## 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. 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 $n=18$. This entry was posted in Scientific ramblings and tagged , , , , , , , , . Bookmark the permalink.