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

(Then desmos stops working.)


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.

enter image description here

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 $\lim\limits_{n\to\infty}(\text{product of areas})=L^2$.

  • 1
    $\begingroup$ It seems that we agree. Look at my edit $\endgroup$ Commented Sep 22, 2022 at 3:49

2 Answers 2


Continuing your calculations

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

Slower would be difficult to find !


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 $ISC$.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Dan
    Commented Sep 21, 2022 at 13:42
  • 1
    $\begingroup$ @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 ? $\endgroup$ Commented Sep 21, 2022 at 13:52
  • 3
    $\begingroup$ Mitochondria? Or Mathematica? I guess you used both :) $\endgroup$
    – Dan
    Commented Sep 21, 2022 at 14:44

The answer is


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.





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

(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


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),


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



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .