Euler’s squares

On account of our fascination with the geometry of origami (albeit not well-endowed in mathematical capacity) we discovered for ourselves shortly after our father had taught us trigonometry that,
\arctan(1)+\arctan(2)+\arctan(3)=\pi
We had earlier shown the origami proof for that. But it was only a little later while drifting away from one of those trigonometric identities that you routinely faced in those annoying college exams we stumbled upon a beautiful relationship that was apparently first discovered by the great Leonhard Euler. This is a relationship parallel to the above one:
\arctan\left(\frac{1}{2}\right)+\arctan\left(\frac{1}{3}\right)=\arctan\left(1\right)=\frac{\pi}{4}
The proof for this, like the origami proof for the above, can be achieved from a self-evident construction of Euler — what the Hindus of yore would have called an upapatti or mathematicians today term “wordless” proof. It is illustrated below but I add several words for the benefit of the non-geometrically oriented reader.

Euler_squares

1) Draw square ABCD and triplicate it so that the three squares share a side.
2) Draw diagonals AC and CG of first two squares and use them to draw square ACGF and duplicate it.
3) Draw \overline{AE}: from the construction it is apparent that \angle GAE=\arctan\left(\frac{1}{3}\right)
4) From the construction it is clear that \angle EAC=\arctan\left(\frac{1}{2}\right)
5) We thus see: \angle GAE+ \angle EAC = \angle BAC= \frac{\pi}{4}= \arctan(1)=\arctan\left(\frac{1}{2}\right)+\arctan\left(\frac{1}{3}\right)

This relationship is one of a class of strange trigonometric relationships that interestingly bring in the meru-średhī (called in western literature as Fibonacci sequence):

\arctan\left(\frac{1}{M_{2n}}\right)=\arctan\left(\frac{1}{M_{2n+1}}\right)+\arctan\left(\frac{1}{M_{2n+2}}\right)

M=1,1,2,3,5,8,13,21..., the meru-średhī; thus for n=1, we get M_2=1; M_3=2; M_4=3.
This leads us to a formula for \pi based on the odd terms of meru-średhī starting from M_3:

\pi=4\displaystyle \sum_{n=1}^\infty \arctan \left(\frac{1}{M_{2n+1}}\right)

Shown below is the convergence of the above series to \pi: We reach an accuracy of 6 decimal places for n=18.

pi_meru

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~ by mAnasa-taraMgiNI on April 20, 2017.

 
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