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ganeshie8
 one year ago
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
 one year ago
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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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2They're the roots of \(x^n  1\) so... using Vieta's Formulas?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Clever! that is still algebra...

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2this is also helpful :O \(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2It's pretty easy when you geometrically see it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[e^{i \pi.\frac{2k}{n}}=e^{i.n'.\pi}=\cos(n' \pi)+i.\sin(n' \pi)\]\[\implies \cos(\frac{2k}{n}\pi)+i.\sin(\frac{2k}{n}\pi)\] can we use this? or is the formula stricly for positive integers??

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2@Nishant_Garg au can and its also work

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0sure euler formula works for ALL \(\theta\), need not be an integer..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0tell me about the geometric method @ParthKohli

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Eh, so far, it's only about seeing. It's not a strict proof. dw:1439296375015:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2These can be looked at as vectors and rearranged to give you a regular polygon.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\cos^2(\frac{k.\pi}{n})\sin^2(\frac{k.\pi}{n}))+2.i.\sin(\frac{k.\pi}{n}).\cos(\frac{k.\pi}{n})\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I 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 selfsufficient condition for \(z_1\) and \(z_2\) to form an equilateral triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wish that n was not there XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Multiplying 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2An ugly method to do this problem is of course computing\[\sum_{k=0}^{n1} \cos \dfrac{2\pi k}{n} \]and\[\sum_{k=0}^{n1} \sin \dfrac{2\pi k}{n}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2For some weird reason, `\frac` isn't working but `\dfrac` does.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0working for me, I use it like this: frac{3}{4} output: \[\frac{3}{4}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439297117562:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2\( \begin{align} \sum_{k=1}^{n} \cos (k\theta) &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2). \end{align} \) so this is solvable if u continued @Nishant_Garg i wanna continue ur method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Feel free to continue, and here I'm trying to study my course and getting distracted!!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2OK, so try seeing it this way: place 2 at the ending of 1, place 3 at the ending of 2, and so on.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, the formula for the sum of sines in AP:\[\frac{\sin n d /2}{\sin d /2} \sin \left(\frac{2a + (n1 ) d }{2}\right) \]

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2either 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
 one year ago
Best ResponseYou've already chosen the best response.2Just wanted to share this interesting oneliner proof as to why \(\sin(x)\) cannot be represented as a polynomial. Proof: It has infinite roots.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.0what about \[\sin(x)=x\frac{x^3}{3!}+\frac{x^5}{5!}\dots\]

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2that's my point Garg :3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, that's a polynomial now isnt it :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just a very long one

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2We want to prove that it cannot be represented as a polynomial. You just gave me another representation of sin(x).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That's a series, not a polynomial. A polynomial has a defined degree.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it's not a polynomial as it goes on infinitely?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2its a polynomial series or as we love to say infinite polynomial function :3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0lets 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
 one year ago
Best ResponseYou've already chosen the best response.0we can approximate... :p \[\sin(x) \approx x\frac{x^3}{3!}+\frac{x^5}{5!}\] I'll call it a pseudopolynomial XD

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2i'll go anyway, bbye.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0P(x) = 0 has infinite roots

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes. If we get a regular polygon, or even a polygon, then the vectorsum 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
 one year ago
Best ResponseYou've already chosen the best response.2By symmetry, each angle of this polygon is \(\pi  \theta = \pi  \frac{2\pi}{n}\). The sum of all angles is \(n\pi  2\pi = (n2)\pi\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2It's of course a regular polygon as all sides are equal.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2And so are all interior angles.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2A 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
 one year ago
Best ResponseYou've already chosen the best response.0Has this not been proved?? like is it an unsolved question?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2What kind of proof do you have with you? @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439298571809:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0multuplying 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
 one year ago
Best ResponseYou've already chosen the best response.0same idea as exterior angles adding up to 360 in a regular polygon

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've the best proof... multiply both sides by... ZERO XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Haha 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
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0i kno you're just kidding, but i wanted to clarify it anyways :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, 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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