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Implement a Matlab function called newtons Method that uses Newton's method to solve f(x) = 0 given an initial guess o. The inputs to your function should include the initial guess To, a convergence tolerence €, and a maximum number of steps (since Newton's Method does not always converge), and can use the function f(x) and the Jacobian J(r). The output should include a flag indicating success or failure, the solution æ* that satisfies ||f(x*)|| ≤ e in the event of success, and the number of steps required for convergence. Test your function with € = 10-10 by finding the four solutions to the example from lecture 15, that is, to f(x) = 0 where f(x) [f₁(x) [f2(x)] = [x²+x₁x-9 3x1x₂-x-4 in the set S = {(₁, x₂) € R² : -5 ≤ x₁ ≤ 5, −5 ≤ x₂ ≤ 5}. Use the following plot from Module 8 to obtain your initial guesses.

Implement a Matlab function called newtons Method that uses Newton's method to solve f(x) = 0 given an initial guess o. The inputs to your function should include the initial guess To, a convergence tolerence €, and a maximum number of steps (since Newton's Method does not always converge), and can use the function f(x) and the Jacobian J(r). The output should include a flag indicating success or failure, the solution æ* that satisfies ||f(x*)|| ≤ e in the event of success, and the number of steps required for convergence. Test your function with € = 10-10 by finding the four solutions to the example from lecture 15, that is, to f(x) = 0 where f(x) [f₁(x) [f2(x)] = [x²+x₁x-9 3x1x₂-x-4 in the set S = {(₁, x₂) € R² : -5 ≤ x₁ ≤ 5, −5 ≤ x₂ ≤ 5}. Use the following plot from Module 8 to obtain your initial guesses.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Implement a Matlab function called newtons Method that uses Newton's method to solve f(x) = 0 given an initial guess To. The inputs to your function should include the initial guess To, a convergence tolerence €, and a maximum number of steps (since Newton's Method does not always converge), and can use the function f(x) and the Jacobian J(x). The output should include a flag indicating success or failure, the solution æ* that satisfies ||f(x*)|| ≤ e in the event of success, and the number of steps required for convergence. Test your function with € = 10-10 by finding the four solutions to the example from lecture 15, that is, to f(x) = 0 where € 50 4 3 in the set S = {(₁, ₂) € R² : -5 ≤ x ≤ 5, -5 ≤ x2 ≤ 5}. Use the following plot from Module 8 to obtain your initial guesses. 2 1 0 -1 -2 -3 -4 5 f(x) = -5 4 [f₁(x)] x² + x₁x²-9 [3x²x₂x² - 4 [f₂(x)] -3 • z}+z₂+x² -2 = -1 0 1 N +3₂ 3 4 5
Transcribed Image Text:Implement a Matlab function called newtons Method that uses Newton's method to solve f(x) = 0 given an initial guess To. The inputs to your function should include the initial guess To, a convergence tolerence €, and a maximum number of steps (since Newton's Method does not always converge), and can use the function f(x) and the Jacobian J(x). The output should include a flag indicating success or failure, the solution æ* that satisfies ||f(x*)|| ≤ e in the event of success, and the number of steps required for convergence. Test your function with € = 10-10 by finding the four solutions to the example from lecture 15, that is, to f(x) = 0 where € 50 4 3 in the set S = {(₁, ₂) € R² : -5 ≤ x ≤ 5, -5 ≤ x2 ≤ 5}. Use the following plot from Module 8 to obtain your initial guesses. 2 1 0 -1 -2 -3 -4 5 f(x) = -5 4 [f₁(x)] x² + x₁x²-9 [3x²x₂x² - 4 [f₂(x)] -3 • z}+z₂+x² -2 = -1 0 1 N +3₂ 3 4 5
Expert Solution
Step 1

 

 

clc%clears screen

clear all%clears history

close all%closes all files

format long

p=@(x1,x2) [x1^2+x1*x2^3-9;3*x1^2*x2-x2^3-4];

f=@(x) p(x(1),x(2));

J=@(x) [x(2)^3+2*x(1),3*x(1)*x(2)^2;6*x(1)*x(2),3*x(1)^2-3*x(2)^2];

x0=[-1;1];

l=[-1,1,-3,3];

u=[1,1,0.5,0];

disp('Solutions are');

for j=1:4

 fprintf('Solution %d is\n\n',j);

x0=[l(j);u(j)];

for i=1:100

x1=x0-inv(J(x0))*f(x0);

 if(norm(f(x1))<1e-10)

break;

 end

0=x1;

end

disp(x1)

end

steps

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