Encontre as primeiras 5 iterações dos sistemas de equações lineares abaixo utilizando os métodos de GaussJacobi e Gauss-Seidel. Utilize como aproximação inicial o vetor nulo. Compute os erros absolutos e relativos em cada iteração para cada método aplicado. Para todos os métodos empregados utilize todas as casas decimais da calculadora. Finalmente, em cada sistema analisado pelos diferentes métodos, apresente uma conclusão baseando-se no que foi possível observar, e nos aspectos que foram abordados em sala de aula.
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Passo 1
Temos o sistema linear
10 x₁ – x₂ = 9,
– x₁ + 10 x₂ – 2 x₃ = 7,
– 2 x₂ + 10 x₃ = 6
O conteúdo aqui é métodos iterativos de Gauss–Jacobi e Gauss–Seidel e vamos precisar derivar as fórmulas de iteração, aplicar 5 passos a partir de x⁽⁰⁾ = (0,0,0) e calcular os erros absoluto e relativo em cada iteração.
10 x₁ – x₂ = 9,
– x₁ + 10 x₂ – 2 x₃ = 7,
– 2 x₂ + 10 x₃ = 6
O conteúdo aqui é métodos iterativos de Gauss–Jacobi e Gauss–Seidel e vamos precisar derivar as fórmulas de iteração, aplicar 5 passos a partir de x⁽⁰⁾ = (0,0,0) e calcular os erros absoluto e relativo em cada iteração.
Passo 2
As fórmulas do método de Gauss–Jacobi são obtidas isolando cada variável em cada equação. Temos
Passo 3
Aplicando Jacobi com x⁽⁰⁾=(0,0,0) obtemos até k=5
k=1: x⁽¹⁾=(0.9, 0.7, 0.6)
k=2: x⁽²⁾=(0.97, 0.91, 0.74)
k=3: x⁽³⁾=(0.991, 0.945, 0.782)
k=4: x⁽⁴⁾=(0.9945, 0.9555, 0.789)
k=5: x⁽⁵⁾=(0.99555, 0.95725, 0.7911)
k=1: x⁽¹⁾=(0.9, 0.7, 0.6)
k=2: x⁽²⁾=(0.97, 0.91, 0.74)
k=3: x⁽³⁾=(0.991, 0.945, 0.782)
k=4: x⁽⁴⁾=(0.9945, 0.9555, 0.789)
k=5: x⁽⁵⁾=(0.99555, 0.95725, 0.7911)
Passo 4
Erros do Jacobi (absoluto e relativo em cada componente):
k=1: |Δx|=(0.9,0.7,0.6), ε_rel=(1.000,1.000,1.000)
k=2: |Δx|=(0.07,0.21,0.14), ε_rel=(0.07216,0.23077,0.18919)
k=3: |Δx|=(0.021,0.035,0.042), ε_rel=(0.02118,0.03704,0.05371)
k=4: |Δx|=(0.0035,0.0105,0.007), ε_rel=(0.00352,0.01099,0.00887)
k=5: |Δx|=(0.00105,0.00175,0.0021), ε_rel=(0.001055,0.00183,0.00265)
k=1: |Δx|=(0.9,0.7,0.6), ε_rel=(1.000,1.000,1.000)
k=2: |Δx|=(0.07,0.21,0.14), ε_rel=(0.07216,0.23077,0.18919)
k=3: |Δx|=(0.021,0.035,0.042), ε_rel=(0.02118,0.03704,0.05371)
k=4: |Δx|=(0.0035,0.0105,0.007), ε_rel=(0.00352,0.01099,0.00887)
k=5: |Δx|=(0.00105,0.00175,0.0021), ε_rel=(0.001055,0.00183,0.00265)
Passo 5
As fórmulas do método de Gauss–Seidel são semelhantes, mas usamos valores mais recentes à medida que os calculamos:
Passo 6
Aplicando Gauss–Seidel com x⁽⁰⁾=(0,0,0) até k=5
k=1: x⁽¹⁾=(0.9, 0.79, 0.758)
k=2: x⁽²⁾=(0.979, 0.9495, 0.7899)
k=3: x⁽³⁾=(0.99495, 0.957475, 0.791495)
k=4: x⁽⁴⁾=(0.9957475, 0.95787375, 0.79157475)
k=5: x⁽⁵⁾=(0.995787375, 0.9578936875, 0.7915787375)
k=1: x⁽¹⁾=(0.9, 0.79, 0.758)
k=2: x⁽²⁾=(0.979, 0.9495, 0.7899)
k=3: x⁽³⁾=(0.99495, 0.957475, 0.791495)
k=4: x⁽⁴⁾=(0.9957475, 0.95787375, 0.79157475)
k=5: x⁽⁵⁾=(0.995787375, 0.9578936875, 0.7915787375)
Passo 7
Erros do Gauss–Seidel (absoluto e relativo em cada componente):
k=1: |Δx|=(0.9,0.79,0.758), ε_rel=(1.000,1.000,1.000)
k=2: |Δx|=(0.079,0.1595,0.0319), ε_rel=(0.08069,0.1680,0.0404)
k=3: |Δx|=(0.01595,0.007975,0.001595), ε_rel=(0.01603,0.00833,0.002015)
k=4: |Δx|=(0.0007975,0.00039875,0.00007975), ε_rel=(0.000800,0.000416,0.0001007)
k=5: |Δx|=(0.000039875,0.0000199375,0.0000039875), ε_rel=(0.0000400,0.0000208,0.00000503)
k=1: |Δx|=(0.9,0.79,0.758), ε_rel=(1.000,1.000,1.000)
k=2: |Δx|=(0.079,0.1595,0.0319), ε_rel=(0.08069,0.1680,0.0404)
k=3: |Δx|=(0.01595,0.007975,0.001595), ε_rel=(0.01603,0.00833,0.002015)
k=4: |Δx|=(0.0007975,0.00039875,0.00007975), ε_rel=(0.000800,0.000416,0.0001007)
k=5: |Δx|=(0.000039875,0.0000199375,0.0000039875), ε_rel=(0.0000400,0.0000208,0.00000503)
Passo 8
Concluindo: após 5 iterações Jacobi dá x⁽⁵⁾≈(0.99555,0.95725,0.7911) e Gauss–Seidel dá x⁽⁵⁾≈(0.9957874,0.9578937,0.7915787). O método de Gauss–Seidel apresenta erros absolutos e relativos menores a cada iteração e, portanto, converge mais rápido.
Resposta
Jacobi até k=5: x⁽⁵⁾=(0.99555, 0.95725, 0.7911);
Gauss–Seidel até k=5: x⁽⁵⁾=(0.995787375, 0.9578936875, 0.7915787375).
Conclusão: Gauss–Seidel converge mais rápido que Jacobi, apresentando menores erros em cada iteração.
Gauss–Seidel até k=5: x⁽⁵⁾=(0.995787375, 0.9578936875, 0.7915787375).
Conclusão: Gauss–Seidel converge mais rápido que Jacobi, apresentando menores erros em cada iteração.