I was playing around a little with approximating $\pi$ by calculating the perimeter of a regular polygon both inscribing and circumscribing a circle.

When using trigonometry then it can be shown that $\pi$ can be approximated using the perimeters of these polygons with $n$ sides as follows

$$ \pi_i(n) = n\,\sin\left(\frac{180^\circ}{n}\right), $$

$$ \pi_c(n) = n\,\tan\left(\frac{180^\circ}{n}\right), $$

with $\pi_i(n)$ and $\pi_c(n)$ the approximation of $\pi$ from an inscribing and circumscribing polygon respectively.

But $\pi_i(n)$ and $\pi_c(n)$ are always a lower and upper bound respectively for $\pi$. Therefore I thought that taking the average should give a better result, which indeed reduced the error. However the rate at which the error gets smaller as $n$ increases was the same as the previous two. When I started using different weights I found the following expression gives the best result

$$ \pi_w(n) = \frac23\,\pi_i(n) + \frac13\,\pi_c(n). $$

Graphing the errors of these approximations gives:

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The rate at which the error of $\pi_i(n)$ and $\pi_c(n)$ goes down seems to be proportional to $n^{-2}$, while the rate at which the error of $\pi_w(n)$ goes down seems to be proportional to $n^{-4}$.

These converges rates can be explained by looking at the Taylor series of each expression. However I am curious if this could also be explained with pure geometry.

  • $\begingroup$ Related question: How can one derive the third-order terms of the Taylor series of $\sin$ and $\tan$ with pure geometry? $\endgroup$
    – user856
    Sep 8, 2017 at 16:39
  • 1
    $\begingroup$ Curiously, if one approximates $\pi$ using the area of inscribed and circumscribed polygons, the best approximation weighs the circumscribed polygon twice as much as the inscribed polygon $\endgroup$ Sep 8, 2017 at 19:02
  • $\begingroup$ Unfortunately it is not possible to get the Taylor series without using calculus. Before Newton and Leibniz, these series were discovered by Indian mathematician Madhava who used integration (as limit of a Riemann sum) to obtain these series. So all of this is possible without differentiation or Fundamental Theorem of Calculus but still the arguments are far from being purely geometric. $\endgroup$
    – Paramanand Singh
    Sep 9, 2017 at 1:22
  • $\begingroup$ @Julian Rosen Your result is obtained by using the method given in my answer below. The function to be minimized by the weight $\alpha$ is$$ \tfrac{1-\alpha}{2}(1-\cos\theta)(\sin\theta+\tan\theta)-(1-\cos\theta)\sin\theta+\tfrac{1}{6}\sin^3\theta$$ which yelds $\alpha=1/3$ and an error $\sim n^{-4}$. $\endgroup$
    – Paul Enta
    Nov 18, 2017 at 9:12

1 Answer 1


Parts of the two polygons and notations

By symmetry, we can discuss the approximation of $\pi$ on an angular sector $\theta=\tfrac{2\pi}{2n}$ of the inscribing and circumscribing polygons (see the figure).

Denoting $H_i$ and $H_c$ the half-side of these polygons, \begin{align} H_i&=R\sin\theta\\ H_c&=R\tan\theta \end{align} Their length are related to the lower and upper approximations of $\pi$ by \begin{align} \pi_i(n)&=2n\frac{H_i}{R}\\ \pi_c(n)&=2n\frac{H_c}{R} \end{align} They approximate the length arc of circle $IC$, noted $S$, which would give the exact value of $\pi$. The question is then to find the value of the weight $\alpha$ which minimize the difference between the lengths of the arc and the weighted approximation: $\varepsilon=\alpha H_i+(1-\alpha)H_c-S$ for $n\to\infty$.

For small angles $\theta$, the arc $IC$ can be approximated by a parabolic shape, with a radius of curvature $R$, its length being \begin{align} S&=\int_0^{H_I}\sqrt{1+\frac{x^2}{R^2}}\,dx\\ &\sim H_I+\frac{1}{2}\int_0^{H_I}\frac{x^2}{R^2}\,dx\\ &\sim R\sin\theta\left( 1+\frac{1}{6}\sin^2\theta \right) \end{align} The error is then \begin{equation} \varepsilon\sim R\left( \alpha\sin\theta+\left( 1-\alpha \right)\tan\theta- \sin\theta\left( 1+\frac{1}{6}\sin^2\theta \right)\right) \end{equation} For $\theta\to 0$ (or $n\to\infty$), \begin{equation} \varepsilon\sim \left( \frac{1}{3}-\frac{\alpha}{2} \right)\theta^3+\left( \frac{5}{24}-\frac{\alpha}{8} \right)\theta^5+O(\theta^7) \end{equation} Taking $\alpha=\frac{2}{3}$ reduces the error: $\varepsilon\sim \frac{\theta^5}{8}+O(\theta^7)$. \begin{align} \frac{2}{3}\pi_i(n)+\frac{1}{3}\pi_c(n)-\pi\sim 2n\frac{1}{8}\left( \frac{\pi}{n} \right)^5=\frac{\pi^5}{4}n^{-4} \end{align} A more exact calculation, taking into account the exact equation of the circle ($H=R\left( 1-\sqrt{1-\frac{x^2}{R^2}} \right)$) shows that the error due to the parabolic approximation is also $\sim \theta^5$. The overall error for the $\pi$ approximation when $\alpha=2/3$ is thus $\sim n^{-4}$.

  • $\begingroup$ Very nice analysis. $\endgroup$ Nov 17, 2017 at 23:23

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