Finding Circle Circumference from Inscribed N-Sided Polygen

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SUMMARY

The perimeter P of a regular polygon with n sides inscribed in a circle of radius r is defined by the equation P = 2nr sin(180° / n). As n approaches infinity, the limit of this perimeter can be expressed using L'Hôpital's rule to derive the circumference of a circle, leading to the conclusion that C = 2πr. The transformation involves recognizing the limit of sin(π/n)/(π/n) as n approaches infinity, which equals 1, thus confirming the relationship between polygonal approximation and circular circumference.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's rule
  • Knowledge of trigonometric functions and their limits
  • Basic concepts of polygon geometry
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  • Study the application of L'Hôpital's rule in calculus
  • Explore the properties of trigonometric limits, particularly sin(x)/x
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  • Learn about the derivation of π and its significance in mathematics
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Mathematicians, calculus students, geometry enthusiasts, and anyone interested in the relationship between polygons and circles.

logan3
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The perimeter P of a regular polygon of n sides inscribed in a circle of radius r is given by P = 2nr sin (180^o / n).

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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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.
 
P = 2nr \sin (180^o / n) = \frac {2nr} {\csc (180^o / n)} = \frac {2nr} {\csc (180^o / n)}
\displaystyle \lim_{n\rightarrow \infty} {\frac {2nr} {\csc (180^o / n)}} = \frac {\infty} {\infty}

L'Hôpital's rule:
\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)}}

Sorry, this is about as far as I got right now. Also, I'm not sure where the \pi is going to come in for the C = 2\pi r, though I assume it will have to come from the trig functions somehow.
 
\lim_{n->\infty}\frac {sin(\frac{\pi}{n})}{\frac{\pi}{n}}=1.
Therefore \lim_{n->\infty}2nrsin(\frac{\pi}{n})=2\pi r.
 
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Where's the denominator of pi / n come from in the second expression?
 
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
 
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micromass said:
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
 
logan3 said:
Where's the denominator of pi / n come from in the second expression?
P=2nrsin(π/n)=2πr(sin(π/n)/(π/n))
 
  • #10
I don't understand, sorry.

But it seems like my thoughts about this relationship were in the right direction.
 
  • #11
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
And, of course, ##\lim_{x\to 0}\frac{\sin x}{x}=1## only if ##x## is measured in radians.
 

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