Math Problem Statement

Solve system of equations using Newton's method

Solve system of equations using Newton's method

Solution

To solve this system of equations using Newton's method, follow these steps:

Step 1: Define the system of equations

The given equations are:

  1. f1(x1,x2)=6−4x1x2f_1(x_1, x_2) = 6 - 4x_1x_2
  2. f2(x1,x2)=3−x13−x22f_2(x_1, x_2) = 3 - x_1^3 - x_2^2

We aim to solve F(x)=0F(\mathbf{x}) = \mathbf{0}, where: F(x)=[f1(x1,x2)f2(x1,x2)].F(\mathbf{x}) = \begin{bmatrix} f_1(x_1, x_2) \\ f_2(x_1, x_2) \end{bmatrix}.

Step 2: Compute the Jacobian matrix

The Jacobian matrix JJ is:

\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix}.$$ Compute the partial derivatives: - $$\frac{\partial f_1}{\partial x_1} = -4x_2$$ - $$\frac{\partial f_1}{\partial x_2} = -4x_1$$ - $$\frac{\partial f_2}{\partial x_1} = -3x_1^2$$ - $$\frac{\partial f_2}{\partial x_2} = -2x_2$$ Thus: $$J(\mathbf{x}) = \begin{bmatrix} -4x_2 & -4x_1 \\ -3x_1^2 & -2x_2 \end{bmatrix}.$$ ### Step 3: Apply Newton's iteration formula The Newton's method update is: $$\mathbf{x}^{(k+1)} = \mathbf{x}^{(k)} - J(\mathbf{x}^{(k)})^{-1} F(\mathbf{x}^{(k)}).$$ Given the starting point $$\mathbf{x}^{(0)} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$$, compute one iteration. ### Step 4: Evaluate $$F(\mathbf{x}^{(0)})$$ $$F(\mathbf{x}^{(0)}) = \begin{bmatrix} 6 - 4(2)(-2) \\ 3 - 2^3 - (-2)^2 \end{bmatrix} = \begin{bmatrix} 6 - (-16) \\ 3 - 8 - 4 \end{bmatrix} = \begin{bmatrix} 22 \\ -9 \end{bmatrix}.$$ ### Step 5: Evaluate $$J(\mathbf{x}^{(0)})$$ $$J(\mathbf{x}^{(0)}) = \begin{bmatrix} -4(-2) & -4(2) \\ -3(2)^2 & -2(-2) \end{bmatrix} = \begin{bmatrix} 8 & -8 \\ -12 & 4 \end{bmatrix}.$$ ### Step 6: Solve for $$\Delta \mathbf{x} = J^{-1} F(\mathbf{x}^{(0)})$$ The inverse of $$J(\mathbf{x}^{(0)})$$ is computed as: $$J^{-1} = \frac{1}{\text{det}(J)} \begin{bmatrix} 4 & 8 \\ 12 & 8 \end{bmatrix},$$ where \( \text{det}(J) = (8)(8)-(3*

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Math Problem Analysis

Mathematical Concepts

Newton's method
Nonlinear equations
Linear algebra
Jacobian matrix

Formulas

Newton's iteration formula: x^(k+1) = x^(k) - J(x^(k))^-1 * F(x^(k))
Jacobian matrix: J(x) = [[∂f1/∂x1, ∂f1/∂x2], [∂f2/∂x1, ∂f2/∂x2]]
Determinant and inverse of 2x2 matrix

Theorems

Newton's method for solving nonlinear systems

Suitable Grade Level

Undergraduate - Junior/Senior (CS/Mathematics/Engineering students)

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