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Topic: Chords on the Unit Circle (Read 2386 times) |
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william wu
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Chords on the Unit Circle
« on: Jan 22nd, 2003, 1:20pm » |
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Pick K equidistant points on the unit circle. Choose one of the points and call it P. Draw line segments connecting P to all the other points on the circle. What is the surprising product of the lengths of these line segments? Prove it.
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Icarus
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Re: Chords on the Unit Circle
« Reply #1 on: Jan 30th, 2003, 7:27pm » |
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The product is K
Proof:
Let r be a primative Kth root of unity, so rK - 1 = 0. All the Kth roots of unity are rn for n=0...K-1. On the unit circle in the complex plane, these roots are K equidistant points. Choose 1 for the point P.
The distance from 1 to rn is |rn-1|. So the product is |(1-r)(1-r2)...(1-rK-1)|
now xK-1 = (x-1)(x-r)(x-r2)...(x-rK-1) = (x-1)(xK-1+xK-2+...+x+1)
Divide both sides by x-1, then set x=1 in the remaining equation, and you get
(1-r)(1-r2)...(1-rK-1) = K
QED
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wowbagger
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Re: Chords on the Unit Circle
« Reply #2 on: Jan 31st, 2003, 2:37am » |
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on Jan 22nd, 2003, 1:20pm, william wu wrote:| Pick K equidistant points on the unit circle. Choose one of the points and call it P. |
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If you weren't allowed to choose your points, but to prove the result holds for any K equidistant points (any one of which may be P) - which is quite obviously true -, I would have argued that one should point to choosing the Kth roots of unity (and 1 as P) can be done without loss of generality.
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| « Last Edit: Jan 31st, 2003, 3:29am by wowbagger » |
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Icarus
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Re: Chords on the Unit Circle
« Reply #3 on: Feb 3rd, 2003, 8:19pm » |
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I actually started to include that point (map the unit circle of the complex plane with 1 mapping to P ...), but left it out because I thought it detracted from the main argument, and should be obvious enough to anyone with sufficient math background to snooping around this forum.
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"Pi goes on and on and on ...
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I wonder: Which is larger
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harpanet
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Re: Chords on the Unit Circle
« Reply #4 on: Mar 25th, 2003, 1:42pm » |
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Quote:Divide both sides by x-1, then set x=1 in the remaining equation, and you get...
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Just idly browsing before bed-time and came across this one. Now, please correct me if I am wrong, but if you divide by x-1 and x equals 1 then you are dividing by 0. I was always taught to look out for these when doing algebraic proofs as they can easily catch you out (or 'prove' non-sensical things  )
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Icarus
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Re: Chords on the Unit Circle
« Reply #5 on: Mar 25th, 2003, 5:05pm » |
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That is often a problem, but it does not occur here. Dividing by x-1 proves the equality
(x-r)(x-r2)...(x-rK-1) = (xK-1+xK-2+...+x+1)
for all x except 1. Extending the equality to x=1 is simply a matter of noting that both sides are continuous functions that are defined at 1 as well. Taking the limits as x-->1 shows equality at x = 1.
This is so familiar a fact to those experienced in higher mathematics, that we usually take it for granted, just as one might go from (x-1)2=0 to x=1 without showing any intervening steps. Thus it did not occur to me to explain it at the time.
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"Pi goes on and on and on ...
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harpanet
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Re: Chords on the Unit Circle
« Reply #6 on: Mar 26th, 2003, 7:06am » |
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Thanks for the info Icarus.
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