Intuition why a combined inscribed and circumscribed polygon converge faster to $\pi$?

Question

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:

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.

Answer 1

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 $\sim n^{-4}$.