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.

π=4√55.2√2+4√5.2√2+√2+4√5.2√2+√2+√2+4√5…+2√105.2√2+6√10.2√2+√2+6√10.2√2+√2+√2+6√10… \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 π \pi

How to prove this identity?

=================

4

Long story short: through the cosine halving formula cos(θ/2)=√1+cosθ2\cos(\theta/2)=\sqrt{\frac{1+\cos\theta}{2}} and the telescopic product cos(x)cos(2x)⋯cos(2Nx)=sin(2Nx)2Nsin(x)\cos(x)\cos(2x)\cdots\cos(2^N x) = \frac{\sin(2^N x)}{2^N \sin(x)}

– Jack D’Aurizio

Oct 20 at 22:38

1

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\pi, since it is so close.)

– Mark Fischler

Oct 20 at 22:41

@SimpleArt: sure. That is exactly the principle behind Vieta’s formula, not just “something similar”.

– Jack D’Aurizio

Oct 20 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 at 22:46

=================

1 Answer

1

=================

Let \theta=\arctan\frac{1}{2}\theta=\arctan\frac{1}{2}. Then \cos\theta = \frac{2}{\sqrt{5}}\cos\theta = \frac{2}{\sqrt{5}}, hence \sec\theta=\frac{\sqrt{5}}{2}\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}}\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)} \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}, \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\pi.

3

There’s a nice proof without words (also well known) of that last fact: i.stack.imgur.com/4j3SN.png

– Brian Tung

Oct 20 at 22:58

I see you’ve added in a link to some others. 🙂

– Brian Tung

Oct 20 at 23:09

@BrianTung Here is another variation on the same theme 😉

– dxiv

Oct 21 at 1:10