function newton_v_secant_v_fpos;
% p is the approximatoin of newtons's method
% p_2, p_1 are the approximations of the secant method
% q_2, q_1 are the approximations of the false position method
% Two initial approximation for 
p_1=-0.5; p_2=0.1; % initial approximation of the secant method,
p=0.5;  %initial approximation of Newton's method, 
q_1=-0.5; q_2=0.1;  %initial approximation of Method of False Position (root is inside [q_1, q_2])
% n is the number of iterations
n=40;
% the same loop is for the both methods
fprintf(1, '\n\n Compare the three methods \n');
for i=1:1:n
p=p-f(p)/der_f(p); % Newton's 
p_secant=p_2-f(p_2)*(p_2-p_1)/(f(p_2)-f(p_1)); p_1=p_2; p_2=p_secant;  %Secant method
q_secant=q_2-f(q_2)*(q_2-q_1)/(f(q_2)-f(q_1)); % false position, root has to be always between q_2 and q_1 
if f(q_secant)*f(q_2)<0
 q_1=q_2; q_2=q_secant;  %next point is computed from q_2 and q_secant 
else 
 q_1=q_1; q_2=q_secant;  %next point is computes from q_1 and q_secant  
end;
fprintf(1,'Iteration: %3i Newton`s: %13.10f Secant: %13.10f False position: %13.10f \n', i,p,p_secant,q_secant);
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Add graphics, just to have a 
%% picture of the function near 
%% the initial approximation
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1);
%x=[2:0.01:4];
%plot(x,pi+0.5*sin(x/2)-x,3.6269420149,pi+0.5*sin(3.6269420149/2)-3.6269420149,'ro');
x=[-1:0.01:1];
%plot(x,x.^2,0,0,'ro');
%plot(x,-x.^3-cos(x), -0.8654740331,-(-0.8654740331)^3-cos(-0.8654740331),'ro');
plot(x,sign(x).*sqrt(abs(x)),0,0,'ro');
grid on
%%% END OF GRAPHICS 


function y=f(x);
% use example 2.2#7 solve as root finding problem by Newton's method
%y=pi+0.5*sin(x/2)-x;
% Another example: solve x^2=0 by Newton's method
%y=x^2;
% One more: se example 2.3#2 solve as root finding problem by Newton's method
%y=-x^3-cos(x);
% a funny example: the function has infinite derivative at $x=0$
y=sign(x)*sqrt(abs(x));

function y=der_f(x);
% use example 2.2#7 solve as root finding problem by Newton's method
%y=cos(x/2)/4-1;
% Another example: solve x^2=0 by Newton's method
%y=2*x;
% One more: se example 2.3#2 solve as root finding problem by Newton's method
%y=-3*x^2+sin(x);
% a funny example: the function has infinite derivative at $x=0$
y=1/(2*sqrt(abs(x)));