Euler's Approximation of pi. I recently stumbled across the formula:
$$\pi=20\arctan\frac{1}{7}+8\arctan\frac{3}{79}$$
developed by Euler, for approximating pi. I evaluated it to several thousand decimal places and up to that point, it accurately represented pi.
When does this equation break down in its representation of pi?
 A: That formula doesn't break down.  It can be proven that the right-hand side is equal to $\pi$.  One reason it comes up in connection with approximation is that the arctangent terms can be approximated well by adding terms from the Taylor expansion of the arctangent function, 
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\cdots.$$
This is valid for $|x|\leq 1$, but the series converges more rapidly the closer $x$ is to $0$, so with the relatively small $\frac{1}{7}$ and $\frac{3}{79}$ as inputs it gives a good approximation without having to take too many terms.  This can be contrasted with the formula $$\pi=4\arctan(1)=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots\right),$$ where the series converges much more slowly.  

In 1950 H.C. Schepler published a 3 part "Chronology of pi" in Mathematics Magazine, available to those with access to JSTOR here, here, and here. In the second part there is the following excerpt indicating that Hutton may have suggested using the formula before Euler:

Schepler's chronology can also be found in L. Berggren, J. Borwein, and P. Borwein's Pi: A Source Book. 
A: The Soviet 8th grade students usually solve this problem as follows. First, divide by 4. Then a portable arc tangent in the opposite direction. Then calculate the tangent of the right and left part using very simple formula tg(A+B).
