Question

Apply Newton’s method to?(?)=?−2sin(?)=0. Compare the number of iterations required for convergence to that of the fixed-point iteration method with formulation ?=?(?)=2sin(?). Solve for x0 = pi/4, piand -piwith14-digit accuracy [i.e., tol = 10-(14+1)= 10-15]. Compare your solutions to those obtained using Matlab’s fsolve and fzero.

Answer 1

Newtons Method : Matlab code

f = @(x) x-2*Sin(x);

fp = @(x) 1-2Cos(x);

x0 = pi/4;

N = 10;

tol = 1E-14;

x(1) = x0; % Set initial guess

n = 2;

nfinal = N + 1;

while (n <= N + 1)

fe = f(x(n - 1));

fpe = fp(x(n - 1));

x(n) = x(n - 1) - fe/fpe;

if (abs(fe) <= tol)

nfinal = n;

break;

end

n = n + 1;

end

plot(0:nfinal - 1,x(1:nfinal),'o-')

title('Solution:')

xlabel('Iterations')

ylabel('X')

Fixed Point Iteration: Matlab Code

function [ x ] = fixedpoint(g,I,y,tol,m)

% input: g, I, y, tol, max

% g - function

% I - interval

% y - starting point

% tol - tolerance (error)

% m - maximal number of iterations

% x - approximate solution

a=I(1);b=I(2);

if(y<a | y>b)

error('The starting iteration does not lie in I.')

end

x=y;

gx=g(y);

while(abs(x-gx)>tol & m>0)

if(gx<a | gx>b)

error('The point g(x) does not lie in I.')

end

y=x;

x=g(y);

m=m-1;

end