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\( f(x)=x^{3}+x-1=0 \) Solve this by using bisection methes fixed poind and Mewton's Method. code this with Matlab

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Feb 02,2025

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Here's how to solve the equation \( f(x) = x^3 + x - 1 = 0 \) using three different methods in MATLAB: 1. **Bisection Method:** - Choose two points where the function changes sign. - MATLAB code provided to find the root. 2. **Fixed Point Iteration:** - Rearrange the equation to \( x = 1 - x^3 \). - MATLAB code provided to find the root. 3. **Newton's Method:** - Use the derivative \( f'(x) = 3x^2 + 1 \). - MATLAB code provided to find the root. Run each MATLAB function to find the root of the equation.

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To solve the equation \( f(x) = x^3 + x - 1 = 0 \) using the Bisection Method, Fixed Point Iteration, and Newton's Method, we can outline the steps and provide MATLAB code for each method. ### 1. Bisection Method The Bisection Method is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie. **Steps:** - Choose two initial points \( a \) and \( b \) such that \( f(a) \) and \( f(b) \) have opposite signs. - Compute the midpoint \( c = \frac{a + b}{2} \). - Check the sign of \( f(c) \): - If \( f(c) = 0 \), then \( c \) is a root. - If \( f(a) \) and \( f(c) \) have opposite signs, set \( b = c \). - Otherwise, set \( a = c \). - Repeat until the desired accuracy is achieved. ### MATLAB Code for Bisection Method ```matlab function root = bisection_method(f, a, b, tol) if f(a) * f(b) >= 0 error('f(a) and f(b) must have opposite signs'); end while (b - a) / 2 > tol c = (a + b) / 2; % Midpoint if f(c) == 0 break; % Found exact root elseif f(a) * f(c) < 0 b = c; % Root is in left half else a = c; % Root is in right half end end root = (a + b) / 2; % Approximate root end % Example usage f = @(x) x^3 + x - 1; root_bisection = bisection_method(f, 0, 1, 1e-6); disp(['Bisection Method Root: ', num2str(root_bisection)]); ``` ### 2. Fixed Point Iteration In Fixed Point Iteration, we rearrange the equation into the form \( x = g(x) \) and then iterate. **Rearrangement:** From \( f(x) = 0 \), we can rearrange to \( x = 1 - x^3 \). ### MATLAB Code for Fixed Point Iteration ```matlab function root = fixed_point_iteration(g, x0, tol, max_iter) for i = 1:max_iter x1 = g(x0); if abs(x1 - x0) < tol root = x1; return; end x0 = x1; end error('Max iterations reached without convergence'); end % Example usage g = @(x) 1 - x^3; root_fixed_point = fixed_point_iteration(g, 0.5, 1e-6, 100); disp(['Fixed Point Iteration Root: ', num2str(root_fixed_point)]); ``` ### 3. Newton's Method Newton's Method uses the derivative of the function to find successively better approximations to the roots. **Formula:** \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] ### MATLAB Code for Newton's Method ```matlab function root = newton_method(f, df, x0, tol, max_iter) for i = 1:max_iter x1 = x0 - f(x0) / df(x0); if abs(x1 - x0) < tol root = x1; return; end x0 = x1; end error('Max iterations reached without convergence'); end % Example usage f = @(x) x^3 + x - 1; df = @(x) 3*x^2 + 1; % Derivative of f root_newton = newton_method(f, df, 0.5, 1e-6, 100); disp(['Newton''s Method Root: ', num2str(root_newton)]); ``` ### Summary - The Bisection Method requires an interval where the function changes sign. - The Fixed Point Iteration requires a rearrangement of the function. - Newton's Method requires the derivative of the function. You can run each of these functions in MATLAB to find the root of the equation \( f(x) = x^3 + x - 1 = 0 \).

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To solve the equation \( f(x) = x^3 + x - 1 = 0 \) using the bisection method, fixed-point iteration, and Newton's method, we can outline the methods and provide a sample MATLAB code. ### Bisection Method The bisection method works by narrowing down the interval where the function changes sign. You first need to identify an interval \([a, b]\) such that \(f(a)\) and \(f(b)\) have opposite signs. Then, repeatedly bisect the interval until you find the root. ### Fixed-Point Iteration In fixed-point iteration, you rewrite the equation in the form \(x = g(x)\) and then iteratively apply \(g\) until convergence. ### Newton's Method Newton’s method requires an initial guess and uses the derivative of the function to find the roots. You update your guess using the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). ### Sample MATLAB Code ```matlab function root_finding_methods() % Define the function and its derivative f = @(x) x^3 + x - 1; % Function df = @(x) 3*x^2 + 1; % Derivative % Bisection method a = 0; b = 1; % Interval [0, 1] tol = 1e-6; % Tolerance while (b - a) / 2 > tol c = (a + b) / 2; % Midpoint if f(c) == 0 % Found exact solution break; elseif f(a) * f(c) < 0 b = c; % Root in left half else a = c; % Root in right half end end root_bisection = (a + b) / 2; % Fixed-point iteration g = @(x) (1 - x^3)^(1/3); % g(x) = (1 - x^3)^(1/3) x0 = 0.5; % Initial guess tol_fixed = 1e-6; % Tolerance x_fixed = x0; for i = 1:1000 % Max iterations x_next = g(x_fixed); if abs(x_next - x_fixed) < tol_fixed break; end x_fixed = x_next; end root_fixed_point = x_fixed; % Newton's method x_newton = 0.5; % Initial guess tol_newton = 1e-6; % Tolerance for i = 1:1000 % Max iterations x_newton = x_newton - f(x_newton)/df(x_newton); if abs(f(x_newton)) < tol_newton break; end end root_newton = x_newton; % Display results fprintf('Bisection root: %f\n', root_bisection); fprintf('Fixed-point root: %f\n', root_fixed_point); fprintf('Newton root: %f\n', root_newton); end ``` Run this function in MATLAB to find the roots using the three methods. Each method allows you to see a different way of tackling the problem, illustrating the strengths and specific use cases of each approach. Have fun experimenting with different initial guesses and intervals!

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