Borwein's algorithm

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In mathematics, Borwein's algorithm is an algorithm devised by Jonathan and Peter Borwein to calculate the value of 1/π. They devised several other algorithms. They published a book: Jonathon M. Borwein, Peter B. Borwein, Pi and the AGM - A Study in Analytic Number Theory and Computational Complexity, Wiley, New York, 1987. Many of their results are available in: Jorg Arndt, Christoph Haenel, Pi Unleashed, Springer, Berlin, 2001, ISBN 3-540-66572-2.

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. This is an example of a Ramanujan–Sato series.

Cubic convergence (1991)

Start out by setting

\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.

Quartic algorithm (1985)

Start out by setting[1]

\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.

Quadratic convergence (1984)

Start out by setting[2] [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 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 of π.

Quintic convergence

Start out by setting

\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:

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

[4]

Nonic convergence

Start out by setting

\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.

References

1. ^ Mak, Ronald (2003). The Java Programmers Guide to Numerical Computation. Pearson Educational. p. 353. ISBN 0-13-046041-9.
2. ^ Arndt, Jörg; Haenel, Christoph (1998). π Unleashed. Springer-Verlag. p. 236. ISBN 3-540-66572-2.
3. ^ Template:Pi Unleashed
4. ^ http://www.cecm.sfu.ca/organics/papers/garvan/paper/html/node12.html