%% ME 2016
%% Fall 2007
%%
%% AUTHOR:  Chris Paredis
%%
%% PURPOSE: plot the function Et(x,n) as defined in Task2 of HW1 for values of
%% x=50 and n ranging from 0 through 200.
%%
%% ASSUMPTIONS: 
%%

%%
%% Problem formulation:
%%

% it is sometimes a good idea to clear the workspace to make sure you are
% not inadvertently using old results
clear all;  % deletes all the variables in the workspace
close all;  % closes all the open figures

% define the problem and initialize the memory
n = 0:200;
x = 50;
% create a vector of the same size as x
Et = zeros(size(n));  

%%
%% Problem solution:
%%

% Since the function taylor_error_cjp only works for scalars, we have to evaluate
% it inside a loop
for i=1:length(n)
    Et(i)=taylor_error_cjp(x,n(i));
end

%%
%% Problem interpretation:
%%

% plot the results in a graph
semilogy(n,abs(Et));
grid on;
title('ME2016: Function from Task 2, HW1 -- Chris Paredis');
xlabel('n');
ylabel('|Et|');

% display the results in a table
% you could use a simple display command: 
format long e
disp([n', Et']);

% or you could create a more elegantly formatted table with headings:
fprintf('+-----------+-------------------------+\n');
fprintf('| n (order) |           Et            |\n');
fprintf('+-----------+-------------------------+\n');
for i=1:length(Et)
    fprintf('|   %3d     | %23.15e |\n',n(i),Et(i));
end
fprintf('+-----------+-------------------------+\n');

%%
%% solution to questions in Task 4
%%

%% Question 4.1:  Why does the graph stop around n=180?
% at n=182 to be exact, the computation of the power term becomes infinite:
% 50^182 is larger than realmax and therefore results in Inf
% However, notice that the result is not Inf but NaN.  That is because the
% denominator, factorial(n), is also larger than realmax, so that
% Inf/Inf = NaN.
% (note: factorial(n) already overflows (>realmax) at n=171, so that the
% last 10 terms in the summation are zero: power_term/Inf = 0

%% Question 4.2:  Why does the graph become flat at n=120?
% We compute the absolute error, Et, by taking the difference between two
% large numbers exp(50) and the Taylor series summation.  Once we reach
% n=115, the difference between exp(50) and the Taylor series approximation
% becomes so small that there are no significant digits left after
% computing the difference.  All that is left is some round-off error.
% Notice that the result is negative!  This is surprising because the
% truncated Taylor series is an underestimate of exp(x), so that Et should
% always be positive.  The computation of some of the the factorials must
% have been underestimated (chopped off) so that some terms of the
% summation were too large.
% The curve remains flat after n=118 because the terms n=119 and above are
% completely chopped off and do not contribute at all to the result.

%% Question 4.3: Why are the values for n=0 through 4 exactly equal?
% For very small values of n, the Taylor series summation result is such a
% small value that when subtracted from exp(50) it is completely chopped
% off.  This is an inherent limitation of the floating point
% representation.  Unlike the two previous artifacts, it cannot be avoided.