This question already has an answer here:

We know as we increase the number of sides in regular polygon, after infinite repetition it will give us a circle. So, is there any way to find a function which approaches to value of $\pi$ when we consider $\lim\limits_{n\to \infty}$?


marked as duplicate by Did, Jyrki Lahtonen, Claude Leibovici, N. F. Taussig, mfl Aug 23 '18 at 8:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ There are a lot of such functions. Do you seek one in general or one that uses your polygon approach? $\endgroup$ – user526015 Aug 23 '18 at 7:59

The area of a regular $n$-gon circumscribed by the circle of radius $r$ is $$A_n = \frac{1}{2}nr^2\sin\left(\frac{2\pi}{n}\right).$$ You can get the proof here. Taking limit, we will get circle when $n\to\infty$. We can use this formula for $\pi$, $$\pi=\frac{\text{Area of circle with radius r}}{\text{radius}^2}=\frac{1}{r^2}\cdot \lim_{n\to\infty}A_n=\lim_{n\to \infty}\frac{n}{2}\sin\left(\frac{2\pi}{n}\right).$$

Using this result you can tend to $\pi$ putting bigger and bigger values of $n$ in $\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$. Here I have plotted $\pi -\frac{x}{2}\sin\left(\frac{2\pi}{x}\right)$, for first few values of $x$ very fast, then slowly approaches $0$, means $\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)\to\pi$ very slowly but, the methods explained in the given link in comments tend $\pi$ much faster than this.


Not the answer you're looking for? Browse other questions tagged or ask your own question.