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Help with proof (Roots of unity question)

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/kno(It n data

    I need to prove that [itex]\prod_{k=1}^{n-1}2\sin\tfrac{k\pi}{n}=n[/itex]
    I just don't know where to start and what type of proof to do.

    Is there any help to get? :D

    2. Relevant equations

    I know this has nothing to do with an equation.
    But i have come up with the equation i have to prove after investigating lenths between roots x^n=1 where x=ℂ.

    I wonder if I can somehow connect the factorization of x^n=1 to the proof.

    3. The attempt at a solution

    I have no idea what to do, this is why im asking for help :D
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2

    HallsofIvy

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    One way of defining "sine" is to say that sin(t) is the y coordinate of the point on the circumference of the unit circle at distance t, around the circumference, from (1, 0). Here, t is [itex]k\pi/n[/itex] so these points are equally spaced about the unit circle and, I am sure you understand, those points are the complex roots of the equation [itex]z^n= 1[/itex]. Now, what can you say about the product of the imaginary parts of those roots?

    (Note [itex](x- a)(x- b)= x^2- (a+ b)x+ ab[/itex], [itex](x- a)(x- b)(x- c)= x^3- (a+ b+ c)x^2+ (ab+ ac+ bc)x- abc[/itex], etc.)
     
  4. Feb 6, 2012 #3

    What if [itex]k=1 [/itex] --- this point isn't a root of [itex]z^n= 1[/itex]? As far as i understand?

    __

    Now to your question I am also a little confused.

    I take [itex]z^3=1 [/itex] as an example. It has the roots [itex]e^{ik2\pi/3}[/itex]. If I exclude the root at (1,0) I have two imaginary parts left: [itex]i\sin2\pi/3[/itex] and [itex]i\sin4\pi/3[/itex]. So: [itex]\pm \sqrt{3}/2[/itex]. The product of these being [itex]3/4[/itex] --> not sure what you are getting at. But i might misunderstand it all.
     
  5. Feb 6, 2012 #4

    HallsofIvy

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    Hmmm- you know I wondered briefly about this but did not follow up on it- what you are trying to prove is not true!

    If n= 3, then k= 1 and 2 so that the product is [itex]2sin(\pi/3)sin(2\pi/3)= 2(sqrt{3}/2)^2= 3/2[/itex], not 3!
     
  6. Feb 6, 2012 #5

    --> [itex]2sin(\pi/3)2sin(2\pi/3)= 3 [/itex]

    Did i write it wrong in my first post since you forgot the two in front of [itex]sin(2\pi/3[/itex])?
     
  7. Feb 6, 2012 #6
    I hope you didnt give up on me :tongue:
     
  8. Feb 7, 2012 #7
    Anyone willing to help here?
     
  9. Feb 7, 2012 #8

    Dick

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    It's a bit of a trick. [itex]x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)[/itex]. You can also write [itex]x^n-1=(x-1)(x-x_1)(x-x_2)...(x-x_{n-1})[/itex] where the [itex]x_k[/itex] are the other roots [itex]x^n-1[/itex] that aren't 1. So define [itex]f(x)=(x^{n-1}+x^{n-2}+...+x+1)=(x-x_1)(x-x_2)...(x-x_{n-1})[/itex]. So [itex]f(1)=n[/itex] and it's also equal to [itex](1-x_1)(1-x_2)...(1-x_{n-1})=|1-x_1| |1-x_2| ... |1-x_{n-1}| [/itex]. Now work out what those absolute values are.
     
  10. Feb 8, 2012 #9

    Yes i see somewhat of a connection here.

    If i do it for n=3,4,5:

    n=3 ---> (x-1) (x^2-x(2cos(2π/3)+1)
    n=4 ---> (x-1)(x+1)(x^2-x(2cos(π/2)+1)
    n=5---> (x-1)(x^2-x(2cos(2π/5)+1)(x^2-x(2cos(4π/5)+1

    The connection is see is that when x=1 then if (x-1) is excluded it equals to n. I am not excactly sure why though.
     
  11. Feb 8, 2012 #10
    Hint:
    [tex]
    \prod_{k = 1}^{n - 1}{2 \, \sin \left( \frac{k \, \pi}{n} \right)} = \st{2 \, \Im \prod_{k = 1}^{n - 1}{\exp \left( i \, \frac{k \, \pi}{n} \right)}} = \ldots
    [/tex]
    What is the sum in the exponential? What is [itex]\Im \exp \left( i \, \alpha \right), \ \alpha \in \mathbb{R}[/itex]? Then, use the values of the sine.

    EDIT:

    The 2nd step does not follow from the first step. Namely: [itex]\mathrm{Im}{z_1} \, \mathrm{Im}{z_2} \neq \mathrm{Im}(z_1 \, z_2)[/itex]
     
    Last edited: Feb 8, 2012
  12. Feb 8, 2012 #11
    :( -- im not completely familiar with all the notation here.

