function newton_n2;
% newtons's method
p=-.1;
% n is the number if iterations
n=20;
fprintf(1, 'Examples of work of Newton"s Method \n');
for i=1:1:n
p=p-f(p)/der_f(p); 
fprintf(1,'Iteration: %3i Newton`s: %12.10f \n', i,p);
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Add graphics, just to have a 
%% picture of the function near 
%% the initial approximation
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1);
x=[-1:0.01:1];
plot(x,-x.^3-cos(x));
%plot(x,x.^2);
%plot(x,sign(x).*sqrt(abs(x)));
grid on
%%% END OF GRAPHICS 

function y=f(x);
% use example 2.3#2 solve as root finding problem by Newton's method
y=-x^3-cos(x);
% An exampl eof slower convergence: y=x^2
%y=x^2;
% a funny example: the function has infinite derivative at $x=0$
%y=sign(x)*sqrt(abs(x));

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