Another interesting property of $y=2^{n-1}\prod_{k=0}^n \left(x-\cos{\frac{k\pi}{n}}\right)$: product of arc lengths converges, but to what?

Question

Here is the curve $y=2^{n-1}\prod\limits_{k=0}^n \left(x-\cos{\frac{k\pi}{n}}\right)$, shown with example $n=8$, together with the unit circle centred at the origin.

Call the arc lengths between neighboring roots $l_1, l_2, l_3, ..., l_n$.

What is the exact value of $L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k$ ?

Desmos suggests that $L$ exists and is approximately $2.94$. Maybe $\frac{8}{e}$ ?

Context

I have studied this curve, and found that it has several interesting properties.

• The curve is tangent to the unit circle at $n$ points, which are uniformly spaced around the circle.

• The magnitude of the gradient at each root inside the circle is $n$; the magnitude of the gradient at $x=\pm1$ is $2n$.

• The total area of the regions enclosed by the curve and the x-axis is $1$.

• As $n\to\infty$, the volume of revolution of those regions about the x-axis approaches $\frac{1}{2}$ of the volume of the unit sphere, and the volume of revolution of those regions about the y-axis approaches $\frac{1}{\pi}$ of the volume of the unit sphere.

• As $n\to\infty$, if the curve is magnified so that the average area of those regions is always $2$, then the product of those areas approaches $4\cosh^2{\left(\frac{\sqrt{\pi^2-8}}{2}\right)}\approx6.18$, as shown here.

I recently discovered that the product of arc lengths between neighboring roots seems to converge to a positive number as $n\to\infty$. Hence, my question.

(If you know any other interesting properties of this curve, feel free to add them in the comments.)

My attempt

The part of the curve inside the circle can be expressed as $y=-\sqrt{1-x^2}\sin{(n\arccos{x})}$. So

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \int_{\cos{\frac{k\pi}{n}}}^{\cos{\frac{(k-1)\pi}{n}}}\sqrt{1+\left(n\cos{(n\arccos{x})+\frac{x\sin{(n\arccos{x})}}{\sqrt{1-x^2}}}\right)^2}dx$$

I do not know how to evaluate this limit. I tried taking the log of the product, without success. I tried to approximate each integral as areas of triangles (hoping that that approximation would become equality with the limit) and a rectangle at the bottom, multiplying each triangle's area by $\frac{4}{\pi}$ (which is the ratio of areas under sine or cosine to the area of an inscribed triangle), but that resulted in a different limit.

EDIT

Further numerical analysis strongly suggests that $L=\frac{8}{e}$. I noticed that when $n$ doubles, the ratio of the two products is a certain number (which is close to $1$), and when $n$ is doubled again, the ratio's distance to $1$ is approximately halved. So then I projected that the product indeed approaches $\frac{8}{e}$. (I don't have Mathematica; anyone who has it is welcome to confirm this.)

I have simplified the expression of $L$. Letting $x=\cos{\frac{u}{n}}$, and ignoring the $1$ in the $\sqrt{1+(...)^2}$ (I think this is OK since $n\to\infty$), we get

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \int_{k\pi}^{(k-1)\pi}\sqrt{\left(n\cos{u}+(\sin{u})\cot{\frac{u}{n}}\right)^2}\left(-\frac{1}{n}\sin{\frac{u}{n}}\right)du$$

$$\space{}=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \int_{(k-1)\pi}^{k\pi}\left|(\cos{u})\sin{\frac{u}{n}}+\frac{1}{n}(\sin{u})\cos{\frac{u}{n}}\right|du$$

So why is this equal to $\frac{8}{e}$ ?

Answer 1

This question is closely related to computing the discriminant of Chebyshev polynomials of the second kind, found in literature (e.g., see G. Szegő Orthogonal polynomials, theorem $6.71$).

