Borwein's algorithm

Borwein's algorithm was devised by Jonathan and Peter Borwein to calculate the value of $$1 / \pi$$. This and other algorithms can be found in the book Pi and the AGM – A Study in Analytic Number Theory and Computational Complexity.

Ramanujan–Sato series
These two are examples of a Ramanujan–Sato series. The related Chudnovsky algorithm uses a discriminant with class number 1.

Class number 2 (1989)
Start by setting


 * $$ \begin{align} A & = 212175710912 \sqrt{61} + 1657145277365 \\

B & = 13773980892672 \sqrt{61} + 107578229802750 \\ C & = \left(5280\left(236674+30303\sqrt{61}\right)\right)^3 \end{align} $$

Then


 * $$\frac{1}{\pi} = 12\sum_{n=0}^\infty \frac{ (-1)^n (6n)!\, (A+nB) }{(n!)^3(3n)!\, C^{n+\frac12}}$$

Each additional term of the partial sum yields approximately 25 digits.

Class number 4 (1993)
Start by setting


 * $$\begin{align}

A = {} & 63365028312971999585426220 \\ & {} + 28337702140800842046825600\sqrt{5} \\ & {} + 384\sqrt{5} \big(10891728551171178200467436212395209160385656017 \\   & {} + \left. 4870929086578810225077338534541688721351255040\sqrt{5}\right)^\frac12 \\ B = {} & 7849910453496627210289749000 \\ & {} + 3510586678260932028965606400\sqrt{5} \\ & {} + 2515968\sqrt{3110}\big(6260208323789001636993322654444020882161 \\   & {} + \left. 2799650273060444296577206890718825190235\sqrt{5}\right)^\frac12 \\ C = {} & -214772995063512240 \\ & {} - 96049403338648032\sqrt{5} \\ & {} - 1296\sqrt{5}\big(10985234579463550323713318473 \\   & {} + \left. 4912746253692362754607395912\sqrt{5}\right)^\frac12 \end{align}$$

Then


 * $$\frac{\sqrt{-C^3}}{\pi} = \sum_{n=0}^{\infty} {\frac{(6n)!}{(3n)!(n!)^3} \frac{A+nB}{C^{3n}}}$$

Each additional term of the series yields approximately 50 digits.

Quadratic convergence (1984)
Start by setting


 * $$ \begin{align} a_0 & = \sqrt{2} \\

b_0 & = 0 \\ p_0 & = 2 + \sqrt{2} \end{align} $$

Then iterate


 * $$ \begin{align} a_{n+1} & = \frac{\sqrt{a_n} + \frac{1}\sqrt{a_n}}{2} \\

b_{n+1} & = \frac{(1 + b_n) \sqrt{a_n}}{a_n + b_n} \\ p_{n+1} & = \frac{(1 + a_{n+1})\, p_n b_{n+1}}{1 + b_{n+1}} \end{align} $$

Then pk converges quadratically to π; that is, each iteration approximately doubles the number of correct digits. The algorithm is not self-correcting; each iteration must be performed with the desired number of correct digits for π's final result.

Cubic convergence (1991)
Start by setting


 * $$ \begin{align} a_0 & = \frac13 \\

s_0 & = \frac{\sqrt{3} - 1}{2} \end{align} $$

Then iterate


 * $$ \begin{align} r_{k+1} & = \frac{3}{1 + 2\left(1-s_k^3\right)^\frac13} \\

s_{k+1} & = \frac{r_{k+1} - 1}{2} \\ a_{k+1} & = r_{k+1}^2 a_k - 3^k\left(r_{k+1}^2-1\right) \end{align} $$

Then ak converges cubically to $1⁄\pi$; that is, each iteration approximately triples the number of correct digits.

Quartic convergence (1985)
Start by setting


 * $$ \begin{align} a_0 & = 2\left(\sqrt{2}-1\right)^2 \\

y_0 & = \sqrt{2}-1 \end{align} $$

Then iterate


 * $$ \begin{align} y_{k+1} & = \frac{1-\left(1-y_k^4\right)^\frac14}{1+\left(1-y_k^4\right)^\frac14} \\

a_{k+1} & = a_k\left(1+y_{k+1}\right)^4 - 2^{2k+3} y_{k+1} \left(1 + y_{k+1} + y_{k+1}^2\right) \end{align} $$

Then ak converges quartically against $1⁄\pi$; that is, each iteration approximately quadruples the number of correct digits. The algorithm is not self-correcting; each iteration must be performed with the desired number of correct digits for π's final result.

One iteration of this algorithm is equivalent to two iterations of the Gauss–Legendre algorithm. A proof of these algorithms can be found here:

Quintic convergence
Start by setting


 * $$ \begin{align} a_0 & = \frac12 \\

s_0 & = 5\left(\sqrt{5} - 2\right) = \frac{5}{\phi^3} \end{align} $$

where $$\phi = \tfrac{1+\sqrt5}{2}$$ is the golden ratio. Then iterate


 * $$ \begin{align} x_{n+1} & = \frac{5}{s_n} - 1 \\

y_{n+1} & = \left(x_{n+1} - 1\right)^2 + 7 \\ z_{n+1} & = \left(\frac12 x_{n+1}\left(y_{n+1} + \sqrt{y_{n+1}^2 - 4x_{n+1}^3}\right)\right)^\frac15 \\ a_{n+1} & = s_n^2 a_n - 5^n\left(\frac{s_n^2 - 5}{2} + \sqrt{s_n\left(s_n^2 - 2s_n + 5\right)}\right) \\ s_{n+1} & = \frac{25}{\left(z_{n+1} + \frac{x_{n+1}}{z_{n+1}} + 1\right)^2 s_n} \end{align} $$

Then ak converges quintically to $1⁄\pi$ (that is, each iteration approximately quintuples the number of correct digits), and the following condition holds:


 * $$0 < a_n - \frac{1}{\pi} < 16\cdot 5^n\cdot e^{-5^n}\pi\,\!$$

Nonic convergence
Start by setting


 * $$ \begin{align} a_0 & = \frac13 \\

r_0 & = \frac{\sqrt{3} - 1}{2} \\ s_0 & = \left(1 - r_0^3\right)^\frac13 \end{align} $$

Then iterate


 * $$ \begin{align} t_{n+1} & = 1 + 2r_n \\

u_{n+1} & = \left(9r_n \left(1 + r_n + r_n^2\right)\right)^\frac13 \\ v_{n+1} & = t_{n+1}^2 + t_{n+1}u_{n+1} + u_{n+1}^2 \\ w_{n+1} & = \frac{27 \left(1 + s_n + s_n^2\right)}{v_{n+1}} \\ a_{n+1} & = w_{n+1}a_n + 3^{2n-1}\left(1-w_{n+1}\right) \\ s_{n+1} & = \frac{\left(1 - r_n\right)^3}{\left(t_{n+1} + 2u_{n+1}\right)v_{n+1}} \\ r_{n+1} & = \left(1 - s_{n+1}^3\right)^\frac13 \end{align} $$

Then ak converges nonically to $1⁄\pi$; that is, each iteration approximately multiplies the number of correct digits by nine.