Attempting Proof by Induction with Pi

**Capricious** · April 15th 2012, 12:32 PM

Sorry if this is in the wrong section, this is apparently University level mathematics (I'm in HS), so I don't really know what topic it belongs to. Anyways, the problem is an attachment (I have no idea how to write that in LaTeX format).

I tried proving it by induction, but I've only ever done that with summation, sigma, and don't know if it even works for Pi ? If someone could show me the proof for this problem, or tell me what way to go, I'd be forever grateful.

Also, $n\geq3$

**jens** · April 15th 2012, 03:08 PM

Hi,
I don't think you need induction. Remember that the modulus of a complex number $z = a+ib$ is $|z| = \sqrt{a^2+b^2}$ . Then keep in mind that $\cos^2(x)+\sin^2(x)=1$ . See what you can do with that.

**Capricious** · April 16th 2012, 01:45 AM

I'm not quite sure how that helps me? Yes, it allows me to simplify the expression to sqrt(2-2cos(2πr/n)), however this expression is still within the multiplication. Is there someway to simplify that expression further to make it work?

Someone please help, I'm really not making any progress on this.

**Deveno** · April 17th 2012, 12:43 PM

i have a totally different suggestion, consider instead the expression:

$\prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = \left|\prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x)\right|$

note that for each r, we have $\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right)$ is a root of $1 - x^n$ ,

and we have every root except the root 1 represented in our product of factors, so:

$\prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x) = \frac{1-x^n}{1-x} = 1 + x + \dots + x^{n-1}$ ,

hence $\prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = |1 + x + \dots + x^{n-1}|$ .

now just let x = 1, and voila!

(edit: by the way, the formula still makes sense for n = 2, and it is true for n = 2...it's just not very interesting, there it boils down to |(-1) - 1| = 2, which is fairly obvious).

**Capricious** · April 17th 2012, 01:02 PM

Originally Posted by Deveno

i have a totally different suggestion, consider instead the expression:

$\prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = \left|\prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x)\right|$

note that for each r, we have $\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right)$ is a root of $1 - x^n$ ,

and we have every root except the root 1 represented in our product of factors, so:

$\prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x) = \frac{1-x^n}{1-x} = 1 + x + \dots + x^{n-1}$ ,

hence $\prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = |1 + x + \dots + x^{n-1}|$ .

now just let x = 1, and voila!

(edit: by the way, the formula still makes sense for n = 2, and it is true for n = 2...it's just not very interesting, there it boils down to |(-1) - 1| = 2, which is fairly obvious).

It's very late here and I'm not at all capable of thinking so I just have to ask you.. does this prove that the expression on the left will equal n?

**Deveno** · April 17th 2012, 01:40 PM

that, and more....the trick is to realize that the expresssions involving cosine and sine really don't have anything to do with trig, but are just one way of expressing n-th roots of unity in the complex numbers (via de moivre's formula). so all i have done is factor the polynomial $1-x^n$ over the complex numbers (which we can do, since any real polynomial factors completely over the complex numbers).

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