I'm going an elementary way here. Denote $x_{n,k}=\cos(k\pi/n)$ for $0\leqslant k\leqslant n$, so that $$S_n(x)=2^{n-1}\prod_{k=0}^n(x-x_{n,k})$$ is our polynomial. Then "ignoring the $1$ in the $\sqrt{1+(\dots)^2}$" amounts to saying that $l_k:=l_{n,k}$ is approximately twice the supremum of $\big|S_n(x)\big|$ on $x_{n,k-1}<x<x_k$; more precisely, we have $$L=\lim_{n\to\infty}\prod_{k=1}^n l_{n,k}=\lim_{n\to\infty}\prod_{k=1}^n\color{LightGray}{\Big[}2\sup_{x_{n,k-1}<x<x_{n,k}}\big|S_n(x)\big|\color{LightGray}{\Big]}=\lim_{n\to\infty}2^n\prod_{k=1}^n \big|S_n(x_{n,k}')\big|,$$ where $x_{n,k}'$ (for $1\leqslant k\leqslant n$) are the roots of $S_n'(x)$.

From properties of resultants, we know that if $f$ is a polynomial of degree $d$, with leading coefficient $a$, roots $x_{d,k}$ for $1\leqslant k\leqslant d$, and roots of its derivative $x'_{d,k}$ for $0<k<d$, then $$\prod_{k=1}^{d-1}f(x_{d,k}')=\frac1{d^d a}\prod_{k=1}^d f'(x_{d,k}).$$

In our case $d=n+1$ and $a=2^{n-1}$, so that $$L=\lim_{n\to\infty}\frac{2L_n}{(n+1)^{n+1}},\qquad L_n=\prod_{k=0}^n\big|S_n'(x_{n,k})\big|.$$

From $S_n(\cos t)=-\sin t\sin nt$ we get $S_n'(\cos t)=n\cos nt+\cot t\sin nt$, hence $$S_n'(1)=2n,\quad S_n'(-1)=2n(-1)^n,\quad S_n'(x_{n,k})=n(-1)^k\quad(0<k<n)$$ and $L_n=4n^{n+1}$, thus $L=\lim\limits_{n\to\infty}8/(1+1/n)^{n+1}=8/e$ as claimed.

Answer 2

A simple parameterization of the curve is given by

$$x = \cos \frac tn, \quad y = -{\sin t \sin \frac tn}, \quad 0 < t < nπ.$$

If we could approximate the arc length of the $k$th lobe $(k - 1)π < t < kπ$ by twice its height at $t = \frac{(k - 1/2)π}{n}$ (the point of tangency with the circle), we’d get

$$\prod_{k = 1}^n l_k ≈ \prod_{k = 1}^n 2\sin \frac{(k - \frac12)π}{n}.$$

That product equals $2$. (Proof: observe that $z^n + 1 = \prod_{k = 1}^n (z - e^{2i(k - 1/2)π/n})$, because these monic polynomials have the same roots; substitute $z = 1$ and take absolute values.)

Unfortunately, as I think you’ve already observed, this approximation isn’t good enough. For $k$ close to $1$ or $n$, the lobe shoots significantly past the tangency point before turning back toward the $x$ axis. However, we can try to correct the approximation as follows.

If we translate the $k$th lobe by $(-1, 0)$ and scale it by a factor of $n$, we get

$$x = n\left(\cos \frac tn - 1\right), \quad y = -n \sin t \sin \frac tn.$$

As $n → ∞$, this approaches

$$x = 0, \quad y = -t \sin t,$$

with the arc length converging as $O(n^{-2})$. This limiting arc length is

$$\max_{(k - 1)π < t < kπ} 2\lvert t \sin t\rvert = 2\lvert r_k\sin r_k\rvert,$$

where $r_k$ is the $k$th positive root of $r \cos r + \sin r = 0$. But we’ve underestimated it as

$$2n \sin \frac{\left(k - \frac12\right)π}{n} \to 2\left(k - \frac12\right)π,$$

also converging as $O{(n^{-2})}$. So we’ll need to correct the approximation by a factor of

$$\frac{\lvert r_k\sin r_k\rvert}{\left(k - \frac12\right)π}.$$

The corrections in lobes $k$ and $n + 1 - k$ are the same, so the overall correction tends to

$$L = 2\prod_{k = 1}^\infty \left(\frac{r_k \sin r_k}{\left(k - \frac12\right)π}\right)^2,$$

with overall error $O(n^{-1}) → 0$. We can compute $L ≈ 2.94303552937$, whose proximity to $\frac 8e$ is extremely striking, although it’s hard to confirm since $r_k$ don’t seem to have a closed form.