Why the sums of cos(x) over 2*pi range not zero?

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when evaluating the following codes:
t = linspace(-pi,pi,128); s = sin(t); c = cos(t); sum(s), sum(c)
ans =
ans =
Q: should not both be zero?
and try
ans =
also quad(@cos,-pi,pi)
ans =
Q: Why the discrepancy?
Thank you for your input.
Thanks J

Accepted Answer

Greg Heath
Greg Heath on 11 Jul 2012
T = fundamental period
N = number of samples
dt = T/N sampling interval
f0 = 1/T fundamental frequency
(n-1)*f0 harmonics 1<=n <= N
Orthogonality interval
t = t0:dt:t0+T-dt;
t = t0+[0:dt:T-dt];
t = t0+dt*[0:N-1];
help fft
doc fft
Hope this helps

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More Answers (3)

Luffy on 11 Jul 2012
In matlab,
sin(pi) = 1.2246e-16
The expression sin(pi) is not exactly zero because pi is not exactly π

Wayne King
Wayne King on 11 Jul 2012
Edited: Wayne King on 11 Jul 2012
If you're trying to establish some equivalence between the integral of cos(t) from -pi and pi and the sum of cos(t), you're forgetting a very important part and that is the dt
t = linspace(-pi,pi,1000);
dt = (pi-(-pi))/length(t);
Using other integration routines in MATLAB is more robust than what I've done, but you see it gets you much closer to zero. Think about the formula for a Riemann sum.
  1 Comment
Tianyou Chen
Tianyou Chen on 11 Jul 2012
Hi Wayne,
Thanks for the answer. But I don't think dt is the issue. try this and nyou will know what I mean: t1 = linspace(-pi,0,64); t2 = linspace(0,pi,64); c1 = cos(t1); c2 = cos(t2); sum(c1) sum(c2)
ans =
ans =
or t1 = linspace(0,pi,64); t2 = linspace(pi,2*pi,64);
c1 = cos(t1); c2 = cos(t2); sum(c1) sum(c2)
ans =
ans =
Thus with or without dt, the sum of a sin or cos over a period of 2pi should be zero. Here the sin function gives the correct answer. I suspect that the even nature of the cos fuction may have something to do with its suming over (-pi, pi) is -1 while over (0,2*pi) is +1.

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Wayne King
Wayne King on 11 Jul 2012
I don't think you can say that simply summing cos(t) on an arbitrary grid should be zero. You have to be careful how the grid is constructed. For example
k = 1;
N = 100;
t = 0:99;
are both zero, because I used a Fourier frequency and a specific discrete-time vector. This has to do with the orthogonality of the N-th roots of unity.
  1 Comment
Tianyou Chen
Tianyou Chen on 11 Jul 2012
Thank Wayne,
You are correct and I should have caught that - the discrete nature of the signal. Cheers, J

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