## 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

1. ParthKohli

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

2. ganeshie8

Clever! that is still algebra...

3. ikram002p

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

4. ParthKohli

It's pretty easy when you geometrically see it.

5. 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??

6. ikram002p

@Nishant_Garg au can and its also work

7. ganeshie8

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

8. ganeshie8

tell me about the geometric method @ParthKohli

9. ParthKohli

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

10. ParthKohli

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

11. anonymous

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

12. 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.

13. anonymous

I wish that n was not there XD

14. 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...

15. 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

16. ParthKohli

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

17. ParthKohli

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

18. anonymous

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

19. ParthKohli

|dw:1439297117562:dw|

20. 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...

21. 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

22. anonymous

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

23. 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.

24. ParthKohli

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

25. 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 .

26. 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.

27. 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)

28. anonymous

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

29. ikram002p

that's my point Garg :3

30. anonymous

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

31. anonymous

just a very long one

32. ParthKohli

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

33. ParthKohli

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

34. anonymous

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

35. ikram002p

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

36. 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.

37. anonymous

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

38. ikram002p

i'll go anyway, bbye.

39. ikram002p

lol xD

40. ganeshie8

P(x) = 0 has infinite roots

41. 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|

42. ParthKohli

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

43. ParthKohli

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

44. ParthKohli

And so are all interior angles.

45. 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.

46. anonymous

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

47. ParthKohli

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

48. ganeshie8

|dw:1439298571809:dw|

49. ParthKohli

Yeah nice.

50. ganeshie8

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

51. 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

52. ganeshie8

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

53. anonymous

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

54. 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 $$0$$

55. 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

56. ganeshie8

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

57. 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