# Product of a geomretic progression

• Apr 8th 2011, 10:26 AM
zhangvict
Product of a geomretic progression
Hey guys, I am new to this forum and I hope to maybe become a regular user.

The following problem is actually from my high school investigation coursework, but it seems too hard to belong to the pre-college forum. By investigating the product of line segments between points of roots of unity, I managed to form the flowing conjecture:

http://latex.codecogs.com/gif.latex?...7D&space;%29=n

n = integer greater than or equal to 2

I tried to prove it by induction, but I don't know how with this type of trigonometric function. By any method, can someone help me prove it?

EDIT: sorry, I mislabeled the thread. This is not actually a geometric progression even though is kind of like one
ALSO, even though this is part of schoolwork, my teacher stated that outside research and discussion is allowed, even among classmates doing the same investigation. As long as we understand why and how the proof works, we are not considered cheating if we learn do the question ourselves, so feel free to provide as much help as you want without feeling like your helping me cheat.
• Apr 8th 2011, 10:44 AM
veileen
$\prod_{i=1}^{n-1}sin(\frac{i\pi }{n})=\frac{n}{2}$

Multiply by $sin (\frac{\pi}{2n})$ and see what happens.

By the way, this is not a geometric progression.
• Apr 8th 2011, 11:01 AM
zhangvict
I'm not sure you can just divide the thing by 2. I plugged in some numbers for n and it didn't work for your modified equation.
• Apr 8th 2011, 11:08 AM
veileen
o.o It's not necessary to do that and it changes nothing.

You know what to do from here?
• Apr 8th 2011, 11:15 AM
zhangvict
Quote:

Originally Posted by veileen
$\prod_{i=1}^{n-1}sin(\frac{i\pi }{n})=\frac{n}{2}$

Multiply by $sin (\frac{\pi}{2n})$ and see what happens.

By the way, this is not a geometric progression.

My original equation was:
http://latex.codecogs.com/gif.latex?...7D&space;%29=n

You changed it to:
$\prod_{i=1}^{n-1}sin(\frac{i\pi }{n})=\frac{n}{2}$

I was saying I don't think you can just simply divide both sides by 2. I think you will need to divide by 2^(n-1) because the 2 was inside the capital pi function.
My original equation worked when I inputted numbers for n, but when I tried your equation, it did not work.

I also don't get how multiplying by $sin (\frac{\pi}{2n})$ will help. Sorry if I misunderstood anything or appear ignorant, I am just a high school student so I don't know as much stuff.
• Apr 8th 2011, 07:42 PM
roninpro
I wrote $\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$. If you think about this for a little bit, you realise that the product can be written

$\displaystyle \prod_{j=1}^{n-1} \frac{e^{i j\pi/n}-e^{-i j\pi/n}}{i}$

where $i=\sqrt{-1}$ and $j$ is the index, to avoid confusion. But we can write $\frac{j\pi}n}=j\frac{2\pi}{2n}$. So we are really looking at

$\displaystyle \prod_{j=1}^{n-1} \frac{\zeta_j-\zeta_j^{-1}}{i}$

where $\zeta_j$ is one of the $2n$-th roots of unity. It looks like we can call on some symmetry or look at the minimal polynomial to complete the problem, but I don't have time to think about it right now. Maybe you can investigate this on your own.
• Apr 8th 2011, 09:06 PM
zhangvict
Quote:

Originally Posted by roninpro
I wrote $\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$. If you think about this for a little bit, you realise that the product can be written

$\displaystyle \prod_{j=1}^{n-1} \frac{e^{i j\pi/n}-e^{-i j\pi/n}}{i}$

where $i=\sqrt{-1}$ and $j$ is the index, to avoid confusion. But we can write $\frac{j\pi}n}=j\frac{2\pi}{2n}$. So we are really looking at

$\displaystyle \prod_{j=1}^{n-1} \frac{\zeta_j-\zeta_j^{-1}}{i}$

where $\zeta_j$ is one of the $2n$-th roots of unity. It looks like we can call on some symmetry or look at the minimal polynomial to complete the problem, but I don't have time to think about it right now. Maybe you can investigate this on your own.

Thanks for your comment. It seems very useful, but I don't have the experience to fully understand the maths in it. Can you tell me what what theoreoms/lemma/formulars you used so I can look them up?

eg. how did you get $\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and why can it be written as $\displaystyle \prod_{j=1}^{n-1} \frac{e^{i j\pi/n}-e^{-i j\pi/n}}{i}$ ?

