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

when evaluating the following codes:

t = linspace(-pi,pi,128); s = sin(t); c = cos(t); sum(s), sum(c)

ans =

     6.811558403281303e-15

ans =

-0.999999999999978

Q: should not both be zero?

and try

 sum(c(2:end))

ans =

   2.5313e-14

also quad(@cos,-pi,pi)

ans =

   4.0143e-09

Q: Why the discrepancy?

Thank you for your input.

Thanks J

Answer 1

[Greg Heath]

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

Greg

[Greg Heath]

t = t0 + linspace(0,T-dt,N);

t = t0 + dt*linspace(0,N-1,N);

[Tianyou Chen]

Thanks Greg. It makes sense now. J

Answer 2

[Luffy]

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

Answer 3

[Wayne King]

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);
    sum(cos(t))*dt

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.

[Tianyou Chen]

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 =

-4.4409e-16

ans =

   5.7732e-15

or t1 = linspace(0,pi,64); t2 = linspace(pi,2*pi,64);

c1 = cos(t1); c2 = cos(t2); sum(c1) sum(c2)

ans =

   5.7732e-15

ans =

-1.1102e-14

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.

Answer 4

[Wayne King]

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;
sum(cos(2*pi*k/N*t))
sum(sin(2*pi*k/N*t))

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.

[Tianyou Chen]

Thank Wayne,

You are correct and I should have caught that - the discrete nature of the signal. Cheers, J