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If we construct regular polygons with larger and larger numbers of sides, they will look more and more like circles. That is intuitively true. I hope you will help me to express and prove it mathematically.

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    $\begingroup$ There is no such thing as a polygon with "largest number of sides". There is also no such thing as a polygon with an infinite number of sides. $\endgroup$ – mrf Aug 28 '13 at 8:35
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    $\begingroup$ First of all, you can't "construct" a regular polygon with an infinite number of sides. By definition a polygon only has a finite number of sides. The statement is mathematically imprecise and certainly not "logically true". You can show by inscribing a regular $n$-gon in the unit circle that letting $n$ go to infinity you can arbitrarily approximate a circle by a regular $n$-gon, but that is somewhat different from your claim. $\endgroup$ – walcher Aug 28 '13 at 8:38
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    $\begingroup$ I find all these comments and the answer by dfeur by far too sarcastic. It is clear that the real meaning of this question is the following. Intuitively, the regular $n$-gon tends to a circle. What would be the appropriate mathematical setting to get this intuition formally correct? $\endgroup$ – J.-E. Pin Aug 28 '13 at 9:27
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    $\begingroup$ @J.-E.Pin: You make a very good point. I'm tired and in a foul mood, so I probably should keep my mouth shut. However, I probably couldn't resist the logic bomb bit even in better circumstances. $\endgroup$ – dfeuer Aug 28 '13 at 9:31
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    $\begingroup$ Be aware that "to become more like a circle" is a weak statement; eg, that property is not enough, in itself, to compute a perimeter. See this (which also "tends to a circle") math.stackexchange.com/questions/12906/is-value-of-pi-4 $\endgroup$ – leonbloy Aug 28 '13 at 11:37
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For Polygons, as number of sides increases,the length of each side is reducing

so at ∞, the length will be equal to a point.

Each Side can be assumed to be shortened to the vertex(ie a point)

Now, the polygon becomes a locus of points which are actually the vertices of the polygon.

However, each vertex is equidistant from the centre of polygon, so the resulting polygon

becomes a locus of points, equidistant from a common centre point, which is a circle.

After Edit: Now, what remains, is to prove that at n->∞, length of side tends to zero, ie a point.

It can be proved by contradiction. 1) Suppose at n->∞, length of each side is not equal to a point, but is L.for n-> ∞, total length = n*L = ∞.

Now, total length is infinite,however, polygon is a closed figure, so its total length is obviously finite.

Thus this contradicts that length is infinite, which in turn contradicts our original supposition.

Therefore, at n-> ∞, length of each side tends to 0, ie each side becomes a point.

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  • $\begingroup$ This at best (disregarding some... inaccuracies, as "at $\infty$, length will be equal to a point") shows that the polygon curve tends to the circle, but this does not prove that its length tends to that of the circle. See this math.stackexchange.com/questions/12906/is-value-of-pi-4 and math.stackexchange.com/a/229269/312 $\endgroup$ – leonbloy Aug 28 '13 at 11:31
  • $\begingroup$ Pl check my solution again, I've edited it. $\endgroup$ – Sumedh Aug 28 '13 at 13:39
  • $\begingroup$ "what remains, is to prove that at $n\to\infty$, length of side tends to zero, ie a point" "What remains" for proving... what? You don't state that. In the linked examples above the length of the small segments all tends to zero. And your statement "polygon is a closed figure, so its total length is obviously finite" is also wrong. ("obviously" ? are you aware of fractals? a closed figure can have infinite perimeter. $\endgroup$ – leonbloy Aug 28 '13 at 13:55
  • $\begingroup$ What remains means that it was left to prove in my first attempt at the solution. However,"obviously finite" was a wrong thing on my part.I was thinking about Koch Snowflake when I thought about the solution.I had a misconception that Koch Snowflake suggests that inspite of having infinite perimeter,length is finite. Anyway, I was wrong, & thanks for clearing that. $\endgroup$ – Sumedh Aug 28 '13 at 14:18
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Here's one serious approach: Let $f_n\colon [0,2\pi]\to \Bbb R_+$ be the function whose graph, in polar coordinates, is the regular $n$-gon centered at the origin with a vertex at $(1,0)$. Then $(f_n)$ converges uniformly to a constant function mapping any angle to $1$, whose graph is a circle.

We can also look at the limit of the area, a la Archimedes, and the perimeter.

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  • $\begingroup$ can you please construct and upload the graph you mentioned in your answer $\endgroup$ – NefinAbraham Oct 3 '13 at 15:30
  • $\begingroup$ @NefinAbraham: I could, but so could you! The GeoGebra program (which is free) is great for this sort of thing. $\endgroup$ – dfeuer Oct 3 '13 at 15:45
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The regular polygon approaches the circle in the following sense:

  • All vertices of the polygon are on the circle.

  • The maximal distance of the polygon to the circle is given by $2R\sin^2(\frac{\pi}{2n})$, which goes to zero as $n$ goes to $\infty$.

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    $\begingroup$ It might be worth to warn that this does NOT guarantee that the perimeter converges to that of the circle. $\endgroup$ – leonbloy Aug 29 '13 at 1:59
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It seems worth emphasizing that "look more and more like circles" admits numerous interpretations. The answers and comments currently visible say that the polygons converge to the circle in several ways: They eventually lie within arbitrarily narrow annuli just within the circle. Their areas converge to the circle's area. Their perimeters converge to the circle's circumference. One could add more; for example, for almost all rays $R$ emanating from the origin, the direction in which the $n$-gon crosses $R$ converges to the direction in which the circle crosses $R$ (namely, perpendicular to $R$). The "almost" here refers to the unpleasantness that a few (countably many) $R$'s pass through a vertex of one of the polygons, so the direction of crossing is undefined there, but even these $R$'s are OK if one uses the average of the directions just to the left and just to the right of $R$. I suspect there are lots of other convergence properties that one could state and prove in this situation. An interesting but non-mathematical question would be to determine which of the many notions of convergence cause people to say that the $n$-gons for large $n$ "look like circles".

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There is nothing wrong with this question. All you have to do to formulate your assumption rigorously is to use the Hausdorff distance. Then you can show, that for a sequence of regular Polygons $P_n$ of the inner radius $r$ and and the circle $S^1_r$ of radius $r$ the Hausdorff distance $d_H (P_n , S^1_r ) $ tends to $0$. More information about Regular polygons and Hausdorff distance can be found in Richard Gardner's book called "Geometric Tomography".

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The half-angle formula is sin(t/2)

S = 2*sin(t/2) Arc length = 2^54*sin(t/2^54) = pi/2 at 90 degree S = 2*sin(90 degree/2) = 2^(1/2) approximate 1.4141 Arc length = pi/2 approximate 1.5707 Sn = 2^54*sin(90 degree/2^54) Sn = 2^53 sides circle total sides = 4*(2^53) = 2^55 sides Tanks Giuseppe Stagno

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I have proven that the circle is a binary polygon and has 2^55 SIDES

The formulas are :

      I1 is the First Increment = (2+2x) ^ (1/2).
     arc lenght =[(2-I53) ^(1/2)]*2^53=pi/2 at 90 degree.
    t= degree

   Arc length=2^54*sin(t/2^54)=pi/2 at 90 degree.

I Giuseppe Stagno the author: my email gstagno31@gmail.com thanks

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