    Especially the (exp) and the imaginary part. Im not sure how it all works together.
    Any chance you can elaborate a little?

    Also.. in the product it goes from k=1 to k=n-1 right? Thats not what you have written above -- is that on purpose or a typo?
     
  13. Feb 8, 2012 #12
    I really need help with this ASAP!

    This proof is only PART of my huge assignment that is due in two days and i can't get it down. I really need some big time help here.
     
  14. Feb 8, 2012 #13

    Dick

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    I'll do the case n=3. [itex]x^3-1=(x-1)(x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3})[/itex]. Also [itex]x^3-1=(x-1)(x^2+x+1)[/itex]. Therefore [itex](x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3})=(x^2+x+1)[/itex]. Put x=1 in that last expression and take the absolute value. Now try showing e.g. that [itex] | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3})[/itex].
     
    Last edited: Feb 8, 2012
  15. Feb 8, 2012 #14
    Yes yes i have done this!

    When n=3 --> then x^2+x+1 = [itex]x^2-x(2cos(3π/2))+1[/itex]
    Pluggin in one --> 3 = [itex]2-2cos(3π/2)[/itex]
    Using DA formula--> 3 = [itex]2(1-(1-sin^2(π/3))[/itex]

    So 3 = 4sin^2 (π/3)

    Now in my assignment i have already shown that [itex] | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3})[/itex] .. A quick way (for now) is to say that [itex]e^\frac{2 \pi i}{3}[/itex] is the complex conjugate of [itex]e^\frac{4 \pi i}{3}[/itex] hence they have the same distance..

    so they are both 2sin(π/3) long.

    But my whole dilemma in all this is that i have no understanding how this all fits together.
    I think it is cool that when taking out (x-1) the remaining is equal to n when putting in x=1.. but i don't understand why.. would it work putting in any other root? And why do we take out (x-1) --

    i need some kind of explanation so i can get understanding of the phenomenon! :D haha
     
  16. Feb 8, 2012 #15

    Dick

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    The whole thing works the same way for n not equal to 3. In general you will need to prove that [itex] | 1 - e^\frac{2 \pi i k}{n} | = 2 \sin \frac{k \pi}{n} [/itex]. Those are the factors in your product. Or show [itex] | 1 - e^{i \theta} | = 2 \sin \frac{ \theta }{2}[/itex] for [itex] 0 \le \theta \le 2 \pi [/itex].
     
    Last edited: Feb 8, 2012
  17. Feb 8, 2012 #16
    I think i get it! I appreciate it! :D
     
  18. Feb 8, 2012 #17
    Man, i can't get it for general term n.. i don't know why

    Any other hint that can help help me.

    This is what i have found for n=3,4,5

    n=3 : 4sin^2(π/3) ..
    n=4 : 8sin^2(π/4)..
    n=5 : 16sin^2(π/5)sin^2(2π/5)

    I noticed this pattern:

    2^n-1 sin^2(π/n) sin^2(2π/n)... sin^2(mπ/n) where m is the number of complex conjugatess..
     
  19. Feb 8, 2012 #18

    Dick

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    It looks like you are ignoring the hint I have you and trying to work it out case by case. That's a lot of work and you don't have to do it. The basic part of the hint was that [itex](x^{n-1}+x^{n-2}+...+x+1)=(x-e^\frac{2 i \pi}{n})(x-e^\frac{i 4 \pi}{n})...(x-e^\frac{i 2 (n-1) \pi}{n})[/itex]. Don't you believe it?
     
    Last edited: Feb 8, 2012
  20. Feb 8, 2012 #19
    I do believe it! Because when x=1 the absolute values of the brackets on the RHS is basically the value of each distance and hence when multiplied together equals to n. But i still don't understand how to prove it.

    i don't know if im just stupid here.. but i can't see it at all.. :(


    btw, on the RHS i think you mean pi instead of x?
     
  21. Feb 8, 2012 #20

    Dick

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    If you know the polynomial [itex]x^n-1[/itex] has n roots [itex]r_1, r_2, ... r_n[/itex], then you can factor the polynomial into [itex]x^n-1 = (x-r_1)(x-r_2)...(x-r_n)[/itex]. But you do know those n roots. They are [itex]1, e^\frac{2 i \pi}{n}, e^\frac{4 i \pi}{n}, ...[/itex]. They are the nth roots of unity. BTW I missing a '2' in the roots in my previous post. I'll fix it.

    But now there is another way to factor [itex]x^n-1 = (x-1) (x^{n-1}+x^{n-2}+...+x+1)[/itex]. Now just compare the two factorizations and 'cancel' the x-1 part.
     
    Last edited: Feb 8, 2012
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