$\lim\limits_{n\to+\infty}\frac{1}{n}\sum_{k=1}^n|\overrightarrow{A_1A_k}|$ for $A_1,..,A_n$ vertices of a regular n sided polygon of radius one

Question

Consider the following problem : Let $A_1,..,A_n$ be the vertices of a regular $n$-sided polygon inscribed in a circle of radius one. Evaluate the limit $$ \lim\limits_{n \to +\infty} \frac{1}{n} \sum_{k=1}^n |\overrightarrow{A_1A_k}|. $$

This can easily be solved by passing to complex numbers indeed we find that this is limit is equal to $$ \lim\limits_{n \to +\infty} \frac{1}{n}\sum_{k=0}^{n-1} |1-e^{\frac{2ik\pi}{n}}| = \lim\limits_{n \to +\infty} \frac{1}{n} \frac{2\cos(\frac{\pi}{2n})}{\sin(\frac{\pi}{2n})} = \boxed{\frac{4}{\pi}}. $$

My question is, is there a geometric interpretation of this result that might make it "obvious" ?

A colleague suggested looking into the cycloid and this seems very promising. Indeed, if I add the vectors $\frac{2\pi}{n}\overrightarrow{A_1A_k}$ one by one starting from the point $(2\pi,0)$ in geogebra it almost perfectly fits the cycloid of radius one and the fits get better with n. Since the cycloid of radius one has length $8$ I obtain the result.

My problem is : I can't explain why the points (after suitable normalization) seem to almost land on the cycloid.

Any ideas ?

I link an image from geogebra where I take n=17.

here is a link to the geogebra file (not cleaned up so I don'y know if it will be helpful) if you want to experiment with different values of $n$ and see the polygon rotate: here

Answer 1

Maybe this should be rather a long comment than an answer.

We can handvawe that the points of the vector sum tend to the cycloid as follows. For each nonegative integer $k\le n-1$ let the vertex $A_{k+1}$ is at $e^{\frac{2ik\pi}{n}}$. Then $$\sum_{j=1}^{k+1} \overrightarrow{A_1A_j}=\sum_{j=0}^k \left(e^{\frac{2ij\pi}{n}}-1\right)=\frac{e^{\frac{2i(k+1)\pi}{n}}-1}{e^{\frac{2i\pi}{n}}-1}-k.$$

Now suppose that $n$ tends to infinity and $\frac{k}{n}$ tends to $\lambda\in [0,1]$. Then

$$\lim 2\pi+\frac{2\pi}{n}\left(\frac{e^{\frac{2i(k+1)\pi}{n}}-1}{e^{\frac{2i\pi}{n}}-1}-k\right)=$$ $$2\pi(1-\lambda)+\lim\frac{2\pi}{n}\cdot \frac{e^{2i\lambda\pi}-1}{e^{\frac{2i\pi}{n}}-1}=$$ $$2\pi(1-\lambda)+\left(e^{2i\lambda\pi}-1\right)\lim\frac{2\pi}{n}\cdot \frac{1}{e^{\frac{2i\pi}{n}}-1}=$$ $$2\pi(1-\lambda)+\left(e^{2i\lambda\pi}-1\right)\frac 1i=$$ $$2\pi(1-\lambda)+\sin 2\lambda\pi+(1-\cos 2\lambda\pi)i=$$ $$2\pi(1-\lambda)-\sin 2(1-\lambda)\pi+(1-\cos 2(1-\lambda)\pi)i.$$

Nevertheless, although I expect that the above arguments are essentially correct, the conclusion should be too weak to imply that the sum $ \frac{1}{n} \sum_{k=1}^n |\overrightarrow{A_1A_k}|$ tends to the length of the cycloid. For instance, we can similarly approximate the diagonal of the unit square by zigzags with sides parallel to the sides of the square, such that each zigzag has length $2$, whereas the diagonal has length $\sqrt 2$.

Answer 2

As you’ve implied this is straightforward with calculus and complex numbers. Here’s another way of going about it. If your circle is $e^{ix}$, then $1-e^{ix}$ is the chord at $x$, then the continuous limit of the sum of the chords up to $x$ is $$\int^x_0 (1-e^{iy})dy=x+i(e^{ix}-1)=x-\sin(x)+i(\cos(x)-1)$$ which is a cycloid. Also, note that since $1-e^{ix}$ is continuous the curve varies smoothly, the integral is well defined, so any sufficiently fine approximation of the integral (i.e. your discrete sum) would approach the length of the curve.

Let’s consider this geometrically. If instead of looking at vectors from a vertex of the polynomial to the other vertices, consider vectors from the center of the polygon to all the vertices. These are $n$ vectors of equal length with constant angle $2\pi/n$ between adjacent ones, so summing them up gives a $n$-sided polygon of side length $1$. The difference between this and the original vectors is that each is shifted by the same vector of length one from the start vertex to the center of the circle. Thus, summing up the vectors from a boundary point consists of a combination of moving around a polygon and moving in a straight line where the speed of movement along both is the same. In the limit, this is the movement along a circle as it moves in a line i.e. a cycloid.