Square wave function Mean Squared Error differing by a factor of 2?

Below are 2 scripts that compute Mean Squared Error, E, of a square wave with the jump discontinuities occurring @ -Tau/2 & +Tau/2. It is computed 2 different ways & the 2 ways differ by a factor of 2. I am not sure why. Maybe someone could enlighten me & which way is most appropriate?

A = 1; % Peak-to-peak amplitude of square wave
Tau = 1/2; % Total range in which the square wave is defined (here -5 to 5)
T0 = 1; % Period (time of repeatation of square wave), here 10 
C = 1000; % Coefficients (sinusoids) to retain 
N = 2001; % Number of points to consider
t = linspace(-(T0-Tau),(T0-Tau),N); % Time axis
X = zeros(1,N);
X(t>=-Tau/2 & t<=Tau/2) = A; % Original signal
% --------go to A/2 at edge of pulse, or not
X(t==-Tau/2 | t==Tau/2) = A/2; % Adjustment in JUMP discontinuity to accommodate the Fourier series y-intercept @ t=0.
%sum(find(X==A/2)) - (N+1)             % should equal 0
% ----------
R = zeros(size(X)); % Initialize the approximated signal
for n = -C:C % entire range of fourier coefficients 
if n~=0
   Sinc = (sin(pi*n*Tau/T0)/((pi*n*Tau/T0))); % At n NOTEQUAL to 0
else
   Sinc = 1; % At n EQUAL to 0
end
Cn = (A*Tau/T0)*Sinc; % Actual Fourier series coefficients
R = R + Cn*exp(1j*n*2*pi/T0.*t); % Sum all the coefficients
end
R = real(R); % So as to get rid of 0.000000000i (imaginary) factor
% E = sum((X(t>-Tau/2 & t<Tau/2)-R(t>-Tau/2 & t<Tau/2)).^2)
E = sum((X-R).^2)
figure(1)
plot(t,X,'o-',t,R,'o-')
xlim([.245 .255])

Below differs in lines 11 & 25 being commented out & removing the comment on line 26.

A = 1; % Peak-to-peak amplitude of square wave
Tau = 1/2; % Total range in which the square wave is defined (here -5 to 5)
T0 = 1; % Period (time of repeatation of square wave), here 10 
C = 1000; % Coefficients (sinusoids) to retain 
N = 2001; % Number of points to consider
t = linspace(-(T0-Tau),(T0-Tau),N); % Time axis
X = zeros(1,N);
X(t>=-Tau/2 & t<=Tau/2) = A; % Original signal
% --------go to A/2 at edge of pulse, or not
% X(t==-Tau/2 | t==Tau/2) = A/2; % Adjustment in JUMP discontinuity to accommodate the Fourier series y-intercept @ t=0.
%sum(find(X==A/2)) - (N+1)             % should equal 0
% ----------
R = zeros(size(X)); % Initialize the approximated signal
for n = -C:C % entire range of fourier coefficients 
if n~=0
   Sinc = (sin(pi*n*Tau/T0)/((pi*n*Tau/T0))); % At n NOTEQUAL to 0
else
   Sinc = 1; % At n EQUAL to 0
end
Cn = (A*Tau/T0)*Sinc; % Actual Fourier series coefficients
R = R + Cn*exp(1j*n*2*pi/T0.*t); % Sum all the coefficients
end
R = real(R); % So as to get rid of 0.000000000i (imaginary) factor
E = sum((X(t>-Tau/2 & t<Tau/2)-R(t>-Tau/2 & t<Tau/2)).^2)
% E = sum((X-R).^2)
figure(1)
plot(t,X,'o-',t,R,'o-')
xlim([.245 .255])

Answer 1

[David Goodmanson]

Hi Jeff,

It depends on how you want to look at it. For the entire square wave there are overshoots at the four locations below. The first method catches the squared error at all four locations, and the second way catches location 2 and 3 so you only get half as much. If the goal is to get the squared error due to a single discontinuity then catching 1 and 2 would make sense although for this particular waveform, numerically you will get the answer you just got.

         2    3
        |------|      
        |      |        
   ------      ------
       1        4

[Jeff]

OK, that makes sense.