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Consider the sum of linearly spaced points in one cycle of a sine wave with arbitrary phase:

$$\sum_{x=t-n+1}^t \sin(\frac{\tau x}{n} +p) $$

When n is even, the result is always 0 as the points cancel each other out symmetrically:

\begin{align*} \sum_{x=t-n+1}^t \sin(\frac{\tau x}{n} +p) &= \sum_{x=t-n/2+1}^t \sin(\frac{\tau x}{n}+p) + \sum_{x=t-n+1}^{t-n/2} \sin(\frac{\tau x}{n}+p) \\ &= \sum_{x=t-n/2+1}^t \sin(\frac{\tau x}{n}+p) + \sin(\frac{\tau(x-n/2)}{n}+p) \\ &= \sum_{x=t-n/2+1}^t \sin(\frac{\tau x}{n}+p) + \sin(\frac{\tau x}{n}-\pi+p) \\ &= 0 \end{align*}

When n is odd, the result is also always 0 if $\frac{\tau}{n}|p$ as one of the points are 0 and the remaining points mirror each other:

\begin{align*} \sum_{x=t-n+1}^t \sin(\frac{\tau x}{n} + \frac{\tau k}{n}) &= \sin(\frac{-\tau k}{n} + \frac{\tau k}{n}) + \sum_{x=1}^{\lfloor n/2 \rfloor} \bigg( \sin(\frac{\tau x}{n}) + \sin(\frac{-\tau x}{n}) \bigg) \\ &= 0 \end{align*}

Does this still hold if n is odd and phase is arbitrary? If so, how can we prove it?

Poh Zi How
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1 Answers1

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Answer based on the link provided by @Blue:

We can employ a trick where we multiply the summation by half of its frequency:

$$ \frac{1}{\sin(\frac{\tau}{2n})} \sum_{x=0}^{n-1} \sin(\frac{\tau x}{n}+p) \sin(\frac{\tau}{2n}) $$

Which simplifies to:

$$ \frac{1}{2\sin(\frac{\tau}{2n})} \sum_{x=0}^{n-1} \bigg( \cos(\frac{\tau x}{n} + p - \frac{\tau}{2n}) - \cos(\frac{\tau x}{n} + p + \frac{\tau}{2n}) \bigg) $$

Which cancels out to 0.

There is a more general form of this at How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? where the frequency doesn't need to be a multiple of $\tau$

Poh Zi How
  • 109