How to prove the summation of cosine or sine of average divided angle in 2*pi*integer
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I guess this question belongs to arithmetic or algebra,although I prefer arithmetic.Thanks for any helpful hints with my question.
What are your thoughts?
Please share your work with us ...even if you know it is wrong
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Harry_the_cat
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Aren't they the same thing?
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That is my interpretation too - that is why I wanted OP to show some work and may be clarify!
pka
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If this is correctly quoted, it is undoubtedly one of the poorest put questions I have ever seen.
I guess this question belongs to arithmetic or algebra,although I prefer arithmetic.Thanks for any helpful hints with my question.
I suspect that one of the \(\displaystyle \cos() \) should have been \(\displaystyle \sin() \).
Moreover, the small \(\displaystyle n\) has no place in question. Either that or \(\displaystyle k=n \)
This appears to be the standard question concerning the sum of the complex \(\displaystyle N^{th} \) roots of unity.
Maybe the poster will do us the honor of a classification??
The question about summation of sine of enumerating angle 2*pi*n*k(enumerate k)
Thanks for advice in my last moved post.After improving the description of my question and adding my work has been done,I post again.I should apologize for my careless work.
Given that n is an arbitrary integer, N is an arbitrary positive integer, and k is the integer enumerating from 0 to N - 1, what I want to prove is the following:
. . . . .\(\displaystyle \displaystyle \sum_{n = 0}^{N - 1}\, \cos\left(\dfrac{2\pi n}{N}\, k\right)\, =\, 0\, \mbox{ as well as }\, \sum_{k = 0}^{N - 1}\, \sin\left(\dfrac{2\pi n}{N}\, k\right)\, =\, 0\)
What I have done is through imagining the complex exponent vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) drawn on a complex plane. If k enumerators from 0 to N - 1, and N is an even number, the vectors can be expressed as:
. . . . .\(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\, =\, \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\, \times\, \exp\bigg[i2\pi \dfrac{n}{N}\, \left(k\, -\, \dfrac{N\, -\, 1}{2}\right)\bigg]\)
If k - (N - 1)/2 is a positive number, the vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) can be regarded as the vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\) clockwised an angle \(\displaystyle 2\pi \dfrac{n}{N}\, \left(k\, -\, \dfrac{N\, -\, 1}{2}\right)\) Otherwise, counter-clockwise for a negative value.
Since N is an even integer, vectors defined as \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) would be distributed symmetrically on both sides of \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\) on the complex plane. As a result:
. . . . .\(\displaystyle \displaystyle \sum_{k = 0}\, N\, -\, 1\, \exp\left(i2\pi \dfrac{n}{N}k\right)\, =\, 0\)
However, this is the case only when N is an even integer. For N odd, I have no idea how to prove this.
Thanks in advance for any advice or hints.
Thanks for advice in my last moved post.After improving the description of my question and adding my work has been done,I post again.I should apologize for my careless work.
Given that n is an arbitrary integer, N is an arbitrary positive integer, and k is the integer enumerating from 0 to N - 1, what I want to prove is the following:
. . . . .\(\displaystyle \displaystyle \sum_{n = 0}^{N - 1}\, \cos\left(\dfrac{2\pi n}{N}\, k\right)\, =\, 0\, \mbox{ as well as }\, \sum_{k = 0}^{N - 1}\, \sin\left(\dfrac{2\pi n}{N}\, k\right)\, =\, 0\)
What I have done is through imagining the complex exponent vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) drawn on a complex plane. If k enumerators from 0 to N - 1, and N is an even number, the vectors can be expressed as:
. . . . .\(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\, =\, \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\, \times\, \exp\bigg[i2\pi \dfrac{n}{N}\, \left(k\, -\, \dfrac{N\, -\, 1}{2}\right)\bigg]\)
If k - (N - 1)/2 is a positive number, the vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) can be regarded as the vector \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\) clockwised an angle \(\displaystyle 2\pi \dfrac{n}{N}\, \left(k\, -\, \dfrac{N\, -\, 1}{2}\right)\) Otherwise, counter-clockwise for a negative value.
Since N is an even integer, vectors defined as \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}k\right)\) would be distributed symmetrically on both sides of \(\displaystyle \exp\left(i2\pi \dfrac{n}{N}\, \dfrac{N\, -\, 1}{2}\right)\) on the complex plane. As a result:
. . . . .\(\displaystyle \displaystyle \sum_{k = 0}\, N\, -\, 1\, \exp\left(i2\pi \dfrac{n}{N}k\right)\, =\, 0\)
However, this is the case only when N is an even integer. For N odd, I have no idea how to prove this.
Thanks in advance for any advice or hints.
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Alright, so if I'm understanding your post correctly, you're asking how to prove that \(\displaystyle \displaystyle \sum _{k=0}^{N-1} \text{cos} \left(\frac{2\pi n}{N}k \right)=0\) for any and all integer values of N and n. But if that's the correct problem statement, you'll never be able to prove it, because it's not true for all values. Consider the case N=n=2:
\(\displaystyle \displaystyle \sum _{k=0}^{2-1} \text{cos} \left(\frac{2\pi 2}{2}k \right)=\sum _{k=0}^{1} \text{cos} (2 \pi k)=cos(2 \pi \cdot 0) + cos(2 \pi \cdot 1)=1+1=2\)
Whoops. So, clearly the given equation is only true for some values of N and n. Figuring out which values make the equation true can be done, if that's what you're after.
\(\displaystyle \displaystyle \sum _{k=0}^{2-1} \text{cos} \left(\frac{2\pi 2}{2}k \right)=\sum _{k=0}^{1} \text{cos} (2 \pi k)=cos(2 \pi \cdot 0) + cos(2 \pi \cdot 1)=1+1=2\)
Whoops. So, clearly the given equation is only true for some values of N and n. Figuring out which values make the equation true can be done, if that's what you're after.
pka
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I know that you have miss-stated a very well-known question.
It is known that \(\displaystyle \sum\limits_{n = 0}^{N - 1} {\cos \left( {\frac{{2n\pi }}{N}} \right)} = 0\;\& \;\sum\limits_{n = 0}^{N - 1} {\sin \left( {\frac{{2n\pi }}{N}} \right)} = 0\) You seem to have too many indexes.
The sum of the \(\displaystyle N,~ N^{th}\) roots of unity is zero.
If \(\displaystyle \zeta=\exp\left(\dfrac{2\pi i}{N}\right) \) then define \(\displaystyle \sigma_n=\zeta^n \). The collection \(\displaystyle \mathscr{R}=\left\{\sigma_n:~n=0,~1\cdots,N-1\right\}\) is the set of the \(\displaystyle N,~ N^{th}\) roots of unity.
Because each of those points is located on the unit circle. As you suggest, using the sum of vectors you get the results.
\(\displaystyle \bf{s} = \sum\limits_{n = 0}^{N - 1} {{\sigma _n}} = 0\). \(\displaystyle \text{Re}({\bf{s}})=0~\&~\text{Im}({\bf{s}})=0\)