Finding Circle Circumference from Inscribed N-Sided Polygen

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  • #1
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Main Question or Discussion Point

The perimeter P of a regular polygon of n sides inscribed in a circle of radius r is given by [itex]P = 2nr sin (180^o / n)[/itex].

I was curious whether it's possible to approximate the circumference of a circle by taking the limit as n goes to infinity of the above perimeter equation is some way?

Thank-you
 

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  • #2
BiGyElLoWhAt
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I would think it should be possible. As it stands, applying that limit would give you infinite*0.
Is there anyway you could express the same function in terms of other trig functions that would give you inf/inf or 0/0? Then you could potentially apply l'hopital's rule and get something useful.
 
  • #3
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[itex]P = 2nr \sin (180^o / n) = \frac {2nr} {\csc (180^o / n)} = \frac {2nr} {\csc (180^o / n)}[/itex]
[itex]\displaystyle \lim_{n\rightarrow \infty} {\frac {2nr} {\csc (180^o / n)}} = \frac {\infty} {\infty}[/itex]

L'Hôpital's rule:
[itex]\displaystyle \lim_{n\rightarrow \infty} {\frac {f'(n)} {g'(n)}} = \displaystyle \lim_{n\rightarrow \infty} {\frac {2r} {-\csc (180^o / n) \cot (180^o / n)}}[/itex]

Sorry, this is about as far as I got right now. Also, I'm not sure where the [itex]\pi[/itex] is going to come in for the [itex]C = 2\pi r[/itex], though I assume it will have to come from the trig functions somehow.
 
  • #4
mathman
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[itex]\lim_{n->\infty}\frac {sin(\frac{\pi}{n})}{\frac{\pi}{n}}=1[/itex].
Therefore [itex]\lim_{n->\infty}2nrsin(\frac{\pi}{n})=2\pi r[/itex].
 
  • #5
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Where's the denominator of pi / n come from in the second expression?
 
  • #6
Svein
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It works in this case. But you need to be careful with polygonal approximation of lengths, it might not always give the right answers:
troll_mathemathics_pi.jpg
 
  • #8
BiGyElLoWhAt
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It works in this case. But you need to be careful with polygonal approximation of lengths, it might not always give the right answers:
Hehe. :D
 
  • #9
mathman
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Where's the denominator of pi / n come from in the second expression?
P=2nrsin(π/n)=2πr(sin(π/n)/(π/n))
 
  • #10
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I don't understand, sorry.

But it seems like my thoughts about this relationship were in the right direction.
 
  • #11
BiGyElLoWhAt
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Its just algebra that he did to get it into a new form. Multiply by pi/pi, and kick the n into the denominator of the denominator. Now you have sin (x)/x.
 
  • #12
LCKurtz
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And, of course, ##\lim_{x\to 0}\frac{\sin x}{x}=1## only if ##x## is measured in radians.
 

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