(Don’t be alarmed by the title; this is a question about mathematics, not programming.)

In the documentation for the

`decimal`

module in the Python Standard Library, an example is given for computing the digits of $\pi$ to a given precision:`def pi(): """Compute Pi to the current precision. >>> print(pi()) 3.141592653589793238462643383 """ getcontext().prec += 2 # extra digits for intermediate steps three = Decimal(3) # substitute "three=3.0" for regular floats lasts, t, s, n, na, d, da = 0, three, 3, 1, 0, 0, 24 while s != lasts: lasts = s n, na = n+na, na+8 d, da = d+da, da+32 t = (t * n) / d s += t getcontext().prec -= 2 return +s # unary plus applies the new precision`

I was not able to find any reference for what formula or fact about $\pi$ this computation uses, hence this question.

Translating from code into more typical mathematical notation, and using some calculation and observation, this amounts to a formula for $\pi$ that begins like:

$$\begin{align}\pi

&= 3+\frac{1}{8}+\frac{9}{640}+\frac{15}{7168}+\frac{35}{98304}+\frac{189}{2883584}+\frac{693}{54525952}+\frac{429}{167772160} + \dots\\

&= 3\left(1+\frac{1}{24}+\frac{1}{24}\frac{9}{80}+\frac{1}{24}\frac{9}{80}\frac{25}{168}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}\frac{121}{624}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}\frac{121}{624}\frac{169}{840}+\dots\right)

\end{align}$$or, more compactly,

$$\pi = 3\left(1 + \sum_{n=1}^{\infty}\prod_{k=1}^{n}\frac{(2k-1)^2}{8k(2k+1)}\right)$$

Is this a well-known formula for $\pi$? How is it proved? How does it compare to other methods, in terms of how how quickly it converges, numerical stability issues, etc? At a glance I didn’t see it on the Wikipedia page for List of formulae involving π or on the MathWorld page for Pi Formulas.

**Answer**

That is the Taylor series of $\arcsin(x)$ at $x=1/2$ (times 6).

**Attribution***Source : Link , Question Author : ShreevatsaR , Answer Author : mlerma54*