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Use Newton's method to find a solution to the following nonlinear systems in the given domain. Iterate until \(\left\|\mathbf{x}^{(k)}-\mathbf{x}^{(k-1)}\right\|_{\infty}<10^{-6}\). a. \(\quad 3 x_{1}^{2}-x_{2}^{2}=0\), b. \(\quad \ln \left(x_{1}^{2}+x_{2}^{2}\right)-\sin \left(x_{1} x_{2}\right)=\ln 2+\ln \pi\), \(3 x_{1} x_{2}^{2}-x_{1}^{3}-1=0\) \(e^{x_{1}-x_{2}}+\cos \left(x_{1} x_{2}\right)=0\) Use \(\mathbf{x}^{(0)}=(1,1)^{t}\) Use \(\mathbf{x}^{(0)}=(2,2)^{t}\) c. \(\quad x_{1}^{3}+x_{1}^{2} x_{2}-x_{1} x_{3}+6=0\), d. $$ \begin{aligned} e^{x_{1}}+e^{x_{2}}-x_{3} &=0 \\ x_{2}^{2}-2 x_{1} x_{3} &=4 \end{aligned} $$ $$ \begin{aligned} 9 x_{2}+\sqrt{x_{1}^{2}+\sin x_{3}+1.06}+0.9 &=0 \\ 60 x_{3}+3 e^{-x_{1} x_{2}}+10 \pi-3 &=0 \end{aligned} $$ Use \(\mathbf{x}^{(0)}=(-1,-2,1)^{t}\) Use \(\mathbf{x}^{(0)}=(0,0,0)^{t}\)

Short Answer

Expert verified
The solution set of the equations consist of the final iteration points derived from Newton's method for each given initial point and non-linear system.

Step by step solution

01

Formulate the Function and Jacobian matrix

The first step is to define our non-linear function(s) and calculate the Jacobian matrix of the system. The Jacobian matrix is the matrix composed of all first-order partial derivatives of a vector-valued function. It's often used in the Newton's method.
02

Initialize the Solution

Use the given initial guess \(\mathbf{x}^{(0)}\) as the starting point to our iterative method.
03

Calculate Next Iteration Point

Use the formulas for Newton's method to find the next point. For a system of equations, Newton's method is defined as: \(\mathbf{x}^{(k+1)} = \mathbf{x}^{(k)} - [J_f(\mathbf{x}^{(k)})]^{-1} f(\mathbf{x}^{(k)})\), where \(f\) is our function, \(J_f\) is the Jacobian matrix of our function and \(\mathbf{x}^{(k)}\) is the current point (vector). We calculate the next point by evaluating the function and its Jacobian at the current point, inverting the Jacobian and subtracting the result from the current point.
04

Convergence Test

Calculate the differences between two successive points \(\left\|\mathbf{x}^{(k)}-\mathbf{x}^{(k-1)}\right\|_{\infty}\) and test if it's less than our tolerance level \(10^{-6}\). If not, repeat step 3.
05

Repeat the Process

The steps 3 and 4 are to be repeated until the changes become smaller than a certain tolerance (here \(10^{-6}\)).

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Key Concepts

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

Nonlinear Systems
In mathematics, a nonlinear system is a collection of equations involving multiple variables where at least one equation contains a nonlinear component, such as a variable being raised to a power or a variable involved in a logarithmic or trigonometric function.
These systems can be quite complex to solve, as they do not follow a linear path and can have multiple solutions or none at all.
Newton's method is often used for solving nonlinear systems by finding successively better approximations to the roots (solutions) of a real-valued function. To apply Newton's method to nonlinear systems, we express the problem in terms of functions equating to zero. For example, each of the functions in the problem acts as a constraint that our solutions must satisfy.
The key challenge here is to find a common solution that satisfies all given equations simultaneously. Thus, understanding the nature of the nonlinear equations helps in setting up the system correctly.
Jacobian Matrix
The Jacobian matrix is a crucial tool for solving systems of nonlinear equations using Newton’s Method. It is constructed by taking the first partial derivatives of the functions in the system with respect to each variable.
For a system of functions \( f_1(x_1, x_2, ..., x_n), f_2(x_1, x_2, ..., x_n), ..., f_m(x_1, x_2, ..., x_n) \), the Jacobian matrix \( J \) would be: \[ J(x) = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \ \vdots & \ddots & \vdots \ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n} \end{bmatrix} \]This matrix provides a linear approximation of the system of equations in the neighborhood of a given point.
Its role in Newton's method is to adjust the current guess by determining how changes in the inputs should result in changes in the outputs.
To advance to the next iteration, we must solve the linear system resulting from the matrix equation: \( J(x^{(k)}) \Delta x^{(k)} = -f(x^{(k)}) \) where \( \Delta x^{(k)} \) represents the vector of changes that needs to be applied to improve the solution. The next point \( x^{(k+1)} \) is updated by adding \( \Delta x^{(k)} \) to the current guess \( x^{(k)} \).
Convergence Criteria
Convergence criteria are essential for determining when the iterative process of Newton's method should stop.
In contexts involving nonlinear systems, convergence is generally achieved when the successive approximations change by an insignificantly small amount.
This ensures that we are extremely close to a solution of the equation.For this exercise, the criterion specified is that the difference between the current and previous solution in terms of the infinity norm \( \| x^{(k)} - x^{(k-1)} \|_\infty \) should be less than \( 10^{-6} \).
The infinity norm measures the largest component-wise difference between the current and former solution vectors.
This strict threshold ensures high precision in the solution, which is critical for many applications.When setting up these criteria:
  • Ensure the tolerance level is appropriate for the problem domain.
  • Confirm the system behavior near the solution is smooth since Newton's method can converge slowly or diverge with poor initial guesses or solutions involving discontinuities.
Using these criteria helps ensure the reliability and accuracy of Newton's method for solving nonlinear systems.

