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Wikipedia

Borwein's algorithm

                   
This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. (April 2009)

In mathematics, Borwein's algorithm is an algorithm devised by Jonathan and Peter Borwein to calculate the value of 1/π.

Contents

  Jonathan Borwein and Peter Borwein's Version (1993)

Start out by setting[citation needed]

\begin{align} A &= 63365028312971999585426220 \\ &\quad + 28337702140800842046825600\sqrt{5} \\ &\quad + 384\sqrt{5} (10891728551171178200467436212395209160385656017 \\ &\qquad + 4870929086578810225077338534541688721351255040\sqrt{5})^{1/2} \\ B &= 7849910453496627210289749000 \\ &\quad + 3510586678260932028965606400\sqrt{5} \\ &\quad + 2515968\sqrt{3110}(6260208323789001636993322654444020882161 \\ &\qquad + 2799650273060444296577206890718825190235\sqrt{5})^{1/2} \\ C &= -214772995063512240 \\ &\quad - 96049403338648032\sqrt{5} \\ &\quad - 1296\sqrt{5}(10985234579463550323713318473 \\ &\qquad + 4912746253692362754607395912\sqrt{5})^{1/2} \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.

  Cubic convergence (1991)

Start out by setting[citation needed]

\begin{align} a_0 & = \frac{1}{3} \\ s_0 & = \frac{\sqrt{3} - 1}{2} \end{align}

Then iterate

\begin{align} r_{k+1} & = \frac{3}{1 + 2(1-s_k^3)^{1/3}} \\ s_{k+1} & = \frac{r_{k+1} - 1}{2} \\ a_{k+1} & = r_{k+1}^2 a_k - 3^k(r_{k+1}^2-1) \end{align}

Then ak converges cubically to 1/π; that is, each iteration approximately triples the number of correct digits.

  Another formula for π (1989)

Start out by setting[citation needed]

\begin{align} A & = 212175710912 \sqrt{61} + 1657145277365 \\ B & = 13773980892672 \sqrt{61} + 107578229802750 \\ C & = (5280(236674+30303\sqrt{61}))^3 \end{align}

Then

1 / \pi = 12\sum_{n=0}^\infty \frac{ (-1)^n (6n)!\, (A+nB) }{(n!)^3(3n)!\, C^{n+1/2}}\,\!

Each additional term of the partial sum yields approximately 31 digits.

  Quadratic convergence (1987)

Start out by setting[1]

\begin{align} x_0 & = \sqrt2 \\ y_1 & = \sqrt[4]2 \\ p_0 & = 2+\sqrt2 \end{align}

Then iterate

\begin{align} x_k & = \frac{1}{2}(x_{k-1}^{1/2} + x_{k-1}^{-1/2}) \\ y_k & = \frac{y_{k-1}x_{k-1}^{1/2} + x_{k-1}^{-1/2}} {y_{k-1}+1} \\ p_k & = p_{k-1}\frac{x_k+1}{y_k+1} \end{align}

Then pk converges monotonically to π; with pk - π ≈ 10−2k+1 for k ≥ 2.s

  Borwein's algorithm (1985)

Start out by setting[2]

\begin{align} a_0 & = 6 - 4\sqrt{2} \\ y_0 & = \sqrt{2} - 1 \end{align}

Then iterate

\begin{align} y_{k+1} & = \frac{1-(1-y_k^4)^{1/4}}{1+(1-y_k^4)^{1/4}} \\ a_{k+1} & = a_k(1+y_{k+1})^4 - 2^{2k+3} y_{k+1} (1 + y_{k+1} + y_{k+1}^2) \end{align}

Then ak converges quartically against 1/π; that is, each iteration approximately quadruples the number of correct digits.

  Quartic convergence (1984)

Start out by setting[3]

\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} + 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 quartically to π; 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 of π.

  Quintic convergence

Start out by setting[citation needed]

\begin{align} a_0 & = \frac{1}{2} \\ s_0 & = 5(\sqrt{5} - 2) \end{align}

Then iterate

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

Then ak converges quintically to 1/π (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 out by setting[citation needed]

\begin{align} a_0 & = \frac{1}{3} \\ r_0 & = \frac{\sqrt{3} - 1}{2} \\ s_0 & = (1 - r_0^3)^{1/3} \end{align}

Then iterate

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

Then ak converges nonically to 1/π; that is, each iteration approximately multiplies the number of correct digits by nine.

  See also

  References

  1. ^ Jean-Luc Chabert et al (1999). A History of Algorithms: From the Pebble to the Microchip. Springer-Verlag. p. 166. ISBN 3-540-63369-3. 
  2. ^ Mak, Ronald (2003). The Java Programmers Guide to Numerical Computation. Pearson Educational. p. 353. ISBN 0-13-046041-9. 
  3. ^ Arndt, Jörg; Haenel, Christoph (1998). π Unleashed. Springer-Verlag. p. 236. ISBN 3-540-66572-2. 
   
   
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