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Consider the system $$ \begin{gathered} x_{1}-1=0 \\ x_{1} x_{2}-1=0 \end{gathered} $$ It is trivial to solve this system immediately, you will surely agree, but suppose we apply Newton's method anyway. For what initial guesses will the method fail? Explain.

Short Answer

Expert verified
Answer: Newton's method will fail for initial guesses (0, x₂) where x₂ can be any real number.

Step by step solution

01

Identify the system of equations

The given system of equations is: $$ \begin{gathered} x_{1}-1=0 \\ x_{1}x_{2}-1=0 \end{gathered} $$
02

Define the functions for the Newton's method

In order to apply Newton's method, we need to represent the system of equations as functions: $$ \begin{gathered} f_{1}(x_{1},x_{2}) = x_{1} - 1 \\ f_{2}(x_{1},x_{2}) = x_{1}x_{2} - 1 \end{gathered} $$
03

Compute the Jacobian matrix

For the Newton's method, we need to compute the Jacobian matrix of the given functions: $$ \mathbf{J}(x_{1},x_{2}) = \begin{bmatrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} \\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ x_{2} & x_{1} \end{bmatrix} $$
04

Apply the Newton's method iteration formula

The iteration formula for Newton's method for a system of equations is given by: $$ \mathbf{x}_{n+1} = \mathbf{x}_{n} - \mathbf{J}^{-1}(\mathbf{x}_{n})\mathbf{F}(\mathbf{x}_{n}) $$ To find the inverse of the Jacobian matrix, we have: $$ \mathbf{J}^{-1}(x_{1},x_{2}) = \frac{1}{\text{det}(\mathbf{J})} \begin{bmatrix} x_{1} & 0 \\ -x_{2} & 1 \end{bmatrix} $$ where \(\text{det}(\mathbf{J}) = x_{1}\).
05

Analyze the inverse of the Jacobian matrix

The inverse of the Jacobian matrix exists when \(x_1 \neq 0\), which implies that for initial guesses where \(x_1 = 0\), the Newton's method will fail. Apart from this case, the inverse exists for all other initial guesses, and Newton's method will work. In conclusion, Newton's method will fail for initial guesses \((0, x_{2})\) where \(x_2\) can be any real number.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Jacobian Matrix
The Jacobian matrix is a crucial component in employing Newton's Method for solving systems of equations. It contains the partial derivatives of each function in a system with respect to each variable. This matrix provides insight into how small changes in the input variables will affect the output of the functions. For a system of equations like the one given, the Jacobian matrix is defined as:\[\mathbf{J}(x_{1},x_{2}) =\begin{bmatrix}\frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} \\frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}}\end{bmatrix} =\begin{bmatrix}1 & 0 \x_{2} & x_{1}\end{bmatrix}\]This matrix captures the sensitivity of the functions to changes in their input variables. The determinant of the Jacobian, \( \text{det}(\mathbf{J}) \), is vital because it influences the feasibility of computing an inverse. Having a non-zero determinant ensures that the Jacobian matrix is invertible, which is essential for progressing in Newton's Method steps. If the determinant is zero, the method's formula cannot proceed, causing the process to potentially fail.
System of Equations
A system of equations consists of multiple equations that share common variables. Solving such a system involves finding values for these variables that satisfy all the different equations simultaneously. In our exercise, the system \[\begin{gathered}x_{1} - 1 = 0 \x_{1} x_{2} - 1 = 0\end{gathered}\] aims to determine simultaneous solutions for \( x_1 \) and \( x_2 \).

To systematically solve systems of equations, especially when they are non-linear, iterative methods like Newton's Method are often used. Converting these equations into functions that can be manipulated by methods like Newton's Method allows for systematic approximations towards the solutions. Furthermore, whether a system's equations are dependent or independent can significantly affect solvability. In our case, if \( x_1 = 0 \), the two equations become inconsistent with each other, failing to provide a solution in practical terms.
Iterative Method
Iterative methods are techniques commonly used to find approximate solutions to complex problems, often non-linear systems, where direct analytical solutions are difficult or impossible. Newton's Method is a popular iterative method that uses the idea of approaching a solution through repeated approximations.
  • It starts with an initial guess, which is iteratively refined through successive calculations.
  • The process continues until the approximations yield a satisfactory level of accuracy for a solution.
  • Each iteration involves using a formula to improve the guess. For Newton's Method in systems of equations, this involves the Jacobian matrix and function values at the given guess.
However, Newton's Method is not fail-proof. For the method to work, some conditions must be met, such as the existence of an inverse of the Jacobian matrix. In our exercise, if the initial guess sets \( x_1 = 0 \), the determinant of the Jacobian matrix becomes zero, making it impossible to compute the inverse. As a result, the iterative method would fail to converge towards the solution, highlighting the importance of choosing suitable initial guesses.