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Most popular questions from this chapter

The nonlinear system $$ \begin{array}{ll} E_{1}: 4 x_{1}-x_{2}+x_{3}=x_{1} x_{4}, & E_{2}:-x_{1}+3 x_{2}-2 x_{3}=x_{2} x_{4} \\ E_{3}: x_{1}-2 x_{2}+3 x_{3}=x_{3} x_{4}, & E_{4}: x_{1}^{2}+x_{2}^{2}+x_{3}^{2} & =1 \end{array} $$ has six solutions. a. Show that if \(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)^{t}\) is a solution then \(\left(-x_{1},-x_{2},-x_{3}, x_{4}\right)^{t}\) is a solution. b. Use Newton's method three times to approximate all solutions. Iterate until \(\left\|\mathbf{x}^{(k)}-\mathbf{x}^{(k-1)}\right\|_{\infty}<10^{-5}\).

Use Broyden's method to approximate solutions to the following nonlinear systems. Iterate until \(\left\|\mathbf{x}^{(k)}-\mathbf{x}^{(k-1)}\right\|_{\infty}<10^{-6}\) a. \(\begin{aligned} x_{1}\left(1-x_{1}\right)+4 x_{2} &=12 \\\\\left(x_{1}-2\right)^{2}+\left(2 x_{7}-3\right)^{2} &=25 \end{aligned}\) b. \(\begin{aligned} 5 x_{1}^{2}-x_{2}^{2} &=0, \\ x_{7}-0.25\left(\sin x_{1}+\cos x_{2}\right) &=0 . \end{aligned}\) c. \(\quad 15 x_{1}+x_{2}^{2}-4 x_{3}=13\), d. \(10 x_{1}-2 x_{2}^{2}+x_{2}-2 x_{3}-5=0\), $$ \begin{aligned} x_{1}^{2}+10 x_{2}-x_{3} &=11 \\ x_{2}^{3}-25 x_{3} &=-22 \end{aligned} $$ $$ \begin{aligned} 8 x_{2}^{2}+4 x_{3}^{2}-9 &=0 \\ 8 x_{2} x_{3}+4 &=0 \end{aligned} $$.

The nonlinear system $$ 3 x_{1}-\cos \left(x_{2} x_{3}\right)-\frac{1}{2}=0, \quad x_{1}^{2}-625 x_{2}^{2}-\frac{1}{4}=0, \quad e^{-x_{1} x_{2}}+20 x_{3}+\frac{10 \pi-3}{3}=0 $$ has a singular Jacobian matrix at the solution. Apply Broyden's method with \(\mathbf{x}^{(0)}=(1,1-1)^{t}\). Note that convergence may be slow or may not occur within a reasonable number of iterations.

Use Newton's method with \(\mathbf{x}^{(0)}=\mathbf{0}\) to compute \(\mathbf{x}^{(2)}\) for each of the following nonlinear systems. a. \(\quad 3 x_{1}-\cos \left(x_{2} x_{3}\right)-\frac{1}{2}=0\), b. \(\quad x_{1}^{2}+x_{2}-37=0\) $$ \begin{array}{r} 4 x_{1}^{2}-625 x_{2}^{2}+2 x_{2}-1=0 \\ e^{-x_{1} x_{2}}+20 x_{3}+\frac{10 \pi-3}{3}=0 \end{array} $$ \(x_{1}+x_{2}+x_{3}-3=0\) c. \(\quad 15 x_{1}+x_{2}^{2}-4 x_{3}=13\), d. \(\quad 10 x_{1}-2 x_{2}^{2}+x_{2}-2 x_{3}-5=0\) $$ \begin{aligned} x_{1}^{2}+10 x_{2}-x_{3} &=11 \\ x_{2}^{3}-25 x_{3} &=-22 \end{aligned} $$ $$ \begin{aligned} 8 x_{2}^{2}+4 x_{3}^{2}-9 &=0 \\ 8 x_{2} x_{3}+4 &=0 \end{aligned} $$

The nonlinear system $$ f_{1}\left(x_{1}, x_{2}\right)=x_{1}^{2}-x_{2}^{2}+2 x_{2}=0, \quad f_{2}\left(x_{1}, x_{2}\right)=2 x_{1}+x_{2}^{2}-6=0 $$ has two solutions, \((0.625204094,2.179355825)^{t}\) and \((2.109511920,-1.334532188)^{2}\). Use the continuation method and Euler's method with \(N=2\) to approximate the solutions where a. \(\quad \mathbf{x}(0)=(0,0)^{r}\) b. \(\quad \mathbf{x}(0)=(1,1)^{\mathrm{r}}\) c. \(\quad \mathbf{x}(0)=(3,-2)^{t}\)

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