# Circle from $n$-gon circumference${}=1$

To get a better intuition why: $$\pi$$ is irrational and transcendental I came up with the following analyses. I am a amateur but I give it my best to describe:

Imagine a piece of very flexible thin robe. The robe length is set to 1, while the robe is continuous all point between robe start $$0$$ and $$1$$ exists. Tie the robe start and end together, creating a loop. Now we can create polygons with $$n$$ sides ($$n$$-gon) with total circumference equal to 1.

To construct the $$n$$-gon, we require the side length $$l(n)$$ and the enclosed angle $$\alpha(n)$$ between $$3$$ edge points.

$$l(n)=\frac{1}{n}$$ $$\alpha(n)=\frac{2\pi}{n}$$

The $$n$$-gon's can be constructed with both formulas below. All $$n$$-gon's have an edge point in the origin:

$$x(n)=\frac{1}{n} \cdot \sum_{k=0}^{n} cos \left( \frac{(2k+1) \ \pi}{n} \right)$$ $$y(n)=\frac{1}{n} \cdot \sum_{k=0}^{n} sin \left( \frac{(2k+1) \ \pi}{n} \right)$$

### Analysis.

With both equations the $$n$$-gon's can be drawn [Github]. It is possible to compute the $$n$$-gon's with $$n$$ as a continuous number interpolation the edge points. 3 plots have been made:

• Top: The polygons with discrete $$n$$ (red polygons) and continuous $$n$$.
• Middle: A finite number of continuous $$n$$.
• Bottom: An infinite “integrated like” continuous $$n$$.

The circle diameter can be determined from the $$y$$-axis, this way we can determine $$\pi$$ see: [Youtube].

### Intuition.

The side lengths $$l$$ can be constructed with compass and straight edge for rational numbers only, irrationals take infinate time. The angle $$\alpha$$ can only be determined for constructible polygons [wiki]. Creating a continuous circle edge would require an iterative/statistical process thus, I would say (intuition): we require a number $$\pi$$ to be irrational, transcendental and random/normal to fill in the gaps.

### Question.

• Is there formal math describing the method of creating a circle with polygons with constant circumference?
• How must I interpret the graph with a continuous $$n$$-gon? What type of math describes that?

• That $\pi$ is transcendental is a very deep result. Even that $\pi$ is irrational is far from trivial. In general, irrationality proofs are incredibly difficult. $\pi$ is known a very long time, but even the irrationality proof occured rather late (1760) . Not earlier than 1882 was it proven that $\pi$ is transcendental. – Peter Feb 23 at 11:12

The following method describes how to derive the circle found from $$n$$-gons with circumference $$1$$. The $$n$$-gons can be described as:

$$x(n)=\frac{1}{n} \cdot \sum_{k=0}^{p} \ cos \left( \frac{2k+1}{n}\ \pi \right)$$

$$y(n)=\frac{1}{n} \cdot \sum_{k=0}^{p} \ sin \left( \frac{2k+1}{n} \ \pi \right)$$

As demonstrated in the question $$n$$ can be substituted for non integer continuous values, interpolating the edge points of the $$n$$-gons. Normally the summation will be till $$p=n$$ and must remain an integer.

Though, by taking the integral also $$k$$ can be a continuous number. This will give us the limit values of the $$n$$-gons:

$$x_c(n)=\frac{1}{n} \int cos \left( \frac{2k+1}{n}\ \pi \right)dk= \frac{1}{2 \pi} \cdot sin\left( \frac{2k+1}{n}\ \pi \right)+a$$

$$y_c(n)=\frac{1}{n} \int sin \left( \frac{2k+1}{n}\ \pi \right)dk=- \frac{1}{2 \pi} \cdot cos \left( \frac{2k+1}{n}\ \pi \right)+b$$

This parametrized function gives us a circle with radius: $$R=\frac{1}{2 \pi}$$:

$$R=\frac{1}{2\pi} =\sqrt{x^{2}+y^{2}}$$

The circumference $$C$$ of the circle is:

$$C=2 \pi R=2 \pi \cdot \frac{1}{2\pi}=1$$

We defined the circumference of the $$n$$-gons $$1$$ this matches the circumference of the found circle.

This method is intuitive and I am not sure if it's valid. Especially substituting the integers for continuous values.