Question

Consider the function f(x) = x - xcosx, which has a root at x = 0. Write a program to compare the rates of convergence of the bisection method (starting with a = -1, b = 1) and Newton’s method (starting with x = 1). Which method converges faster? Why?

Answer 1

Solution:

Newton's method converges faster

In Bisection method more iteration then converge real root but Newton's method less iteration and converge real roots

%% Bisection method
clc
clear all


%f= input('enter the function f \n')
f = @(x) [x-x*cos(x)];
a=input('enter the first initial guesse point \n')

b=input('enter the second initial guesse point b \n')

%% provide the equation you want to solve with R.H.S = 0 form.
%% Write the L.H.S by using inline function
%% Give initial guesses.
%% Solves it by method of bisection.
%% A very simple code. But may come handy

if f(a)*f(b)>0
    disp('Wrong choice Sp')
else
    p = (a + b)/2;
    err = abs(f(p));
    while err > 1e-8
   if f(a)*f(p)<0
       b = p;
   else
       a = p;         
   end
    p = (a + b)/2;
   err = abs(f(p));
    end
end

out_put = p  
   

%% matlab code ==newton Raphson methed

clc
clear all
f = @(x) [x-x*cos(x)]
g = @(x) [1+x*sin(x)-cos(x)] %% diff of function f
%g = input('enter the diff of function f = \n')
n = input('enter the number of iterations n = \n')
x0 = input('enter the inital approximation x0 \n')
for i=1:n   %% increase iterations converge real root
    x(1) = x0;
    x(i+1) = x(i) - (f(x(i))/g(x(i)));
    error_tolerance = abs(x(i) - x(i+1));
end
out_put= x(n)

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