Perform three iterations of the Newton Raphson method to solve the system of equations \{\begin{matrix}x^{2}=3xy-1\\ y=2(x+1)\end{matrix}\} by taking X^{(0)}=[0.5~0.5]^{T}
To solve the system of equations using the Newton Raphson method, we need to compute the Jacobian matrix, the function values, and their derivatives at each iteration.
Given the system of equations:
x^2 = 3xy - 1 ----(1)
y = 2(x + 1) ----(2)
Let X^(0) = [0.5, 0.5]^T
Step 1:
Compute the Jacobian matrix:
Jacobian matrix = [∂F₁/∂x ∂F₁/∂y]
[∂F₂/∂x ∂F₂/∂y]
where F₁ = x^2 - 3xy + 1 and F₂ = y - 2(x + 1)
∂F₁/∂x = 2x - 3y
∂F₁/∂y = -3x
∂F₂/∂x = -2
∂F₂/∂y = 1
Jacobian matrix at X^(0) = [2(0.5) - 3(0.5), -3(0.5)]
[-2, 1]
Jacobian matrix at X^(0) = [0, -1]
[-2, 1]
Step 2:
Compute the function values at X^(0):
F₁(X^(0)) = (0.5)^2 - 3(0.5)(0.5) + 1
= 0.25 - 0.75 + 1
= 0.5
F₂(X^(0)) = 0.5 - 2(0.5 + 1)
= 0.5 - 2(1.5)
= 0.5 - 3
= -2.5
Step 3:
Compute the derivatives of the functions at X^(0):
∂F₁/∂x (X^(0)) = 2(0.5) - 3(0.5)
= 1 - 1.5
= -0.5
∂F₁/∂y (X^(0)) = -3(0.5)
= -1.5
∂F₂/∂x (X^(0)) = -2
∂F₂/∂y (X^(0)) = 1
Step 4:
Update the X^(0) using the formula:
X^(k+1) = X^(k) - [Jacobian(matrix)]^(-1) * [Function values]
where [Jacobian(matrix)]^(-1) is the inverse of the Jacobian matrix.
Jacobian(matrix) = [0, -1]
[-2, 1]
Inverse of Jacobian(matrix) = [(1/(-2)), (-1/(-2))]
[(2/(-2)), (0/(-2))]
= [-0.5, 0.5]
[-1, 0]
[X^(1)] = [0.5, 0.5]^T - [-0.5, 0.5] * [0.5, -2.5]
= [0.5, 0.5]^T - [0.25 + 1.25]
= [0.5, 0.5]^T - [1.5]
= [-1, -1]
Step 5:
Compute the function values at X^(1):
F₁(X^(1)) = (-1)^2 - 3(-1)(-1) + 1
= 1 - 3 + 1
= -1
F₂(X^(1)) = -1 - 2((-1) + 1)
= -1 - 2(0)
= -1
Step 6:
Repeat steps 2-5 until convergence or for the desired number of iterations.
To perform three iterations, we repeat steps 2-5 using X^(1) as the new X^(0):
Step 2:
Compute the function values at X^(1):
F₁(X^(1)) = (-1)^2 - 3(-1)(-1) + 1
= 1 - 3 + 1
= -1
F₂(X^(1)) = -1 - 2((-1) + 1)
= -1 - 2(0)
= -1
Step 3:
Compute the derivatives of the functions at X^(1):
∂F₁/∂x (X^(1)) = 2(-1) - 3(-1)
= -2 + 3
= 1
∂F₁/∂y (X^(1)) = -3(-1)
= 3
∂F₂/∂x (X^(1)) = -2
∂F₂/∂y (X^(1)) = 1
Step 4:
Update X^(1) using the formula:
X^(2) = X^(1) - [Jacobian(matrix)]^(-1) * [Function values]
where [Jacobian(matrix)]^(-1) is the inverse of the Jacobian matrix.
= [-1, -1]^T - [-0.5, 0.5] * [-1, -1]
= [-1, -1]^T - [-0.5 + 0.5, 0.5 - 0.5]
= [-1, -1]^T - [0, 0]
= [-1, -1]^T
Step 5:
Compute the function values at X^(2):
F₁(X^(2)) = (-1)^2 - 3(-1)(-1) + 1
= 1 - 3 + 1
= -1
F₂(X^(2)) = -1 - 2((-1) + 1)
= -1 - 2(0)
= -1
Step 6:
Repeat steps 2-5 using X^(2) as the new X^(0):
Step 2:
Compute the function values at X^(2):
F₁(X^(2)) = (-1)^2 - 3(-1)(-1) + 1
= 1 - 3 + 1
= -1
F₂(X^(2)) = -1 - 2((-1) + 1)
= -1 - 2(0)
= -1
Step 3:
Compute the derivatives of the functions at X^(2):
∂F₁/∂x (X^(2)) = 2(-1) - 3(-1)
= -2 + 3
= 1
∂F₁/∂y (X^(2)) = -3(-1)
= 3
∂F₂/∂x (X^(2)) = -2
∂F₂/∂y (X^(2)) = 1
Step 4:
Update X^(2) using the formula:
X^(3) = X^(2) - [Jacobian(matrix)]^(-1) * [Function values]
where [Jacobian(matrix)]^(-1) is the inverse of the Jacobian matrix.
= [-1, -1]^T - [-0.5, 0.5] * [-1, -1]
= [-1, -1]^T - [-0.5 + 0.5, 0.5 - 0.5]
= [-1, -1]^T - [0, 0]
= [-1, -1]^T
Step 5:
Compute the function values at X^(3):
F₁(X^(3)) = (-1)^2 - 3(-1)(-1) + 1
= 1 - 3 + 1
= -1
F₂(X^(3)) = -1 - 2((-1) + 1)
= -1 - 2(0)
= -1
After three iterations using the Newton Raphson method, X^(3) = [-1, -1]^T.