Borwein's algorithm

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

Start by setting{{cite journal | last=Bailey | first=David H | title=Peter Borwein: A Visionary Mathematician | journal=Notices of the American Mathematical Society | volume=70 | issue=4 | date=2023-04-01 | issn=0002-9920 | doi=10.1090/noti2675 | pages=610–613}}

:

B & = 13773980892672 \sqrt{61} + 107578229802750 \\

C & = \left(5280\left(236674+30303\sqrt{61}\right)\right)^3

\end{align}

Then

:$\backslash frac\{1\}\{\backslash pi\}\; =\; 12\backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\{\; (-1)^n\; (6n)!\backslash ,\; (A+nB)\; \}\{(n!)^3(3n)!\backslash ,\; C^\{n+\backslash frac12\}\}$

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

Start by setting{{cite journal | last1=Borwein | first1=J.M. | last2=Borwein | first2=P.B. | title=Class number three Ramanujan type series for 1/π | journal=Journal of Computational and Applied Mathematics | volume=46 | issue=1–2 | date=1993 | doi=10.1016/0377-0427(93)90302-R | pages=281–290| doi-access=free }}

: $\backslash begin\{align\}$

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

: $\backslash frac\{\backslash sqrt\{-C^3\}\}\{\backslash pi\}\; =\; \backslash sum\_\{n=0\}^\{\backslash infty\}\; \{\backslash frac\{(6n)!\}\{(3n)!(n!)^3\}\; \backslash frac\{A+nB\}\{C^\{3n\}\}\}$

Each additional term of the series yields approximately 50 digits.

Start by setting{{cite book| title={{pi}} Unleashed |first1=Jörg |last1=Arndt |first2=Christoph |last2=Haenel |publisher=Springer-Verlag |isbn=3-540-66572-2 |year=1998 |page=236}}

:

b_0 & = 0 \\

p_0 & = 2 + \sqrt{2}

\end{align}

Then iterate

:

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 p_k converges quadratically to {{pi}}; 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 {{pi}}'s final result.

Start by setting

:

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

\end{align}

Then iterate

:

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 a_k converges cubically to {{sfrac|1|{{pi}}}}; that is, each iteration approximately triples the number of correct digits.

Start by setting{{cite book | title=The Java Programmers Guide to Numerical Computation |last=Mak |first=Ronald |publisher=Pearson Educational |year=2003 |isbn=0-13-046041-9 |page=353}}

:

y_0 & = \sqrt{2}-1

\end{align}

Then iterate

:

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 a_k converges quartically against {{sfrac|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 {{pi}}'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:{{citation|title=Easy Proof of Three Recursive {{pi}}-Algorithms|last1=Milla|first1=Lorenz|arxiv=1907.04110|year=2019}}

Start by setting

:

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

\end{align}

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

:

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 a_k converges quintically to {{sfrac|1|{{pi}}}} (that is, each iteration approximately quintuples the number of correct digits), and the following condition holds:

:

Start by setting

:

r_0 & = \frac{\sqrt{3} - 1}{2} \\

s_0 & = \left(1 - r_0^3\right)^\frac13

\end{align}

Then iterate

:

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 a_k converges nonically to {{sfrac|1|{{pi}}}}; that is, each iteration approximately multiplies the number of correct digits by nine.{{cite web|url=http://www.hvks.com/Numerical/Downloads/HVE%20Practical%20implementation%20of%20PI%20Algorithms.pdf|title=Practical implementation of π Algorithms|author=Henrik Vestermark|date=4 November 2016|access-date=29 November 2020}}

{{Portal|Mathematics}}

• Bailey–Borwein–Plouffe formula
• Chudnovsky algorithm
• Gauss–Legendre algorithm
• Ramanujan–Sato series

• [http://mathworld.wolfram.com/PiFormulas.html Pi Formulas] from Wolfram MathWorld

Category:Pi algorithms