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Possible Duplicate:
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

How is it possible to show for integer $m$:

$\frac{1}{M}\sum_{k=1}^{M}\sin(m\cdot y_{k})=0$

Thank you very much

Interval $[-\pi,\pi]$ split into $M$ equal intervals, with the mid point of interval is $y_{k}$

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    @Gerry: while abstractly I agree it is a duplicate, in this particular case the question is about summing an odd function at points symmetric about zero.2012-06-20

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The sine function is odd function, meaning that in general $\sin(-t)=-\sin(t)$.

The interval $[-\pi,\pi]$ is symmetrical about $0$. So if $x$ is one of the $x_k$, then $-x$ also is one of the $x_k$, and the values of $\sin(mx)$ and $\sin(-mx)$ add up to $0$. (If $N$ is odd, there isn't the perfect twinning, but the untwinned point is $0$, and $\sin(0)=0$.)

Exactly the same argument works for any odd function and any interval symmetric about the origin.

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    There is a kind of general technique for sine and cosine. But it uses complex numbers.2012-06-17
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When you sum over the $y_k$'s backwards, you get the both the same value and its negative:

$\frac{1}{N}\sum_{k=1}^N \sin(m y_k) = \frac{1}{N} \sum_{k=1}^N \sin(m y_{N+1-k}) = \frac{1}{N} \sum_{k=1}^N \sin(-m y_k) = - \frac{1}{N} \sum_{k=1}^N \sin(m y_k)$

so the sum is zero.

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    The $y_k$ are symmetric about 0 and $\sin(-x) = -\sin(x)$ for any $x$.2012-06-17