show that \[\sum\limits_{k=1}^n e^{i2k\pi/n} = 0\] w/o using geometric series http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5En+e%5E%7Bi2k%5Cpi%2Fn%7D

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show that \[\sum\limits_{k=1}^n e^{i2k\pi/n} = 0\] w/o using geometric series http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5En+e%5E%7Bi2k%5Cpi%2Fn%7D

Discrete Math
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They're the roots of \(x^n - 1\) so... using Vieta's Formulas?
Clever! that is still algebra...
this is also helpful :O \(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

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It's pretty easy when you geometrically see it.
\[e^{i \pi.\frac{2k}{n}}=e^{i.n'.\pi}=\cos(n' \pi)+i.\sin(n' \pi)\]\[\implies \cos(\frac{2k}{n}\pi)+i.\sin(\frac{2k}{n}\pi)\] can we use this? or is the formula stricly for positive integers??
@Nishant_Garg au can and its also work
sure euler formula works for ALL \(\theta\), need not be an integer..
tell me about the geometric method @ParthKohli
Eh, so far, it's only about seeing. It's not a strict proof. |dw:1439296375015:dw|
These can be looked at as vectors and rearranged to give you a regular polygon.
\[(\cos^2(\frac{k.\pi}{n})-\sin^2(\frac{k.\pi}{n}))+2.i.\sin(\frac{k.\pi}{n}).\cos(\frac{k.\pi}{n})\]
I really love the use of complex numbers looked at as rotation operators. Beautiful stuff. For example using that analogy I could easily discover that \(z_2 = z_1 e^{i\pi/3}\) is a self-sufficient condition for \(z_1\) and \(z_2\) to form an equilateral triangle.
I wish that n was not there XD
Multiplying a complex number by another complex number adds up the arguments, so you get rotation. But what you had there are position vectors which happen to lie on a circle. I don't see a regular polygon yet...
@Nishant_Garg yeah we may show that the individual components are 0 http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5En+cos%282*k*%5Cpi%2Fn%29 http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5En+sin%282*k*%5Cpi%2Fn%29
An ugly method to do this problem is of course computing\[\sum_{k=0}^{n-1} \cos \dfrac{2\pi k}{n} \]and\[\sum_{k=0}^{n-1} \sin \dfrac{2\pi k}{n}\]
For some weird reason, `\frac` isn't working but `\dfrac` does.
working for me, I use it like this: frac{3}{4} output: \[\frac{3}{4}\]
|dw:1439297117562:dw|
@ParthKohli your diagram just shows that multiplying \(e^{i2\pi/n}\) to itself for \(n\) times gives \(1\). I don't see how it is related to the sum of roots...
\( \begin{align} \sum_{k=1}^{n} \cos (k\theta) &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2). \end{align} \) so this is solvable if u continued @Nishant_Garg i wanna continue ur method
Feel free to continue, and here I'm trying to study my course and getting distracted!!
OK, so try seeing it this way: place 2 at the ending of 1, place 3 at the ending of 2, and so on.
Yes, the formula for the sum of sines in AP:\[\frac{\sin n d /2}{\sin d /2} \sin \left(\frac{2a + (n-1 ) d }{2}\right) \]
either way cant get rid of geometric series :P which we need to prove this some its only a twist sounds like the origin proof to me so ive better check out something else .
Just wanted to share this interesting one-liner proof as to why \(\sin(x)\) cannot be represented as a polynomial. Proof: It has infinite roots.
so cosine do also any other periodic function, but there is always away to approximate i would say Taylor series is an infinite polynomial that might represent sin x (just an example there is many other ways)
what about \[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\]
that's my point Garg :3
yeah, that's a polynomial now isnt it :P
just a very long one
We want to prove that it cannot be represented as a polynomial. You just gave me another representation of sin(x).
That's a series, not a polynomial. A polynomial has a defined degree.
So it's not a polynomial as it goes on infinitely?
its a polynomial series or as we love to say infinite polynomial function :3
lets get back to this after finishing that polynomial thingy OK, so try seeing it this way: place 2 at the ending of 1, place 3 at the ending of 2, and so on.
we can approximate... :p \[\sin(x) \approx x-\frac{x^3}{3!}+\frac{x^5}{5!}\] I'll call it a pseudopolynomial XD
i'll go anyway, bbye.
lol xD
P(x) = 0 has infinite roots
Yes. If we get a regular polygon, or even a polygon, then the vector-sum is zero. Now it makes a lot of sense as to why we should get a regular polygon without completing the process. |dw:1439298174946:dw|
By symmetry, each angle of this polygon is \(\pi - \theta = \pi - \frac{2\pi}{n}\). The sum of all angles is \(n\pi - 2\pi = (n-2)\pi\).
It's of course a regular polygon as all sides are equal.
And so are all interior angles.
A simpler way to look at it is that the sum of exterior angles is also \(2\pi/n \times n = 2\pi\). None of this constitutes a proof but it's enough to convince.
Has this not been proved?? like is it an unsolved question?
What kind of proof do you have with you? @ganeshie8
|dw:1439298571809:dw|
Yeah nice.
I like this analytic proof in particular say \(\omega = e^{i2\pi/n}\), then \(S=\sum\limits_{k=1}^{n}e^{i2k\pi/n} = \sum\limits_{k=1}^{n} \omega^k \) \(\implies \omega S = \sum\limits_{k=1}^{n} \omega^{k+1} =S\) \(\implies S=0\)
multuplying each root by \(\omega\) rotates the number by\(\arg(\omega)\), but we still have exact same terms in the sum.. so the sum before and after rotation is same
same idea as exterior angles adding up to 360 in a regular polygon
I've the best proof... multiply both sides by... ZERO XD
Haha that doesn't work because \(ab=0 \land a\ne 0 \implies b=0\) but we don't know the value of \(b\) if \(a\) is also \(0\)
In the earlier proof I defined \(\omega = e^{i2\pi/n}\), it is nonzero. and we have \(\omega S = S \) multiplying a nonzero number by \(S\) is not changing its value, that means \(S\) must be 0. because 0 is the only number with this property
i kno you're just kidding, but i wanted to clarify it anyways :)
Actually, a quick google search revealed that you can't multiply both sides by 0 because unlike other numbers where u can divide to get back the original equation, division by zero is not defined

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