You are such a nerd.

a geek, cait. i'm a geek. i used to be very secretive about it and everything, but i'm quite out of the closet now.

we're here, we're geeky, get used to it. there would be some big demonstration about it on the mall in washington except all the geeks are busy with math problems.

More power to ya! I might have said "geek," but feared you might be offended by its original meaning. (I admire you as a word-geek also.)

Hot holy damn, that's awesome. I love all those little coincidences in mathematics. Freakin' crazy.

So what was the proof on this, anyways? I worked through the trig to get it as a product of sines but had no idea how to solve that product. Thanks.

hint for a proof: complex roots of unity.

Setting up a problem with so much as this example, it's not too surprising that you get a result that's so closely linked to the setting up of the problem.

Setting up a problem with so much symmetry as this example, it's not too surprising that you get a result that's so closely linked to the setting up of the problem. (Heh, I missed out the important word)

huh?

Not to sound dumb, but when will I use Commutative, Associative, or Identity properties in real life?

put the unit circle on the complex plane with p_0=1. then p_0, p_1,...,p_(n-1) are roots of x^n -1=0. for the moment we will calculate the distance between a point x and all those point, not from p_0. this is abs((x-p_1)(x-p_2)(x-p_3)...(x-p_(n-1)). by fundamental theorem of algebra, this factors to abs((x^n -1)/(x-p_0)) or (x^n -1)/(x-1)). when divided out this becomes abs(x^(n-1)+x^(n-2)+x^(n-3)+...+x+1). now since p_0=1 we can plug in x=1 to find the distance we want. this will be abs(n)=n.
QED

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