Also what branch of maths uses the functions in $\displaystyle \prod_{j=1}^{n-1} \frac{\zeta_j-\zeta_j^{-1}}{i}$ (i.e the name of the methods) ?

When you talk about symmetry, do you mean symmetry in the sine curve? and what does minimal polynomial mean?
• Apr 8th 2011, 10:41 PM
Drexel28
Quote:

Originally Posted by zhangvict
Thanks for your comment. It seems very useful, but I don't have the experience to fully understand the maths in it. Can you tell me what what theoreoms/lemma/formulars you used so I can look them up?

eg. how did you get $\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and why can it be written as $\displaystyle \prod_{j=1}^{n-1} \frac{e^{i j\pi/n}-e^{-i j\pi/n}}{i}$ ?

Also what branch of maths uses the functions in $\displaystyle \prod_{j=1}^{n-1} \frac{\zeta_j-\zeta_j^{-1}}{i}$ (i.e the name of the methods) ?

When you talk about symmetry, do you mean symmetry in the sine curve? and what does minimal polynomial mean?

I'll let roninpro answer all your questions.

That said, my immediate response is similar to his. Namely, note that since your product, call it $P$, is positive an real you have that $|P|=P$ where $|\cdot|$ is the modulus function. But, a little manipulation (ask if it isn't clear) shows that $\displaystyle |P|=2^{1-n}\left|\prod_{k=1}^{n-1}\left(1-e^{\frac{2\pi i k}{n}\right)\right|$ but the second term in the product is easily (by considering it as a factorized form of $1+z+\cdots+z^{n-1}$) seen to be equal to $n$. Thus, $P=2^{1-n}n$.
• Apr 8th 2011, 10:52 PM
zhangvict
Quote:

Originally Posted by Drexel28
I'll let roninpro answer all your questions.
But, a little manipulation (ask if it isn't clear) shows that $\displaystyle |P|=2^{1-n}\left|\prod_{k=1}^{n-1}\left(1-e^{\frac{2\pi i k}{n}\right)\right|$ but the second term in the product is easily (by considering it as a factorized form of $1+z+\cdots+z^{n-1}$) seen to be equal to $n$. Thus, $P=2^{1-n}n$.

Yes, can you explain it? I understand why 2^(n-1) can be taken out of the capital pi fuction, but I don't get why the sine function can be converted into the 1-e form.

How is $P=2^{1-n}n$ significant to prove the conjecture?

Oh yeah, in the conjecture, the product is stated to be equal to n. Does that mean that $n=2^{1-n}n$, implying that 2^(1-n) = 1?
• Apr 9th 2011, 01:35 AM
Deveno
Quote:

Originally Posted by zhangvict
Thanks for your comment. It seems very useful, but I don't have the experience to fully understand the maths in it. Can you tell me what what theoreoms/lemma/formulars you used so I can look them up?

eg. how did you get $\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and why can it be written as $\displaystyle \prod_{j=1}^{n-1} \frac{e^{i j\pi/n}-e^{-i j\pi/n}}{i}$ ?

Also what branch of maths uses the functions in $\displaystyle \prod_{j=1}^{n-1} \frac{\zeta_j-\zeta_j^{-1}}{i}$ (i.e the name of the methods) ?

When you talk about symmetry, do you mean symmetry in the sine curve? and what does minimal polynomial mean?

$\displaystyle \sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ comes from the fact that:

$\displaystyle e^{i\theta}=\cos \theta+i\sin \theta$ and $\displaystyle e^{-i\theta}=\cos \theta-i\sin \theta$ (this is known as euler's formula)

so each term in your product $\displaystyle \prod_{k=1}^{n-1} 2\sin(\frac{k\pi}{n})$ is of the form: $\displaystyle \frac{e^{i k\pi/n}-e^{-i k\pi/n}}{i}$

and if we set $\displaystyle \zeta_k = e^{i2k\pi/2n}$ we get the product $\displaystyle \prod_{k=1}^{n-1} \frac{\zeta_k-\zeta_k^{-1}}{i}$

complex numbers of the form $\displaystyle e^{i2k\pi/m}$ have a special name, m-th roots of unity, because they form the m solutions to $\displaystyle x^{m}-1 = 0$.

it turns out that $\displaystyle \zeta_k$ and $\displaystyle \zeta_k^{-1}$ are conjugates, so their difference is $\displaystyle 2Im(\zeta_k)i$ (this is really just another way of saying the y-coordinate of a point on a circle is sin of the angle between the ray going through that point, and the x-axis).
• Apr 9th 2011, 02:37 AM
zhangvict
From your information, I can deduce that the original becomes:

