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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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    what is m, N? did you mean "for integer m"? if so, what about m=1?2012-06-17
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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

2 Answers 2

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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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    but how to proove this?2012-06-17
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    How to prove that $\sin x$ is an odd function?2012-06-17
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    thank you very much but problem I try to apply concept to new problem but cosine function not symmmetric and so don't know how to show the statements truth.$$ \frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})=\begin{cases} 1, & \ m \equiv 0\pmod{M}\\ 0, & \text{else} \end{cases}$$2012-06-17
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    @nanme: Different functions may require different techniques. My answer used a technique that works for all odd functions. The cosine problem above is not yet clear enough for me to give an answer.2012-06-17
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    I understand. it follows from the same description as the original sin question. I try to find technique that is general and so can be used in many cases but am not sure if there is one that does exist. thank you again very much2012-06-17
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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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    hi and thank you for response. however i don't seem to follow your logic thank you again2012-06-17
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    The $y_k$ are symmetric about 0 and $\sin(-x) = -\sin(x)$ for any $x$.2012-06-17