Question

Please use MATLAB to solve the question

Computer Problem. Solve fi(x) = x – cos(x) = 0 and f2(x) = x2 – 2xcos(x) + cos²(x) = 0 both with initial guess xo = 0 by using Newton's mehtod and fill the following table. Stopping tolerance 10-5 fi(x) = 0 # of iterations Root $2(x) = 0 # of iterations Root 10-6 10-8 10-10

Answer 1

MATLAB Script:

close all
clear
clc

syms x
f1 = x - cos(x);
f2 = x^2 - 2*x*cos(x) + (cos(x))^2;
f1d = diff(f1); % f1'(x)
f2d = diff(f2); % f2'(x)

tols = 10 .^ -[5 6 8 10];
x0 = 0;
results_1 = zeros(1, length(tols));
results_2 = zeros(1, length(tols));
num_iters_1 = zeros(1, length(tols));
num_iters_2 = zeros(1, length(tols));
for i = 1:length(tols)
tol = tols(i);
[results_1(i), num_iters_1(i)] = my_newton(f1, f1d, x0, tol);
[results_2(i), num_iters_2(i)] = my_newton(f2, f2d, x0, tol);
end

fprintf('Tolerance\t\tIterations (f1)\t\tRoot (f1)\t\tIterations (f2)\t\tRoot (f2)\n')
for i = 1:length(tols)
fprintf('%e\t%d\t\t\t\t\t%.10f\t%d\t\t\t\t\t%.10f\n', tols(i), num_iters_1(i), results_1(i), num_iters_2(i), results_2(i))
end

function [res, num_iter] = my_newton(func,func_d,ic,tol)
res = ic;
num_iter = 0;
while true
num_iter = num_iter + 1;
x_ = res; % Previous value of approximate root
res = double(res - subs(func, res)/subs(func_d, res)); % Newton update rule
if abs(x_ - res) < tol
break;
end
end
end

Output:

Iterations (f1) Command Window Tolerance 1.000000e-955 1.000000-06 1.000000e-08 1.000000e-10 fx >> Root (fi) 0.7390851332 0.7390851332 0.7390851332 0.7390851332 Iterations (f2) 16 19 5 5 6 Root (f2) 0.7390786791 0.7390843265 0.7390851269 0.7390851331 26