function fixedpoint_n6
% 0.5 is the Initial approximation
%p=0.5
p=1
% calculate the number of iterations for #6
%%
n=2*log(10)/log(3*(4^(1/3)))
% calculate the number of iterations for #7
n=(2*log(10)+log(pi))/log(4)
n=ceil(n);
for i=1:1:n
p=f(p)
end
p



function y=f(x)
% #6
%y=(x+1)^(1/3);
% #7
y=pi+0.5*sin(x/2);