Problem #2 A numerical algorithm is used to solve the following equation The alg

Question

Problem #2 A numerical algorithm is used to solve the following equation The algorithm starts with an initial guess x(1)-1 It can be seen that this is not the solution (f(1)--I # 0). An update term is then used to improve the solution. x(2) = x(1) + ??(1) v here ??(1) = and f'(x) = 3x2-2x This process will be repeated until If (x(N)ls 0.0001 (the solution has been found) or 1) Ax(N)I s 0.0001 (the solution cannot be improved). f (x(k)) x(k + 1) x(k) + ??(k) Write a Matlab script file for the following tasks (I) Use "while-end" loop to find the solution of the equation and the number of iteration required for finding the solution. (To make sure the program can stop even a s olution cannot be found please ensure the number of iteration is restricted to 50). (2) Find the solution using the command "izero" and compare the result with the result obtained

Explanation / Answer

MATLAB code

close all
clear
clc

f = @(x) (x^3 - x^2 - 1);
fd = @(x) (3*x^2 - 2*x);

max_iter = 50;
x = 1; % initial guess
iter = 0;
while 1
    x_prev = x;
    deltax = -double(f(x))/double(fd(x));
    x = x + deltax;
    iter = iter + 1;
    if abs(double(f(x))) <= 0.0001 || abs(deltax) <= 0.0001 || iter == max_iter
        break
    end
end
fprintf('Number of iterations: %d ', iter);
fprintf('Solution: %.8f ', x);
fprintf('f(Solution): %.8f ', double(f(x)));
x = fzero(f,1);
fprintf(' Using ''fzero'': ');
fprintf('Solution: %.8f ', x);
fprintf('f(Solution): %.8f ', double(f(x)));

Output

Number of iterations: 5
Solution: 1.46557137
f(Solution): 0.00000050

Using 'fzero':
Solution: 1.46557123
f(Solution): 0.00000000

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