# How can I show that a sequence of regular polygons with $n$ sides becomes more and more like a circle as $n \to \infty$?

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.

• 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. – mrf Aug 28 '13 at 8:35
• 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. – walcher Aug 28 '13 at 8:38
• 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? – J.-E. Pin Aug 28 '13 at 9:27
• @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. – dfeuer Aug 28 '13 at 9:31
• 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 – leonbloy Aug 28 '13 at 11:37

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.

• 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 – leonbloy Aug 28 '13 at 11:31
• Pl check my solution again, I've edited it. – Sumedh Aug 28 '13 at 13:39
• "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. – leonbloy Aug 28 '13 at 13:55
• 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. – Sumedh Aug 28 '13 at 14:18

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.

• @NefinAbraham: I could, but so could you! The GeoGebra program (which is free) is great for this sort of thing. – dfeuer Oct 3 '13 at 15:45

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 $\infty$.

• It might be worth to warn that this does NOT guarantee that the perimeter converges to that of the circle. – leonbloy Aug 29 '13 at 1:59

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".

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".

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

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