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I'm a little stuck at the moment and wondered if someone could point me in the direction of the theory I need to read.

I have a $2\pi$-periodic function, $f:\mathbb{R}\rightarrow\mathbb{R}$ which I have sampled at $N\in\mathbb{N}$ and $3N$ regularly spaced intervals, beginning at zero, over 1 period. I have proved using Fourier series that,

\begin{equation} \sum_{p=1}^{3N}f\left(\frac{2\pi(p-1)}{3N}\right)=3\sum_{p=1}^Nf\left(\frac{2\pi(p-1)}{N}\right) \end{equation}

The $N$ samples of the function are present in the $3N$ samples of the function, which means that at least one $\sum_{p=1}^Nf\left(\frac{2\pi(p-1)}{N}\right)$ can be found easily from within $\sum_{p=1}^{3N}f\left(\frac{2\pi(p-1)}{3N}\right)$. I need to understand how the remaining terms come together to form $2\sum_{p=1}^Nf\left(\frac{2\pi(p-1)}{N}\right)$.

Is there a theorem detailing this result? I have a more complicated sum I need to deal with and I'm hoping that the ideas from this theorem could help me. Thanks.

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Your statement is most certainly false. Take $N=1$, then the $2 \pi$ periodicity doesn't matter, so you're making a claim about all functions defined on a $2\pi$ interval. –  Calvin Lin Dec 31 '12 at 11:35
    
Thanks for pointing that out Calvin. I think $N$ needs to be sufficiently large, but I will have another think. –  Alanaj5 Dec 31 '12 at 12:16
    
What you want is the infinite sum. The statement is almost never true (except in very rare cases like the constant function) in the finite case, since $N/3 \pmod{2\pi}$ are all distinct values. –  Calvin Lin Dec 31 '12 at 18:39
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