^{1-1}c_{1j}x_{j}^{(k+1)}+\sum _{j=1}^{3}c_{1j}x_{j}^{(k)}+d_{1}}$, то есть ${\displaystyle x_{1}^{(k+1)}=c_{11}x_{1}^{(k)}+c_{12}x_{2}^{(k)}+c_{13}x_{3}^{(k)}+d_{1},}$

${\displaystyle x_{2}^{(k+1)}=\sum _{j=1}^{2-1}c_{2j}x_{j}^{(k+1)}+\sum _{j=2}^{3}c_{2j}x_{j}^{(k)}+d_{2}}$, то есть ${\displaystyle x_{2}^{(k+1)}=c_{21}x_{1}^{(k+1)}+c_{22}x_{2}^{(k)}+c_{23}x_{3}^{(k)}+d_{2},}$

${\displaystyle x_{3}^{(k+1)}=\sum _{j=1}^{3-1}c_{3j}x_{j}^{(k+1)}+\sum _{j=3}^{3}c_{3j}x_{j}^{(k)}+d_{3}}$, то есть ${\displaystyle x_{3}^{(k+1)}=c_{31}x_{1}^{(k+1)}+c_{32}x_{2}^{(k+1)}+c_{33}x_{3}^{(k)}+d_{3}.}$

1. метод Гаусса-Зейделя сходится;
2. скорость сходимости метода равна скорости сходимости геометрической прогрессии со знаменателем ${\displaystyle q=\|\mathrm {A} _{2}\|}$;
3. верна оценка погрешности: ${\displaystyle \|{\vec {x}}^{(k)}-{\vec {x}}\|=q^{k}\,\|{\vec {x}}^{(0)}-{\vec {x}}\|}$.

${\displaystyle \parallel x^{(k+1)}-x^{(k)}\parallel \leq \varepsilon }$

${\displaystyle \parallel Ax^{(k)}-b\parallel \leq \varepsilon }$

import numpy as np

def seidel(A, b, eps):
    n = len(A)
    x = np.zeros(n)  # zero vector

    converge = False
    while not converge:
        x_new = np.copy(x)
        for i in range(n):
            s1 = sum(A[i][j] * x_new[j] for j in range(i))
            s2 = sum(A[i][j] * x[j] for j in range(i + 1, n))
            x_new[i] = (b[i] - s1 - s2) / A[i][i]

        converge = np.linalg.norm(x_new - x) <= eps
        x = x_new

• Геометрическая прогрессия
• Метод простой итерации
• Метод Якоби
• Теорема Банаха
• Метод итерации

• https://www.maa.org/press/periodicals/loci/joma/iterative-methods-for-solving-iaxi-ibi-gauss-seidel-method
• https://www3.nd.edu/~zxu2/acms40390F12/Lec-7.3.pdf
• http://mathworld.wolfram.com/Gauss-SeidelMethod.html