(UPDATED) Why didn't Archimedes further approximate $\pi$ this way (or did he)?

Solution 1:

This is not an answer, just an alternative exposition of your findings as far as I was able to reproduce them.

First, the video describes Archimedes' method for approximating $\pi$ via the following recurrence for the side length of a regular $3\cdot2^n$-gon inscribed in a circle, which I'll call $\ell_n$: $$ \ell_1=1 \text{ ,}\qquad \ell_{n+1} = 2\sqrt{2-\sqrt{4-\ell_n^2}} \tag1 $$ Then we approximate $\pi$ as half the perimeter of that $3\cdot2^n$-gon, that is, $$ h_n = 3\cdot 2^{n-1} \ell_n $$

What you've noticed numerically is that $$ \frac{h_n-h_{n-1}}{h_{n-1}-h_{n-2}} \approx \frac14 \tag2 $$ (Nice observation, by the way. If you do the trigonometry, it turns out that $$ \frac{h_n-h_{n-1}}{h_{n-1}-h_{n-2}} = \frac1{2\cos\theta_n(1+\cos\theta_n)} \text{ ,}\quad\text{ where } \theta_n = \frac{\pi}{3\cdot2^n} $$ which confirms your finding because $\cos\theta\to 1$ as $\theta\to 0$. This also shows that the LHS is always $\ge\frac14$.)

Next, you started approximating $h_n$ using (2) instead of the Archimedean recurrence (1). Actually, you say you have something more precise than $\frac14$ on the RHS of (2), but you don't say what it is, so I'll just proceed as if you used $\frac14$.

Let's call the values obtained by this method $\hat h_n$. Since you started using this method after the $96$-gon, the sequence is given by the rules $$ \hat h_4 = h_4 \text{ ,}\quad \hat h_5 = h_5 \text{ ,}\quad \frac{\hat h_n-\hat h_{n-1}}{\hat h_{n-1}-\hat h_{n-2}} = \frac14 $$ By telescoping product, we get $$ \hat h_n - \hat h_{n-1} = (\hat h_5 - \hat h_4) \prod_{k=6}^n \frac{\hat h_k-\hat h_{k-1}}{\hat h_{k-1}-\hat h_{k-2}} = \frac{\hat h_5 - \hat h_4}{4^{n-5}} $$ and then, by telescoping sum, $$ \hat h_n = \hat h_5 + \sum_{k=6}^n (\hat h_k-\hat h_{k-1}) = \hat h_5 + (\hat h_5-\hat h_4)\sum_{k=6}^n \frac1{4^{k-5}} = \hat h_5 + \frac13 (\hat h_5-\hat h_4) \Big(1-\frac1{4^{n-4}}\Big) $$ Now you can see that as $n\to\infty$, we have $$ \hat h_n \to \hat h_5 + \frac13 (\hat h_5-\hat h_4) = \frac{4h_5-h_4}{3} \approx 3.1415925335 $$ So, this sequence is not actually tending to $\pi$, but to something a bit smaller.

Update: I don't know whether Archimedes knew your method, but I will note that the summation of the infinite geometric series used here was absolutely something he knew — he did that as part of his computation of the area of a parabolic sector.

Update: It occurs to me that another way to use your idea is to compute $h_n$ via (1) as usual, but rather than report that as your estimate, report $(4h_n-h_{n-1})/3$. This is as if you had used (2) starting from $n$ instead of starting from $6$, and it gives a sharper lower bound for $\pi$. (It's a lower bound because, as noted above, in (2) we have $\ge$.) You still have to use (1), unfortunately.