# What is a closed form of this limit? (product of areas in circle with parabolas)

I am looking for a closed form of

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(\left(6+\frac{4n}{\pi}\left(\sin{\frac{\pi}{2n}}\right)\left(\cos{\frac{(2k-1)\pi}{2n}}\right)\right)^2-\left(\frac{32n}{\pi}\left(\sin{\frac{\pi}{4n}}\right)\left(\cos{\frac{(2k-1)\pi}{4n}}\right)\right)^2\right)$$

The limit seems to exist. Calling the product $$P_n$$, we have

$$P_1\approx10.062$$
$$P_{10}\approx18.409$$
$$P_{100}\approx19.788$$
$$P_{200}\approx19.869$$
$$P_{300}\approx19.896$$
$$P_{400}\approx19.909$$
$$P_{500}\approx19.917$$
$$P_{570}\approx19.919$$
(Then desmos stops working.)

Context

I came up with the following question. $$4n$$ $$(n\in\mathbb{Z}^+)$$ points are uniformly distributed around a circle. $$2n$$ parabolas are drawn in the manner shown below with example $$n=3$$. The parabolas share a vertex, that is one of the designated points, and a line of symmetry. Each parabola passes through two more of the designated points on the circle. One of the parabolas is degenerate and looks like a line segment.

Assume that, for every value of $$n$$, the average (arithmetic mean) area of the $$4n$$ regions is exactly $$6$$.

Find $$\lim\limits_{n\to\infty}(\text{product of areas})$$.

I have worked out that the answer is $$L^2$$ where $$L$$ is the limit shown above. So I am looking for a closed form of $$L$$.

If the average area is $$5.999$$ the product approaches $$0$$; if the average area is $$6.001$$ the product diverges. I have tested my expression for $$P_{n}$$ with some small values of $$n$$, and I think it is correct.

This question was inspired by other questions about the product of areas in a circle, such as this, this and this. I have tried to apply the techniques from the answers to those questions, to no avail.

For the limits in those other questions, the closed forms (if not $$1$$) involve $$\text{cosh}$$, so I would not be surprised if the closed form of $$L$$ in this question also involves cosh. Maybe $$L=2\cosh{\left(\pi e^{\frac{1}{\pi}-\frac{1}{e}}\right)}\approx19.929$$ ?

UPDATE: From @Claude Leibovici's answer, it seems that $$\left(\dfrac{P_{4n}}{P_{2n}}-1\right)\to\dfrac{1}{2}\left(\dfrac{P_{2n}}{P_{n}}-1\right)$$ as $$n\to\infty$$. Then based on his value of $$P_{60000}$$, I project that $$L\approx 19.950955$$.

UPDATE2: I believe the number of points around the circle does not have to be a multiple of $$4$$. Any positive integer will do. If the average area of the regions is $$6$$, then $$\lim\limits_{n\to\infty}(\text{product of areas})=L^2$$.

• It seems that we agree. Look at my edit Commented Sep 22, 2022 at 3:49

$$\left(\begin{array}{cc} n & P_n \\ 1000 & 19.93455668774900154012061 \\ 2000 & 19.94275333857622878761460 \\ 3000 & 19.94548671811436159136837 \\ 4000 & 19.94685362596789631782790 \\ 5000 & 19.94767384047956738149078 \\ 6000 & 19.94822067924005947638804 \\ 7000 & 19.94861129260183551122982 \\ 8000 & 19.94890426041492105340228 \\ 9000 & 19.94913212888703196970719 \\ 10000 & 19.94931442657381322499540 \\ 20000 & 19.95013479816578198778725 \\ 30000 & 19.95040826700053826290505 \\ 40000 & 19.95054500360003303052213 \\ 50000 & 19.95062704625801974143739 \\ 60000 & 19.95068174165430121991535 \\ \end{array}\right)$$

Slower would be difficult to find !

Edit

Using the last six values of the above table and the simplistic $$P_n=a+\frac b n$$ a curve fit predicts an asymptotic value of $$19.9509552$$ which is not recognized by the $$ISC$$.

