ganeshie8
  • ganeshie8
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ParthKohli
  • ParthKohli
They're the roots of \(x^n - 1\) so... using Vieta's Formulas?
ganeshie8
  • ganeshie8
Clever! that is still algebra...
ikram002p
  • ikram002p
this is also helpful :O \(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

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More answers

ParthKohli
  • ParthKohli
It's pretty easy when you geometrically see it.
anonymous
  • 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
  • ikram002p
@Nishant_Garg au can and its also work
ganeshie8
  • ganeshie8
sure euler formula works for ALL \(\theta\), need not be an integer..
ganeshie8
  • ganeshie8
tell me about the geometric method @ParthKohli
ParthKohli
  • ParthKohli
Eh, so far, it's only about seeing. It's not a strict proof. |dw:1439296375015:dw|
ParthKohli
  • ParthKohli
These can be looked at as vectors and rearranged to give you a regular polygon.
anonymous
  • 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
  • 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
  • anonymous
I wish that n was not there XD
ganeshie8
  • 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
  • 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
  • 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
  • ParthKohli
For some weird reason, `\frac` isn't working but `\dfrac` does.
anonymous
  • anonymous
working for me, I use it like this: frac{3}{4} output: \[\frac{3}{4}\]
ParthKohli
  • ParthKohli
|dw:1439297117562:dw|
ganeshie8
  • 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
  • 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
  • anonymous
Feel free to continue, and here I'm trying to study my course and getting distracted!!
ParthKohli
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
what about \[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\]
ikram002p
  • ikram002p
that's my point Garg :3
anonymous
  • anonymous
yeah, that's a polynomial now isnt it :P
anonymous
  • anonymous
just a very long one
ParthKohli
  • ParthKohli
We want to prove that it cannot be represented as a polynomial. You just gave me another representation of sin(x).
ParthKohli
  • ParthKohli
That's a series, not a polynomial. A polynomial has a defined degree.
anonymous
  • anonymous
So it's not a polynomial as it goes on infinitely?
ikram002p
  • ikram002p
its a polynomial series or as we love to say infinite polynomial function :3
ganeshie8
  • 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
  • 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
  • ikram002p
i'll go anyway, bbye.
ikram002p
  • ikram002p
lol xD
ganeshie8
  • ganeshie8
P(x) = 0 has infinite roots
ParthKohli
  • 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
  • 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
  • ParthKohli
It's of course a regular polygon as all sides are equal.
ParthKohli
  • ParthKohli
And so are all interior angles.
ParthKohli
  • 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
  • anonymous
Has this not been proved?? like is it an unsolved question?
ParthKohli
  • ParthKohli
What kind of proof do you have with you? @ganeshie8
ganeshie8
  • ganeshie8
|dw:1439298571809:dw|
ParthKohli
  • ParthKohli
Yeah nice.
ganeshie8
  • 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
  • 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
  • ganeshie8
same idea as exterior angles adding up to 360 in a regular polygon
anonymous
  • anonymous
I've the best proof... multiply both sides by... ZERO XD
ganeshie8
  • 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
  • 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
  • ganeshie8
i kno you're just kidding, but i wanted to clarify it anyways :)
anonymous
  • 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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