$\pi$ and rapid series for the inverse tangent

Question

I design an educational program to demonstrate convergence rate in computation of $\pi$ by using the Machin like formula. According to Weisstein, the Hwang’s equation (see eq. (32) in http://mathworld.wolfram.com/Machin-LikeFormulas.html): $$\begin{align} \frac{1}{4}\pi = & 183{{\cot }^{-1}}239+32{{\cot }^{-1}}1023-68{{\cot }^{-1}}5832 \\ & +12{{\cot }^{-1}}110443-12{{\cot }^{-1}}4841182-100{{\cot }^{-1}}6826318 \\ \end{align}$$ is the most efficient Machin like formula to compute pi. I used three series for the inverse tangent ($\cot^{-1}x = \tan^{-1}\frac{1}{x}$): $$\tag{1}\tan^{-1}x=\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{x}^{2n+1}}}{2n+1}},$$

$$\tag{2}\tan^{-1}x=\sum\limits_{n=1}^{\infty }{\frac{{{2}^{2n}}{{\left( n! \right)}^{2}}}{\left( 2n+1 \right)!}\frac{{{x}^{2n+1}}}{{{\left( 1+{{x}^{2}} \right)}^{n+1}}}},$$

$$\tag{3}\tan^{-1}x=2\sum\limits_{n=1}^{\infty }{\frac{1}{2n-1}\frac{{{a}_{n}}\left( x \right)}{a_{n}^{2}\left( x \right)+b_{n}^{2}\left( x \right)}}, $$ where $$\begin{align} & {{a}_{1}}\left( x \right)=2/x, \\ & {{b}_{1}}\left( x \right)=1, \\ & {{a}_{n}}\left( x \right)={{a}_{n-1}}\left( x \right)\left( 1-4/{{x}^{2}} \right)+4{{b}_{n-1}}\left( x \right)/x, \\ & {{b}_{n}}\left( x \right)={{b}_{n-1}}\left( x \right)\left( 1-4/{{x}^{2}} \right)-4{{a}_{n-1}}\left( x \right)/x. \\ \end{align}$$

Eq. (1) is the Maclaurin series, eq. (2) is the Euler’s series, eq. (3) is from the paper Abrarov & Quine arXiv:1706.08835. It appears to be that eqs. (2) and (3) are more rapid than eq. (1). I want to ask you two questions. Is eq. (3) more efficient than eq. (2)? Are there other series for the inverse tangent with rapid convergence?

Answer 1

The Hwang equation is not the most efficient Machin like formula to compute $\pi$. The following formula

$${\pi \over 4} = 32 \, \hbox{arctan}({1\over 40}) - \hbox{arctan}\biggl( {38035138859000075702655846657186322249216830232319 \over 2634699316100146880926635665506082395762836079845121}\biggr) $$ has a Lehmer's measure around $1.16751$ thus beating Hwang's formula (with Lehmer's measure $1.51240$).

Abrarov and Quine gave a formula with Lehmer's measure $0.245319$ last summer, together with the relevant algorithms, in their paper An iteration procedure for a two-term Machin-like formula for pi with small Lehmer’s measure. That formula provides 16 new digits of $\pi$ per term increment thus beating the famous Chudnovsky formula.

At $k=27$ the Abrarov-Quine formula gives a fraction that has a numerator with $522\,185\,807$ digits and a denominator with $522\,185\,816$ digits. This type of fractions with huge numbers can also be obtained from the Borwein’s integral as shown in the preprint by Uwe Bäsel and Robert Baillie Sinc integrals and tiny numbers such that a formula for $\pi$ can be written with $453\,130\,145$ and $453\,237\,170$ digits in numerator and denominator, respectively.

Answer 2

This is not exactly an answer, but R. P. Brent has an excellent survey of computing everything under the Sun (including $\arctan.$)

Answer 3

The efficiency of such series depends not just on the rate of convergence but on the bit size of the terms.

If $\sum_{k=0}^{\infty} T(k)$ converges linearly, you can always rewrite it as $\sum_{k=0}^{\infty} U(k)$ where $U(k) = \sum_{j=0}^{N-1} T(N k + j)$ to get a series that converges $N$ times faster. But nothing has really been gained since the terms now cost $N$ times more to compute.

Answer 4

Arctangent formula (3) that can also be written as $$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right)$$ is more rapid in convergence than the equation (2). Therefore, it is more efficient in computation. Here is a link: How to compare convergence of two equations for the arctangent function?