• Thanks for the numbers. So my guess that $L=2\cosh{\left(\pi e^{\frac{1}{\pi}-\frac{1}{e}}\right)}$ was wrong. Out of curiosity, what did you use to find these?
– Dan
Commented Sep 21, 2022 at 13:42
• @Dan. Let us see how good you are with riddles : the name starts with M, ends with a and there is an h somewhere. Then ? Commented Sep 21, 2022 at 13:52
• Mitochondria? Or Mathematica? I guess you used both :)
– Dan
Commented Sep 21, 2022 at 14:44

$$L=2\cosh{\left(\pi\sqrt{\dfrac{\phi}{\sqrt5}}\right)}+2\cosh{\left(\pi\sqrt{\dfrac{1}{\phi\sqrt5}}\right)}=19.9505522212724487...$$

where $$\phi=\dfrac{1+\sqrt5}{2}$$, the golden ratio.

We have

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(\left(6+\dfrac{4n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\left(\cos{\dfrac{(2k-1)\pi}{2n}}\right)\right)^2-\left(\dfrac{32n}{\pi}\left(\sin{\dfrac{\pi}{4n}}\right)\left(\cos{\dfrac{(2k-1)\pi}{4n}}\right)\right)^2\right)$$

Let $$\theta=\dfrac{(2k-1)\pi}{2n}$$, and replace all $$n$$ with $$\frac{n}{2}$$. (If $$n$$ is odd, replace all $$n$$ with $$\frac{n-1}{2}$$.)

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2} \left(\left(6+\dfrac{2n}{\pi}\left(\sin{\dfrac{\pi}{n}}\right)\left(2\cos^2{\theta}-1\right)\right)^2-\left(\dfrac{16n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\cos{\theta}\right)^2\right)$$

Factorise the difference of squares. I will use $$A$$ and $$B$$ for shorthand.

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)(A-B)$$

$$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)\prod\limits_{k=1}^{n/2}(A-B)$$

$$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)\prod\limits_{k=n/2+1}^{n}(A+B)$$

$$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}(A+B)$$

$$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}\left(6+\dfrac{2n}{\pi}\left(\sin{\dfrac{\pi}{n}}\right)\left(2\cos^2{\theta}-1\right)+\dfrac{16n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\cos{\theta} \right)$$

(In fact, we could have arrived at this last expression for $$L$$ more directly by using another method to express the areas of the regions in the circle, i.e. on each side of the common line of symmetry, let $$f(k)=$$ area enclosed by the $$k$$th parabola from the bottom and its outer arc, then the areas enclosed by neighboring parabolas and the circle are $$f(k)-f(k-1)$$.)

Factorise the quadratic in $$\cos{\theta}$$, and use the Maclaurin series for sine. After a lot of simplification, we get

$$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}\left(2\cos\theta-2-\dfrac{\dfrac{5+2\sqrt5}{20}\pi^2}{n^2}+O\left(\dfrac{1}{n^4}\right)\right)\left(2\cos\theta-2-\dfrac{\dfrac{5-2\sqrt5}{20}\pi^2}{n^2}+O\left(\dfrac{1}{n^4}\right)\right)$$

Using the fact that $$\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(2\cos\theta-2-\dfrac{\alpha^2}{n^2} \right)=2\cosh\alpha$$ for all $$\alpha\in\mathbb{R}$$ (which can be proved using the technique shown in @metamorphy's answer here),

$$L=4\cosh{\left(\pi\sqrt{\dfrac{5+2\sqrt5}{20}}\right)}\cosh{\left(\pi\sqrt{\dfrac{5-2\sqrt5}{20}}\right)}$$

Using the identity for $$\cosh{A}+\cosh{B}$$,

$$L=2\cosh{\left(\pi\sqrt{\dfrac{\phi}{\sqrt5}}\right)}+2\cosh{\left(\pi\sqrt{\dfrac{1}{\phi\sqrt5}}\right)}$$