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Most popular questions from this chapter

Consider minimizing the function \(\phi(\mathbf{x})=\mathbf{c}^{T} \mathbf{x}+\frac{1}{2} \mathbf{x}^{T} H \mathbf{x}\), where \(\mathbf{c}=(5.04,-59.4,146.4,-96.6)^{T}\) and $$ H=\left(\begin{array}{cccc} .16 & -1.2 & 2.4 & -1.4 \\ -1.2 & 12.0 & -27.0 & 16.8 \\ 2.4 & -27.0 & 64.8 & -42.0 \\ -1.4 & 16.8 & -42.0 & 28.0 \end{array}\right) $$ Try both Newton and BFGS methods, starting from \(\mathbf{x}_{0}=(-1,3,3,0)^{T}\). Explain why the BFGS method requires significantly more iterations than Newton's.

Use Newton's method to solve a discretized version of the differential equation $$ y^{\prime \prime}=-\left(y^{\prime}\right)^{2}-y+\ln x, \quad 1 \leq x \leq 2, y(1)=0, y(2)=\ln 2 $$ The discretization on a uniform mesh, with the notation of Example \(9.3\), can be $$ \frac{y_{i+1}-2 y_{i}+y_{i-1}}{h^{2}}+\left(\frac{y_{i+1}-y_{i-1}}{2 h}\right)^{2}+y_{i}=\ln (i h), \quad i=1,2, \ldots, n $$ The actual solution of this problem is \(y(x)=\ln x .\) Compare your numerical results to the solution \(y(x)\) for \(n=8,16,32\), and 64 . Make observations regarding the convergence behavior of Newton's method in terms of the iterations and the mesh size, as well as the solution error.

(a) Suppose Newton's method is applied to a linear system \(A \mathbf{x}=\mathbf{b} .\) How does the iterative formula look and how many iterations does it take to converge? (b) Suppose the Jacobian matrix is singular at the solution of a nonlinear system of equations. Speculate what can occur in terms of convergence and rate of convergence. Specifically, is it possible to have a situation where the Newton iteration converges but convergence is not quadratic?

Consider the saddle point matrix $$ K=\left(\begin{array}{cc} H & A^{T} \\ A & 0 \end{array}\right) $$ where the matrix \(H\) is symmetric positive semidefinite and the matrix \(A\) has full row rank. (a) Show that \(K\) is nonsingular if \(\mathbf{y}^{T} H \mathbf{y} \neq 0\) for all \(\mathbf{y} \in \operatorname{null}(A), \mathbf{y} \neq \mathbf{0}\). (b) Show by example that \(K\) is symmetric but indefinite, i.e., it has both positive and negative eigenvalues.

Consider the nonlinear least square problem of minimizing $$ \phi(\mathbf{x})=\frac{1}{2}\|\mathbf{g}(\mathbf{x})-\mathbf{b}\|^{2} $$ (a) Show that $$ \nabla \phi(\mathbf{x})=A(\mathbf{x})^{T}(\mathbf{g}(\mathbf{x})-\mathbf{b}) $$ where \(A\) is the \(m \times n\) Jacobian matrix of \(\mathbf{g}\). (b) Show that $$ \nabla^{2} \phi(\mathbf{x})=A(\mathbf{x})^{T} A(\mathbf{x})+L(\mathbf{x}) $$ where \(L\) is an \(n \times n\) matrix with elements $$ L_{i, j}=\sum_{k=1}^{m} \frac{\partial^{2} g_{k}}{\partial x_{i} \partial x_{j}}\left(g_{k}-b_{k}\right) $$ You may want to check first what \(\frac{\partial \phi}{\partial x_{j}}\) looks like for a fixed \(i\); later on, look at \(\frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}}\) for a fixed \(j .]\)

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