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Finding Circle Circumference from Inscribed N-Sided Polygen

  1. Feb 9, 2015 #1
    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
     
  2. jcsd
  3. Feb 9, 2015 #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.
     
  4. Feb 9, 2015 #3
    [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.
     
  5. Feb 9, 2015 #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].
     
  6. Feb 9, 2015 #5
    Where's the denominator of pi / n come from in the second expression?
     
  7. Feb 10, 2015 #6

    Svein

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  8. Feb 10, 2015 #7

    micromass

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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
     
  9. Feb 10, 2015 #8

    BiGyElLoWhAt

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    Hehe. :D
     
  10. Feb 10, 2015 #9

    mathman

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    P=2nrsin(π/n)=2πr(sin(π/n)/(π/n))
     
  11. Feb 10, 2015 #10
    I don't understand, sorry.

    But it seems like my thoughts about this relationship were in the right direction.
     
  12. Feb 11, 2015 #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.
     
  13. Feb 19, 2015 #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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