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

- ganeshie8

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- ParthKohli

They're the roots of \(x^n - 1\) so... using Vieta's Formulas?

- ganeshie8

Clever! that is still algebra...

- ikram002p

this is also helpful :O
\(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

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## More answers

- ParthKohli

It's pretty easy when you geometrically see it.

- anonymous

\[e^{i \pi.\frac{2k}{n}}=e^{i.n'.\pi}=\cos(n' \pi)+i.\sin(n' \pi)\]\[\implies \cos(\frac{2k}{n}\pi)+i.\sin(\frac{2k}{n}\pi)\]
can we use this?
or is the formula stricly for positive integers??

- ikram002p

@Nishant_Garg au can and its also work

- ganeshie8

sure euler formula works for ALL \(\theta\), need not be an integer..

- ganeshie8

tell me about the geometric method @ParthKohli

- ParthKohli

Eh, so far, it's only about seeing. It's not a strict proof. |dw:1439296375015:dw|

- ParthKohli

These can be looked at as vectors and rearranged to give you a regular polygon.

- anonymous

\[(\cos^2(\frac{k.\pi}{n})-\sin^2(\frac{k.\pi}{n}))+2.i.\sin(\frac{k.\pi}{n}).\cos(\frac{k.\pi}{n})\]

- ParthKohli

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.

- anonymous

I wish that n was not there XD

- ganeshie8

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...

- ganeshie8

@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

- ParthKohli

An ugly method to do this problem is of course computing\[\sum_{k=0}^{n-1} \cos \dfrac{2\pi k}{n} \]and\[\sum_{k=0}^{n-1} \sin \dfrac{2\pi k}{n}\]

- ParthKohli

For some weird reason, `\frac` isn't working but `\dfrac` does.

- anonymous

working for me, I use it like this:
frac{3}{4}
output:
\[\frac{3}{4}\]

- ParthKohli

|dw:1439297117562:dw|

- ganeshie8

@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...

- ikram002p

\( \begin{align}
\sum_{k=1}^{n} \cos (k\theta)
&=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2).
\end{align} \)
so this is solvable if u continued @Nishant_Garg i wanna continue ur method

- anonymous

Feel free to continue, and here I'm trying to study my course and getting distracted!!

- ParthKohli

OK, so try seeing it this way: place 2 at the ending of 1, place 3 at the ending of 2, and so on.

- ParthKohli

Yes, the formula for the sum of sines in AP:\[\frac{\sin n d /2}{\sin d /2} \sin \left(\frac{2a + (n-1 ) d }{2}\right) \]

- ikram002p

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 .

- ParthKohli

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.

- ikram002p

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)

- anonymous

what about
\[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\]

- ikram002p

that's my point Garg :3

- anonymous

yeah, that's a polynomial now isnt it :P

- anonymous

just a very long one

- ParthKohli

We want to prove that it cannot be represented as a polynomial. You just gave me another representation of sin(x).

- ParthKohli

That's a series, not a polynomial. A polynomial has a defined degree.

- anonymous

So it's not a polynomial as it goes on infinitely?

- ikram002p

its a polynomial series or as we love to say infinite polynomial function :3

- ganeshie8

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.

- anonymous

we can approximate... :p
\[\sin(x) \approx x-\frac{x^3}{3!}+\frac{x^5}{5!}\]
I'll call it a pseudopolynomial XD

- ikram002p

i'll go anyway, bbye.

- ikram002p

lol xD

- ganeshie8

P(x) = 0 has infinite roots

- ParthKohli

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|

- ParthKohli

By symmetry, each angle of this polygon is \(\pi - \theta = \pi - \frac{2\pi}{n}\).
The sum of all angles is \(n\pi - 2\pi = (n-2)\pi\).

- ParthKohli

It's of course a regular polygon as all sides are equal.

- ParthKohli

And so are all interior angles.

- ParthKohli

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.

- anonymous

Has this not been proved?? like is it an unsolved question?

- ParthKohli

What kind of proof do you have with you? @ganeshie8

- ganeshie8

|dw:1439298571809:dw|

- ParthKohli

Yeah nice.

- ganeshie8

I like this analytic proof in particular
say \(\omega = e^{i2\pi/n}\), then
\(S=\sum\limits_{k=1}^{n}e^{i2k\pi/n} = \sum\limits_{k=1}^{n} \omega^k \)
\(\implies \omega S = \sum\limits_{k=1}^{n} \omega^{k+1} =S\)
\(\implies S=0\)

- ganeshie8

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

- ganeshie8

same idea as exterior angles adding up to 360 in a regular polygon

- anonymous

I've the best proof... multiply both sides by...
ZERO XD

- ganeshie8

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 \(0\)

- ganeshie8

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

- ganeshie8

i kno you're just kidding, but i wanted to clarify it anyways :)

- anonymous

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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