We have,

I. Liu Hui (c. 300 AD) $$\pi \approx 3\cdot2^{\color{red}8}\times \underbrace{\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\sqrt{\color{blue}1}}}}}}}}}}}_{\color{red}{10}\text{ square roots}}$$

II. Viete (c. 1590 AD)

$$\pi = \lim_{k\to\infty} 4\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}2}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$

III. (yours truly)

$$\pi \approx 6\cdot2^{\color{red}8}\times \underbrace{\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\sqrt{\color{blue}3}}}}}}}}}}}_{\color{red}{10}\text{ square roots}}$$

For the blue numbers, I tried $\sqrt{5}$, $\sqrt{6}$, and other $\sqrt{n}$, nothing worked, until I remembered $\sqrt{2}$ and $\sqrt{3}$ has a trigonometric context.

Question: Given,


then is it true that,

$$\pi = \lim_{k\to\infty} \color{brown}{\alpha}\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}\beta}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$

So the first three are simply the cases $\alpha = 3,4,6$, though if the general form is true, then one can use other positive integers.

P.S. For example, note that if $\alpha = 5$, then we relate $\pi$ to our old friend the golden ratio $\phi$,

$$\pi = \lim_{k\to\infty} \color{brown}{5}\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}{\phi^2}}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$

  • 1
    $\begingroup$ Though I don't have time to write a full answer right now, I can tell you that if this identity is true, then it can almost certainly be obtained by repeatedly applying the cosine half-angle formula, followed by the sin half-angle formula, and noting that: $$\pi=\lim_{x\rightarrow\infty}x\sin\left(\frac{\pi}{x}\right)$$ $\endgroup$
    – C-RAM
    Nov 21, 2022 at 4:11
  • 1
    $\begingroup$ @C-RAM Feel free to answer it when you have time. We will wait. $\endgroup$ Nov 21, 2022 at 4:42

1 Answer 1


Firstly, we define the recursive sequence \begin{equation} \begin{split} A_{k+1}&=\sqrt{2+A_k}\\ A_0&=\sqrt{\beta}=2\cos\left(\frac{\pi}{\alpha}\right)\\ \end{split} \end{equation} and let $B_k=\sqrt{2-A_k}$. Your claim is that $$\pi=\lim_{k\rightarrow\infty}2^k\alpha B_k$$ To prove this claim, we shall first show by induction that $A_k$ admits the simple closed form $$A_k=2\cos\left(\frac{\pi}{2^k\alpha}\right)$$ By definition, the above equality is satisfied for $k=0$. Suppose that the above formula is true for $k$, then by the cos half-angle formula $2\cos(x/2)=\sqrt{2+2\cos(x)}$, we have that $$A_{k+1}=\sqrt{2+A_k}=\sqrt{2+2\cos\left(\frac{\pi}{2^k\alpha}\right)}=2\cos\left(\frac{\pi}{2^{k+1}\alpha}\right)$$ and therefore, by induction $A_k=2\cos\left(\frac{\pi}{2^k\alpha}\right)$ for all $k\geq 0$. Now, applying the sin half-angle formula $2\sin(x/2)=\sqrt{2-2\cos(x)}$ gives us that $$B_k=\sqrt{2-A_k}=\sqrt{2-2\cos\left(\frac{\pi}{2^k\alpha}\right)}=2\sin\left(\frac{\pi}{2^{k+1}\alpha}\right)$$ Finally, applying the fact that $\pi=\lim\limits_{x\rightarrow 0}\frac{\sin(\pi x)}{x}$, we have that $$\lim_{k\rightarrow\infty}2^k\alpha B_k=\lim_{k\rightarrow\infty}2^{k+1}\alpha\sin\left(\frac{\pi}{2^{k+1}\alpha}\right)=\pi$$ as desired.

  • $\begingroup$ As a note: Most identities involving similar nested radicals of $2$ can be evaluated using the same methods. This is where identities like Viéte's formula come from. I too have played around and created a few identities of this sort in the past; they certainly make for odd looking formulas. : ) $\endgroup$
    – C-RAM
    Nov 25, 2022 at 17:04
  • $\begingroup$ When I first found $(1), (2), (3)$, I naturally wondered what other "kernel" ($\sqrt1, \sqrt2,\sqrt3$, respectively) could be used. It was quite satisfying to find the general form. :) $\endgroup$ Nov 26, 2022 at 3:14

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