% Unlimited growth of a single roundoff error 
% 
% calculate I(n)=e\int_{0}^{1} t^n e^{-t}dt 
% 
% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  USE I(n)=nI(n-1)-1:
format long 
I=exp(1)-1;
fprintf(1,'Iteration #0: I(0)= %13.10f \n', I);
for i=1:20
    I=i*I-1;
    fprintf(1,'Iteration: #%3i: I(%3i)= %13.10f \n', i,i,I);
end

fprintf(1,'\n \n');

% Now let us try compute things backwards 
% 
% USE I(n-1)=(I(n)+1)/n
I=0.123456789;
fprintf(1,'Iteration #20: I(20)= %13.10f \n', I);
for i=20:-1:1
    I=(I+1)/i;
    fprintf(1,'Iteration: #%3i: I(%3i)= %13.10f \n', i,i,I);
end