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Borwein's algorithm (others)

From Wikipedia, the free encyclopedia

Jonathan and Peter Borwein devised various algorithms to calculate the value of π. The most prominent and oft-used one is explained under Borwein's algorithm and Bailey-Borwein-Plouffe formula. Other algorithms found by them include the following:

  1. Quadratic convergence, 1987:
    • Start out by setting
      x_0 = \sqrt2
      y_0 = \sqrt[4]2
      p_0 = 2+\sqrt2
    • Then iterate
      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}

    Then pk converges monotonically to π; with

    p_k - \pi = 10^{-2^{k+1}}

    for k > = 2

  2. Cubical convergence, 1991:
    • Start out by setting
      a_0 = \frac{1}{3}
      s_0 = \frac{\sqrt{3} - 1}{2}
    • Then iterate
      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)

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

  3. Quartical convergence, 1984:
    • Start out by setting
      a_0 = \sqrt{2}
      b_0 = 0\,\!
      p_0 = 2 + \sqrt{2}
    • Then iterate
      a_{n+1} = \frac{\sqrt{a_n} + 1/\sqrt{a_n}}{2}
      b_{n+1} = \frac{\sqrt{a_n} (1 + b_n)}{a_n + b_n}
      p_{n+1} = \frac{p_n b_{n+1} (1 + a_{n+1})}{1 + b_{n+1}}

    Then pk converges quartically against 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.

  4. Quintical convergence:
    • Start out by setting
      a_0 = \frac{1}{2}
      s_0 = 5(\sqrt{5} - 2)
    • Then iterate
      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}

    Then ak converges quintically against 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
  5. Nonical convergence:
    • Start out by setting
      a_0 = \frac{1}{3}
      r_0 = \frac{\sqrt{3} - 1}{2}
      s_0 = (1 - r_0^3)^{1/3}
    • Then iterate
      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}

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

  6. Another formula for π, 1989:
    • Start out by setting
      A = 212175710912 \sqrt{61} + 1657145277365
      B = 13773980892672 \sqrt{61} + 107578229802750
      C = (5280(236674+30303\sqrt{61}))^3
    • 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 series yields approximately 31 digits.

  7. Jonathan Borwein and Peter Borwein, 1993:
    • Start out by setting
      \,A = 63365028312971999585426220
      + 28337702140800842046825600\sqrt5
      + 384\sqrt5(10891728551171178200467436212395209160385656017
      + 4870929086578810225077338534541688721351255040\sqrt5)^{1/2}
      \,B = 7849910453496627210289749000
      + 3510586678260932028965606400\sqrt5
      + 2515968\sqrt{3110}(6260208323789001636993322654444020882161
      + 2799650273060444296577206890718825190235\sqrt5)^{1/2}
      \,C = -214772995063512240
      - 96049403338648032\sqrt5
      - 1296\sqrt5(10985234579463550323713318473
      + 4912746253692362754607395912\sqrt5)^{1/2}
    • 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.

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