function [ x, k, fx ] = bisect( a,b,tol)
%UNTITLED2 Summary of this function goes here
%   Detailed explanation goes here
% find a zero for the function f(x) on [a,b]
% using bisection. tol is the tolerance, say, 1e-10
%
% x is the answer if any, k is the number of iterations, and
% fx = f(x)

% f(x) is defined at the end

if f(a)*f(b)>0, disp('f(a) and f(b) have the same sign. quit'); return; end
if abs(f(a))<tol, x=a; return; end
if abs(f(b))<tol, x=b; return; end

c=(a+b)/2;

k=0;

while (b-a)>tol   % abs(f(c))>tol
    if f(a)*f(c)>0
        a=c;
    else
        b=c;
    end
    c = (a+b)/2;
    k = k+1;
end

x = c;
fx = f(x);

% define the function f(x)
function [y] = f(x)
y = x + sin(x) - 2;

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function [ x, k, fx ] = newtonbacktrack( x0, tol, kmax )
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here

% Newton with backtracking to find a zero for f(x)
%
% x = x - f(x) / df(x) where df is f'(x)

x = x0;
k = 0;

while abs(f(x)) > tol && k < kmax
    if abs(df(x))<tol, disp('df(x) is close to zero. Try a different x');
       return;
    end
    
    xtrial = x - f(x) / df(x);  % calculate xtrial by Newton;
    
    % backtrack if necessary
    while abs(f(xtrial)) >= abs(f(x))
        xtrial = (x + xtrial) / 2;
    end
    
    x = xtrial,
    
    k = k + 1,
end

fx = f(x);

% define f(x)
function [y] = f(x)
y = x^2 + 1;

% define df(x)
function [y] = df(x)
y = 2*x;

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