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Recently, I found the following handwritten expression in an old math book of my family. Probably it belonged to my great grandfather Boris, who had a P.D. in mathematics.

$ \pi = \frac{4\sqrt{5}}{5}.\frac{2}{\sqrt{2 + \frac{4}{\sqrt{5}}}}.\frac{2}{\sqrt{2 + \sqrt{2 +\frac{4}{\sqrt{5}}}}}.\frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 +\frac{4}{\sqrt{5}}}}}}... +\frac{2\sqrt{10}}{5}.\frac{2}{\sqrt{2 + \frac{6}{\sqrt{10}}}}.\frac{2}{\sqrt{2 + \sqrt{2 +\frac{6}{\sqrt{10}}}}}.\frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 +\frac{6}{\sqrt{10}}}}}}... $

I found this identity extremely interesting, and, in fact, I had never seen it. It's similar, but different, from Vieta's formula for $ \pi $

How to prove this identity?

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    Long story short: through the cosine halving formula $$\cos(\theta/2)=\sqrt{\frac{1+\cos\theta}{2}}$$ and the telescopic product $$\cos(x)\cos(2x)\cdots\cos(2^N x) = \frac{\sin(2^N x)}{2^N \sin(x)}$$ – Jack D'Aurizio Oct 20 '16 at 22:38
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    The convergence is remarkably rapid; 2 parts in a million accuracy after just 7 terms, and the error seems to fall by a factor of about 5 each subsequent term.. Have you tried nesting similar but differently oriented 72-72-36 triangles in a circle? [Of course, the REALLY astonishing result would be to show that this converges to something other than $\pi$, since it is so close.) – Mark Fischler Oct 20 '16 at 22:41
  • @SimpleArt: sure. That is exactly the principle behind Vieta's formula, not just "something similar". – Jack D'Aurizio Oct 20 '16 at 22:42
  • @Jack D'Aurizio Usually I can immediately see solutions given one of your hints, but this one has me stuck. – Mark Fischler Oct 20 '16 at 22:46

1 Answers1

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Let $\theta=\arctan\frac{1}{2}$. Then $\cos\theta = \frac{2}{\sqrt{5}}$, hence $\sec\theta=\frac{\sqrt{5}}{2}$ and $$\cos\frac{\theta}{2}=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}},\qquad \cos\frac{\theta}{4}=\sqrt{\frac{1+\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}}{2}}$$

and so on. Perform a bit of maquillage, then recall that $$ \cos(\theta)\cos(2\theta)\cdots\cos(2^N\theta)=\frac{\sin(2^{N+1}\theta)}{2^N\sin(\theta)}$$ by a telescopic product, and you will recognize that your identity is just asserting $$ \arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4},$$ pretty well-known. Your grandfather just combined a Machin-like formula with the principle behind Vieta's formula for $\pi$.

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