$\displaystyle \prod_{k=1}^{n-1} \frac{\zeta_k-\zeta_k^{-1}}{i}$ = n

$\displaystyle \prod_{k=1}^{n-1} \frac{\2Im(\zeta_k)i}{i}$ = n

$\displaystyle \prod_{k=1}^{n-1}{\2Im(\zeta_k)}$ = n

assuming that the m you mean in $\displaystyle \2Im(\zeta_k)$ is the mth power in $\displaystyle x^{m}-1 = 0$:

$\displaystyle \prod_{k=1}^{n-1}{\2In(\zeta_k)}$ = n
(because n also represents the power of x in the roots of unity)

I still don't know how this changed version can help me prove it though. What is the standard method used to prove this type of thing and can you show me how to carry out this method?. Also, what does capital I represent?
• Apr 9th 2011, 06:03 AM
Opalg
Quote:

Originally Posted by zhangvict
even though this is part of schoolwork, my teacher stated that outside research and discussion is allowed, even among classmates doing the same investigation. As long as we understand why and how the proof works, we are not considered cheating if we learn do the question ourselves, so feel free to provide as much help as you want without feeling like your helping me cheat.

It looks as though the first stage of your project should be to find out something about complex numbers. A complex number $z=x+iy$ has two parts, a real part x and an imaginary part y. The notation $y = \text{Im}\,z$ is used to denote the fact that y is the imaginary part of z. That is the meaning of the abbreviation Im in previous comments in this thread.

You need to study enough of the basic theory of complex numbers to cover the fact that $e^{i\theta} = \cos\theta+i\sin\theta$, and to know what is meant by complex roots of unity. You should by then be able to follow the neat argument of Drexel28 that $\displaystyle \prod_{k=1}^{n-1}\left(z-e^{2\pi i k/n}\right) = 1+z+z^2+\ldots+z^{n-1}$ from which, by putting z=1, you get $\displaystyle \prod_{k=1}^{n-1}\left(1-e^{2\pi i k/n}\right) = n.$
• Apr 9th 2011, 06:24 AM
zhangvict
I have already studied the basic theory of complex numbers; my whole investigation is based around plotting complex numbers on an argand diagram and drawing lines between the points of roots of unity. I was confused mainly because of the notation and the fact that people here tend to post the next step of the equation without referencing to the formula or theorem they use, so it seems like things came out of nowhere. I don't blame them though, the basic stuff of complex numbers they probably already know inside out and have used so many times it seems pointless to them to state what they are doing.

Now that things have cleared up and I actually get the maths going on, Opalg, I think your reminder about Drexel's idea is key now. If you equate the two equations you provided, then you just basically get 1+1+1+1+1+1+1.... until you get n.

Just one thing though, why can you put z=1?

is it because since z^n=1, then z=1? why are you allowed to put z=1 and not z = the other roots?
• Apr 9th 2011, 07:01 AM
Opalg
Quote:

Originally Posted by zhangvict
I have already studied the basic theory of complex numbers; my whole investigation is based around plotting complex numbers on an argand diagram and drawing lines between the points of roots of unity. I was confused mainly because of the notation and the fact that people here tend to post the next step of the equation without referencing to the formula or theorem they use, so it seems like things came out of nowhere. I don't blame them though, the basic stuff of complex numbers they probably already know inside out and have used so many times it seems pointless to them to state what they are doing.

I didn't intend to be patronising. It's just that it wasn't clear from your previous comments how much you already knew.

Quote:

Originally Posted by zhangvict
Now that things have cleared up and I actually get the maths going on, Opalg, I think your reminder about Drexel's idea is key now. If you equate the two equations you provided, then you just basically get 1+1+1+1+1+1+1.... until you get n.

Just one thing though, why can you put z=1?

is it because since z^n=1, then z=1? why are you allowed to put z=1 and not z = the other roots?

The equation $\displaystyle \prod_{k=1}^{n-1}\left(z-e^{2\pi i k/n}\right) = 1+z+z^2+\ldots+z^{n-1}$ is an identity, true for all values of z and hence in particular when z=1. It comes from factorising $z^n-1$ in two different ways. Firstly, as the product of all the linear factors, which are given by the roots of the equation $z^n-1=0$, namely z=1 together with all the complex n'th roots of unity. Secondly, as $z^n-1 = (z-1)(1+z+z^2+\ldots+z^{n-1})$. When you cancel the factor z–1 on both sides, you are left with the required identity.