function Et = taylor_error_cjp(x,n)

%TAYLOR_ERROR_CJP computes the expression in Equation 1 of HW1
%   Et = TAYLOR_ERROR_CJP(X,N) computes Et as a function of the scalar
%   input X and the non-negative integer input N according to the equation
%   given in Task2 of HW1:  Et is the true absolute error of an N-th order
%   Taylor series approximation of exp(x).  If N is not an integer, the
%   function will truncate it to the nearest integer smaller than N.
%
%   Inputs:
%      X: any scalar
%      N: the order of the Taylor approximation; must be a non-negative
%         integer
%
%   Outputs:
%      Et: a scalar representing the true absolute error of the Taylor
%          series approximation
%
%   Assumptions: 
%         - this function only provides accurate results when the inputs X
%         and N are relatively small.

%%
%% ME 2016 Section
%% Fall 2007
%%
%% AUTHOR:  Chris Paredis
%%

% start with some error checking
% the input n must be a non-negative integer scalar
% NOTE: it is important that the function stop executing when an error is
% detected.  The best way to accomplish this is by using the error()
% function
if length(n)>1
    error('the input must be a scalar');
end
if n<0
    error('the input must be a non-negative integer');
end
% make sure the input is an integer
% NOTE: it is equally acceptable to generate an error if n is not an
% integer
n = floor(n);

% this is a straightforward implementation of Equation 1 in HW1.  It works
% well for when x and n are relatively small, but runs into floating point
% problems when x and n are large.
taylor_sum = 1;
power_term = 1;
fact = 1; % factorial (abbreviated to avoid conflict with factorial() function
for i=1:n
    % computing the power term and factorial based on the terms from the
    % previous iteration is slightly more efficient
    power_term = power_term * x;
    fact = fact * i;
    taylor_sum = taylor_sum + power_term/fact;
end

Et = exp(x) - taylor_sum;

% An experienced Matlab programmer would probably avoid using the for-loop
% as follows: (uncomment the lines below if you want to test this)

%sum_range = 1:n;
%Et = exp(x) - sum(x.^sum_range./factorial(sum_range));

% one can argue as to which of these implementations is most readable or
% maintainable.  The second implementation is somewhat more elegant and
% concise; but for somebody who is not experienced using the colon operator
% it can appear somewhat convoluted.
% Both approaches are acceptable 
return;